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DPOSVX(3) uses the Cholesky factorization A = U**T*U or A = L*L**T to compute the solution to a real system of linear equations A * X = B,
## SYNOPSIS
SUBROUTINE DPOSVX(
FACT, UPLO, N, NRHS, A, LDA, AF, LDAF, EQUED, S, B, LDB, X, LDX, RCOND, FERR, BERR, WORK, IWORK, INFO )
CHARACTER EQUED, FACT, UPLO
INTEGER INFO, LDA, LDAF, LDB, LDX, N, NRHS
DOUBLE PRECISION RCOND
INTEGER IWORK( * )
DOUBLE PRECISION A( LDA, * ), AF( LDAF, * ), B( LDB, * ), BERR( * ), FERR( * ), S( * ), WORK( * ), X( LDX, * )
## PURPOSE
DPOSVX uses the Cholesky factorization A = U**T*U or A = L*L**T to compute the solution to a real system of linear equations
A * X = B, where A is an N-by-N symmetric positive definite matrix and X and B are N-by-NRHS matrices.
Error bounds on the solution and a condition estimate are also provided.
## DESCRIPTION
The following steps are performed:
1. If FACT = 'E', real scaling factors are computed to equilibrate
the system:
diag(S) * A * diag(S) * inv(diag(S)) * X = diag(S) * B
Whether or not the system will be equilibrated depends on the
scaling of the matrix A, but if equilibration is used, A is
overwritten by diag(S)*A*diag(S) and B by diag(S)*B.
2. If FACT = 'N' or 'E', the Cholesky decomposition is used to
factor the matrix A (after equilibration if FACT = 'E') as
A = U**T* U, if UPLO = 'U', or
A = L * L**T, if UPLO = 'L',
where U is an upper triangular matrix and L is a lower triangular
matrix.
3. If the leading i-by-i principal minor is not positive definite,
then the routine returns with INFO = i. Otherwise, the factored
form of A is used to estimate the condition number of the matrix
A. If the reciprocal of the condition number is less than machine
precision, INFO = N+1 is returned as a warning, but the routine
still goes on to solve for X and compute error bounds as
described below.
4. The system of equations is solved for X using the factored form
of A.
5. Iterative refinement is applied to improve the computed solution
matrix and calculate error bounds and backward error estimates
for it.
6. If equilibration was used, the matrix X is premultiplied by
diag(S) so that it solves the original system before
equilibration.
## ARGUMENTS
FACT (input) CHARACTER*1
Specifies whether or not the factored form of the matrix A is supplied on entry, and if not, whether the matrix A should be equilibrated before it is factored. = 'F': On entry, AF contains the factored form of A. If EQUED = 'Y', the matrix A has been equilibrated with scaling factors given by S. A and AF will not be modified. = 'N': The matrix A will be copied to AF and factored.
= 'E': The matrix A will be equilibrated if necessary, then copied to AF and factored.
UPLO (input) CHARACTER*1
= 'U': Upper triangle of A is stored;
= 'L': Lower triangle of A is stored.
N (input) INTEGER
The number of linear equations, i.e., the order of the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of columns of the matrices B and X. NRHS >= 0.
A (input/output) DOUBLE PRECISION array, dimension (LDA,N)
On entry, the symmetric matrix A, except if FACT = 'F' and EQUED = 'Y', then A must contain the equilibrated matrix diag(S)*A*diag(S). If UPLO = 'U', the leading N-by-N upper triangular part of A contains the upper triangular part of the matrix A, and the strictly lower triangular part of A is not referenced. If UPLO = 'L', the leading N-by-N lower triangular part of A contains the lower triangular part of the matrix A, and the strictly upper triangular part of A is not referenced. A is not modified if FACT = 'F' or 'N', or if FACT = 'E' and EQUED = 'N' on exit. On exit, if FACT = 'E' and EQUED = 'Y', A is overwritten by diag(S)*A*diag(S).
LDA (input) INTEGER
The leading dimension of the array A. LDA >= max(1,N).
AF (input or output) DOUBLE PRECISION array, dimension (LDAF,N)
If FACT = 'F', then AF is an input argument and on entry contains the triangular factor U or L from the Cholesky factorization A = U**T*U or A = L*L**T, in the same storage format as A. If EQUED .ne. 'N', then AF is the factored form of the equilibrated matrix diag(S)*A*diag(S). If FACT = 'N', then AF is an output argument and on exit returns the triangular factor U or L from the Cholesky factorization A = U**T*U or A = L*L**T of the original matrix A. If FACT = 'E', then AF is an output argument and on exit returns the triangular factor U or L from the Cholesky factorization A = U**T*U or A = L*L**T of the equilibrated matrix A (see the description of A for the form of the equilibrated matrix).
LDAF (input) INTEGER
The leading dimension of the array AF. LDAF >= max(1,N).
EQUED (input or output) CHARACTER*1
Specifies the form of equilibration that was done. = 'N': No equilibration (always true if FACT = 'N').
= 'Y': Equilibration was done, i.e., A has been replaced by diag(S) * A * diag(S). EQUED is an input argument if FACT = 'F'; otherwise, it is an output argument.
S (input or output) DOUBLE PRECISION array, dimension (N)
The scale factors for A; not accessed if EQUED = 'N'. S is an input argument if FACT = 'F'; otherwise, S is an output argument. If FACT = 'F' and EQUED = 'Y', each element of S must be positive.
B (input/output) DOUBLE PRECISION array, dimension (LDB,NRHS)
On entry, the N-by-NRHS right hand side matrix B. On exit, if EQUED = 'N', B is not modified; if EQUED = 'Y', B is overwritten by diag(S) * B.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= max(1,N).
X (output) DOUBLE PRECISION array, dimension (LDX,NRHS)
If INFO = 0 or INFO = N+1, the N-by-NRHS solution matrix X to the original system of equations. Note that if EQUED = 'Y', A and B are modified on exit, and the solution to the equilibrated system is inv(diag(S))*X.
LDX (input) INTEGER
The leading dimension of the array X. LDX >= max(1,N).
RCOND (output) DOUBLE PRECISION
The estimate of the reciprocal condition number of the matrix A after equilibration (if done). If RCOND is less than the machine precision (in particular, if RCOND = 0), the matrix is singular to working precision. This condition is indicated by a return code of INFO > 0.
FERR (output) DOUBLE PRECISION array, dimension (NRHS)
The estimated forward error bound for each solution vector X(j) (the j-th column of the solution matrix X). If XTRUE is the true solution corresponding to X(j), FERR(j) is an estimated upper bound for the magnitude of the largest element in (X(j) - XTRUE) divided by the magnitude of the largest element in X(j). The estimate is as reliable as the estimate for RCOND, and is almost always a slight overestimate of the true error.
BERR (output) DOUBLE PRECISION array, dimension (NRHS)
The componentwise relative backward error of each solution vector X(j) (i.e., the smallest relative change in any element of A or B that makes X(j) an exact solution).
WORK (workspace) DOUBLE PRECISION array, dimension (3*N)
IWORK (workspace) INTEGER array, dimension (N)
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
> 0: if INFO = i, and i is
<= N: the leading minor of order i of A is not positive definite, so the factorization could not be completed, and the solution has not been computed. RCOND = 0 is returned. = N+1: U is nonsingular, but RCOND is less than machine precision, meaning that the matrix is singular to working precision. Nevertheless, the solution and error bounds are computed because there are a number of situations where the computed solution can be more accurate than the value of RCOND would suggest.
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# Garland
Time Limit: 1 second
Memory Limit: 256 megabytes
## Description
Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.
## Input
The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.
## Output
Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.
## Sample Input
InputaaabbacaabbccacOutput6InputazOutput-1
## Sample Output
None
## Hint
In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z.
None
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## Constrained Integer Behavior
May 26, 2014
The wheels go round and round, round and round ...
Integer arithmetic is ubiquitous in digital hardware implementations, it's prolific in the control and data-paths. When using fixed width (constrained) integers, overflow and underflow is business as usual.
Building with Integers
The subtitle of this post mentions a wheel - before I get to the wheel I want to look at an example. The recursive-windowed-averager (rwa, a.k.a moving average)...
## DSP Related Math: Nice Animated GIFs
April 24, 20143 comments
I was browsing the ECE subreddit lately and found that some of the most popular posts over the last few months have been animated GIFs helping understand some mathematical concepts. I thought there would be some value in aggregating the DSP related gifs on one page.
The relationship between sin, cos, and right triangles: Constructing a square wave with infinite series (see this...
## Signed serial-/parallel multiplication
February 16, 2014
Keywords: Binary signed multiplication implementation, RTL, Verilog, algorithm
Summary
• A detailed discussion of bit-level trickstery in signed-signed multiplication
• Algorithm based on Wikipedia example
• Includes a Verilog implementation with parametrized bit width
Signed serial-/parallel multiplication
A straightforward method to multiply two binary numbers is to repeatedly shift the first argument a, and add to a register if the corresponding bit in the other argument b is set. The...
## Finding the Best Optimum
November 4, 2013
When I was in school learning electrical engineering I owned a large mental pot, full of simmering resentment against the curriculum as it was being taught.
It really started in my junior year, when we took Semiconductor Devices, or more accurately "how to build circuits using transistors". I had been seduced by the pure mathematics of sophomore EE courses, where all the circuit elements (resistors, capacitors, coils and -- oh the joy -- dependent sources) are ideally modeled, and the labs...
## Is It True That j is Equal to the Square Root of -1 ?
September 16, 20136 comments
A few days ago, on the YouTube.com web site, I watched an interesting video concerning complex numbers and the j operator. The video's author claimed that the statement "j is equal to the square root of negative one" is incorrect. What he said was:
He justified his claim by going through the following exercise, starting with:
Based on the algebraic identity:
the author rewrites Eq. (1) as:
If we assume
Eq. (3) can be rewritten...
## Python scipy.signal IIR Filtering: An Example
May 19, 2013
Introduction
In the last posts I reviewed how to use the Python scipy.signal package to design digital infinite impulse response (IIR) filters, specifically, using the iirdesign function (IIR design I and IIR design II ). In this post I am going to conclude the IIR filter design review with an example.
Previous posts:
## A Quadrature Signals Tutorial: Complex, But Not Complicated
April 12, 201364 comments
Introduction Quadrature signals are based on the notion of complex numbers and perhaps no other topic causes more heartache for newcomers to DSP than these numbers and their strange terminology of j operator, complex, imaginary, real, and orthogonal. If you're a little unsure of the physical meaning of complex numbers and the j = √-1 operator, don't feel bad because you're in good company. Why even Karl Gauss, one the world's greatest mathematicians, called the j-operator the "shadow of...
## FIR sideways (interpolator polyphase decomposition)
September 12, 20129 comments
An efficient implementation of a symmetric-FIR polyphase 1:3 interpolator that doesn't follow the usual tapped delay line-paradigm. The example exploits the impulse response symmetry and avoids four multiplications out of 10. keywords: symmetric polyphase FIR filter implementation ASIC Matlab / Octave implementation
Introduction
An interpolating FIR filter can be implemented with a single tapped delay line, possibly going forwards and backwards for a symmetric impulse response. To...
## Design of an anti-aliasing filter for a DAC
August 18, 2012
Overview
• Octaveforge / Matlab design script. Download: here
• weighted numerical optimization of Laplace-domain transfer function
• linear-phase design, optimizes vector error (magnitude and phase)
• design process calculates and corrects group delay internally
• includes sinc() response of the sample-and-hold stage in the ADC
• optionally includes multiplierless FIR filter
Problem Figure 1: Typical FIR-DAC-analog lowpass line-up
Digital-to-analog conversion connects digital...
## Understanding the 'Phasing Method' of Single Sideband Demodulation
August 8, 201230 comments
There are four ways to demodulate a transmitted single sideband (SSB) signal. Those four methods are:
• synchronous detection,
• phasing method,
• Weaver method, and
• filtering method.
Here we review synchronous detection in preparation for explaining, in detail, how the phasing method works. This blog contains lots of preliminary information, so if you're already familiar with SSB signals you might want to scroll down to the 'SSB DEMODULATION BY SYNCHRONOUS DETECTION'...
## Least-squares magic bullets? The Moore-Penrose Pseudoinverse
October 24, 20109 comments
Hello,
the topic of this brief article is a tool that can be applied to a variety of problems: The Moore-Penrose Pseudoinverse.While maybe not exactly a magic bullet, it gives us least-squares optimal solutions, and that is under many circumstances the best we can reasonably expect.
I'll demonstrate its use on a short example. More details can be found for example on Wikipedia, or the Matlab documentation...
## Filter a Rectangular Pulse with no Ringing
May 12, 201610 comments
To filter a rectangular pulse without any ringing, there is only one requirement on the filter coefficients: they must all be positive. However, if we want the leading and trailing edge of the pulse to be symmetrical, then the coefficients must be symmetrical. What we are describing is basically a window function.
Consider a rectangular pulse 32 samples long with fs = 1 kHz. Here is the Matlab code to generate the pulse:
N= 64; fs= 1000; % Hz sample...
## Algebra's Laws of Powers and Roots: Handle With Care
September 25, 202318 comments
Recently, for entertainment, I tried to solve a puzzling algebra problem featured on YouTube [1]. In due course I learned that algebra’s $$(a^x)^y=a^{xy}\qquad\qquad\qquad\qquad\qquad(1)$$
Law of Powers identity is not always valid (not always true) if variable a is real and exponents x and y are complex-valued.
The fact that Eq. (1) can’t reliably be used with complex x and y exponents surprised me. And then I thought, “Humm, …what other of algebra’s identities may also...
## There's No End to It -- Matlab Code Plots Frequency Response above the Unit Circle
October 23, 20179 comments
Reference [1] has some 3D plots of frequency response magnitude above the unit circle in the Z-plane. I liked them enough that I wrote a Matlab function to plot the response of any digital filter this way. I’m not sure how useful these plots are, but they’re fun to look at. The Matlab code is listed in the Appendix.
This post is available in PDF format for easy...
## Compute Images/Aliases of CIC Interpolators/Decimators
November 1, 20202 comments
Cascade-Integrator-Comb (CIC) filters are efficient fixed-point interpolators or decimators. For these filters, all coefficients are equal to 1, and there are no multipliers. They are typically used when a large change in sample rate is needed. This article provides two very simple Matlab functions that can be used to compute the spectral images of CIC interpolators and the aliases of CIC decimators.
1. CIC Interpolators
Figure 1 shows three interpolate-by-M...
## Approximating the area of a chirp by fitting a polynomial
November 15, 20158 comments
Once in a while we need to estimate the area of a dataset in which we are interested. This area could give us, for example, force (mass vs acceleration) or electric power (electric current vs charge).
## Polar Coding Notes: Channel Combining and Channel Splitting
October 19, 2018
Channel Combining
Channel combining is a step that combines copies of a given B-DMC $W$ in a recursive manner to produce a vector channel $W_N : {\cal X}^N \to {\cal Y}^N$, where $N$ can be any power of two, $N=2^n, n\le0^{[1]}$.
The notation $u_1^N$ as shorthand for denoting a row vector $(u_1, \dots , u_N)$.
The vector channel $W_N$ is the virtual channel between the input sequence $u_1^N$ to a linear encoder and the output sequence $y^N_1$ of $N$...
## Simulink-Simulation of SSB demodulation
June 13, 20211 comment
≥≥≥ Simulink-Simulation of SSB demodulation or modulation from the article “Understanding the ‘Phasing Method’ of Single Sideband Demodulation” by Richard Lyons Josef Hoffmann
The article “Understanding the ‘Phasing Method’ of Single Sideband Demodulation” by Richard Lyons is a very good description of this topic. The block representation from the figures are clear and easy to understand. They are predestined for a simulation in Simulink. The simulation can help...
## The Zeroing Sine Family of Window Functions
August 16, 20202 comments
Introduction
This is an article to hopefully give a better understanding of the Discrete Fourier Transform (DFT) by introducing a class of well behaved window functions that the author believes to be previously unrecognized. The definition and some characteristics are displayed. The heavy math will come in later articles. This is an introduction to the family, and a very special member of it.
This is one of my longer articles. The bulk of the material is in the front half. The...
## DFT Bin Value Formulas for Pure Complex Tones
March 17, 2017
Introduction
This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by deriving an analytical formula for the DFT of pure complex tones and an alternative variation. It is basically a parallel treatment to the real case given in DFT Bin Value Formulas for Pure Real Tones. In order to understand how a multiple tone signal acts in a DFT it is necessary to first understand how a single pure tone acts. Since a DFT is a linear transform, the...
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0
# What is 121 over 99 in simplest form?
Updated: 9/19/2023
Wiki User
13y ago
Factor the numerator and denominator and cancel out like factors:
121/99 = (11*11)/(9*11) = 11/9
Wiki User
13y ago
Wiki User
12y ago
9/11
Anonymous
Lvl 1
3y ago
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9/11
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99/121 = 9/11
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### What is 99 over 32 in simplest form?
99/32 is in its simplest form.
### What is the simplest form for 99 over 125?
99/125 is the simplest form.
### What is 14 over 99 in its simplest form?
14/99 is in its simplest form.
### What is 89 over 99 in simplest form?
89/99 is in its simplest form.
### What is 53 over 99 in simplest form?
53/99 is in its simplest form.
### Is 99 over 100 in simplest form?
99/100 is in its simplest form
### What is 85 over 99 in simplest form?
85/99 is the simplest form as a fraction. In decimal form it is 0.858685/99 is the simplest form.
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Roman Numerals
# Roman Numerals 1548 - 1554
## The numbers 1548 to 1554 in roman numerals
The right column shows how each roman numeral adds up to the total.
1548 = MDXLVIII = 1000 + 500 + 50 − 10 + 5 + 1 + 1 + 1 1549 = MDXLIX = 1000 + 500 + 50 − 10 + 10 − 1 1550 = MDL = 1000 + 500 + 50 1551 = MDLI = 1000 + 500 + 50 + 1 1552 = MDLII = 1000 + 500 + 50 + 1 + 1 1553 = MDLIII = 1000 + 500 + 50 + 1 + 1 + 1 1554 = MDLIV = 1000 + 500 + 50 + 5 − 1
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### Problem 1 : Matched Brackets
We consider sequences of opening and closing brackets with two types of brackets, () and []. A bracket sequence is well-bracketed if we can pair up each opening bracket with a matching closing bracket in the usual sense. For instance, the sequences (), [] ([]) and []([]) are well-bracketed, while (, ()], (], )( and [(]) are not well-bracketed. In the last case, each opening bracket has a matching closing bracket and vice versa, but the intervals spanned by the different types of brackets intersect each other instead of being contained one within the other.
The alternating depth of a well-bracketed sequence tells us the maximum number of times we switch between the two types of brackets when we have inner matched brackets enclosed within outer matched brackets. For instance, the alternating depth of (), [[[]]] and ()[][] is 1, the alternating depth of [()] and ()([]) is 2, the alternating depth of ([()]) and [()][(([]))] is 3, and so on.
Given a well-bracketed sequence, we are interested in computing three quantities.
• The alternating depth of the sequence.
• The maximum number of symbols between any pair of matched brackets of the type ( and ), including both the outer brackets.
• The maximum number of symbols between any pair of matched brackets of the type [ and ], including both the outer brackets.
For instance, the alternating depth of (([]))[[[()]]] is 2, the maximum number of symbols between a matched pair () is 6 and the maximum number of symbols between a matched pair [] is 8.
### Input format
The input consists of two lines. The first line is a single integer N, the length of the bracket sequence. Positions in the sequence are numbered 1,2,…,N. The second line is a sequence of N space-separated integers that encode the bracket expression as follows: 1 denotes an opening bracket (, 2 denotes a closing bracket ), 3 denotes an opening bracket [ and 4 denotes a closing bracket ]. Nothing other than 1, 2, 3 or 4 appears in the second line of input and the corresponding expression is guaranteed to be well-bracketed.
### Output format
Your program should print 3 space-separated integers in a line, denoting the three quantities asked for in the following order: alternating depth, length of the maximum sequence between matching () brackets and length of the maximum sequence between matching [] brackets.
### Testdata
You may assume that 2 ≤ N ≤ 105. In 30% of the test cases, 2 ≤ N ≤ 103.
### Sample Input
```14
1 1 3 4 2 2 3 3 3 1 2 4 4 4
```
```2 6 8
```
### Time and memory limits
The time limit for this task is 1 second. The memory limit is 32MB.
Copyright (c) IARCS 2003-2022; Last Updated: 23 Sep 2012
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https://brilliant.org/problems/everything-in-the-us-has-to-be-supersized/
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Everything In The US Has To Be Supersized
Calculus Level 4
During Expo '74 in Spokane, Washington, an IMAX screen that measured 27 meters by 20 meters was featured in the US Pavilion. For those that stood in front of it, it appears that the total vision field was completely filled. This created a sensation of motion in most viewers, which resulted in motion sickness in some people.
Suppose that the floor of the US pavilion is perfectly flat, and that the bottom of the screen is located 5 meters above of Susie's eye level. The viewing angle is defined to be the angle between the two lines which connect the top and the bottom of the screen to Susie's eyes.
To one decimal place, what is the measure (in degrees) of the largest viewing angle that Susie can have as she walks around the US pavilion?
×
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http://www.governmentadda.com/ibps-rrb-syllabus-for-po-officer-scale-1/
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IBPS RRB Complete Syllabus for PO Officer Scale 1 Prelims | Mains Exam 2018 : Read Now | | GovernmentAdda
Saturday , September 22 2018
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Home / Exam Syllabus / IBPS RRB Complete Syllabus for PO Officer Scale 1 Prelims | Mains Exam 2018 : Read Now
# IBPS RRB Complete Syllabus for PO Officer Scale 1 Prelims | Mains Exam 2018 : Read Now
SSC CGL Study Material Book Free PDF >>
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## IBPS RRB PO (Prelims & Mains) Syllabus 2018
Hello Aspirants,
We are providing the Complete Syllabus For IBPS RRB Po Officer 2018 exam here for the candidates who are preparing for IBPS RRB PO 2018 examination. Candidates are advised to study all topics of each section thoroughly.The questions will be asked in IBPS RRB Exam will be based on latest IBPS RRB Exam Syllabus – Officer Scale I (Prelims & Mains). So here we are providing you the Detailed Syllabus for IBPS RRB PO Scale 1 Officer 2018 Exam.
We are providing you full syllabus for IBPS RRB Po 2018. Read all section wise syllabus from below.
## IBPS RRB PO Syllabus for Data Analysis & Interpretation & Quantitative Aptitude (Prelims & Mains)
• Number Series.
• Missing DI sets
• Double Diagram DI
• Caselet DI
• Data Sufficiency
• Mathematical inequality
• Simplification / Approximation.
• Data Interpretation.
• Average.
• Profit & Loss.
• Simple and Compound Interest.
• Time & Work.
• Time & Distance.
• Partnership.
• Ratio & Proportion.
• Problem on Ages.
• Probability.
• Mensuration.
• Mixture & Allegation.
• Percentages.
• Pipes & Cisterns.
• Averages.
## IBPS RRB PO Syllabus for English Language (Mains)
• Cloze Test.
• Spotting Errors.
• Sentence Improvement/ Correction.
• Para Jumbles.
• Sentence/ Para Completion.
• Sentence Connectors.
• Fill in the blanks.
• Vocabulary based questions (Synonym-Antonyms, Misspelt words, Idioms & Phrases etc).
## IBPS RRB PO Syllabus for Reasoning Ability & Computer Aptitude (Prelims & Mains)
• Seating Arrangement.
• Puzzles.
• Inequality.
• Syllogism.
• Input-Output.
• Data Sufficiency.
• Logical Reasoning.
• Blood Relations.
• Distance & Direction.
• Series based questions (Alphanumeric/ Numeric/ Alphabet).
• Alphabet Test.
• Order and Ranking
• 1) Number System
• 2) History Of Computers
• 3) Hardware
• 4) Software
• 5) Database ( introduction)
• 6) Communication (Basic Introduction)
• 7) Networking( LAN, WAN,)
• 8) Internet (Concept, History, working environment, Application)
• 9) Security Tools, Virus, Hacker
• 10) MS Windows, MS Office
• 11) Logic Gates
## IBPS RRB PO Syllabus for General/ Banking/Economy/Financial Awareness (Mains)
• Current Affairs of last 4-5 months.
• Banking Awareness (Static as well as Current Banking Awareness questions)
• Static General Awareness
• History of Indian banking system
• Overview of Indian Financial System
• Recent credit and monetary policies
• Abbreviations and Economic terminologies & Banking Terms
• Important Government Schemes, Important Summits, Recent developments in banking sector, Important banking terms etc.
• Introduction to National financial institutions like RBI, SEBI, IRDA, FSDC etc and of International organizations like IMF, World Bank, ADB, UN etc.
• Important Dates and Dates.
• Countries and their Capital & Currencies.
• Important National & International Organizations and their Headquarters.
• National Parks and wildlife sanctuaries.
## IBPS RRB PO Syllabus for Hindi Language (Mains)
• अपठित गद्यांश
• वाक्यक्रम व्यवस्थापन
• वाक्यों में त्रुटियाँ
• वाक्यों में रिक्त-स्थानों की पूर्ति
• वर्तनी अशुद्धियाँ
• विपरीतार्थक शब्द
• पर्यायवाची शब्द
• मुहावरे तथा लोकोक्तियां
• गुढार्थी (क्लोज टाइप) प्रकार का गद्यांश
• वाक्यांश के लिए एक शब्द
#### Preparation
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https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_7&oldid=114278
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2013 AMC 10B Problems/Problem 7
Problem
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$
Solution 1
$[asy] unitsize(72); draw((0,0)--(1/2,sqrt(3)/2)); draw((1/2,sqrt(3)/2)--(3/2,sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(2,0)); draw((2,0)--(3/2,-sqrt(3)/2)); draw((3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((1/2,-sqrt(3)/2)--(0,0)); draw((3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2)); label("2",(3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2),NE); [/asy]$
If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$, so the side opposite the thirty degree angle in the triangle is also $1$. From the properties of $30-60-90$ triangles, the area is $1\cdot\sqrt3/2$=$\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$
Solution 2—Similar to Solution 1
As every point on the circle is evenly spaced, the length of each arc is $\frac{\pi}{3}$, because the circumference is $2\pi$. Once we draw the triangle (as is explained in solution 1), we see that one angle in the triangle subtends one such arc. Thus, the measure of that angle is thirty degrees. Similarly, another angle in the triangle subtends an arc of twice the length, and thus equals 60 degrees. The last angle is equal to 90 degrees and the triangle is a $30-60-90$ triangle. We know that as the diameter, the length of the hypotenuse is 2, and thus, the other sides are 1 and $\sqrt{3}$. We then find the area to be $\boxed{\textbf{(B) } \frac{\sqrt{3}}{2} }$.
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http://www.purplemath.com/learning/viewtopic.php?p=4342
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## Writing number in index form with a prime number base?
Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Trillian
Posts: 2
Joined: Tue Jul 20, 2010 12:01 pm
Contact:
### Writing number in index form with a prime number base?
Hi, I desperately need help in knowing how to solve a problem that requires a number to be written in index form with a prime number base. I can see the answers in the back of the text, but I have no idea how they get those answers. This is for both whole numbers and fractions.
An example of the questions are;
Write each number in index form with a prime number base
a) 8 b) 64 c) 81 d) 1/13 e) 1/49 f) 1/13^3
Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:
### Re: Writing number in index form with a prime number base?
Hi, I desperately need help in knowing how to solve a problem that requires a number to be written in index form with a prime number base. I can see the answers in the back of the text, but I have no idea how they get those answers. This is for both whole numbers and fractions.
An example of the questions are;
Write each number in index form with a prime number base
a) 8 b) 64 c) 81 d) 1/13 e) 1/49 f) 1/13^3
do you mean....
$2^3$
$2^6$
$3^4$
$13^{-1}$
$7^{-2}$
$13^{-3}$
Trillian
Posts: 2
Joined: Tue Jul 20, 2010 12:01 pm
Contact:
### Re: Writing number in index form with a prime number base?
Yes! Those are the answers but I don't know how they get those answers - I need to know how to work it out for myself. Thanks for replying
Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:
### Re: Writing number in index form with a prime number base?
Yes! Those are the answers but I don't know how they get those answers - I need to know how to work it out for myself. Thanks for replying
$8=2\times4=2\times2\times2=2^3$
stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
I don't know how they get those answers - I need to know how to work it out for myself.
To learn how to factor numbers, try this lesson.
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-8-section-8-1-the-square-root-property-and-completing-the-square-8-1-exercises-page-512/79
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## Intermediate Algebra (12th Edition)
$x=\left\{ \dfrac{1-2i\sqrt{2}}{6},\dfrac{1+2i\sqrt{2}}{6} \right\}$
$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $(6x-1)^2=-8 ,$ take the square root of both sides (Square Root Property). Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} 6x-1=\pm\sqrt{-8} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to\begin{array}{l}\require{cancel} 6x-1=\pm\sqrt{-1}\cdot\sqrt{8} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} 6x-1=\pm i\sqrt{8} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} 6x-1=\pm i\sqrt{4\cdot2} \\\\ 6x-1=\pm i\sqrt{(2)^2\cdot2} \\\\ 6x-1=\pm i(2)\sqrt{2} \\\\ 6x-1=\pm 2i\sqrt{2} .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 6x=1\pm 2i\sqrt{2} \\\\ x=\dfrac{1\pm 2i\sqrt{2}}{6} .\end{array} Hence, $x=\left\{ \dfrac{1-2i\sqrt{2}}{6},\dfrac{1+2i\sqrt{2}}{6} \right\} .$
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https://www.oddscoach.com/arbitrage-calculator/
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# Arbitrage Calculator
The arbitrage calculator gives you betting strategies for series of bets where the odds have slipped in the bettors favor. Arbitrage in sports betting is a series of bets where the sportsbooks edge is below zero meaning that regardless of outcome, the bettor can break even or make a profit at no risk. Arbitrage opportunities don’t happen every day so be sure to calculate out your best outcome and make sure you aren’t going to be the one losing when you don’t have to.
Selection
Include
Bias
Highest Odds
Selection 1
Selection 2
Selection 3
Selection 4
Selection 5
### Betting Outcomes
#### Strategies (Bet on Selection 1)
ProfitsArbitrage
NoneNo BiasBiased
Selection 1
Wager
Profit
Selection 2
Wager
Profit
Selection 3
Wager
Profit
Selection 4
Wager
Profit
Selection 5
Wager
Profit
Usually, you find arbitrage opportunities by maintaining betting accounts at multiple sportsbooks and as the odds shift you get odds on one site that haven’t moved quickly enough. You place a bet on both sides but across two sites to guarantee profit. Sometimes, as in futures betting, you spread the risk across a number picks and sportsbooks.
The arbitrage calculator is going to give you a couple of pieces of information. First of all, you are going to need to enter the american odds offered for a single betting event. You will need at least two but ca enter a number of outcomes (i.e. team 1 win, team 2 win, tie). Also you are going to have to choose a favorite. This is the pick that you would bet on if you had to choose one.
Once you have entered your odds and favorite hit calculate and get some important information: the fact that an arbitration opportunity exists. If your edge over the sportsbook isn’t in the positive numbers, then no amount of calculating is going to give you a sure thin. Providing the arbitrage calculator tells that you you have an opportunity, the chart below will show you your options. Mostly for illustration the first column just shows you the outcome if you placed your bet your favorite. The second column gives you the options if you spread the risk between each of the betting options. This gives you the best chance for an outcome from each of the outcomes. The final column shows the best outcome if your favorite is right, but limits you to break-even if the other options occur.
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https://math.stackexchange.com/questions/2996840/mapping-of-a-complex-function-from-z-plane-to-w-plane
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# Mapping of a complex function from z-plane to w-plane
I have been tasked to find the image of a set $$U=\{z \in \mathbb{C} \mid \frac{-\pi}{2} \lt \Re(z) \lt \frac{\pi}{2} \}$$ under the function $$f(z)=\sin(z)$$ which I have been asked to do so answering a series of sub-questions that go as follows:
(a): What is the image of the line segment $$L_1=(\frac{-\pi}{2}, \frac{\pi}{2})$$ (the real axis) under $$f$$?
(b): What is the image of the Imaginary Axis $$L_2=\{iy \mid y \in \mathbb{R} \}$$ under $$f$$?
(c): What is the image of the vertical line $$L_3=\{\frac{-\pi}{2}+ iy \mid y \in \mathbb{R}\}$$ under $$f$$?
(d): What is the image of the vertical line $$L_3=\{\frac{\pi}{2}+iy \mid y \in \mathbb{R} \}$$ under $$f$$?
(e): Given your observations in the previous steps what do you guess the image of the set $$U$$ is under $$f$$?
I am done figuring out the image of the invidual parts from (a)-(d) but ultimately cannot combine the results to judge the image of the set $$U$$ under $$f$$.
For parts (a)-(d), I got the following results:
(a): $$(-1, 1)$$
(b): $$\{\iota \sinh(y) \mid y \in \mathbb{R}\}$$
(c): $$\{-\cosh(y) \mid y \in \mathbb{R} \}$$
(d): $$\{\cosh(y) \mid y \in \mathbb{R}\}$$
How do I combine all the information obtained above into finding the image of the set $$U$$ under $$f$$. I also replaced $$z$$ with $$x+\iota y$$ and substituted it in the definition formula of the complex sine function and the addition formula and obtained the equivalent of $$\sin(x+\iota y)$$ as $$\sin(x)\cosh(y)+\iota \cos(x)\sinh(y)$$. How can I restrict the $$x$$ i.e. the $$\Re(z)$$ to being in the set $$U$$ so that I can plug in that information in the obtained equivalent of the complex sine function to obtain something meaningful?
Thanks
• In particular, what's happening here is that the image $f(U)$ is the entire complex plane, except for "most of" the real axis; e.g. $\Bbb C-\{z\in\Bbb C : |\Re(z)|\geq 1\}$. [The part of the real axis that is included is the part calculated in (a), and parts (c) and (d) calculate the parts excluded.] – aleph_two Dec 19 '18 at 4:27
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Welcome! On this website you can find information about each number.
# Number 240403678
## Number 240403678 basic info
Number 240403678 has 9 digits. Number 240403678 can be formatted as 240,403,678 or 240.403.678 or 240 403 678 or in case this was a phone number 240-403-678 or 24-040-3678 to be easier to read. Number 240403678 in English words is "two hundred and fourty million, four hundred and three thousand, six hundred and seventy-eight". Number 240403678 can be read by triplets (groups of 3 digits) as "two hundred and fourty, four hundred and three, six hundred and seventy-eight". Number 240403678 can be read digit by digit as "two four zero four zero three six seven eight". Number 240403678 is even. Number 240403678 is divisible by: two. Number 240403678 is a composite number (non-prime number).
## Number 240403678 conversions
Number 240403678 in binary code is 1110010101000100010011011110. Number 240403678 in octal code is: 1625042336. Number 240403678 in hexadecimal (hexa): e5444de.
The sum of all digits of this number is 34. The digital root (repeated digital sum until you get single-digit number) is 7. Number 240403678 divided by two (halved) equals 120201839. Number 240403678 multiplied by two (doubled) equals 480807356. Number 240403678 multiplied by ten equals 2404036780. Number 240403678 raised to the power of 2 equals 5.7793928395928E+16. Number 240403678 raised to the power of 3 equals 1.389387295245E+25. The square root (sqrt) of 240403678 is 15504.956562338. The sine (sin) of 240403678 degree is -0.034899497361311. The cosine (cos) of 240403678 degree is 0.99939082699609. The base-10 logarithm of 240403678 equals 8.3809411077687. The natural logarithm of 240403678 equals 19.297830060009. The number 240403678 can be encoded to characters as BDJDJCFGH. The number 240403678 can be encrypted to chemical element names as helium, beryllium, neon, beryllium, neon, lithium, carbon, nitrogen, oxygen.
## Numbers simmilar to 240403678
Numbers simmilar to number 240403678 (one digit altered): 140403678340403678230403678250403678241403678240303678240503678240413678240402678240404678240403578240403778240403668240403688240403677240403679
Possible variations of 240403678 with a digit pair swapped: 420403678204403678244003678240043678240430678240406378240403768240403687
Number 240403678 typographic errors with one digit missing: 404036782040367824403678240036782404367824040678240403782404036824040367
Number 240403678 typographic errors with one digit doubled: 224040367824404036782400403678240440367824040036782404033678240403667824040367782404036788
Previous number: 240403677
Next number: 240403679
## Several randomly selected numbers:
5538635242931543043889912818794874498902771624582505137452790854150872514592034922466958752398039984451593841198588330544660843255602885843970591167893900186308531306595198478059525387595133922038948099245365486396688322955767789894881179268762796174543741265056354590533620205242096528872379866004234745907388736608649923712451664317216174949791294989754571355368130715151757324114793333277.
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Index: Xⁿ starts with n | Scoins.net | DJS
## Index: Xⁿ starts with n
Often said in my lessons are that 2^6 is 64 and 2^10 is ten twenty four, 1024, and that these are the only 2ⁿ that start with n. [Prove!] This page explores the relationship between integers to the power of another integer, looking especially at those results which, like 2⁶ = 64, begin with the same digits as the index used.
For example, 42⁵⁹ is 5.9x10⁹⁵ or 42⁵⁹ = 59x10⁹⁴, in a non-standard format.
Sometimes we will need general terms to describe this computation; I choose to refer, should I need to and using the example 42⁵⁹, such that 42 is the base and 59 is the index.
1. Can you point straightaway to some powers of 20 that will fit this property of 20ⁿ starting with n? Demonstrate that you're right and make a statement about other cases where the base number is a multiple of ten.
2. Thinking of easy squares —those you know well— spot the square that starts with a two. Write this down and also the similar result for ten times as much, which demonstrates the property you wrote down in Q1.
3. Using your calculator, show that there is a run of (two-digit) squares as the base number rises by one that also start with the digit two.
4. 2^10 is given at the top and is generally assumed knowledge (by me in my classes), so you might easily recognise that 8⁴ (= 2¹², = 4x1024) starts with a 4. More than this, you should now recognise that 80⁴ starts with a 4 and, quite possibly, 81⁴ does. Show that you can extend this towards 85 and show that 84 works but 85 does not.
5. Quite soon your calculator complains. By writing the values down in standard form, show that these calculations fit the requirement:
31⁸, 58⁵⁶, 87⁵⁴, 99⁵⁶.
For some calculators, those last two will prove too hard. Explain why this is, and use what you learned in Q1 and Q2 to avoid the problem and still find the answer. Whatever solution you found, show it. One is enough.
6. Now you see a way to recognise whether you've found combinations that work, find which of these fit the property of having the first few digits (or non-zero digits if you're using numbers less than unity) matching the index: 86⁶⁴, 88⁵⁰, 470³², 850⁴, 840¹², 81⁸⁸, 82¹¹, 99855⁸⁸
The result you discovered in Q4 suggests that if a single digit number like 8 raised to a power like 4 fits, then it is likely that 8.1 and 8.2 will fit until some limit is reached.
7. Find the simple 0.7ⁿ such the first non-zero digit is n and find the biggest number you can, quite close to 0.7, that still starts with that n.
8. 0.55⁵=0.050, so 0.551⁵ begins 0.5. The first number bigger than 0.55, X, that fails to fit the pattern here (numbers close to 0.55 that fit the property) is such that X⁵=0.06. [Obviously; if it doesn't start with 5 and it is a bit bigger, it must start with 6.] Find this X and consequently show the biggest integer close to 550 such that its fifth power will begin with the digit 5. Does 549 work?
Around about now you begin to see that using numbers between zero and one is going to be an awful lot easier. So we might change 'fit the pattern' to 'the truncated decimal matches the index' and call our effect something like a 'truncated index match'. In class, we might call that a 'tim', especially if there was a Tim present.
9. The fifth power produces more numbers of this sort. 0.9⁵= 0.59049. By looking at the fifth root of 0.5 and of 0.6, find the range of integers close to 900 (above and below) whose fifth power starts with a 5.
Looking for more numbers where X⁵ begins with a 5 but not a 6, Q8&Q9 suggest that looking at the fifth root of 0.5, 0.05, 0.005 etc and 0.6, 0.06, 0.006, etc will reveal a range of three digit numbers that fit the pattern.
10. By writing the fifth root of 5x10⁻ⁿ and 6x10⁻ⁿ for various n, find the five groups of numbers below 10000 which will fit the pattern.
You now have a techniques for discovering a lot of numbers that fit this pattern. I decided to be interested in the fifteenth power, which produced a table like this on Excel.
The bottom row, which I'll call row 15, suggests to me that, of numbers below 1000, 103¹⁵=0.15.....
11. Use the table given ( or make your own) to write down which integers¹⁵ below 1000 will begin 0.15.
Here are the formulas I put into EXCEL, where I could have Named cell R4 as 'index', in which case it would look a lot tidier, like this:-
I've decided to be interested in the index being 37, as you can see.
12. Produce your own spreadsheet (this is now well into GCSE-rated project-work) and investigate.
I say the digit integers such that X³⁷ begins 37... when truncated (i.e. fitting the 'Xⁿ starts with an n' pattern we've been investigating) include 63 and 86, but also 338, 383 and nine more three digit integers.
The same spreadsheet showed that 0.9856^137 =0.137(0866), 0.9692^137 = 0.0137(057), 0.9063^137= 137.94x10⁻⁴ and so on. 766^137 works, but I think it is the only three digit solution.
I found it quite difficult to persuade my spreadsheet to calculate or show which integer solutions would work without it being me doing the decision-making. Maybe I was being thick, but at the time I found this hard.
Few students will complete all twelve questions. This might make up a week of lessons for some classes, a independent piece of coursework, or just a weekend homework.
This page as a whole could be seen as chasing numbers around on a calculator. It could also be seen as having a little problem that deserves just enough attention to cause a class collectively or individually to hunt for answers. It may serve to educate a class into the use of the xʸ button and its inverse, ˣ√y. Not least, it shows that while there may be 'answers' there is also a region of grey not-quite answers in the immediate neighbourhood. The more maths one does the less the subject is black and white and it becomes seen as more filled with many shades of grey. I would suggest that a whole-class exercise gives the feature to be pursued its own name (Tim, perhaps)— as I've written before, possession is nine-tenths of the learning.
If we did some maths first (rather than play with arithmetic) then all the exploration turns into preparation for understanding what comes next - the maths. For an index of 37,
we want 0.37≤X³⁷<0.38, so ³⁷√0.37≤X< ³⁷√0.38,
which in turn means 0.973486≤X<0.974188 so that the integer 'solutions' are 974, 9735-41, 97349-97418 and so on. That approach (eventually) produced this next spreadsheet, where the index is 5 and the 3 indicates that solutions will be recorded from 1 to 999. The bold 5 is cell AC1 and is now called Nind, so the leftmost column is AB. I have a figure 3 in cell AE1 which is the number of digits (I thought of it as 'size') I want to have shown before a decimal point, as the next example shows. Beware filling downwards, for I've used pairs of lines (columns AC and AF) to achieve my objectives.
That same spreadsheet now shows all of the results for X⁵ = 5....., where X<1000. The top two 871 and 902 (on the right in Col AF) indicate that all the integers from 871 to 902 inclusive succeed in beginning with a five. While that 871-902 range implies 90 and 9 will also fit (because 900 does), filling the table downwards produces these answers too.
I swap the top left 5 to a 37 and instantly have the results for that different problem in descending order: 974, 915, 860, 808, 759, 630, 592, 491, 383, 338 and presumably 86 and 63. And no others. 1036 appeared in my table and works. I subsequently found some larger numbers where it is more difficult to show the valid integer range, such as where X³⁷ = 3.7x10⁵, for which 1415 works
I switch the index to 137 and only 766 shows. I change the size setting from 3 to 4 and get this quite long list:- 9856, 9692, 9062, 8616, 8056, 7660, 7162, 6925, 6475, 6367, 6054, 5953, 5566, 5473, 5382, 5204, 5032, 4948, 4399, 3977, 2560, 2402, 2362, 1805, 1166 and of course 766.
DJS 20210531
1. Surely 20^6 = (2x10)⁶ = 2⁶x10⁶. Similarly 20^6 = 1024x10⁶ = 10.24x10¹¹
5. 87⁵⁴ = 8.7⁵⁴x 10⁵⁴ = 5.4x10⁵⁰x10⁵⁴ = 5.4x10¹⁰⁴. Extending this 0.87⁵⁴ = 0.00054(2050) =5.4x10⁻⁴ so 87⁵⁴ = 5.4x10⁻⁴x10¹⁰⁸ = 5.4x10¹⁰⁴. Some will go to a bigger calculator, such as within Excel. That gets the answer but doesn't understand why the calculator won't do it.
6. 8.4⁴ = 4978 so 84 and 840 both fit the property but 8.5 = 5220, so doesn't. Therefore 850⁴ doesn't either. You might prefer to work with 0.84⁴=0.4(9787) and 0.85⁴=0.5(2200625).
7. 0.7³ = 0.343, 0.73³=0.389, 0.736806300 is just too big. Think about how to persuade the calculator to tell you that. This isn't about chasing answers, it is about understanding how numbers fit together.
8. ⁵√0.06 = 0.569679052 so 569 works but not 570. Also ⁵√0.05 = 0.54928 so 55 and 56 both work, making the whole range here the integers from 550 to 569.
9. ⁵√0.6 = 0.90288 so 901 and 902 will work but 903 will not. ⁵√0.5 = 0.87055 so 870 won't work but 871 will, and many bigger integers. So the range of three digit numbers here that work is 871 to 902. We could write this as 871 ≤ X ≤ 902.
10. ⁵√0.6 = 0.90288 and ⁵√0.5 = 0.87055 so 871 to 902 from Q9
⁵√0.06 = 0.5696 and ⁵√0.05 = 0.5492. So 550 to 569 (not 549, not 570) from Q8
⁵√0.006 = 0.3594, ⁵√0.005 = 0.34657. So 35⁵ works, and 3466⁵ to 3593⁵ will work
⁵√0.0006 = 0.22679, ⁵√0.0005 = 0.21867 ; 22⁵ works and 2187 to 2267 will work
⁵√0.00006 = 0.143096; ⁵√0.00005 = 0.13797; 14⁵ works and 1380 to 1430 will work.
Inserting yet another zero causes the pattern to repeat. Some will say "Of course it does" and others will need to hunt for why this is true. I suggest you write some of the numbers in standard form.
11. Working from the bottom row upwards gives the values in number order; 103, 120, 140, 163, 190, 222, 259, 301-2, 351-2, 410, 477-8, 556-8, 649-51, 756-9, 882-4 and, removing zeroes, 12,14 and 19 also work..
In November 2021 I threw this at a Y13 already known to be a good mathematician (and Oxbridge aspirant) and he found this hard. But then he had only had 20 minutes of horrible, off-the-wall me and not the backgrouind that any PMC Y8 top set student would have become entirely familiar with. So perhaps problems like this are harder than I imagine them to be. Alternatively maybe a whole classful of bright Y8 are actually sharing enough working brain cells (called teamwork if I'm polite) to produce some relatively brilliant results?
Covid Email: David@Scoins.net © David Scoins 2021
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > clwwlkn2 Structured version Visualization version GIF version
Theorem clwwlkn2 26303
Description: In an undirected simple graph, a closed walk of length 2 represented as word is a word consisting of 2 symbols representing vertices connected by an edge. (Contributed by Alexander van der Vekens, 19-Sep-2018.)
Assertion
Ref Expression
clwwlkn2 (𝑉 USGrph 𝐸 → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ ((#‘𝑊) = 2 ∧ 𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
Proof of Theorem clwwlkn2
Dummy variable 𝑖 is distinct from all other variables.
StepHypRef Expression
1 usgrav 25867 . . 3 (𝑉 USGrph 𝐸 → (𝑉 ∈ V ∧ 𝐸 ∈ V))
2 2nn0 11186 . . . . 5 2 ∈ ℕ0
3 isclwwlkn 26297 . . . . 5 ((𝑉 ∈ V ∧ 𝐸 ∈ V ∧ 2 ∈ ℕ0) → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ (𝑊 ∈ (𝑉 ClWWalks 𝐸) ∧ (#‘𝑊) = 2)))
42, 3mp3an3 1405 . . . 4 ((𝑉 ∈ V ∧ 𝐸 ∈ V) → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ (𝑊 ∈ (𝑉 ClWWalks 𝐸) ∧ (#‘𝑊) = 2)))
5 isclwwlk 26296 . . . . 5 ((𝑉 ∈ V ∧ 𝐸 ∈ V) → (𝑊 ∈ (𝑉 ClWWalks 𝐸) ↔ (𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸)))
65anbi1d 737 . . . 4 ((𝑉 ∈ V ∧ 𝐸 ∈ V) → ((𝑊 ∈ (𝑉 ClWWalks 𝐸) ∧ (#‘𝑊) = 2) ↔ ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2)))
74, 6bitrd 267 . . 3 ((𝑉 ∈ V ∧ 𝐸 ∈ V) → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2)))
81, 7syl 17 . 2 (𝑉 USGrph 𝐸 → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2)))
9 3anass 1035 . . . . 5 ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ↔ (𝑊 ∈ Word 𝑉 ∧ (∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸)))
10 oveq1 6556 . . . . . . . . . . . 12 ((#‘𝑊) = 2 → ((#‘𝑊) − 1) = (2 − 1))
1110oveq2d 6565 . . . . . . . . . . 11 ((#‘𝑊) = 2 → (0..^((#‘𝑊) − 1)) = (0..^(2 − 1)))
1211raleqdv 3121 . . . . . . . . . 10 ((#‘𝑊) = 2 → (∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ ∀𝑖 ∈ (0..^(2 − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
1312ad2antlr 759 . . . . . . . . 9 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ ∀𝑖 ∈ (0..^(2 − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
14 2m1e1 11012 . . . . . . . . . . . . 13 (2 − 1) = 1
1514oveq2i 6560 . . . . . . . . . . . 12 (0..^(2 − 1)) = (0..^1)
16 fzo01 12417 . . . . . . . . . . . 12 (0..^1) = {0}
1715, 16eqtri 2632 . . . . . . . . . . 11 (0..^(2 − 1)) = {0}
1817a1i 11 . . . . . . . . . 10 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (0..^(2 − 1)) = {0})
1918raleqdv 3121 . . . . . . . . 9 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (∀𝑖 ∈ (0..^(2 − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ ∀𝑖 ∈ {0} {(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
20 c0ex 9913 . . . . . . . . . 10 0 ∈ V
21 fveq2 6103 . . . . . . . . . . . . 13 (𝑖 = 0 → (𝑊𝑖) = (𝑊‘0))
22 oveq1 6556 . . . . . . . . . . . . . . 15 (𝑖 = 0 → (𝑖 + 1) = (0 + 1))
23 0p1e1 11009 . . . . . . . . . . . . . . 15 (0 + 1) = 1
2422, 23syl6eq 2660 . . . . . . . . . . . . . 14 (𝑖 = 0 → (𝑖 + 1) = 1)
2524fveq2d 6107 . . . . . . . . . . . . 13 (𝑖 = 0 → (𝑊‘(𝑖 + 1)) = (𝑊‘1))
2621, 25preq12d 4220 . . . . . . . . . . . 12 (𝑖 = 0 → {(𝑊𝑖), (𝑊‘(𝑖 + 1))} = {(𝑊‘0), (𝑊‘1)})
2726eleq1d 2672 . . . . . . . . . . 11 (𝑖 = 0 → ({(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
2827ralsng 4165 . . . . . . . . . 10 (0 ∈ V → (∀𝑖 ∈ {0} {(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
2920, 28mp1i 13 . . . . . . . . 9 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (∀𝑖 ∈ {0} {(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
3013, 19, 293bitrd 293 . . . . . . . 8 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ↔ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
3130anbi1d 737 . . . . . . 7 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ((∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ↔ ({(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸)))
32 lsw 13204 . . . . . . . . . . . . . 14 (𝑊 ∈ Word 𝑉 → ( lastS ‘𝑊) = (𝑊‘((#‘𝑊) − 1)))
3332adantl 481 . . . . . . . . . . . . 13 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ( lastS ‘𝑊) = (𝑊‘((#‘𝑊) − 1)))
3410ad2antlr 759 . . . . . . . . . . . . . . 15 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ((#‘𝑊) − 1) = (2 − 1))
3534, 14syl6eq 2660 . . . . . . . . . . . . . 14 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ((#‘𝑊) − 1) = 1)
3635fveq2d 6107 . . . . . . . . . . . . 13 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (𝑊‘((#‘𝑊) − 1)) = (𝑊‘1))
3733, 36eqtr2d 2645 . . . . . . . . . . . 12 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → (𝑊‘1) = ( lastS ‘𝑊))
3837preq2d 4219 . . . . . . . . . . 11 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → {(𝑊‘0), (𝑊‘1)} = {(𝑊‘0), ( lastS ‘𝑊)})
39 prcom 4211 . . . . . . . . . . 11 {(𝑊‘0), ( lastS ‘𝑊)} = {( lastS ‘𝑊), (𝑊‘0)}
4038, 39syl6eq 2660 . . . . . . . . . 10 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → {(𝑊‘0), (𝑊‘1)} = {( lastS ‘𝑊), (𝑊‘0)})
4140eleq1d 2672 . . . . . . . . 9 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ({(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸 ↔ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸))
4241biimpd 218 . . . . . . . 8 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ({(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸 → {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸))
4342pm4.71d 664 . . . . . . 7 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ({(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸 ↔ ({(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸)))
4431, 43bitr4d 270 . . . . . 6 (((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) ∧ 𝑊 ∈ Word 𝑉) → ((∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ↔ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
4544pm5.32da 671 . . . . 5 ((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) → ((𝑊 ∈ Word 𝑉 ∧ (∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸)) ↔ (𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
469, 45syl5bb 271 . . . 4 ((𝑉 USGrph 𝐸 ∧ (#‘𝑊) = 2) → ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ↔ (𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
4746ex 449 . . 3 (𝑉 USGrph 𝐸 → ((#‘𝑊) = 2 → ((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ↔ (𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))))
4847pm5.32rd 670 . 2 (𝑉 USGrph 𝐸 → (((𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸 ∧ {( lastS ‘𝑊), (𝑊‘0)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2) ↔ ((𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2)))
49 3anass 1035 . . . 4 (((#‘𝑊) = 2 ∧ 𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸) ↔ ((#‘𝑊) = 2 ∧ (𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
50 ancom 465 . . . 4 (((#‘𝑊) = 2 ∧ (𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)) ↔ ((𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2))
5149, 50bitr2i 264 . . 3 (((𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2) ↔ ((#‘𝑊) = 2 ∧ 𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸))
5251a1i 11 . 2 (𝑉 USGrph 𝐸 → (((𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸) ∧ (#‘𝑊) = 2) ↔ ((#‘𝑊) = 2 ∧ 𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
538, 48, 523bitrd 293 1 (𝑉 USGrph 𝐸 → (𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘2) ↔ ((#‘𝑊) = 2 ∧ 𝑊 ∈ Word 𝑉 ∧ {(𝑊‘0), (𝑊‘1)} ∈ ran 𝐸)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ∀wral 2896 Vcvv 3173 {csn 4125 {cpr 4127 class class class wbr 4583 ran crn 5039 ‘cfv 5804 (class class class)co 6549 0cc0 9815 1c1 9816 + caddc 9818 − cmin 10145 2c2 10947 ℕ0cn0 11169 ..^cfzo 12334 #chash 12979 Word cword 13146 lastS clsw 13147 USGrph cusg 25859 ClWWalks cclwwlk 26276 ClWWalksN cclwwlkn 26277 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-fzo 12335 df-hash 12980 df-word 13154 df-lsw 13155 df-usgra 25862 df-clwwlk 26279 df-clwwlkn 26280 This theorem is referenced by: numclwwlkovf2 26611
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# Tagged Questions
an asymmetric (e.g. public-key) cryptosystem, based on modular exponentiation with big exponents and modulus. RSA can be used both for signature and encryption.
111 views
### RSA Signature Verification Implementation on Cortex M0
I have to implement RSA Signature Verification procedure (RSA-2048) on a Cortex M0 based MCU. My budget is 15kB Flash and 2-4kB Flash RAM. Is it even possible to do RSA on a low end MCU based on ...
62 views
### public/private key: is it possible to have master public/private key
This may be a very primitive question which might not be possible but I have this question to understand. Assume Yahoo provides a messenger where yahoo users can chat in secure way (end-end ...
120 views
### ECDH or RSA more secure for symmetric key wrapping?
Suppose a message is encrypted with a symmetric block cipher with a random key. RSA is often used to wrap the symmetric key using the recipient's public key. In this case, the size of the message is ...
181 views
### Can an ephemeral RSA key give forward secrecy?
Suppose party A generates an ephemeral RSA key and sends the public key to B. Party B then generates a symmetric key, encrypts it with Party A's public key and sends the Ciphertext to Party A. Party ...
83 views
### Inverting RSA function
I am in high school and I am writing a paper on RSA. I want to show that low values of the public key exponent can make it easy to 'invert' the function so that the encrypted message can be recovered. ...
101 views
### Wanted: Simple ring signature example
I spent all day trying to figure ring signature schemes out but apparently I wasted time. I read the original ring signature paper (Rivest's How to leak a secret). I'm having numerous problems. Let's ...
313 views
### How does RSA compute such enormous numbers?
So I have been reading and learning a lot about cryptography lately and in particular asymmetric ciphers such as RSA. One thing that I am curious about but never seems to be mentioned is how the ...
148 views
### RSA 1024 bit forge a new matching signature from a chosen message
I have an RSA signature scheme with 1024-bit key where I know the following: Public modulus $N$ Public exponent $e=3$ A lot of signatures Summary: To compute a signature, an MD5 hash is ...
155 views
### lcm versus phi in RSA
In textbook RSA, the Euler $\varphi$ function $$\varphi(pq) := (p-1)(q-1)$$ is used to define the private exponent $d$. On the other hand, real-world cryptographic specifications require the ...
59 views
### How is RSA able to prevent brute forcing using the public key?
Bob calculates a private and public key. Bob sends his public key to John. Jeff is a third party unwanted member and manages to snatch the public key mid-transfer. John encrypts his message ...
32 views
### Is it possbile to get same set of public and private keys twice while creating CSR ? If so where does the conflict happens !?
I'm new to cryptography and please clarify if my below assumption is wrong. Lets assume, that two different banks creating CSR and they get identical public and private key pairs of RSA. In such ...
257 views
### generate RSA public key having public modulus and exponent [closed]
i want to generate an RSA public public key file using openssl (or other tools) having public modulus and exponent, so i can use it later to encrypt files i have this: ...
87 views
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cs代写|复杂网络代写complex network代考|TSKS33
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• Foundations of Data Science 数据科学基础
cs代写|复杂网络代写complex network代考|Multiple Lyapunov functions
To proceed, the notion of time dependent switching is introduced.
As a special kind of hybrid dynamic system, switched system has been studied for quite some time by researchers from applied mathematics, systems and control fields. Roughly speaking, a switched system is a dynamic system that consists of a number of subsystems and a switching rule that determines switches among these subsystems. Suppose the switched system is generated by the following family of subsystems
$$\dot{x}(t)=f_{p}(t, x(t)), x(t) \in \mathbb{R}^{n}, p \in{1, \ldots, \kappa},$$
together with a switching signal $\sigma(t):\left[t_{0},+\infty\right) \mapsto{1, \ldots, \kappa}$. Note that $\sigma(t)$ is a piecewise constant function that switches at the switching time instants $t_{1}, t_{2}, \ldots$, and is constant on the time interval $\left[t_{k}, t_{k+1}\right), k=0,1, \ldots$. In this book, we assume $\sigma(t)$ is right continuous, i.e., $\sigma(t)=\lim {\iota} t \sigma(\iota)$, and $\inf {k \in \mathbb{N}}\left(t_{k+1}-t_{k}\right) \geq \tau_{m}$ for some given positive scalar $\tau_{m}$ where inf represents the infimum. Please see Figure $2.2$ for an example. Thus the switched systems with time-dependent switching signal $\sigma(t)$ can be described by the equation
$$\dot{x}(t)=f_{\sigma(t)}(t, x(t)) .$$
According to Theorem 2.1, each subsystem has a unique solution over arbitrary interval $\left[t_{k}, t_{k+1}\right), k=0,1, \ldots$, with arbitrary initial value $x\left(t_{k}\right) \in \mathbb{R}^{n}$ if the function $f_{p}$, for each $p=1, \ldots, \kappa$, is globally Lipschitz in $x(t)$ uniformly over $t$. Thus the switched system (2.10) is well defined for arbitrary switching signal $\sigma(t)$ defined above and any given initial value $x\left(t_{0}\right) \in \mathbb{R}^{n}$. Throughout this chapter, we assume that such a globally Lipschitz condition holds for the subsystems, and thus the well-definedness of the switched system is guaranteed. We further assume that $f_{p}\left(t, \mathbf{0}{n}\right)=\mathbf{0}{n}$ for each $p=1, \ldots, \kappa$. Thus, the zero vector is an equilibrium point of the switched system (2.10). Next, some stability notions for the zero equilibrium point of switched systems are introduced.
cs代写|复杂网络代写complex network代考|CONSENSUS OF LINEAR CNSS WITH DIRECTED SWITCHING TOPOLOGIES
In the past decade, the consensus problem of general linear CNSs has received a lot of attention $[76,146,162,185,186,224]$. Specifically, the consensus problem of linear CNSs under a directed fixed communication topology has been addressed in $[76,224]$. In [162], the robust consensus of linear CNSs with additive perturbations of the transfer matrices of the nominal dynamics was studied. In [163] and a number of subsequent papers, the robust consensus was analyzed from the viewpoint of the $\mathcal{H}_{\infty}$ control theory. Among other relevant references, we mention [146] where, while assuming that the open loop systems are Lyapunov stable, the consensus problem of linear CNSs with undirected switching topologies has been investigated. In the situation where the CNS is equipped with a leader and the topology of the system belongs to the class of directed switching topologies, the consensus tracking problem has been studied in $[185,186]$. One feature of the results in these references is that the open loop agents’ dynamics do not have to be Lyapunov stable. Note that the presence of the leader in the CNSs considered in these references facilitate the derivations and the direct analyses of the consensus error system. However, when the open loop systems are not Lyapunov stable and/or there is no designated leader in the group, the consensus problem for linear CNSs with directed switching topologies remains challenging.
Motivated by the above discussion, this section aims to study the consensus problem for linear CNSs with directed switching topologies. Several aspects of the current study are worth mentioning. Firstly, some of the assumptions in the existing works are dismissed, e.g., the open loop dynamics of the agents do not have to be Lyapunov stable in this chapter. Furthermore, the CNSs under consideration are not required to have a leader. Compared with the consensus problems for linear CNSs with a designated leader, the point of difference here concerns the assumption on the system’s communication topology. In the previous work on the consensus tracking of linear CNSs such as [185], each possible augmented system graph was required to contain a directed spanning tree rooted at the leader. Compared with that work, the switching topologies in this section are allowed to have spanning trees rooted at different nodes. This is a significant relaxation of the previous conditions since it enables the system to be reconfigured if necessary (e.g., to allow different nodes to serve as the formation leader). This also has a potential to make the system more reliable.
cs代写|复杂网络代写complex network代考|Multiple Lyapunov functions
$$\dot{x}(t)=f_{p}(t, x(t)), x(t) \in \mathbb{R}^{n}, p \in 1, \ldots, \kappa$$
$$\dot{x}(t)=f_{\sigma(t)}(t, x(t)) .$$
有限元方法代写
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MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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2015-01-23T15:50:04-05:00
So
x+y = 5
x-y =7
if you add the two equations you get
x+y = 5
x-y = 7
--------------
2x = 12
x = 6
Now plug in 6 into one of the equations to find y
6+y = 5
y = -1
lets check
6+-1 = 5
6 - (-1) = 7
it works
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# Chemistry
posted by .
a student conducts and experiment by reacting 39.6g CuSO4 with an excess of MgS. The reaction yields 9.5599g CuS. What is the percentage yield? MgS+CuSO4=CuS+MgSO4
• Chemistry -
mols CuSO4 = grams/molar mass = ?
Convert mols CuSO4 to mols MgS using the coefficients in the balanced equation.
Convert mols MgS to grams. g = mols x molar mass. This is the theoretical yield(TY). The actual yield (YD) in the problem is 9.5599 g.
Then %yield = (AY/TY)*100 = ?
## Similar Questions
1. ### Chemistry
I have a problem that has been driving me crazy trying to solve, and I was wondering if someone could help. The problem is this: "Given the reaction: CuSO4 + 4 NH3 ----> Cu(NH3)4SO4, if 10 grams of CuSO4 reacts with 30 grams of …
2. ### Chem help
I have the answers for these but can you please explain to me how to get them?
3. ### chemistry
Calculate the mass of the solid you must measure out to prepare 100 mL of 0.025 M CuSO4. Note that this salt is a hydrate, so its formula is CuSO4⋅5H2O. You must include the waters of hydration when calculating the formula weights. …
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Consider the reaction 2Al + 3CuSO4 = Al2(SO4)3 + 3Cu What is the maximum amount of Cu (63.5 g/mol) that could be produced by reacting 20.0 grams of Al (27.0 g/mol) with excess CuSO4?
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I need to find out Molar solubility of CuS. How many grams of CuS are present in 15.0 L of a saturated CuS solution,
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Morecooperative games
# Morecooperative games - More on cooperative games...
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More on cooperative games
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Landowner-worker game, 2 workers possible revolution Let x1,x2,x3 be an allocation of the output f(k) from k people working on landowner’s land. Two workers could revolt, kill landowner, and take land. Output after revolution is less than two workers with no revolt. With no revolution, f(1)=1, f(2)=3, f(3)=4. With a revolution, output with 2 workers is 1.5. What’s in the core? All work, no
One owner two possible buyers Owner (person 1) has an object that is worthless to him, worth \$1 to either of two possible buyers (persons 2 and 3). Persons 2 and 3 each start out with more than \$1. Trade is possible. Two outcomes are in the core. Person 1 sells object to 2 for \$1. Person 1 sells object to 3 for \$1. Why is nothing else in the core?
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Person 2 values object at 1. Person 3 values it at \$v<1. What is in core? Person 2 gets the object and pays person
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https://www.physicsforums.com/threads/mathematica-maximize-element-extraction-from-list.569765/
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Mathematica maximize element extraction from list
1. Jan 22, 2012
parzio
Hi guys,
I think this is a simple question for mathematica experts.
How can I maximize the extracted value from a list given a index that has to respect some constrains?
For example:
S = {4,2,3,5}
Maximize[{Extract[S,x], x<= 3, x>=1},{x}]
I would like 4 is returned instead of this error:
Extract::psl: "Position specification x in Extract[{4,2,3,5},x] is not an integer or a list of integers."
Does someone know like solve this?
Thanks a lot.
2. Jan 22, 2012
Bill Simpson
In[1]:= S={4,2,3,5};Max[Take[S,{1,3}]]
Out[2]= 4
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# Projectile Motion. Regardless of its path, a projectile will always follow these rules:
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## Transcription
1 Projectile Motion What is a projectile? Regardless of its path, a projectile will always follow these rules: 1. A horizontally launched projectile moves both horizontally and vertically and traces out a parabolic trajectory. (Trajectory: The parabolic path of a projectile.) 2. The horizontal and vertical motions of a projectile are completely independent of one another. a. In the absence of air resistance, there is no net horizontal force on the projectile; therefore the projectile travels with a constant horizontal velocity. b. In the absence of air resistance, gravity is the only vertical force on the projectile; therefore the projectile travels with a uniformly accelerated vertical motion. Every second, the vertical velocity of the projectile changes by 9.8 m/s (10 m/s)
2
3
4 3. Horizontal and vertical motion are completely independent of each other. Therefore, the velocity of a projectile can be separated into horizontal (v x ) and vertical components (v y ). v x = velocity in the x-direction v y = velocity in the y-direction 4. For a projectile beginning and ending at the same height, the time it takes to rise to its highest point equals the time it takes to fall from the highest point back to the original position. 5. For a projectile beginning and ending at the same height, the initial speed is equal to its final speed.
5 Questions: 1. A projectile is launched at an angle into the air. If air resistance is negligible, what is the acceleration of its vertical component of motion? Of its horizontal component of motion? 2. At what part of its trajectory does a projectile have minimum speed? 3. Supposing a snowmobile is equipped with a flare launcher that is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)? a. in front of the snowmobile b. behind the snowmobile c. in the snowmobile 4. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane? 5. A hunter fires a gun at the same time a monkey drops a coconut as shown below. Where should the hunter aim to hit the coconut?
6 Horizontal Launched Projectiles d x = v x t d x = horizontal distance v x = horizontal velocity t = time d y = 4.9 x t 2 d y = vertical distance t = time A stone is thrown horizontally at +15 m/s from the top of a cliff that is 44-m high. a. How long does the stone take to reach the bottom of the cliff? b. How far from the base of the cliff does the stone strike the ground? c. Sketch the trajectory of the stone.
7 Horizontally Launched Projectile Practice 1. A ball is projected horizontally at velocity of 10 m/s from the top of a 50 m high cliff. A. How long is the ball in the air? B. How far from the base of the cliff does the ball land? C. What is the vertical velocity of the ball just before it strikes the ground? D. What is the horizontal velocity of the ball just before it strikes the ground? 2. A ball is projected horizontally at velocity of 8.0 m/s from the top of a 125 m high cliff. How long is the ball in the air? A. How long is the ball in the air? B. How far from the base of the cliff does the ball land? C. What is the vertical velocity of the ball just before it strikes the ground? D. What is the horizontal velocity of the ball just before it strikes the ground? 3. A ball is projected horizontally at velocity of 5.0 m/s from the top of a 10 m high cliff. How long is the ball in the air? A. How long is the ball in the air? B. How far from the base of the cliff does the ball land? C. What is the vertical velocity of the ball just before it strikes the ground? D. What is the horizontal velocity of the ball just before it strikes the ground? 4. You are trying to roll a ball off an 1.8 m high table to hit a target on the floor 2.0 m from the table's edge. With what speed should you roll the ball?
8 Projectiles Launched At An Angle Let s do one example to illustrate this concept: A ball is thrown with a velocity of 5 m/s at an angle of 60 above the horizontal. V yf =0 at top v yi = 4.36 m/s v xi Separate into x and y components. v yi 5 m/s 60 o v xi
9 Magic Chart Equations d t v I v F a d y = ½ (v yi + v yf ) t x x x x NO d y = v yi t t 2 x x x NO x V yf = v yi t NO x x x x V yf 2 = v yi d x NO x x x Example: A ball is thrown with a velocity of 8 m/s at an angle of 30 above the horizontal. 1. Calculate the time it takes the ball to reach the apex? (time up ) 2. Determine the total time the ball is in the air? (time total ) 3. How high does the ball go? (maximum height) V xi = initial horizontal velocity V yi = initial vertical velocity V fy = final vertical velocity 4. How far (horizontally) does the ball go? (range)
10 Projectiles Launched at an Angle 1. A ball is projected at a 30 angle above the horizontal at velocity of 12.0 m/s from the ground. A. How long does it take the ball to reach its apex? B. How high does the ball go? C. What is the total time the ball is in the air? D. How far does the ball go? 2. A ball is projected at a 60 angle above the horizontal at velocity of 15 m/s. A. How long does it take the ball to reach its apex? B. How high does the ball go? C. What is the total time the ball is in the air? D. How far does the ball go?
11 3. A scared kangaroo once cleared a fence by jumping with a speed of 4 m/s at an angle of 45 with respect to the ground. A. How long does it take the ball to reach its apex? B. How high does the ball go? C. What is the total time the ball is in the air? D. How far does the ball go? 4. A soccer ball is kicked with a speed of 7.5 m/s at an angle of 20 with respect to the ground. A. How long does it take the ball to reach its apex? B. How high does the ball go? C. What is the total time the ball is in the air? D. How far does the ball go?
12 The Snowball Fight Mr. Grant (aka Mr. Sneaky) throws a snowball 25 m/s at an angle of 70 o at Mr. Menzella. A. Draw a picture depicting the scenario above. B. How long is the snowball in the air? C. How far does the snowball travel? D. Does the snowball hit Mr. Menzella who is standing 41.5 m away?
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# Chapter 13 – financial statement analysis
Chapter 13 – Financial Statement Analysis
Assemble the financial statements prepared for the City of Monroe. These financial statements will be in the solutions to Exercises 5–C, 6–C, 7–C, and 8–C. Assume a population of 30,000 and fair value of property in the amount of \$350 million. Compute the following ratios, following the guidance used for the Village of Elizabeth in this chapter:
(1) Financial Position – Governmental Activities
(2) Financial Position – General Fund.
(3) Quick Ration – Governmental Activities
(4) Leverage – Primary Government
(5) Debt Coverage – Enterprise Funds
(6) Debt Service to Total Expenditures
(7) Debt per Capital – Primary Government
(8) Debt to Assessed Value of Property – Primary Government
I’m attaching the complete assignment but need Chapter 13 completed. I’m also attaching answers for 7-C and 8-C
*Other notes that will explain 2 of the screenshot images
CHAPTER 13 “FINANCIAL RATIOS” OF THE CAPSTONE CONTINUOUS PROBLEM FOR THE CITY OF MONROE IS ACTUALLY BASED UPON CHAPTER 13 IN THE TEXTBOOK. THE DESCRIPTION OF THE PROBLEM MAKES REFERENCE TO THE “VILLAGE OF ELIZABETH” THAT WAS DESCRIBED IN THE 12TH EDITION OF THE TEXTBOOK. IT SHOULD READ THE “VILLAGE OF RIVERSIDE” THAT IS DESCRIBED IN THE CURRENT 13TH EDITION OF THE TEXTBOOK ON PAGE 389.
(550 words)
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# Prove that the partial trace is a quantum operation, finding its Kraus representation
I am referring to Nielsen and Chuang Quantum Computation and Quantum Information 10th Anniversary Edition Textbook, Chapter 8.3.
A linear operator $$E_i:H_{QR}\longrightarrow H_Q$$ is defined by:
$$E_i \bigg(\sum_j \lambda_j |q_j\rangle|j\rangle \bigg)\equiv \lambda_i |q_i\rangle$$
whereby $$|q_j\rangle$$ and $$|j\rangle$$ are arbitrary states of system Q and the basis of system R respectively. Define $$\varepsilon$$ to be the quantum operation with the operation elements {$$E_i$$}:
$$\varepsilon(\rho)\equiv \sum_i E_i \rho E_i^{\dagger}$$
The text went on to say:
$$\varepsilon(\rho\otimes|j\rangle\langle j'|)=\rho \space \delta_{j,j'}=tr_R(\rho\otimes|j\rangle\langle j'|)$$
Question: I do not understand how to arrive at $$\delta_{j,j'}$$, and what form will be the operator representation of $$E_i$$ take? From what I've observed, system Q and R are not entangled in the last equation and $$E_i$$ seems to disregard whatever $$|j\rangle$$ basis of system R. Help will be much appreciated.
## 2 Answers
I think the presentation in N&C is a little confusing because $$\rho$$ is used in two contexts. I'll substitute one of those for a $$\sigma$$.
You can define $$E_i=I\otimes\langle i|,$$ which will certainly achieve the effect stated in your first equation. This lets us define the quantum operation $$\mathcal{E}(\sigma)=\sum_iE_i\sigma E_i^\dagger$$ where $$\sigma$$ is a density matrix on $$QR$$.
Now, let $$\rho$$ be a density matrix on $$Q$$. We have \begin{align} \mathcal{E}(\rho\otimes|j\rangle\langle j'|)&=\sum_iE_i(\rho\otimes|j\rangle\langle j'|)E_i^\dagger\\ &=\sum_i(I\otimes\langle i|)(\rho\otimes|j\rangle\langle j'|)(I\otimes|i\rangle)\\ &=\sum_i\rho\otimes\langle i|j\rangle\langle j'|i\rangle\\ &=\sum_i\rho\otimes\delta_{i,j}\delta_{j',i}\\ &=\rho\otimes\delta_{j,j'}\\ &=\rho\delta_{j,j'} \end{align}
• Got it. Thanks!
– C.C.
May 7, 2020 at 15:16
Say $$\lambda_j=\delta_{j,k}$$ so that the first equation gives:
$$E_i\left|q_k\right>\left|k\right>=\delta_{k,i}\left|q_i\right>$$ Now, we can write $$\rho$$ as:
$$\rho = \sum_k p_k \left|q_k\right>\left so that
$$\varepsilon(\rho\otimes\left|j\right>\left\left|j\right>\left\left This term is non-zero only when both kronecker deltas are 1 which happens only when $$i=j$$ and $$i=j'$$, which is only possible when $$j=j'$$. This gives us the required
$$\varepsilon(\rho\otimes\left|j\right>\left\left
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### Home > CALC > Chapter 6 > Lesson 6.1.4 > Problem6-54
6-54.
1. Examine the integrals below. Consider the multiple tools available for integrating and use the best strategy. After evaluating each integral, write a short description of your method. Homework Help ✎
Remember that the derivative of 3x = (ln 3) · 3x.
Distribute the 'x' then try evaluating the integral.
How can you rewrite elnx?
elnx = x
2m + 2 = 22 · 2m
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# Problems
1 – 12 of 12
Problem Title Likes Solvers Difficulty
#### Problem 701. Play Tic-Tac-Toe: Easy Bots
Created by: Richard Zapor
1
18
#### Problem 141. Solve the Sudoku Row
Created by: @bmtran (Bryant Tran)
Tags sudoku, solve, logic
16
950
#### Problem 114. Check to see if a Sudoku Puzzle is Solved
Created by: @bmtran (Bryant Tran)
Tags sudoku, solve, logic
7
302
Created by: goc3
6
25
#### Problem 3043. Fill-a-pix - Solver (basic)
Created by: goc3
Tags game, logic, solver
2
8
#### Problem 1184. Hangman (strategy)
Created by: Alfonso Nieto-Castanon
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Created by: Richard Zapor
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Created by: Benoit Charles
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Measuring water level in tank.......??
How to measure water level in tank in the cheapest way????
The first thing that came to my mind is pressure sensor..... but it's ruled out because it's expensive.
Can somebody suggest a cheaper mechanism??????
And it is preferred to have a low power mechanism.....
PhysOrg.com engineering news on PhysOrg.com >> Researchers use light projector and single-pixel detectors to create 3-D images>> GPS solution provides 3-minute tsunami alerts>> Single-pixel power: Scientists make 3-D images without a camera
Recognitions: Science Advisor http://www.rainharvesting.com.au/rain_alert.asp That one looks ideal but I guess the price might be too high. Looks like they use ultrasonics, although they don't say so. If you just wanted to know if the tank was full or not, you could use a float switch. If the water is conductive, you can have wires of different lengths coming from the top of the tank and as they get immersed, they conduct to a common wire which is always submerged. You can have a tube in the water with an opening at each end and you apply a range of audio frequencies to the tube and see where it resonates. The air column above the water resonates at a higher frequency as the water height rises. Using the same tube, but with a floating radio transmitter in it. The strength of the signal at the top of the tube is a measure of how deep the water is. It only has to pulse once an hour or so, so power consumption can be minimal. You can have a transparent tube coming from the bottom of the tank. The water riises in this tube and you can use a lot of different methods to work out how high the water is. You can have a floating magnet which operates Hall switches as it passes them. Or you can have an opaque float which interrupts a light beam. Or you can have conductance between electrodes embedded in the tube. As you have an interest in Microcontrollers, some of these methods may lend themselves to micro control.
I used to fill a 100 gallon drum for making chemicals. It took about 15 minutes to fill and I needed to know when the water was reaching the top. I built an ultra-cheap alarm with a 9V battery, a micro-buzzer and two long probes to connect the circuit when the water level reached them. The battery was the most expensive component.
Recognitions:
Gold Member
Measuring water level in tank.......??
You can measure the change in pressure of air being bubbled down a tube inserted in the tank from top to bottom. Cheapest way I can think of.
Bubbler systems are used to measure water level by detecting the pressure required to force air through a submerged tube. The tube is mounted with the end of the tube below the water surface being measured, and the air emerges from the bottom of the tube as a stream of bubbles. The air flow rate is relatively small--just enough to prevent water from backing up into the tube--so the pressure required to push air through the tube is equal to the pressure at the tube's outlet. This pressure is proportional to the water depth above the bottom of the tube.
http://www.usbr.gov/pmts/hydraulics_...bler/index.htm
You could also use a capacitive water level sensor where two insulated coaxial conductors are immersed in the tank. The total measured capacitance will be the sum of the water-filled portion and the water-free portion of the sensor, thus making the capacitance a function of the water level in the tank. You'd also have to correct for the temperature dependency of the water's dielectric constant -- with a thermistor perhaps? I have no idea about the cost of such a setup. It's probably not as cheap as opening the lid and peaking inside.
Quick googling led me to this article http://www.siliconchip.com.au/cms/A_30607/article.html Conversely if you don't want something complex and electric, how about putting a floating ball in the tank attached to a string. The string then pulls a little indicator attached to a spring, and with some calibration you have a fairly accurate measurement.
Mentor Google sump pum float buy: http://www.google.com/products?q=sum...w=dd&scoring=p Easy way, a $22 tethered float switch. http://www.google.com/products/catal...461&sa=title#p Recognitions: Gold Member Science Advisor Quote by russ_watters Google sump pum float buy: http://www.google.com/products?q=sum...w=dd&scoring=p Easy way, a$22 tethered float switch. http://www.google.com/products/catal...461&sa=title#p
Isn't that only for determining there is water present? I think the OP wants to measure how much water there is in the tank.
i need a help in understanding the working of water level indicator
plzzzzzzzzzzzz ...........help me in my project of water level indicator
Sorry, we do not spoonfeed homework here. Show us what you've got and we'll help you along.
0 down vote Another solution (no pun intended); Use a potentiometer . The normal rotation range is 270 degrees. Attach a float to the potentiometer using a boom arm. (length=1 unit) Between full and empty the boom arm will move 90 degrees. The ADC on a PIC is either 256 or 1024 steps (yes, zero is a step). I'll use 256 steps for clarity. 270 degrees = 256 ADC steps. 270/90 = 3 (one third of potentiometer range) 255/3 = 85 ADC steps Program some code to mark the 0 degree point (tank empty) when a button is pushed. This stores an offset point in the PIC eeprom. Now the potentiometer does not have to be exactly at zero because this calibration point can be set. Using trigometry calculate a lookup table (hint: php script) to correspond to each step of the ADC. Hint: each ADC step corresponds to 90/85 = 1.0588 degrees. Yes, you should have paid more attention in maths class. dumb waste of time back then, indespensible now. engage brain. learn trigonometry. teach others. Acronym : The Old Arab Carried A Heavy Sack Of Hay. The hypontenuse is the length of the boom arm. Make it 1 unit long. Lookup table then provides percentage of tank depth. (multiply by 100 of course)
Recognitions: Gold Member 1.Dipstick 2.Completely or partially transparent tank/container arrangement with scale inscribed. 3.Tank sitting on suitably calibrated scales
With a ruler
USound can be used. Here is a concept. Might be helpful for future searches... http://www.open-electronics.org/wate...-with-arduino/
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# 4.8. Matrices III#
## 4.8.1. Orthogonal Matrices#
Definition 4.109
A real square matrix $$\bU$$ is called orthogonal if the columns of $$\bU$$ form an orthonormal set. In other words, let
$\bU = \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix}$
with $$\bu_i \in \RR^n$$. Then we have
$\bu_i \cdot \bu_j = \delta_{i , j}.$
Lemma 4.44
An orthogonal matrix $$\bU$$ is invertible with $$\bU^T = \bU^{-1}$$.
Proof. Let
$\bU = \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix}$
be orthogonal with
$\begin{split} \bU^T = \begin{bmatrix} \bu_1^T \\ \bu_2^T \\ \vdots \\ \bu_n^T \end{bmatrix}. \end{split}$
Then
$\begin{split} \bU^T \bU = \begin{bmatrix} \bu_1^T \\ \bu_2^T \\ \vdots \\ \bu_n^T \end{bmatrix} \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix} = \begin{bmatrix} \bu_i \cdot \bu_j \end{bmatrix} = \bI. \end{split}$
Since columns of $$\bU$$ are linearly independent and span $$\RR^n$$, hence $$\bU$$ is invertible. Thus
$\bU^T = \bU^{-1}.$
Lemma 4.45 (Determinant of an orthogonal matrix)
Determinant of an orthogonal matrix is $$\pm 1$$.
Proof. Let $$\bU$$ be an orthogonal matrix. Then
$\det (\bU^T \bU) = \det (\bI) \implies \left ( \det (\bU) \right )^2 = 1.$
Thus we have
$\det(\bU) = \pm 1.$
## 4.8.2. Unitary Matrices#
Definition 4.110 (Unitary matrix)
A complex square matrix $$\bU$$ is called unitary if the columns of $$\bU$$ form an orthonormal set. In other words, let
$\bU = \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix}$
with $$u_i \in \CC^n$$. Then we have
$\bu_i \cdot \bu_j = \langle \bu_i , \bu_j \rangle = \bu_j^H \bu_i = \delta_{i , j}.$
Lemma 4.46
A unitary matrix $$\bU$$ is invertible with $$\bU^H = \bU^{-1}$$.
Proof. Let
$\bU = \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix}$
be unitary with
$\begin{split} \bU^H = \begin{bmatrix} \bu_1^H \\ \bu_2^H \\ \vdots \\ \bu_n^H \end{bmatrix}. \end{split}$
Then
$\begin{split} \bU^H \bU = \begin{bmatrix} \bu_1^H \\ \bu_2^H \\ \vdots \\ \bu_n^H \end{bmatrix} \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix} = \begin{bmatrix} \bu_i^H \bu_j \end{bmatrix} = \bI. \end{split}$
Since columns of $$\bU$$ are linearly independent and span $$\CC^n$$, hence $$\bU$$ is invertible. Thus
$\bU^H = \bU^{-1}.$
Lemma 4.47 (Determinant of unitary matrices)
The magnitude of determinant of a unitary matrix is $$1$$.
Proof. Let $$\bU$$ be a unitary matrix. Then
$\det (\bU^H \bU) = \det (\bI) \implies \det(\bU^H) \det(\bU) = 1 \implies \overline{\det(\bU)}{\det(\bU)} = 1.$
Thus we have
$|\det(\bU) |^2 = 1 \implies |\det(\bU) | = 1.$
## 4.8.3. F Unitary Matrices#
We provide a common definition for unitary matrices over any field $$\FF$$. This definition applies to both real and complex matrices.
Definition 4.111
A square matrix $$\bU \in \FF^{n \times n}$$ is called $$\FF$$ unitary if the columns of $$\bU$$ form an orthonormal set. In other words, let
$\bU = \begin{bmatrix} \bu_1 & \bu_2 & \dots & \bu_n \end{bmatrix}$
with $$\bu_i \in \FF^n$$. Then we have
$\langle \bu_i , \bu_j \rangle = \bu_j^H \bu_i = \delta_{i , j}.$
We note that a suitable definition of inner product transports the definition appropriately into orthogonal matrices over $$\RR$$ and unitary matrices over $$\CC$$.
When we are talking about $$\FF$$ unitary matrices, then we will use the symbol $$\bU^H$$ to mean its inverse. In the complex case, it will map to its conjugate transpose, while in real case it will map to simple transpose.
This definition helps us simplify some of the discussions in the sequel (like singular value decomposition).
Following results apply equally to orthogonal matrices for real case and unitary matrices for complex case.
Lemma 4.48 (Norm preservation)
$$\FF$$-unitary matrices preserve norm. i.e.
$\| \bU \bx \|_2 = \| \bx \|_2.$
Proof. We have
$\| \bU \bx \|_2^2 = (\bU \bx)^H (\bU \bx) = \bx^H \bU^H \bU \bx = \bx^H \bI \bx = \| \bx \|_2^2.$
Remark 4.11
For the real case we have
$\| \bU \bx \|_2^2 = (\bU \bx)^T (\bU \bx) = \bx^T \bU^T \bU \bx = \bx^T \bI \bx = \| \bx\|_2^2.$
Lemma 4.49 (Preservation of inner products)
$$\FF$$-unitary matrices preserve inner product. i.e.
$\langle \bU \bx, \bU \by \rangle = \langle \bx, \by \rangle.$
Proof. We have
$\langle \bU \bx, \bU \by \rangle = (\bU \by)^H \bU \bx = \by^H \bU^H \bU \bx = \by^H \bx.$
Remark 4.12
For the real case we have
$\langle \bU \bx, \bU \by \rangle = (\bU \by)^T \bU \bx = \by^T \bU^T \bU \bx = \by^T \bx.$
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Class 12
MATHS
Hyperbola
# Let the foci of the hyperbola (X^(2))/(A^(2))-(y^(2))/(B^(2))=1 be the vertices of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and the foci of the ellipse be the vertices of the hyperbola. Let the eccentricities of the ellipse and hyperbola be e_(E) and e_(H), respectively. Then match the following lists. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_CG_C07_E10_001_Q01.png" width="80%">
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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Text Solution
a rarrp; b rarr r, s;c rarr q; drarrs
Solution :
We have <br> A=ae_(E) and a=Ae_(H) <br> "or "e_(E)e_(H)=1 <br> therefore" "e_(E)+e_(H)gt2" "("Using "e_(E)+e_(H)gtsqrt(e_(E)e_(H))) <br> B^(2)=A^(2)(e_(H)^(2)-1)=a^(2)(1-e_(E)^(2))=b^(2) <br> "or "(b)/(B)=1 <br> Also, the angle between the asymptotes is <br> 2tan^(-1).(B)/(A)=(2pi)/(3) <br> Also, (B)/(A)=sqrt3or(b)/(ae_(E))=sqrt3ore_(E)^(2)=(1)/(4) <br> Solving (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(a^(2)e_(E)^(2))-(y^(2))/(b^(2))=1 or(2x^(2))/(a^(2))-(y^(2))/(b^(2))=1 <br> Now, solve.
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Class 12th
Hyperbola
Class 12th
Hyperbola
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# Primeval number
In mathematics, a primeval number is a natural number n for which the number of prime numbers which can be obtained by permuting some or all of its digits (in base 10) is larger than the number of primes obtainable in the same way for any smaller natural number. Primeval numbers were first described by Mike Keith.
The first few primeval numbers are
1, 2, 13, 37, 107, 113, 137, 1013, 1037, 1079, 1237, 1367, ... (sequence A072857 in OEIS)
The number of primes that can be obtained from the primeval numbers is
0, 1, 3, 4, 5, 7, 11, 14, 19, 21, 26, 29, ... (sequence A076497 in OEIS)
The largest number of primes that can be obtained from a primeval number with n digits is
1, 4, 11, 31, 106, ... (sequence A076730 in OEIS)
The smallest n-digit prime to achieve this number of primes is
2, 37, 137, 1379, 13679, ... (sequence A134596 in OEIS)
Primeval numbers can be composite. The first is 1037 = 17×61. A Primeval prime is a primeval number which is also a prime number:
2, 13, 37, 107, 113, 137, 1013, 1237, 1367, 10079, ... (sequence A119535 in OEIS)
The following table shows the first six primeval numbers with the obtainable primes and the number of them.
Primeval number Primes obtained Number of primes
1 none 0
2 2 1
13 3, 13, 31 3
37 3, 7, 37, 73 4
107 7, 17, 71, 107, 701 5
113 3, 11, 13, 31, 113, 131, 311 7
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# smoothing pot input
skyjumper: Currently I am storing the last 20 samples and averaging them, but I have been looking for a way to achieve this without consuming 80 bytes (20 floats)... I am getting speed readings several times each second from a transducer. Each of these readings tends to vary a bit from the prior one, and I just want to get a stable reading to use to present to the driver.
α = 0.25 gives a window that almost decays away at 20 samples and can be implemented without floating-point. I think this will work...
``````unsigned long history;
unsigned short value;
void setup( void )
{
Serial.begin( 250000 );
history = analogRead( 0 ) * 4;
}
void loop( void )
{
history = analogRead( 0 ) + (((3 * history) + 2) / 4);
value = (history + 2) / 4;
Serial.println( value );
delay( 100 );
}
``````
Thanks!!! I’ll play with that…
I think dot-8 rather than dot-2 fixed-point may give slightly more accurate results. Try this one instead...
``````unsigned long history;
unsigned short value;
void setup( void )
{
Serial.begin( 250000 );
history = analogRead( 0 ) * 256;
}
void loop( void )
{
history = (64*analogRead(0)) + (((64*3*history)+128) / 256);
value = (history + 128) / 256;
Serial.println( value );
delay( 100 );
}
``````
This looks very interesting, but I don't seem to follow the formula.
Can you explain what it's doing?
The basic formula is…
v1 = (α * analogRead) + ((1 - α) * v0)
α = 0.25 or 1/4 …
v1 = ((1/4) * analogRead) + ((1 - (1/4)) * v0)
v1 = ((1/4) * analogRead) + ((3/4) * v0)
To make it fixed-point with eight bits for the fraction multiple both sides by 256 (2 to the power of 8)…
256v1 = 256 * { ((1/4) * analogRead) + ((3/4) * v0) }
256
v1 = ((256/4) * analogRead) + ((3256/4) * v0)
256
v1 = ((64) * analogRead) + ((3*64) * v0)
The right-side has “v0” not "256v0" so we have to perform the division when calculating the next value. The multiplication is performed first to preserve the precision…
256
v1 = (64 * analogRead) + ((364v0) / 256)
Finally, to improve the accuracy we need to include rounding…
256v1 = (64 * analogRead) + (((364*v0)+(256/2)) / 256)
So, history is the “actual” value multiplied by 256. Another way to look it: history / 256 is the whole number part and history % 256 is the fractional part.
Nice!
Going back to the original, is the big disavantage that you have to use a float?
Is that much slower? How much?
Float is a lot slower, but I haven't got figures to say exactly how much. At least four times slower.
db2db: Going back to the original, is the big disavantage that you have to use a float?
Is that much slower? How much?
Floats are bigger as well. In my case, unsigned short int is more than enough space for 2 bytes.
Good point about storage. history in Reply #20 can be unsigned short (half the size of float). Assuming the "raw" values are between 0 and 1023, up to six fractional bits are possible with an unsigned short (instead of multiplying both sides by 256, multiply both sides by 64). Which is a very nice compromise: about 1.5 decimals, smoothing, and fast all from just two bytes! Warning: updating history overflows an unsigned short so the equation will have to be cast to an unsigned long before the right-side multiply.
history in Reply #22 has to remain an unsigned long (same size as float).
Note: I updated #20 and #22 to make them a bit more accurate.
One thing I didn’t make clear, is that I get numbers directly from the instrument system. I don’t need to do the A/D conversion.
Anyhow, I have been modeling this in a spreadsheet. I used this formula:
``````=(\$B\$3 * D4) + ((1 - \$B\$3) * D3)
``````
Where \$B\$3 is Alpha. Column D contains the series of input values, so in this example D4 is the current input and D3 is the prior input. I believe I am modeling this correctly from looking at your sample source code.
I like this quite a bit. I plotted the input data, the output data from this filter, and the output data from the strategy of averaging the past 5, 10 and 20 samples. I put in some data the is typical of what I usually see.
I found that an alpha of .25 very closely corresponded to averaging the last 5 samples. Alpha == .15 is very close to the past 10 samples. Alpha == .08 approximates averaging the last 20 samples, although not very well in that case (but not at all badly).
Using .25 seems to provide the best results. I get samples at a rate of 2Hz, so thats 2.5 seconds of data.
Getting rid of the past 10 samples saves 18 bytes (since I now only need to save the more recent result). Of course if the user sets the filter to be slower (a lower alpha) then even more bytes are saved. Since I need to use this filter for about 10 different inputs, this will save at least 180 bytes, which is pretty huge.
Is there a name for this filter?
Thank you very, very much, this is going to be a huge improvement for my project!
I’ll attach the spreadsheet for anyone who wants to play with it. I did it in Libre Office, but saved it as an XLS file since Office can’t read odt. The numbers in the spread sheet are floats, but in the code of course I’ll multiply them by 100.
filter-model.xls (18 KB)
Is there a name for this filter?
It is a non recursive low pass filter.
skyjumper:
I believe I am modeling this correctly from looking at your sample source code.
You are.
Thank you very, very much, this is going to be a huge improvement for my project!
You are welcome!
I’ll attach the spreadsheet for anyone who wants to play with it.
Thank you. I updated it to also model the code from my previous post…
but in the code of course I’ll multiply them by 100.
Which works well if you decide to use the fixed-point version (OUT 5 in the updated workbook).
Is there a name for this filter?
filter-model.xls (34.5 KB)
I'll look that over this evening, thanks again...
I have to admit, I don't fully understand what's going on in post #24. The A/D conversion is a 10 bit conversion. Are you just deciding that the fractional part of that should be 8 of those 10 bits? From looking at your revision to the model spreadsheet, I noticed that making that change degrades how accurately the output of the filter reflects the input (when gain/alpha is set to 1). I assume this reflects the reduced precision? Whether or not this is acceptable of course depends on the application and the range of data.
skyjumper: One thing I didn't make clear, is that I get numbers directly from the instrument system. I don't need to do the A/D conversion.
Anyhow, I have been modeling this in a spreadsheet. I used this formula:
``````=(\$B\$3 * D4) + ((1 - \$B\$3) * D3)
``````
Where \$B\$3 is Alpha. Column D contains the series of input values, so in this example D4 is the current input and D3 is the prior input. I believe I am modeling this correctly from looking at your sample source code.
I like this quite a bit. I plotted the input data, the output data from this filter, and the output data from the strategy of averaging the past 5, 10 and 20 samples. I put in some data the is typical of what I usually see.
Actually I think that I did this wrong. The spread sheet is probably right, but the formula should be:
``````=(\$B\$3 * D4) + ((1 - \$B\$3) * E3)
``````
Where \$B\$3 is Alpha. Column D contains the series of input values and column E has output values, so in this example D4 is the current input and E3 is the prior output.
I noticed some discrepancies as the numbers got bigger and the filter became less effective.
skyjumper: I have to admit, I don't fully understand what's going on in post #24. The A/D conversion is a 10 bit conversion. Are you just deciding that the fractional part of that should be 8 of those 10 bits?
No. The full 10 bits are used as-is.
Imagine you want to add 1/2 to 1/2. Using integer operations the result is not at all what we want: zero.
Let's take a simple 16 bit integer...
bbbb bbbb bbbb bbbb
...pretend there is a decimal point in the middle...
bbbb bbbb . bbbb bbbb
...also pretend that each bit to the right of the decimal is a fraction of a successive power of two...
bbbb bbbb . 1234 5678
"1" is the 1/2 place "2" is the 1/4 place "3" is the 1/8 place "4" is the 1/16 place etcetera
...and pretend each bit to the left of the decimal is just a normal integer...
iiii iiii . 1234 5678
The value 9.00 would be stored as...
0000 1001 . 0000 0000
...or 0x0900. The value 0.50 (1/2) would be stored as...
0000 0000 . 1000 0000
...or 0x0080. Adding 1/2 to 1/2 is just like adding normal integers (remember, the decimal isn't really there; we're just pretending it is)...
0000 0000 . 1000 0000
## 0000 0000 . 1000 0000
0000 0001 . 0000 0000
...or 0x0080 + 0x0080 = 0x0100.
There is an implied / imaginary divided-by-256 always present in our value. So 0x0080 + 0x0080 = 0x0100 can also be viewed as 128 (/256) + 128 (/256) = 256 (/256).
That's essentially what I'm doing in my fixed-point EWMA code. I pretend there is a decimal point to the left of the right-most eight bits (just like the example above). In order to convert from a "normal" integer (like the value returned from analogRead) I have to shift the value left by eight bits so the decimal points are aligned. That's the purpose of multiplying by 256; to shift the whole number into the whole number position.
Does that help?
From looking at your revision to the model spreadsheet, I noticed that making that change degrades how accurately the output of the filter reflects the input (when gain/alpha is set to 1).
I should have mentioned, the alpha for OUT 5 is permanently set at 0.25. No attempt is made to use the value in cell B3.
Just to make certain I understand: You're saying that OUT 1 and OUT 5 are a bit different even though they both have an alpha of 0.25. Correct?
I assume this reflects the reduced precision?
Yes. Excel uses IEEE 64-bit floats which give 11 to 12 total digits (10 to 11 decimals in this case). My code always has 2.4 decimals {log(256)}. The two values will frequently be a bit different but should never diverge for any extended period and should never be more than 0.005 different.
Whether or not this is acceptable of course depends on the application and the range of data.
Exactly. Two more things to consider...
It may not be worth using the fixed-point version because of conversions. In your case, you have the motor speed with two decimals. In order to use the fixed-point version you have to multiply the motor speed by 100 and round to an integer. If you display the smoothed value, you will have to convert the fixed-point value to text. After all that, the floating-point version may actually be more efficient.
The fixed-point version works well when alpha is predetermined and unlikely to change. In your case, you will probably want to experiment with different alpha values which makes the fixed-point version annoying and difficult to use.
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https://www.mathworks.com/help/matlab/ref/ezpolar.html?requestedDomain=www.mathworks.com&nocookie=true
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# Documentation
### This is machine translation
Translated by
Mouseover text to see original. Click the button below to return to the English version of the page.
# ezpolar
Easy-to-use polar coordinate plotter
## Syntax
```ezpolar(fun) ezpolar(fun,[a,b]) ezpolar(axes_handle,...) h = ezpolar(...) ```
## Description
`ezpolar(fun)` plots the polar curve `rho = fun(theta)` over the default domain 0 < `theta` < 2π.
`fun` can be a function handle or a character vector (see the Tips section).
`ezpolar(fun,[a,b])` plots `fun` for `a` < `theta` < `b`.
`ezpolar(axes_handle,...)` plots into the axes with handle `axes_handle` instead of the current axes (`gca`).
`h = ezpolar(...)` returns the handle to a line object in `h`.
## Examples
collapse all
Plot the function over the domain .
```figure ezpolar('1+cos(t)')```
## Tips
### Passing the Function as a Character Vector
Array multiplication, division, and exponentiation are always implied in the expression you pass to `ezpolar`. For example, the MATLAB® syntax for a plot of the expression
`t.^2.*cos(t)`
which represents an implicitly defined function, is written as
`ezpolar('t^2*cos(t)')`
That is, `t^2` is interpreted as `t.^2` in the character vector you pass to `ezpolar`.
### Passing a Function Handle
Function handle arguments must point to functions that use MATLAB syntax. For example, the following statements define an anonymous function and pass the function handle `fh` to `ezpolar`.
```fh = @(t) t.^2.*cos(t); ezpolar(fh)```
Note that when using function handles, you must use the array power, array multiplication, and array division operators (```.^, .*, ./```) since `ezpolar` does not alter the syntax, as in the case with character vector inputs.
If your function has additional parameters, for example `k1` and `k2` in `myfun`:
```function s = myfun(t,k1,k2) s = sin(k1*t).*cos(k2*t);```
then you can use an anonymous function to specify the parameters:
`ezpolar(@(t)myfun(t,2,3))`
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https://calculat.io/en/number/hex-to-dec/28e
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# 28e in decimal
## What is 28e in decimal?
Answer: Hexadecimal Number 28E (0x28E) It Is Decimal: 654
(Six hundred fifty-four)
27f639
280640
281641
642
283643
284644
285645
286646
287647
288648
289649
28a
28b651
28c652
28d653
654
28f
290656
291657
292658
293659
294660
295661
296662
297663
298664
299665
29a
29b667
29c668
## About "Hex to Decimal" Calculator
This calculator will help you to convert hexadecimal numbers to decimal. For example, it can help you find out what is 28e in decimal? (The answer is: 654). Enter hexadecimal number (e.g. '28e') and hit the 'Convert' button.
## FAQ
### What is 28e in decimal?
Hexadecimal Number 28E (0x28E) It Is Decimal: 654
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https://solvedlib.com/how-do-you-solve-9g-285,289009
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# How do you solve 9g = 28.5?
###### Question:
How do you solve 9g = 28.5?
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http://dozenalsociety.org.uk/roses/munro.html
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# Thoughts - Andy Munro
We need to agree on two new digits and some names for numbers.
If we look at the modern words for our numbers we can see that they have been evolving for many years and a certain amount of erosion has taken place.
Four tens and nine has become for-ty five
So our dozenal system should have this contraction factored in
ie. Four dozens and five becomes for-zy five
Thirteen becomes one-zy one *11
Next is the problem of ten and eleven
For ten I think a nod to the Romans is called for here and the amount 'ten' should be re-named 'dec' the symbol being a stylised 'd'.
Ok now for the eleven. Bit more tricky for the name, but surely the symbol has to be something like epsilon.
I think that keeping the stem of eleven will ease recognition so I'm proposing the German word for eleven which is elf.
Therefore the mouthful of 'one hundred and thirty-one' is reduced to *dec-zy elf.
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https://arbourj.wordpress.com/2012/05/05/more-on-finite-simple-extensions/
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Recall that in last post, we saw that a finite simple extension $K(\alpha)/K$ is such that $K(\alpha) \simeq K[x]/(f)$ for $f \in K[x]$ a monic irreducible nonconstant polynomial. This polynomial is called the irreducible polynomial for $\alpha$ over $K.$ However, by the theorem in my post about adjunction of roots in a ring, we know that in general there is not a unique isomorphism $K[x]/(f) \to K(\alpha).$ Indeed, there are as many distinct isomorphisms as there are roots of $f$ in $K(\alpha).$
Now, recall that the construction of the polynomial ring over a ring is functorial : every ring morphism $\varphi : R \to S$ lifts uniquely to a ring morphism $\tilde{\varphi} : R[x] \to S[x]$ agreeing with $\varphi$ on $R.$ ie. such that this diagram commutes :
We would like a similar situation with finite simple field extensions but as remarked, even the identity morphism from a field $K$ to itself doesn’t lift uniquely to a morphism from $K(\alpha_1)$ to $K(\alpha_2)$ fixing $K,$ even if $\alpha_1$ and $\alpha_2$ have the same irreducible polynomial $f(x)$ over $K.$ Indeed, we will certainly have isomorphisms $K(\alpha_1) \simeq K[x]/(f) \simeq K(\alpha_2)$ but there will be as many distinct isomorphisms $\varphi : K(\alpha_1) \to K(\alpha_2)$ fixing $K$ as there are roots of $f(x)$ in $K(\alpha_2).$ We thus need to specify the image of $\alpha_1$ if we want unicity. More generally, here is the situation:
Theorem: Let $K_1(\alpha_1)/K_1$ and $K_2(\alpha_2)/K_2$ be two simple field extensions. Let $f_1(x) \in K_1[x]$ and $f_2(x) \in K_2[x]$ be the respective irreducible polynomials of $\alpha_1$ and $\alpha_2$ respectively over $K_1$ and $K_2.$ Let $\varphi : K_1 \to K_2$ be an isomorphism such that $\tilde{\varphi}(f_1(x)) = f_2(x)$ (See the above diagram for polynomials). Then there exists a unique isomorphism $\phi : K_1(\alpha_1) \to K_2(\alpha_2)$ such that $\phi | _{K_1} = \varphi$ and $\phi(\alpha_1) = \alpha_2.$
Here is a diagram to help picture what has just been said:
Proof: Consider the map
$\psi : K_1[x] \to K_2[x]/(f_2(x))$
defined as $\psi = \pi \circ \tilde{\varphi},$ where $\pi : K_2[x] \to K_2[x]/(f_2(x))$ is the canonical projection. Since $\tilde{\varphi}$ is injective and $\tilde{\varphi}(f_1(x)) = f_2(x),$ we must have $\ker(\psi) = \tilde{\varphi}^{-1}(\ker(\pi)) \supset (f_1(x)).$ On the other hand, since $(f_1(x))$ is a maximal ideal, we have $\ker(\psi) = (f_1(x))$ hence the isomorphism
$K_1[x]/(f_1) \simeq K_2[x]/(f_2).$
Moreover, in the theorem of last post we saw that finite simple extensions $K(\alpha)$ were characterized as quotients of the polynomial ring over $K$ by the ideal generated by the irreducible polynomial of $\alpha$ over $K.$ More explicitly, for $f(x)$ the irreducible polynomial for $\alpha$ over $K,$ we have an isomorphism $K[x]/(f) \simeq K(\alpha)$ identifying the coset of $x$ with $\alpha.$ Composing this all together, we have an isomorphism
$\phi : K_1(\alpha_1) \simeq K_1[x]/(f) \simeq K_2[x](f) \simeq K_2(\alpha_2)$
such that $\phi(\alpha_1) = \alpha_2,$ as desired.
For the unicity of $\phi,$ note that every element of $K_1(\alpha_1)$ is a finite linear combination of powers of $\alpha_1$ over $K_1$ so $\phi$ is completely and uniquely determined by its action on $K_1$ together with its action on $\alpha_1.$ Thus, since $\phi(K_1) = \varphi(K_1)$ and $\phi(\alpha_1) = \alpha_2,$ the isomorphism $\phi$ is unique.
QED
More often than not, we will be concerned with morphisms of field extensions that extend the identity:
Definition: Let $F/K$ be a field extension. A $K$-automorphism, or an automorphism of $K$-extension, is a field automorphism $\varphi : F \to F$ such that $\varphi|_K = \text{Id}_K.$ The group of $K$-automorphism of $F$ will be denoted by $\text{Aut}_K(F).$
A corollary of the theorem hints at the interaction between group theory and field theory:
Corollary: Let $F/K$ be a simple finite extension and let $f(x)$ be the minimal polynomial of $\alpha$ over $K$ (where $F \simeq K(\alpha)).$ Then $|\text{Aut}_K(F)|$ is the number of distinct roots of $f(x)$ in $F.$ In particular
$|\text{Aut}_K(F)| \leq [F : K]$
with equality iff $f(x)$ factors over $K$ in a product of distinct polynomials of degree 1.
Proof: Let $\sigma \in \text{Aut}_K(F).$ Since $\sigma|_K = \text{Id}_K,$ $\sigma$ is completely determined by $\sigma(\alpha).$ But it must send $\alpha$ to a root of $f(x)$ because
$f(\sigma(\alpha)) = \sigma ( f(\alpha)) = 0.$
Hence there are at most the number of roots of $f(x)$ elements in $\text{Aut}_K(F).$ But by the theorem, every choice of a root of $f(x)$ determines a unique lift of the identity $\sigma \in \text{Aut}_K(F)$ hence the equality. Since the number of roots of $f(x)$ is smaller or equal to $\text{deg } f(x) = [K : F],$ the proof is complete.
QED
Reference: P. Aluffi, Algebra : Chapter 0.
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https://eng.kakprosto.ru/how-35513-how-to-find-specific-heat
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Instruction
1
The specific heat capacity of a substance is the quantity of heat required to heat or cool 1 kg of substance by 1 Kelvin. That is, in other words, if for example the specific heat of water is equal to 4.2 kJ/(kg*K) means that in order to heat one kg of water by one degree, you must pass this kg of water is 4.2 kJ of energy. Specific heat of a substance is given by:
C = Q/m(T_2-T_1)
Unit of specific heat capacity has the units in the system SI – (j/kg*K).
2
The specific heat of a body is determined experimentally by using a calorimeter and thermometer. A simple calorimeter consists of a polished metal Cup, then put inside the other metal of glass tubes (for heat insulation) and filled with water or other liquid with known specific heat capacity. Body (solid or liquid), heated to a certain temperature t, is immersed in a calorimeter where temperature is measured. Let the test lowering of the body temperature of the liquid in the calorimeter was equal to t_1, and after the temperature of the water (liquid) and lowered into it the body will be equal, it will be equal ?.
3
From the law of conservation of energy it follows that the heat Q given to the heated body, equal to the amount of heat Q_1 obtained with water, and Q_2 obtained by the calorimeter:
Q=Q_1+Q_2
Q=cm(t- ?), Q_1=c_1 m_1 (?-t_1), Q_2=c_2 m_2(?-t_1)
cm(t- ?)= c_1 m_1 (?-t_1)+ c_2 m_2(?-t_1)
here m_1 and c_1 - specific heat and mass of water in the calorimeter, and c_2 m_2 - specific heat and mass of the material of the calorimeter.
This equation expresses the balance of thermal energy, is called the equation of thermal balance. He will find
c=(Q_1+Q_2)/m(t- ?) =( c_1 m_1 (?-t_1)+ c_2 m_2(?-t_1))/m(t- ?) = (c_1 m_1+m_2 c_2)( ?- t_1)/m(t- ?)
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https://puzzles.mit.edu/2015/puzzle/say_cheese_and_dive/solution/
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Location: Graveyard
Depth: 302
# Solution to Say Cheese and Dive!
### by Shawn Westerdale, Pweaver, and Adam Rosenfield
This puzzle is a text adventure game where the solver is exploring a series of sub-dungeons in the form of holes in a giant Swiss Cheese Dungeon.
At the start of the game, you can choose a hole to explore. At the entrance to the dungeons, you encounter a sign warning you of the perils that await you. This conveniently acronyms to "IGNORE ROOM NAMES"; they're there to give each room a unique identifier and aren't actually otherwise relevant.
As you adventure around, there are a few things you might notice: there are strange objects (weapons) laying around, there are terrifying muensters that you occasionally encounter, some rooms have various amounts of mold, and other rooms smell like bravery.
Rooms that smell like bravery are adjacent to a room containing a weapon. Rooms that contain mold are one or two rooms away from a muenster. The point of these two vaguely Hunt the Wumpus-y features is that they allow you to find things in the hole without having to stumble upon them by pure luck. They also mean that you can be careful and avoid accidentally running into a muenster and dying before you get the weapon you need to kill it.
The solver can begin their adventures by exploring these holes. Each weapon has a power level associated with it ranging from 1 to N (where N is the num- ber of weapons in that hole), and each weapon pairs to exactly one muenster (generally guessable by some sort of terrible pun or wordplay, or other silliness). When you encounter a muenster, you can try to fight it by using a weapon against it. If you use the correct weapon, the muenster dies; if you use the incorrect weapon, the munster eats you. You can also try to run away, in which case you have a 50% chance of escaping into an adjacent room, and otherwise you get eaten. Getting eaten is annoying but doesn't suck that much, since you can just start the hole again, and it doesn't really change.
As you adventure around the holes, you'll eventually find all the weapons and kill all the muensters. Once you defeat all the muensters in the hole, you nd a piece of paper. Do this in each hole and sort the pieces of paper by the hole number and you get the message "CONNECT THE LOCATIONS OF THE MUENSTERS". Might not be so clear what that means yet, so we'll get back to that later.
Now, in your adventures, there is another strange thing you may have no- ticed. If you try to draw a map of these holes, you'll find that you can't do so on a single at sheet of paper. It turns out all of the rooms in the holes are arranged on the surface of a 3D letter. Reconstruct each of the holes and you find that the letters range from A to F.
Now back to the cluephrase we got from the pieces of paper earlier. If you make a 3D map of each of the holes, you can then connect each of the muensters with a string, ordered by the power level of the weapon you used to kill them. Doing so, you'll nd that the string draws out another letter.
Sort the holes by the letter formed by the hole topology and extract the letter you obtain from connecting the muensters and you get the answer CURL UP.
Below are pictures of what the 3D maps of all of the holes look like.
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https://www.freeeconhelp.com/2011/08/what-is-market-equilibrium-how-supply.html
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What is market equilibrium? How supply and demand interact to reach equilibrium price and quantity - FreeEconHelp.com, Learning Economics... Solved!
## 8/18/11
Finding market equilibrium is a pretty easy task once you understand what to look for. The easiest way to find it is by looking for the point at which supply and demand cross, but it is a little difficult trying to understand why the market will converge to this point. This post goes over that process, using the market for educational consultants as an example.
On the graph to the write we have our typical supply and demand graph, with S representing our upward sloping supply curve, and D representing our downward sloping demand curve. We are now going to experiment with three different price levels and discuss what happens in the market when confronted with those three different prices.
The first price level we are going to experiment with will be above the point where the supply and demand curves cross. When prices are higher, we see that the line intersects the demand curve before the supply curve. This means that firms are willing to supply more educational consulting services then will be demanded in the market, so there will be a surplus. You can see this result in the graph to the left, but why is this happening? We have a surplus because at high prices suppliers are willing to produce a lot of a good or service because they can make a lot of money doing so, but consumers on the other hand have to pay the higher prices and therefore buy less of the good or service.
The other extreme is to have price below the point where supply and demand intersect. With this scenario, the price line runs into the supply curve first, and then the demand curve. Because we have more people demanding the product than those willing to supply it, we have a shortage. Considering our example, prices are so low that everyone wants to hire an educational adviser, but because prices are so low, not very many firms are willing to supply the service (maybe they will lose money).
So what happens when we have a price that isn't equal to the price where supply and demand cross? Lets go back to the surplus example first, when there is a surplus there are too many goods on the market. When this happens, firms need to lower their price in order to get rid of the extra merchandise (ever see clearance signs up at your local mall? They are trying to get rid of last years model, or surplus inventory). When the firms lower their price, then more people are willing to buy the good, and firms are less likely to supply it. This happens until the price lowers to the level shown below, where quantity demanded is equal to quantity supplied:
We can also analyze the shortage scenario. Here the pries are so low that more of the good is demanded than is being supplied. This means that market forces direct the price up, so that less people will demand the good, and firms will be more willing to supply it. This market force keeps pushing prices up until we reach equilibrium, and quantity demanded is equal to quantity supplied. For an example of this, think of NFL Super Bowl tickets. If the NFL sells tickets for too low of a price, then demand will be much higher than supply (which in this case is fixed). In order to reach equilibrium, scalpers will buy up tickets at the cheap price and sell them at a much higher price. As this price goes higher and higher we may see season ticket holders enter the market, which is our movement along the supply curve. Eventually we hit price equilibrium and the number of people willing to sell tickets is equal to those willing to buy.
Remember: Market equilibrium occurs at the point where supply and demand intersect. If the price is too high, we will have a surplus. If the price is too low we will have a shortage. In either of this situations, market forces will work to either raise or lower the price until we get to equilibrium, and the number of those willing and able to buy will be equal to those willing and able to sell.
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Skip to content
/ SymmetryBook Public
This book will be an undergraduate textbook written in the univalent style, taking advantage of the presence of symmetry in the logic at an early stage.
### License
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# SymmetryBook
This book will be an undergraduate textbook written in the univalent style, taking advantage of the presence of symmetry in the logic at an early stage.
## Style guide
• Try to be informal. Use as few formulas as possible, especially for the parts about type theory and logic, to ease the entry into group theory.
• We call objects in a type elements of that type even if the type is not a set.
• An element of a proposition can be called a proof.
• Identity types are denoted in general using the macro \eqto, which produces ⥱ (that is an arrow with an = on top). An element of an identity type is called an identification, and otherwise a path. We may say that it shows how to identify two elements. If the type is a set, we may denote its identity types by a = b and call them equations. When a = b has an element we say that a and b are equal.
• Types similar to identity types, like the type of eqivalences from A to B, are also denoted with a macro ending in "to", like \equivto, producing ⥲ (that is an arrow with an equivalence sign on top).
• The type containing the variable in a family is called the "parameter type", not the "index type", nor the "base type".
• Being equal by definiton is denoted with three lines and is called just that, and not definitionally equal or judgmentally equal.
• A synonym of "function" is "map". We don't use "mapping" or "application" as synonyms.
• In the preliminary chapters (up to subgroups), the underlying set map U from groups to sets has to be applied explicitly. Thereafter, it can be a coercion.
• Composition of p: a⥱b and q: b⥱c is denoted by either p∗q (p\ct q), or by q·p (q\cdot p), qp or q∘p (q\circ p). The latter is preferred when p and q come from equivalences.
• In dependent pairs, components having propositional type may be omitted.
• If x is a bound variable and c is less bound, then we prefer c ⥱ x to x ⥱ c. Typically, if c is the center of contraction.
• If k and n are number variables that can be renamed, then we prefer k < n to k > n or n < k.
• up to versus modulo regarding a group action: Up to is the stacky version, the orbit type (typically for us, a groupoid), whereas modulo refers to the set of connected components/the set of orbits. For example, given a group G, we have the groupoid of elements up to conjugation versus the set of elements modulo conjugation.
• Globally defined constants are typeset roman, while variables are italic. One exception is the B construction: The B matches whatever it operates on and joins to it without any space.
• When a structure is introduced and unpacked at the same time, use ≡ to connect the new variable with the unpacked parts. For example: “Let M ≡ (S,ι,μ) be a monoid”.
• Hints to exercises go in footnotes in the margin, with the footnote marker at the end of the exercise.
• Margin notes should usually to be made as footnotes (i.e., with a footnote marker).
• For a G-set X, we also write X for the underlying set, and we may write X^z to mean X twisted by a G-shape z : BG.
• Whenever possible, do not use a letter for a variable when the same letter is being used as an operator. E.g., try to avoid a variable B when the classifying type/map operator B is used in the same paragraph.
• Use macros with mathematical meaning, such as \conncomp, whenever possible, for uniformity of notation.
• Avoid the use of acronyms, such as LEM and LPO.
• Construct sort-order keys for glossary entries this way:
• for unary operators, use 1 followed by something (e.g., for $-y$ use (1-);
• for binary operators, use 2 followed by something (e.g., for $x+y$ use (2+);
• for numbers, use 8 followed by the number (e.g., for $0$ use (80).
• for identifiers in the Greek alphabet use 9 followed by the 2-digit ordinal number of the first letter (for proper alphabetization) and then something (e.g., for $\loops$ use (924Omega):
01 Α α, 02 Β β, 03 Γ γ, 04 Δ δ, 05 Ε ε, 06 Ζ ζ, 07 Η η, 08 Θ θ, 09 Ι ι,
10 Κ κ, 11 Λ λ, 12 Μ μ, 13 Ν ν, 14 Ξ ξ, 15 Ο ο, 16 Π π, 17 Ρ ρ, 18 Σ σ, 19 Τ τ,
20 Υ υ, 21 Φ φ, 22 Χ χ, 23 Ψ ψ, and 24 Ω ω;
• for identifiers in the Roman alphabet use the name (e.g., for $\Ker$ use (Ker) or (ker);
• Given a: A, we refer to elements of a ⥱ a as either symmetries of a, or symmetries in A.
## Compiling the book
To speed up compilation while writing the book, we cache the macros as a TeX format, and we externalize most of the figures, so that they're only compiled once (or when necessary after changes to them). This is managed with the Makefile, so just run make. The first run takes about 5–10 minutes, but subsequent compilations should finish in seconds.
## An icosahedron for your viewing pleasure
Shield:
This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.
## About
This book will be an undergraduate textbook written in the univalent style, taking advantage of the presence of symmetry in the logic at an early stage.
## Releases
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;; ** The Picky interpreter, no explicit types #lang pl #| The grammar: ::= | | { + } | { - } | { = } | { < } | { fun { } } | { call } | { with { } } | { if } The types are no longer part of the input syntax. Evaluation rules: eval(N,env) = N eval(x,env) = lookup(x,env) eval({+ E1 E2},env) = eval(E1,env) + eval(E2,env) eval({- E1 E2},env) = eval(E1,env) - eval(E2,env) eval({= E1 E2},env) = eval(E1,env) = eval(E2,env) eval({< E1 E2},env) = eval(E1,env) < eval(E2,env) eval({fun {x} E},env) = <{fun {x} E}, env> eval({call E1 E2},env1) = eval(Ef,extend(x,eval(E2,env1),env2)) if eval(E1,env1) = <{fun {x} Ef}, env2> = error! otherwise -- but this doesn't happen eval({with {x E1} E2},env) = eval(E2,extend(x,eval(E1,env),env)) eval({if E1 E2 E3},env) = eval(E2,env) if eval(E1,env) is true = eval(E3,env) otherwise Type checking rules (note the ambiguity of the `fun' rule): Γ ⊢ n : Number Γ ⊢ x : Γ(x) Γ ⊢ A : Number Γ ⊢ B : Number ——————————————————————————————— Γ ⊢ {+ A B} : Number Γ ⊢ A : Number Γ ⊢ B : Number ——————————————————————————————— Γ ⊢ {< A B} : Boolean Γ[x:=τ₁] ⊢ E : τ₂ ———————————————————————————— Γ ⊢ {fun {x} E} : (τ₁ -> τ₂) Γ ⊢ F : (τ₁ -> τ₂) Γ ⊢ V : τ₁ —————————————————————————————— Γ ⊢ {call F V} : τ₂ Γ ⊢ C : Boolean Γ ⊢ T : τ Γ ⊢ E : τ ——————————————————————————————————————— Γ ⊢ {if C T E} : τ Γ ⊢ V : τ₁ Γ[x:=τ₁] ⊢ E : τ₂ —————————————————————————————— Γ ⊢ {with {x V} E} : τ₂ |# (define-type PICKY [Num Number] [Id Symbol] [Add PICKY PICKY] [Sub PICKY PICKY] [Equal PICKY PICKY] [Less PICKY PICKY] [Fun Symbol PICKY] ; no types even here [Call PICKY PICKY] [With Symbol PICKY PICKY] [If PICKY PICKY PICKY]) (: parse-sexpr : Sexpr -> PICKY) ;; parses s-expressions into PICKYs (define (parse-sexpr sexpr) (match sexpr [(number: n) (Num n)] [(symbol: name) (Id name)] [(list '+ lhs rhs) (Add (parse-sexpr lhs) (parse-sexpr rhs))] [(list '- lhs rhs) (Sub (parse-sexpr lhs) (parse-sexpr rhs))] [(list '= lhs rhs) (Equal (parse-sexpr lhs) (parse-sexpr rhs))] [(list '< lhs rhs) (Less (parse-sexpr lhs) (parse-sexpr rhs))] [(list 'call fun arg) (Call (parse-sexpr fun) (parse-sexpr arg))] [(list 'if c t e) (If (parse-sexpr c) (parse-sexpr t) (parse-sexpr e))] [(cons 'fun more) (match sexpr [(list 'fun (list (symbol: name)) body) (Fun name (parse-sexpr body))] [else (error 'parse-sexpr "bad `fun' syntax in ~s" sexpr)])] [(cons 'with more) (match sexpr [(list 'with (list (symbol: name) named) body) (With name (parse-sexpr named) (parse-sexpr body))] [else (error 'parse-sexpr "bad `with' syntax in ~s" sexpr)])] [else (error 'parse-sexpr "bad expression syntax: ~s" sexpr)])) (: parse : String -> PICKY) ;; parses a string containing a PICKY expression to a PICKY AST (define (parse str) (parse-sexpr (string->sexpr str))) ;; Typechecker and related types and helpers ;; this is not a part of the AST now, and it also has a new variant ;; for type variables (see `same-type' for how it's used) (define-type TYPE [NumT] [BoolT] [FunT TYPE TYPE] [?T (Boxof (U TYPE #f))]) ;; this is similar to ENV, but it holds type information for the ;; identifiers during typechecking; it is essentially "Γ" (define-type TYPEENV [EmptyTypeEnv] [ExtendTypeEnv Symbol TYPE TYPEENV]) (: type-lookup : Symbol TYPEENV -> TYPE) ;; similar to `lookup' for type environments; note that the ;; error is phrased as a typecheck error, since this indicates ;; a failure at the type checking stage (define (type-lookup name typeenv) (cases typeenv [(EmptyTypeEnv) (error 'typecheck "no binding for ~s" name)] [(ExtendTypeEnv id type rest-env) (if (eq? id name) type (type-lookup name rest-env))])) (: typecheck : PICKY TYPE TYPEENV -> Void) ;; Checks that the given expression has the specified type. ;; Used only for side-effects, so return a void value. There ;; are two side-effects that it can do: throw an error if the ;; input expression doesn't typecheck, and type variables can ;; be mutated once their values are known -- this is done by ;; the `types=' utility function that follows. (define (typecheck expr type type-env) ;; convenient helpers (: type= : TYPE -> Void) (define (type= type2) (types= type type2 expr)) (: two-nums : PICKY PICKY -> Void) (define (two-nums e1 e2) (typecheck e1 (NumT) type-env) (typecheck e2 (NumT) type-env)) (cases expr [(Num n) (type= (NumT))] [(Id name) (type= (type-lookup name type-env))] [(Add l r) (type= (NumT)) (two-nums l r)] ; note that the [(Sub l r) (type= (NumT)) (two-nums l r)] ; order in these [(Equal l r) (type= (BoolT)) (two-nums l r)] ; things can be [(Less l r) (type= (BoolT)) (two-nums l r)] ; swapped... [(Fun bound-id bound-body) (let (;; the identity of these type variables is important! [itype (?T (box #f))] [otype (?T (box #f))]) (type= (FunT itype otype)) (typecheck bound-body otype (ExtendTypeEnv bound-id itype type-env)))] [(Call fun arg) (let ([type2 (?T (box #f))]) ; same here (typecheck arg type2 type-env) (typecheck fun (FunT type2 type) type-env))] [(With bound-id named-expr bound-body) (let ([type2 (?T (box #f))]) ; and here (typecheck named-expr type2 type-env) (typecheck bound-body type (ExtendTypeEnv bound-id type2 type-env)))] [(If cond-expr then-expr else-expr) (typecheck cond-expr (BoolT) type-env) (typecheck then-expr type type-env) (typecheck else-expr type type-env)])) (: types= : TYPE TYPE PICKY -> Void) ;; Compares the two input types, and throw an error if they don't ;; match. This function is the core of `typecheck', and it is used ;; only for its side-effect. Another side effect in addition to ;; throwing an error is when type variables are present -- they will ;; be mutated in an attempt to make the typecheck succeed. Note ;; that the two type arguments are not symmetric: the first type is ;; the expected one, and the second is the one that the code implies ;; -- but this matters only for the error messages. Also, the ;; expression input is used only for these errors. As the code ;; clearly shows, the main work is done by `same-type' below. (define (types= type1 type2 expr) (unless (same-type type1 type2) (error 'typecheck "type error for ~s: expecting ~a, got ~a" expr (type->string type1) (type->string type2)))) (: type->string : TYPE -> String) ;; Convert a TYPE to a human readable string, ;; used for error messages (define (type->string type) (format "~s" type) ;; The code below would be useful, but unfortunately it doesn't ;; work in some cases. To see the problem, try to run the example ;; below that applies identity on itself. It's left here so you ;; can try it out when you're not running into this problem. #| (cases type [(NumT) "Num"] [(BoolT) "Bool"] [(FunT i o) (string-append (type->string i) " -> " (type->string o))] [(?T box) (let ([t (unbox box)]) (if t (type->string t) "?"))]) |#) ;; Convenience type to make it possible to have a single `cases' ;; dispatch on two types instead of nesting `cases' in each branch (define-type 2TYPES [PairT TYPE TYPE]) (: same-type : TYPE TYPE -> Boolean) ;; Compares the two input types, return true or false whether ;; they're the same. The process might involve mutating ?T type ;; variables. (define (same-type type1 type2) ;; the `PairT' type is only used to conveniently match on both ;; types in a single `cases', it's not used in any other way (cases (PairT type1 type2) ;; flatten the first type, or set it to the second if it's unset [(PairT (?T box) type2) (let ([t1 (unbox box)]) (if t1 (same-type t1 type2) (begin (set-box! box type2) #t)))] ;; do the same for the second (reuse the above case) [(PairT type1 (?T box)) (same-type type2 type1)] ;; the rest are obvious [(PairT (NumT) (NumT)) #t] [(PairT (BoolT) (BoolT)) #t] [(PairT (FunT i1 o1) (FunT i2 o2)) (and (same-type i1 i2) (same-type o1 o2))] [else #f])) ;; Evaluator and related types and helpers (define-type ENV [EmptyEnv] [Extend Symbol VAL ENV]) (define-type VAL [NumV Number] [BoolV Boolean] [FunV Symbol PICKY ENV]) (: lookup : Symbol ENV -> VAL) ;; lookup a symbol in an environment, return its value or throw an ;; error if it isn't bound (define (lookup name env) (cases env [(EmptyEnv) (error 'lookup "no binding for ~s" name)] [(Extend id val rest-env) (if (eq? id name) val (lookup name rest-env))])) (: strip-numv : Symbol VAL -> Number) ;; converts a VAL to a Racket number if possible, throws an error if ;; not using the given name for the error message (define (strip-numv name val) (cases val [(NumV n) n] ;; this error will never be reached, see below for more [else (error name "expected a number, got: ~s" val)])) (: arith-op : (Number Number -> Number) VAL VAL -> VAL) ;; gets a Racket numeric binary operator, and uses it within a NumV ;; wrapper (define (arith-op op val1 val2) (NumV (op (strip-numv 'arith-op val1) (strip-numv 'arith-op val2)))) (: bool-op : (Number Number -> Boolean) VAL VAL -> VAL) ;; gets a Racket numeric binary predicate, and uses it ;; within a BoolV wrapper (define (bool-op op val1 val2) (BoolV (op (strip-numv 'bool-op val1) (strip-numv 'bool-op val2)))) (: eval : PICKY ENV -> VAL) ;; evaluates PICKY expressions by reducing them to values (define (eval expr env) (cases expr [(Num n) (NumV n)] [(Id name) (lookup name env)] [(Add l r) (arith-op + (eval l env) (eval r env))] [(Sub l r) (arith-op - (eval l env) (eval r env))] [(Equal l r) (bool-op = (eval l env) (eval r env))] [(Less l r) (bool-op < (eval l env) (eval r env))] [(Fun bound-id bound-body) (FunV bound-id bound-body env)] [(Call fun-expr arg-expr) (let ([fval (eval fun-expr env)]) (cases fval [(FunV bound-id bound-body f-env) (eval bound-body (Extend bound-id (eval arg-expr env) f-env))] ;; `cases' requires complete coverage of all variants, but ;; this `else' is never used since we typecheck programs [else (error 'eval "`call' expects a function, got: ~s" fval)]))] [(With bound-id named-expr bound-body) (eval bound-body (Extend bound-id (eval named-expr env) env))] [(If cond-expr then-expr else-expr) (let ([bval (eval cond-expr env)]) (if (cases bval [(BoolV b) b] ;; same as above: this case is never reached [else (error 'eval "`if' expects a boolean, got: ~s" bval)]) (eval then-expr env) (eval else-expr env)))])) (: run : String -> Number) ;; evaluate a PICKY program contained in a string (define (run str) (let ([prog (parse str)]) (typecheck prog (NumT) (EmptyTypeEnv)) (let ([result (eval prog (EmptyEnv))]) (cases result [(NumV n) n] ;; this error is never reached, since we make sure ;; that the program always evaluates to a number above [else (error 'run "evaluation returned a non-number: ~s" result)])))) ;; tests -- including translations of the FLANG tests (test (run "5") => 5) (test (run "{fun {x} {+ x 1}}") =error> "type error") (test (run "{call {fun {x} {+ x 1}} 4}") => 5) (test (run "{with {x 3} {+ x 1}}") => 4) (test (run "{with {identity {fun {x} x}} {call identity 1}}") => 1) (test (run "{with {add3 {fun {x} {+ x 3}}} {call add3 1}}") => 4) (test (run "{with {add3 {fun {x} {+ x 3}}} {with {add1 {fun {x} {+ x 1}}} {with {x 3} {call add1 {call add3 x}}}}}") => 7) (test (run "{with {identity {fun {x} x}} {with {foo {fun {x} {+ x 1}}} {call {call identity foo} 123}}}") => 124) (test (run "{with {x 3} {with {f {fun {y} {+ x y}}} {with {x 5} {call f 4}}}}") => 7) (test (run "{call {with {x 3} {fun {y} {+ x y}}} 4}") => 7) (test (run "{with {f {with {x 3} {fun {y} {+ x y}}}} {with {x 100} {call f 4}}}") => 7) (test (run "{call {call {fun {x} {call x 1}} {fun {x} {fun {y} {+ x y}}}} 123}") => 124) (test (run "{call {fun {x} {if {< x 2} {+ x 5} {+ x 6}}} 1}") => 6) (test (run "{call {fun {x} {if {< x 2} {+ x 5} {+ x 6}}} 2}") => 8) ;; Note that we still have a language with the same type system, ;; even though it looks like it could be more flexible -- for ;; example, the following two examples work: (test (run "{with {identity {fun {x} x}} {call identity 1}}") => 1) (test (run "{with {identity {fun {x} x}} {if {call identity {< 1 2}} 1 2}}") => 1) ;; but this doesn't, since identity can not be used with different ;; types: (test (run "{with {identity {fun {x} x}} {if {call identity {< 1 2}} {call identity 1} 2}}") =error> "type error") ;; this doesn't work either -- with an interesting error message: (test (run "{with {identity {fun {x} x}} {call {call identity identity} 1}}") =error> "type error") ;; ... but these two work fine: (test (run "{with {identity1 {fun {x} x}} {with {identity2 {fun {x} x}} {+ {call identity1 1} {if {call identity2 {< 1 2}} 1 2}}}}") => 2) (test (run "{with {identity1 {fun {x} x}} {with {identity2 {fun {x} x}} {call {call identity1 identity2} 1}}}") => 1)
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# chapter 2 diffusion
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Chapter 2: Diffusion in Solids2.1 2.2 2.3 2.4 Fundamental Equations of Diffusion Interstitial Diffusion Einstein Relation Atomic Mobility Substitutional Diffusion (Self-diffusion, Vacancy Diffusion, Darken Relation) Solutions to the Diffusion Equations
2.5
MMAT 305
Institute of Materials Science
DIFFUSION IN SOLIDS
WHY STUDY DIFFUSION? Materials often heat treated to improve properties Atomic diffusion occurs during heat treatment Depending on situation higher or lower diffusion rates desired Heat treating temperatures and times, and heating or cooling rates can be determined using the mathematics/physics of diffusion Example: steel gears are case-hardened by diffusing C or N to outer surfaceInstitute of Materials Science
DIFFUSION IN SOLIDSDiffusion: Material transport by atomic or particle transport from region of high to low concentration (???)
What forces the particles to go from left to right? Does each particle know its local concentration? Every particle is equally likely to go left or right! At the interfaces in the above picture, there are more particles going right than left this causes an average flux of particles to the right! Largely determined by probability & statisticsInstitute of Materials Science
DIFFUSION IN SOLIDS Interdiffusion: In an alloy or diffusion couple, atoms tendto migrate from regions of large to lower concentration.Initially (diffusion couple) After some time
Adapted from Figs. 5.1 and 5.2, Callister 6e.
Cu 100%
Ni
100%
0
Concentration Profiles
0 Concentration ProfilesInstitute of Materials Science
Correct Definition of Diffusion
Fig. 2.1 Free energy and chemical potential changes during diffusion. (a) and (b) down-hill diffusion. (c) and (d) up-hill 2 1 diffusion. (e) Q A " Q A therefore A atoms move from (2) to (1), 2 2 Q 1 " Q B therefore B atoms move from (1) to (2). (f) Q 1 " Q A B A 2 therefore A atoms move from (1) to (2), Q B " Q 1 therefore B B atoms move from (2) to (1).
MMAT 305
Institute of Materials Science
Correct Definition of Diffusion
Diffusion: Material transport by atomic transport from regions of high to low CHEMICAL POTENTIALS
Institute of Materials Science
2.1 Fundamental Equations of Diffusion
Institute of Materials Science
2.2 Interstitial Diffusion1 2
a)
atoms of parent lattice B interstitials
b) CB
axC B a ! (C xxX
Interstitial diffusion by random jumps in a concentration gradient.MMAT 305 Institute of Materials Science
2.2 Interstitial Diffusiona) C
0 J b)J1 J2
X X H X
X
0 c)J1
1
2
XJ2
HX area A
Figure 2.2.1 The derivation of Ficks second law.
MMAT 305
Institute of Materials Science
2.3 Einstein Relation Atomic Mobility
G2
(Gm1 3
a
MMAT 305
Institute of Materials Science
2.4 Substitutional Diffusion(Self-diffusion, Vacancy diffusion, Darken Relation)
Interdiffusion and vacancy flow. (a) Composition profile after interdiffusion of A and B. (b) The corresponding fluxes of atoms and vacancies as a function of position x. (c) The rate at which the vacancy concentration would increase or decrease if vacancies were not created or destroyed by dislocation climb.
MMAT 305
Institute of Materials Science
2.4 Substitutional Diffusion Self-diffusion: In an elemental solid, atomsalso migrate.Label some atoms After some time
C C A A D B B D
Institute of Materials Science
2.4 Substitutional DiffusionVacancy Diffusion
Fig. 2.17 (a) before, (b) after: a vacancy is absorbed at a jog on an edge dislocation (positive climb). (b) before, (a) after: a vacancy is created by negative climb of an edge dislocation. (c) Perspective drawing of a jogged edge dislocation.
Fig. 2.16 The jumping of atoms in one direction can be considered as the jumping of vacancies in the other direction. Fig. 2.18 A flux of vacancies causes the atomic planes to move through the specimen. MMAT 305 Institute of Materials Science
2.4 Substitutional DiffusionSubstitutional Diffusion: applies to substitutional impurities atoms exchange with vacancies rate depends on: -- number of vacancies -- temperature -- activation energy to exchange.
i
r
i g l p
d timInstitute of Materials Science
2.4 Substitutional Diffusion
Initial state
Intermediate state
Final state
Energy
Activation energy
Also called energy barrier for diffusionInstitute of Materials Science
2.5 Solutions to the Diffusion Equations Copper diffuses into a bar of aluminum.S rf f e t .,
(x,t)
pr - xi ti
.,
f
b r pper t
to t
t3 t2 position, x
Adapted from Fig. 5.5, Callister 6e.
Boundary conditions: For t = 0, C = C0 at x > 0 For t > 0, C = Cs at x = 0 C = C0 at x =
d2C dC =D 2 dt dxInstitute of Materials Science
2.5 Solutions to the Diffusion Equations Copper diffuses into a bar of aluminum.S rf ce conc., toms s of
Cs
C(x,t)
pre-existing onc.,
b r o of copper toms
o
to t
t3 t2 position, x
Adapted from Fig. 5.5, Callister 6e.
General solution:
x Co ! C(x, t) 1 erf 2 Dt Cs Co"error function"
Institute of Materials Science
2.5 Solutions to the Diffusion Equations Suppose we desire to achieve a specific concentration C1 at a certain point in the sample at a certain time C ( x , t ) C0 x ! 1 erf C s C0 2 Dt becomes C1 C0 x ! constant ! 1 erf Cs C0 2 Dt x2 ! constant Dt
Institute of Materials Science
2.5 Solutions to the Diffusion Equations The experiment: record combinations oft and x that kept C constant.to t1 t2 t3 xo x1 x2 x3
C(x i i Co x i ! 1 erf = (constant here) Cs Co i
Diffusion depth given by:
x i w Dt iInstitute of Materials Science
2.5 Solutions to the Diffusion Equations Copper diffuses into a bar of aluminum. 10 hours at 600C gives desired C(x). How many hours would it take to get the same C(x) if we processed at 500C, given D500 and D600? Key point 1: C(x,t500C) = C(x,t600C). Key point 2: Both cases have the same Co and Cs.
Result: Dt should be held constant. x C(x,t) Co (Dt) 5 = 1 erf 2Dt C C o s 5.3x Answer: 4.8x - 3m2/s 10hrs
C (Dt) 6
C
t5
- 4m2/s
(Dt)6 ! D5
!
hr
Note: values of D are provided here.Institute of Materials Science
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Necessary Topics to Master Before Completing Statistical Modeling Assignments
August 28, 2023
Harry Windsor
🇬🇧 United Kingdom
Statistical Models
Harry Windsor, a distinguished expert in GARCH models, holds a Ph.D. in statistics from the University of San Diego with 13+ years of experience. He provides invaluable assistance to students in completing their statistical assignments.
Key Topics
• Honing the Foundation
• Probability and Distributions
• Descriptive Statistics
• Hypothesis Testing and Inference
• Regression Analysis
• Experimental Design
• Model Validation and Overfitting
• Approaching Statistical Modeling Assignments
• Understand the Assignment
• Data Exploration
• Choose the Right Model
• Feature Selection and Engineering
• Model Building and Training
• Model Evaluation
• Iterate and Optimize
• Practice Regularly
• Seek Resources
• Collaborate and Discuss
• Learn from Feedback
• Real-World Applications
• Conclusion
Statistical modeling is a powerful tool that allows us to extract insights and make informed decisions from data. Whether you're a student diving into the world of statistics or a professional seeking to enhance your analytical skills, understanding the key topics and strategies for tackling assignments on statistical modeling is crucial. In this guide, we'll explore the fundamental topics you should be familiar with before starting a statistical modeling assignment and provide effective techniques for solving statistical modeling assignment successfully.
Honing the Foundation
Before delving into statistical modeling assignments, it's essential to have a solid grasp of the foundational concepts. A strong understanding of probability, distributions, descriptive statistics, hypothesis testing, regression analysis, experimental design, and model validation sets the stage for successful statistical modeling assignments. These foundational topics provide the tools to unravel data intricacies and create robust models that extract meaningful insights from complex datasets. Here are the key topics you should be well-versed in:
Probability and Distributions
Probability forms the bedrock of statistical modeling, enabling us to quantify uncertainty. It's the language through which we express chances and make informed decisions in the face of randomness. Understanding random variables and their distributions is crucial; they're the building blocks of various modeling techniques. Distributions like the normal, binomial, and Poisson distributions help us comprehend how data is spread out and the likelihood of different outcomes.
As you embark on statistical modeling assignments, grasp the significance of these concepts. Probability and distributions offer the lens through which you view data's underlying patterns. Armed with this knowledge, you can effectively assess uncertainty, formulate hypotheses, and develop models that capture the essence of real-world phenomena. A solid foundation here paves the way for more advanced modeling techniques and a deeper understanding of data's hidden stories.
Descriptive Statistics
Descriptive statistics breathes life into raw data, revealing its story in a comprehensible manner. By exploring measures of central tendency like the mean, median, and mode, you gain insights into the data's typical behavior. Variability measures such as the range, variance, and standard deviation highlight data's spread and consistency.
In statistical modeling assignments, these tools are your initial glimpse into the dataset's characteristics. They aid in identifying potential outliers, understanding data distribution, and making informed decisions on preprocessing steps. For effective communication, visualizations like histograms, box plots, and scatter plots paint a clearer picture of the data's landscape.
Incorporate descriptive statistics into your analytical toolkit. They empower you to interpret data, discern patterns, and ultimately construct accurate models. A strong grasp of these fundamentals enhances your ability to extract meaningful insights from the often intricate world of statistical modeling.
Hypothesis Testing and Inference
Hypothesis testing and inference are the bridges connecting sample data to broader conclusions. They provide a systematic way to make sense of observations and draw reliable insights. In statistical modeling, hypothesis testing involves proposing assumptions about populations and then assessing the likelihood of those assumptions based on sample data.
These concepts guide your decision-making process in assignments. By setting up null and alternative hypotheses, calculating p-values, and establishing confidence intervals, you're equipped to quantify the uncertainty associated with your findings. These techniques ensure your results are not merely based on chance but hold true significance.
Mastering hypothesis testing and inference is akin to becoming a detective of data. These tools enable you to uncover hidden relationships, validate assumptions, and communicate your findings with confidence. In the realm of statistical modeling, these skills empower you to transform raw data into actionable insights.
Regression Analysis
Regression analysis is the flashlight illuminating the intricate relationships between variables. It's a versatile tool that helps you understand how one variable influences another. Linear regression, multiple regression, and logistic regression are essential techniques that model these connections.
In assignments, regression analysis plays a pivotal role in predictive modeling. By identifying the underlying patterns and quantifying relationships, you can make informed forecasts and decisions. Understanding coefficients, interpreting results, and assessing model fit are crucial steps in this process.
Equipped with regression analysis skills, you can unveil insights hidden within the data's complexity. These insights inform strategies, policies, and business decisions across various domains. By mastering regression, you position yourself to solve real-world problems and extract valuable knowledge from data.
Experimental Design
Experimental design is the compass guiding you through the realm of data collection and analysis. It's the meticulous planning that ensures your findings are credible and meaningful. Understanding the distinction between observational and experimental studies is crucial. By carefully controlling variables, randomizing conditions, and considering potential confounding factors, you reduce biases and enhance the reliability of your results.
In assignments, a solid grasp of experimental design helps you create a robust foundation for analysis. Whether you're investigating the effects of a new drug, evaluating marketing strategies, or studying social behaviors, proper experimental design ensures your findings accurately reflect reality.
By mastering experimental design, you cultivate the ability to set up controlled studies that provide insights with integrity. This skill is invaluable in the world of statistical modeling, enabling you to contribute to the pool of accurate knowledge and make informed decisions based on sound evidence.
Model Validation and Overfitting
Model validation is the compass steering you toward accurate predictions and reliable insights. It's the process of assessing how well your model performs on unseen data. However, a pitfall to watch out for is overfitting, where a model fits training data too closely and fails to generalize well to new data.
In assignments, model validation is your litmus test for success. Techniques like cross-validation ensure your model doesn't suffer from overfitting. By splitting data into training and validation sets, you're able to fine-tune your model's parameters and evaluate its performance on fresh data.
Understanding model validation and guarding against overfitting is crucial in statistical modeling. It ensures your insights are not artifacts of your training data but rather robust representations of reality. By mastering these aspects, you position yourself to develop models that stand up to the complexities of real-world data.
Approaching Statistical Modeling Assignments
Once you've solidified your understanding of the foundational topics, it's time to tackle statistical modeling assignments with confidence. Here's a step-by-step guide to help you navigate these assignments effectively:
Understand the Assignment
The first step to acing a statistical modeling assignment is deciphering the task at hand. Carefully read the instructions to grasp the objectives, dataset, and required techniques. Clarity is paramount. Seek clarification if needed. Understanding the assignment ensures you channel your efforts in the right direction from the start, setting the stage for a well-executed and relevant analysis that aligns seamlessly with the assignment's requirements.
Data Exploration
Before embarking on any modeling journey, data exploration is your compass. It's the process of unearthing insights, patterns, and anomalies within your dataset. By visualizing distributions, identifying outliers, and understanding variable relationships, you lay the groundwork for informed decisions throughout your assignment. Data exploration not only helps you clean and preprocess data effectively but also guides your modeling choices. It's the initial step that sets the tone for a successful statistical modeling endeavor.
Choose the Right Model
Selecting the appropriate model is akin to choosing the right tool for a task. Just as a carpenter wouldn't use a hammer to paint a canvas, you shouldn't use a model that doesn't fit your data. In assignments, understanding the problem's nature—regression, classification, clustering—guides your choice. Picking the right model increases the chances of obtaining accurate and meaningful results, setting the stage for successful statistical modeling.
Feature Selection and Engineering
Feature selection and engineering are the chisels you use to craft models that capture data's essence. In assignments, identifying the right features to include can significantly impact your model's accuracy. Meanwhile, feature engineering involves transforming or creating variables to better represent relationships.
By understanding the relevance of features and applying creative engineering, you enhance your model's ability to discern patterns and make accurate predictions. These skills empower you to extract maximum value from data, shaping it into a form that magnifies the power of statistical modeling.
Model Building and Training
Model building is the art of transforming data into predictive power. Assignments challenge you to apply theoretical knowledge to construct models tailored to specific problems. This process involves selecting appropriate algorithms, setting hyperparameters, and training the model on the data. Through iterations and fine-tuning, you cultivate the ability to develop models that capture data intricacies and generate accurate predictions. This hands-on experience equips you to bridge the gap between theory and practical application, a crucial skill in the realm of statistical modeling.
Model Evaluation
Model evaluation is the yardstick by which you measure your model's effectiveness. It involves using specific metrics to assess how well your model performs its intended task. For regression, mean squared error quantifies prediction accuracy; for classification, metrics like accuracy, precision, and recall provide nuanced insights. By mastering model evaluation, you gain the ability to objectively determine your model's strengths and weaknesses, refining your approach and delivering more accurate results in statistical modeling assignments.
Iterate and Optimize
Embrace the iterative nature of statistical modeling assignments. Rarely does a model perform perfectly on the first try. Take a systematic approach to refining your work. Tweak parameters, experiment with different algorithms, and reevaluate your data preprocessing steps. Each iteration hones your model's accuracy and enhances your problem-solving skills. Through persistence and adaptation, you'll transform initial concepts into sophisticated models that extract valuable insights from complex data landscapes.
Becoming proficient in statistical modeling takes practice and dedication. Here are additional strategies to enhance your skills and excel in your assignments:
Practice Regularly
Statistical modeling proficiency is honed through consistent practice. Engage with diverse datasets, experiment with various techniques, and tackle a range of problems. Just as athletes train to excel, regular practice builds your intuition and refines your analytical prowess. Each modeling encounter adds to your expertise, bringing you closer to mastering the art of turning data into actionable intelligence.
Seek Resources
Expand your knowledge by tapping into a wealth of online resources. Platforms like Khan Academy, Coursera, and DataCamp offer comprehensive courses on statistical modeling. These resources provide interactive learning experiences, allowing you to grasp complex concepts at your own pace. Supplementing your coursework with these resources empowers you to dive deeper into the world of statistical modeling, fostering a strong foundation for assignments.
Collaborate and Discuss
Engaging in discussions with peers and mentors enriches your understanding of statistical modeling. Explaining your thoughts and learning from others' viewpoints widens your horizons. Collaborative environments foster creativity and expose you to alternative approaches. Through shared insights, you can refine your strategies and conquer challenges more effectively, ultimately elevating your ability to tackle statistical modeling assignments with confidence.
Learn from Feedback
Feedback on your assignments is a treasure trove of learning. Understand where you went wrong, why certain approaches were ineffective, and how to rectify mistakes. Adapting based on feedback sharpens your analytical acumen, enabling you to tackle future challenges with improved strategies and a deeper understanding of statistical modeling nuances.
Real-World Applications
Understanding statistical modeling isn't just theoretical—it has tangible real-world applications. Analyzing case studies, research papers, or business reports showcases how statistical models influence decision-making. Whether it's predicting market trends, optimizing supply chains, or studying disease spread, these models empower you to tackle complex challenges and contribute meaningfully to diverse industries, bridging the gap between data and impactful solutions.
Conclusion
A strong foundation in probability, distributions, descriptive statistics, hypothesis testing, regression analysis, experimental design, and model validation is paramount. Armed with these essential topics and guided by effective strategies, you're well-equipped to solve your statistical modeling assignments with confidence. Remember, practice and persistence are key. By mastering these concepts, you'll not only excel in your assignments but also unlock the power to solve intricate problems and extract meaningful insights from data.
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# In a boat 25 persons were sitting. Their average weight increased one kilogram when One man goes and a new man comes in. The weight of the new man is 70kgs. Find the Weight of the man who is going.
1
by Deleted account
2014-07-29T21:55:05+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
W1 = avg weight initially
25 * W1 = weight of 25 persons in itially
Let weight of the man who leaves = W
remaining 24 persons weight is = 25 W1 - W
When new man comes in total weight = 25 W1 - W + 70
Average weight of the new set of 25 persons = (25 W1 - W +70) / 25
= W1 + 1 one more than previous average
=> 25W1 - W +70 = 25 W1 + 25
=> W = 45 Kg
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# Need help understanding rotary attachment configuration
I purchased an OemTech K40+ machine a year ago, and it came with a roller attachment and Lightburn included. Recently I decided to use the rotary, but I am having many issues with it to the point I am even thinking this attachment is defective.
The physical installation is done, but the Lightburn settings are confusing
To start, the mm per rotation is not clear. Lightburn says it is the circumference of the rollers. The diameter of the roller is 25 mm and the circumference is 78.54 mm. Why is Lightburn asking for it when it can be calculated with the diameter?
Also, once I entered the diameter of the object (60.7 mm) and hit âTESTâ, the roller rotates the object almost one full rotation. Why not a full rotation if I entered what Lightburn asked for? After adjusting the mm per rotation (from 78.54 to 83), it does a complete rotation.
Here is what is even more confusing:
If the TEST is working fine, I would guess the settings are correct and I can go ahead and start engraving, or that is what I thoughtâŚwrong
If the objectâs circumference is 190.695 (based on its diameter), I would understand I could be able to engrave a thin square with a length of 190.695 mm and cover the whole circumference of the object, however, when I frame the movement of the laser and the rotary, it covers twice the circumference of the object, meaning the square would completely overlap. What am I missing here?
I went and changed the mm per rotation to 41.5 so the framing shows one full rotation. Why does this value need to be adjusted if the roller diameter and the mm per rotation should never be adjusted? or that is what I thought.
Once more, by changing the mm per rotation messed up the TEST part in the rotary setup because it now rotates the object half. It needs 83.
Also, if this is a common issue setting up this attachment for every object with different diameter?
I decided to engrave a 40x40 graphic and the result is disastrous. The graphic is completely distorted and deformed.
I am using âCurrent Positionâ instead of âAbsolute Positionâ and is it normal for the object practically rotates briskly before starting to engrave and once finished, rotates the object a few times really fast before stopping? Is there a way to tell Lightburn to start engraving right on the current location and stop and remain in the position where it finished engraving?
If you have a roller rotary, there is no need for an object diameter ⌠you only need this with a chuck type rotary.
When you click on the test button, the roller should rotate one complete rotation and backâŚ
Since a roller drives the surface of the object, not the center, there is no need for an object diameter.
Make sense?
@jkwilborn is correct.
Thanks for your responses, but I am still not sure why I cannot make my rotary attachment work properly.
I understand why the object diameter is not required for the setup, however, the diameter and therefore the circumference are needed to know the maximum area available for engraving, correct?
If I have a bottle with 2.4â in diameter (circumference of 7.5â), I assume I can create a drawing 7.5â long and fit perfectly around the bottle âs surface, but I canât. The drawing is stretched so long that is overlaps a lot as if the drawing were around 12â or 14â long.
Why is it doing that?
Do I need to go every time to the rotary setup and change the âmm per rotation until the âFrameâ tool show the proper rotation? Or just reduce my drawing âs height accordingly?
It would benefit all of us if you would answer the questions put to you. You have added nothing to help us assist you with this issue.
Good luck
I donât see any questions asked? Unless Iâm missing something.
I know how I got my roller rotary to work but the community doesnât agree with my settings.
Here are some troubleshooting I found from lasers123.com.
a. In LightBurn, create a rectangle that is 80mm wide x 5mm high.
b. Rotate the design 90-degrees so that the 80mm dimension runs along the round part of the cup.
c. Engrave the design.
d. Measure the rectangle in mm. (For this example, lets say it measured 95mm.)
e. Divide the measurement by 80. (e.g., 95 / 80 = 1.1875)
f. Find your current steps per rotation. (For this example, lets say it is 5000.)
g. Divide your current steps per rotation by the value obtained in step e. (e.g., 5000 / 1.1875 = 4210.53)
h. Test the value calculated in step g as your steps per rotation.
Here it is a screen shot of my rotary settings
When I hit TEST, the object does a full rotation and it returns to the initial position, which it is what it shoul do, right?
But, when I frame the 155 mm thin square you see next to the rotary setup, it does 2 full rotations.
Lightburn needs to know the diameter of the roller. I assume you have that correct. Is that diameter marked on the rotary or in the docs, or did you measure it?
The mm per rotation is not directly related to the diameter of the roller, it is related to what your controller setting is for your Y axis. You need to experiment with that setting until you get the correct setting, then you should never need to change it again. Since you said itâs making 2 full rotations at 84, change it to 42.
Lets work with easier numbers. Make a 100mm skinny box, cover your tumbler or whatever you are working with in painters tape and set your layer to a low power burn, just enough to mark the tape. Run the test and remove the tape and place on a flat surface so you can measure it. If it now measures 100mm, youâre good. It will probably be off by a little bit at this point. If the test burn is less than 100, increase your mm per rotation and repeat the test. If larger, reduce. It is trial and error until you get it perfect.
Micrololin,
I did what you mentioned and it works, but that is what I wanted to avoid: making calculations everytime I wanted to engrave objects with diferent diameter. I thougth it was only matter of measuring the diameter of the object and adjust my drawings based on the maximum area available on that object.
I guess I will have to get used to that now.
If you now have a correct setting for the rotary, you do not need to calculate anything. Those settings will not change. You only need to know your object diameter for your design. You donât need to change the rotary settings.
Since itâs a roller it is using distance covered by the roller, if it were a chuck rotary you would have to calculate the object diameter every time.
Thelmuth,
I guess as long as I use objects with the same diameter, I will not beed to modify anything, but when I use objects with different diameter I will have to adjust the mm per rotation,correct?
I mention this because when I use a different object, the TEST function does not make a full rotation
No, because it is traveling along the outer circumference of the object. It knows that when the roller travels 1 rotation it covers x distance. It doesnât matter if your object is 1" or 5" diameter.
Forget the test button. Use what works. Try it on something of a different diameter using the tape. You should see the same results.
You might need to adjust your designs SLIGHTLY to get a perfect full wrap, if that 's what youâre going for. Itâs easy to stretch or squeeze a little bit if needed. But donât keep changing the settings once you have them dialed in.
2 Likes
Thank you Thelmuth,
I finally understand now.
2 Likes
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http://www.philipzucker.com/solving-the-xy-model-using-mixed-integer-optimization-in-python/
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There are many problems in physics that take the form of minimizing the energy. Often this energy is taken to be quadratic in the field. The canonical example is electrostatics. The derivative of the potential $\phi$ gives the electric field E. The energy is given as $\int (|\nabla \phi|^2 + \phi \rho) d^3 x$. We can encode a finite difference version of this (with boundary conditions!) directly into a convex optimization modelling language like so.
import cvxpy as cvx
import numpy as np
import matplotlib.pyplot as plt
import scipy.linalg
from mpl_toolkits import mplot3d
N = 10
# building a finite difference matrix. It is rectangle of size N x (N-1). It maps from the vertices of our grid to the lines in between them, where derivatives live.
col = np.zeros(N)
col[0] = -1
col[1] = 1
delta = scipy.linalg.toeplitz(col, np.zeros(N-1)).T
print(delta)
# a variable for our potential
phi = cvx.Variable((N, N))
# vectorization is useful. It flattens out the x-y 2Dness.
phivec = cvx.vec(phi)
constraints = []
# boundary conditions. Dirichlet
constraints += [phi[:,0] == 0, phi[0,:] == 0, phi[:,-1] == 0, phi[-1,:] == 0 ]
# fixed charge density rho
rho = np.zeros((N,N))
rho[N//2,N//2] = 1
print(rho)
# objective is energy
objective = cvx.Minimize(V + cvx.sum(cvx.multiply(rho,phi)))
prob = cvx.Problem(objective, constraints)
res = prob.solve()
print(res)
print(phi.value)
# Plotting
x = np.linspace(-6, 6, N)
y = np.linspace(-6, 6, N)
X, Y = np.meshgrid(x, y)
fig = plt.figure()
ax = plt.axes(projection='3d')
ax.plot_surface(X, Y, phi.value, rstride=1, cstride=1,
cmap='viridis', edgecolor='none')
plt.show()
The resulting logarithm potential
It is noted rarely in physics, but often in the convex optimization world that the barrier between easy and hard problems is not linear vs. nonlinear, it is actually more like convex vs. nonconvex. Convex problems are those that are bowl shaped, on round domains. When your problem is convex, you can’t get caught in valleys or on corners, hence greedy local methods like gradient descent and smarter methods work to find the global minimum. When you differentiate the energy above, it results in the linear Laplace equations $\nabla^2 \phi = \rho$. However, from the perspective of solvability, there is not much difference if we replace the quadratic energy with a convex alternative.
def sum_abs(x):
return cvx.sum(cvx.abs(x))
a = 1
V = cvx.sum(cvx.maximum( -a - dxphi, dxphi - a, 0 )) + cvx.sum(cvx.maximum( -a - dyphi, dyphi - a, 0 ))
Materials do actually have non-linear permittivity and permeability, this may be useful in modelling that. It is also possible to consider the convex relaxation of truly hard nonlinear problems and hope you get the echoes of the phenomenology that occurs there.
Another approach is to go mixed integer. Mixed Integer programming allows you to force that some variables take on integer values. There is then a natural relaxation problem where you forget the integer variables have to be integers. Mixed integer programming combines a discrete flavor with the continuous flavor of convex programming. I’ve previously shown how you can use mixed integer programming to find the lowest energy states of the Ising model but today let’s see how to use it for a problem of a more continuous flavor.
As I’ve described previously, in the context of robotics, the non-convex constraint that variables lie on the surface of a circle can be approximated using mixed integer programming. We can mix this fairly trivially with the above to make a global solver for the minimum energy state of the XY model. The XY model is a 2d field theory where the value of the field is constrained to lie on a circle. It is a model of a number of physical systems, such as superconductivity, and is the playground for a number of interesting phenomenon, like the Kosterlitz-Thouless phase transition. Our encoding is very similar to the above except we make two copies of the field $phi$ and we then force them to lie on a circle. I’m trying to factor out the circle thing into my library cvxpy-helpers, which is definitely a work in progress.
import cvxpy as cvx
import numpy as np
import matplotlib.pyplot as plt
import scipy.linalg
from mpl_toolkits import mplot3d
from cvxpyhelpers import cvxpyhelpers as mip
N = 6
# building a finite difference matrix. It is rectangle of size Nx(N-1). It maps from the vertices of our grid to the lines in between them, where derivatives live.
col = np.zeros(N)
col[0] = -1
col[1] = 1
delta = scipy.linalg.toeplitz(col, np.zeros(N-1)).T
print(delta)
# a variable for our potential
phix = cvx.Variable((N, N))
phiy = cvx.Variable((N, N))
# vectorization is useful. It flattens out the x-y 2Dness.
phixvec = cvx.vec(phix)
phiyvec = cvx.vec(phiy)
def sum_abs(x):
return cvx.sum(cvx.abs(x))
constraints = []
# coundary conditions. Nice and vortexy.
constraints += [phix[:,0] >= 0.9, phiy[0,1:-1] >= 0.9, phix[:,-1] <= -0.9, phiy[-1,1:-1] <= -0.9 ]
for i in range(N):
for j in range(N):
x, y, c = mip.circle(4)
constraints += c
constraints += [phix[i,j] == x]
constraints += [phiy[i,j] == y]
# fixed charge density rho
rho = np.ones((N,N)) * 0.01
rho[N//2,N//2] = 1
print(rho)
# objective is energy
objective = cvx.Minimize(V + cvx.sum(cvx.multiply(rho,phix)))
prob = cvx.Problem(objective, constraints)
print("solving problem")
res = prob.solve(verbose=True, solver=cvx.GLPK_MI)
print(res)
print(phix.value)
# Plotting
x = np.linspace(-6, 6, N)
y = np.linspace(-6, 6, N)
X, Y = np.meshgrid(x, y)
fig = plt.figure()
plt.quiver(X,Y, phix.value, phiy.value)
plt.show()
Now, this isn’t really an unmitigated success as is. I switched to an absolute value potential because GLPK_MI needs it to be linear. ECOS_BB works with a quadratic potential, but it was not doing a great job. The commercial solvers (Gurobi, CPlex, Mosek) are supposed to be a great deal better. Perhaps switching to Julia, with it’s richer ecosystem might be a good idea too. I don’t really like how the solution of the absolute value potential looks. Also, even at such a small grid size it still takes around a minute to solve. When you think about it, it is exploring a ridiculously massive space and still doing ok. There are hundreds of binary variables in this example. But there is a lot of room for tweaking and I think the approach is intriguing.
Musings:
• Can one do steepest descent style analysis for low energy statistical mechanics or quantum field theory?
• Is the trace of the mixed integer program search tree useful for perturbative analysis? It seems intuitively reasonable that it visits low lying states
• The Coulomb gas is a very obvious candidate for mixed integer programming. Let the charge variables on each grid point = integers. Then use the coulomb potential as a quadratic energy. The coulomb gas is dual to the XY model. Does this exhibit itself in the mixed integer formalism?
By absolute value potential, I mean using $|\nabla \phi|$ as compared to a more ordinary quadratic $|\nabla \phi|^2$ .
$|\nabla \phi| = |\nabla \phi_x| + |\nabla \phi_y|$ ? This is the only linear thing I can think of.
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# Peeter Joot's (OLD) Blog.
• ## Archives
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## PHY450H1S. Relativistic Electrodynamics Tutorial 4 (TA: Simon Freedman). Waveguides: confined EM waves.
Posted by peeterjoot on March 14, 2011
# Motivation
While this isn’t part of the course, the topic of waveguides is one of so many applications that it is worth a mention, and that will be done in this tutorial.
We will setup our system with a waveguide (conducting surface that confines the radiation) oriented in the $\hat{\mathbf{z}}$ direction. The shape can be arbitrary
PICTURE: cross section of wacky shape.
## At the surface of a conductor.
At the surface of the conductor (I presume this means the interior surface where there is no charge or current enclosed) we have
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= 0\end{aligned} \hspace{\stretch{1}}(1.1)
If we are talking about the exterior surface, do we need to make any other assumptions (perfect conductors, or constant potentials)?
## Wave equations.
For electric and magnetic fields in vacuum, we can show easily that these, like the potentials, separately satisfy the wave equation
Taking curls of the Maxwell curl equations above we have
\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{E}}}{\partial {{t}}^2} \\ \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{B}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{B}}}{\partial {{t}}^2},\end{aligned} \hspace{\stretch{1}}(1.5)
but we have for vector $\mathbf{M}$
\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{M})=\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{M}) - \Delta \mathbf{M},\end{aligned} \hspace{\stretch{1}}(1.7)
which gives us a pair of wave equations
\begin{aligned}\square \mathbf{E} &= 0 \\ \square \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(1.8)
We still have the original constraints of Maxwell’s equations to deal with, but we are free now to pick the complex exponentials as fundamental solutions, as our starting point
\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{i k^a x_a} = \mathbf{E}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{i k^a x_a} = \mathbf{B}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) },\end{aligned} \hspace{\stretch{1}}(1.10)
With $k_0 = \omega/c$ and $x_0 = c t$ this is
\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) }.\end{aligned} \hspace{\stretch{1}}(1.12)
For the vacuum case, with monochromatic light, we treated the amplitudes as constants. Let’s see what happens if we relax this assumption, and allow for spatial dependence (but no time dependence) of $\mathbf{E}_0$ and $\mathbf{B}_0$. For the LHS of the electric field curl equation we have
\begin{aligned}0 &= \boldsymbol{\nabla} \times \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + \mathbf{E}_0 \times \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + i \mathbf{E}_0 \times \mathbf{k} ) e^{i k_a x^a}.\end{aligned}
Similarly for the divergence we have
\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 ) e^{i k_a x^a}.\end{aligned}
This provides constraints on the amplitudes
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 - i \mathbf{k} \times \mathbf{E}_0 &= -i \frac{\omega}{c} \mathbf{B}_0 \\ \boldsymbol{\nabla} \times \mathbf{B}_0 - i \mathbf{k} \times \mathbf{B}_0 &= i \frac{\omega}{c} \mathbf{E}_0 \\ \boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B}_0 - i \mathbf{k} \cdot \mathbf{B}_0 &= 0\end{aligned} \hspace{\stretch{1}}(1.14)
Applying the wave equation operator to our phasor we get
\begin{aligned}0 &=\left(\frac{1}{{c^2}} \partial_{tt} - \boldsymbol{\nabla}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})} \\ &=\left(-\frac{\omega^2}{c^2} - \boldsymbol{\nabla}^2 + \mathbf{k}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})}\end{aligned}
So the momentum space equivalents of the wave equations are
\begin{aligned}\left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(1.18)
Observe that if $c^2 \mathbf{k}^2 = \omega^2$, then these amplitudes are harmonic functions (solutions to the Laplacian equation). However, it doesn’t appear that we require such a light like relation for the four vector $k^a = (\omega/c, \mathbf{k})$.
# Back to the tutorial notes.
In class we went straight to an assumed solution of the form
\begin{aligned}\mathbf{E} &= \mathbf{E}_0(x, y) e^{ i(\omega t - k z) } \\ \mathbf{B} &= \mathbf{B}_0(x, y) e^{ i(\omega t - k z) },\end{aligned} \hspace{\stretch{1}}(2.20)
where $\mathbf{k} = k \hat{\mathbf{z}}$. Our Laplacian was also written as the sum of components in the propagation and perpendicular directions
\begin{aligned}\boldsymbol{\nabla}^2 = \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(2.22)
With no $z$ dependence in the amplitudes we have
\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(2.23)
# Separation into components.
It was left as an exercise to separate out our Maxwell equations, so that our field components $\mathbf{E}_0 = \mathbf{E}_\perp + \mathbf{E}_z$ and $\mathbf{B}_0 = \mathbf{B}_\perp + \mathbf{B}_z$ in the propagation direction, and components in the perpendicular direction are separated
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 &=(\boldsymbol{\nabla}_\perp + \hat{\mathbf{z}}\partial_z) \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times (\mathbf{E}_\perp + \mathbf{E}_z) \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=( \hat{\mathbf{x}} \partial_x +\hat{\mathbf{y}} \partial_y ) \times ( \hat{\mathbf{x}} E_x +\hat{\mathbf{y}} E_y ) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=\hat{\mathbf{z}} (\partial_x E_y - \partial_z E_z) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z.\end{aligned}
We can do something similar for $\mathbf{B}_0$. This allows for a split of 1.14 into $\hat{\mathbf{z}}$ and perpendicular components
\begin{aligned}\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{E}_z - i \mathbf{k} \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_z - i \mathbf{k} \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= i k E_z - \partial_z E_z \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i k B_z - \partial_z B_z.\end{aligned} \hspace{\stretch{1}}(3.25)
So we see that once we have a solution for $\mathbf{E}_z$ and $\mathbf{B}_z$ (by solving the wave equation above for those components), the components for the fields in terms of those components can be found. Alternately, if one solves for the perpendicular components of the fields, these propagation components are available immediately with only differentiation.
In the case where the perpendicular components are taken as given
\begin{aligned}\mathbf{B}_z &= i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp \\ \mathbf{E}_z &= -i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.31)
we can express the remaining ones strictly in terms of the perpendicular fields
\begin{aligned}\frac{\omega}{c} \mathbf{B}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) + \mathbf{k} \times \mathbf{E}_\perp \\ \frac{\omega}{c} \mathbf{E}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) - \mathbf{k} \times \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= -i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp).\end{aligned} \hspace{\stretch{1}}(3.33)
Is it at all helpful to expand the double cross products?
\begin{aligned}\frac{\omega^2}{c^2} \mathbf{B}_\perp &= \boldsymbol{\nabla}_\perp (\boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp) -{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ &= i \frac{c}{\omega}(i k - \partial_z)\boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp)-{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \end{aligned}
This gives us
\begin{aligned}\left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{B}_\perp &= - \frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ \left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{E}_\perp &= -\frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) - \frac{\omega}{c} \mathbf{k} \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.37)
but that doesn’t seem particularly useful for completely solving the system? It appears fairly messy to try to solve for $\mathbf{E}_\perp$ and $\mathbf{B}_\perp$ given the propagation direction fields. I wonder if there is a simplification available that I am missing?
# Solving the momentum space wave equations.
Back to the class notes. We proceeded to solve for $\mathbf{E}_z$ and $\mathbf{B}_z$ from the wave equations by separation of variables. We wish to solve equations of the form
\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x}}^2} + \frac{\partial^2 {{}}}{\partial {{y}}^2} + \frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \phi(x,y) = 0\end{aligned} \hspace{\stretch{1}}(4.39)
Write $\phi(x,y) = X(x) Y(y)$, so that we have
\begin{aligned}\frac{X''}{X} + \frac{Y''}{Y} = \mathbf{k}^2 - \frac{\omega^2}{c^2}\end{aligned} \hspace{\stretch{1}}(4.40)
One solution is sinusoidal
\begin{aligned}\frac{X''}{X} &= -k_1^2 \\ \frac{Y''}{Y} &= -k_2^2 \\ -k_1^2 - k_2^2&= \mathbf{k}^2 - \frac{\omega^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(4.41)
The example in the tutorial now switched to a rectangular waveguide, still oriented with the propagation direction down the z-axis, but with lengths $a$ and $b$ along the $x$ and $y$ axis respectively.
Writing $k_1 = 2\pi m/a$, and $k_2 = 2 \pi n/ b$, we have
\begin{aligned}\phi(x, y) = \sum_{m n} a_{m n} \exp\left( \frac{2 \pi i m}{a} x \right)\exp\left( \frac{2 \pi i n}{b} y \right)\end{aligned} \hspace{\stretch{1}}(4.44)
We were also provided with some definitions
\begin{definition}TE (Transverse Electric)
$\mathbf{E}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Magnetic)
$\mathbf{B}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Electromagnetic)
$\mathbf{E}_3 = \mathbf{B}_3 = 0$.
\end{definition}
\begin{claim}TEM do not existing in a hollow waveguide.
\end{claim}
Why: I had in my notes
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = 0 & \implies \frac{\partial {E_2}}{\partial {x^1}} -\frac{\partial {E_1}}{\partial {x^2}} = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} = 0 & \implies \frac{\partial {E_1}}{\partial {x^1}} +\frac{\partial {E_2}}{\partial {x^2}} = 0\end{aligned}
and then
\begin{aligned}\boldsymbol{\nabla}^2 \phi &= 0 \\ \phi &= \text{const}\end{aligned}
In retrospect I fail to see how these are connected? What happened to the $\partial_t \mathbf{B}$ term in the curl equation above?
It was argued that we have $\mathbf{E}_\parallel = \mathbf{B}_\perp = 0$ on the boundary.
So for the TE case, where $\mathbf{E}_3 = 0$, we have from the separation of variables argument
\begin{aligned}\hat{\mathbf{z}} \cdot \mathbf{B}_0(x, y) =\sum_{m n} a_{m n} \cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.45)
No sines because
\begin{aligned}B_1 \propto \frac{\partial {B_3}}{\partial {x_a}} \rightarrow \cos(k_1 x^1).\end{aligned} \hspace{\stretch{1}}(4.46)
The quantity
\begin{aligned}a_{m n}\cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.47)
is called the $TE_{m n}$ mode. Note that since $B = \text{const}$ an ampere loop requires $\mathbf{B} = 0$ since there is no current.
Writing
\begin{aligned}k &= \frac{\omega}{c} \sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 } \\ \omega_{m n} &= 2 \pi c \sqrt{ \left(\frac{m}{a} \right)^2 + \left(\frac{n}{b} \right)^2 }\end{aligned} \hspace{\stretch{1}}(4.48)
Since $\omega < \omega_{m n}$ we have $k$ purely imaginary, and the term
\begin{aligned}e^{-i k z} = e^{- {\left\lvert{k}\right\rvert} z}\end{aligned} \hspace{\stretch{1}}(4.50)
represents the die off.
$\omega_{10}$ is the smallest.
Note that the convention is that the $m$ in $TE_{m n}$ is the bigger of the two indexes, so $\omega > \omega_{10}$.
The phase velocity
\begin{aligned}V_\phi = \frac{\omega}{k} = \frac{c}{\sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 }} \ge c\end{aligned} \hspace{\stretch{1}}(4.51)
However, energy is transmitted with the group velocity, the ratio of the Poynting vector and energy density
\begin{aligned}\frac{\left\langle{\mathbf{S}}\right\rangle}{\left\langle{{U}}\right\rangle} = V_g = \frac{\partial {\omega}}{\partial {k}} = 1/\frac{\partial {k}}{\partial {\omega}}\end{aligned} \hspace{\stretch{1}}(4.52)
(This can be shown).
Since
\begin{aligned}\left(\frac{\partial {k}}{\partial {\omega}}\right)^{-1} = \left(\frac{\partial {}}{\partial {\omega}}\sqrt{ (\omega/c)^2 - (\omega_{m n}/c)^2 }\right)^{-1} = c \sqrt{ 1 - (\omega_{m n}/\omega)^2 } \le c\end{aligned} \hspace{\stretch{1}}(4.53)
We see that the energy is transmitted at less than the speed of light as expected.
# Final remarks.
I’d started converting my handwritten scrawl for this tutorial into an attempt at working through these ideas with enough detail that they self contained, but gave up part way. This appears to me to be too big of a sub-discipline to give it justice in one hours class. As is, it is enough to at least get an concept of some of the ideas involved. I think were I to learn this for real, I’d need a good text as a reference (or the time to attempt to blunder through the ideas in much much more detail).
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### Olympiad Math Preparation – Class 2 -Subtraction of 2 and 3 digit numbers ( without regrouping/borrow )
1. The difference between greatest and smallest number is _________________ . Numbers : 178, 608, 578, 801, 999
2. Select the correct option
3. The difference of numbers shown in two abacuses below is ________________
4. If and then what is value of ?
5. If is subtracted from 100, the result is 45. Then is _________.
6. What is difference of largest and smallest 3 digit number ?
7. Duggu has 85 chocolates. Dakshu has 24 chocolates. How many more chocolates Duggu has than Dakshu ?
8. and . Find the value of
RELATED POSTS :
# Olympiad Math Preparation – Class 2
## Addition of 2 and 3 digit numbers ( without regrouping/carry )
1. The sum of greatest and smallest number is _________________ . Numbers : 188, 508, 345, 801, 290
2. Select the correct option
3. The sum of numbers shown in two abacuses below is ________________
4. If and then what is value of ?
5. If is added to 100, the result is 150. Then is _________.
6. What is sum of largest and smallest 3 digit number ?
7. Duggu has 54 chocolates. Dakshu has 45 chocolates. How many chocolates both of them have ?
8. and . Find the value of
RELATED POSTS :
# Olympiad Math Preparation – Class 2
## Expanded Form
1. 8 hundred + 5 tens + 2 ones is an even number ?
2. 5 hundred + 5 tens + 7 ones is an even number ?
3. The next odd number which comes after 6 hundred + 2 tens + 3 ones __________________
4. The next even number which comes after 3 hundred + 9 tens + 8 ones is _____________________
5. 4 hundred + 8 tens = _________________________
6. Expanded form of 346
7. Which abacus represents 7 hundred + 3 tens + 2 ones ?
8. Which option is correct ?
9. Duggu has 55 balls and Dakshu has 47 balls . What is total number of balls in expanded form ?
RELATED POSTS :
# Olympiad Math Preparation – Class 2
## Even and Odd Numbers
1. Which number is an even number ?
2. Which number is an odd number ?
3. The next odd number which comes after 145 is __________________
4. The next even number which comes after 98 is _____________________
5. Find the even number in the list of numbers: 145, 181, 192, 99, 177
6. Find the odd number in the list of numbers: 100, 78, 124, 191, 136
7. Which abacus shows odd number ?
8. Which option is correct ?
9. Which number does not belong to the group of numbers : 98, 114, 778, 277, 88, 190
10. Duggu has 54 chocolates and Dakshu has 47 chocolates. Who has odd number of chocolates ?
RELATED POSTS :
# Olympiad Math Preparation – Class 2
## Abacus and Place Value
1. The place value of 8 in 687 is ____________________
2. The place value of 9 in 975 is ____________________
3. The place value of 4 in 564 is ____________________
4. 2 in 925 is at ________________ place.
5. In 845, the number 8 is at ____________________ place.
6. 6 in 216 is at ________________ place.
7. Which abacus shows 623 ?
8. Duggu has 8 beads. He puts 8 beads in columns H, T and O in abacus. Hundreds column has 3 and ones column has 2 beads. Which of below is Duggu’s abacus with 8 beads.
9. What is number shown in abacus below :
RELATED POSTS:
# Olympiad Math Preparation – Class 2
## Number Sense – Ascending or Descending order
1. Which is second smallest number – 456, 567, 545, 434, 445
2. Arrange the numbers in increasing order – 345, 235, 325, 245, 335
3. Arrange the numbers in decreasing order – 693, 723, 623, 713, 703,
4. Which of following are in descending order ?
5. Which of following are in ascending order ?
6. Find the number in below group of numbers which lies between 400 and 500 and is smallest ?
212, 454, 434, 494, 534
7. Find the number in below group of numbers which is less than 500 and is greatest ?
751, 456, 890, 485, 286
RELATED POSTs:
# Olympiad Math Preparation – Class 2
## Number Sense – Comparing Numbers
1. 7 hundred forty five is ___________________ 7 hundred fifty
2. Which of below boxes has least number ?
3. Largest one digit number is _______________
4. Which of following number has largest value ?
5. Largest two digit number is _______________
6. Which of below balance is correct ?
7. Duggu has three cards . What is least 3 digit number that can be formed using each card only once
8. Which of below coin boxes have greatest number of coins ?
9. Largest three digit number is ________________
10. Which of following number is more greater than 450 and less than 575 in the screen below ?
11. The smallest 3 digit number is ___________
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# Multiples
Home > By Subject > Factors and Multiples > Lesson on Mutliples
This lesson is on multiples. A multiple is a number you say when you "count by" a number, or use skip-counting. When counting by fives, it is like saying only the answers on the 5 times table:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.and so on.
All of these numbers are called multiples of five.
Work through the examples and explanations in this lesson with your children and then try the worksheet that you will find at the bottom of this page.
Multiples are most often thought about as skip counting, or "count by" numbers.
You will find more on skip-counting here.
## Multiples of 5
Let's use a real-world example. If you were going to count up the value of a stack of nickels, you would "count by" fives, since each nickel is worth five cents.
All the numbers you say as you count by fives are the multiples of five. You might also recognize that they are the products, or answers, to the times table for fives.
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20
5 x 5 = 25
5 x 6 = 30
5 x 7 = 35
5 x 8 = 40
5 x 9 = 45
5 x 10 = 50
5 x 11 = 55
5 x 12 = 60
The pattern continues on past the times tables you've seen most often. Just keep on adding another set of five for each multiple of five.
## Multiples of 10
Let’s try another example. This time, we’ll count dimes. Since dimes are worth ten cents, we’ll “count by” tens. This will give us the multiples of 10.
So the multiples of 10 are: 10,20,30,40,50,60,70,80,90,100,110,120,130,140,150, and so on. Just keep on adding another set of 10 for each multiple of 10.
## Multiples of 12
Of course, there are many examples besides money. How about eggs? Eggs come by the dozen, and a dozen is 12. To find the multiples of 12, I count each set of 12.
So the multiples of 12 are: 12, 24, 36, 48, 60, 72, and so on.
## More Multiples
You can also figure out multiples by making your own sets with anything you can count. For example, you can make sets of three candies to figure out the multiples of three.
So the multiples of three are: 3, 6, 9, 12, and so on.
### Finding Multiples By Dividing
Since multiplication and division are inverse operations, meaning that they are related to one another and “un-do” each other, division can also be used to determine whether or not a given number is a multiple of another. See the examples below:
Since 12 can be divided evenly by 3 (12 ÷ 3 = 4), 12 is a multiple of 3. 4 sets of 3 would equal 12. Since 77 can be divided evenly by 11 (77 ÷ 11 = 7), 77 is a multiple of 11. 7 sets of 11 would equal 77. Since 56 can be divided evenly by 8 (56 ÷ 8 = 7), 56 is a multiple of 8. 7 sets of 8 would equal 56.
## Multiples Worksheets
Click the links below and get your children to try the worksheets that provide practice questions on multiples.
### Prevent Bullying
Click the links below for information and help on dealing with bullying .
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Sample Arkansas Math Word Problems Worksheet
Arkansas Theme Unit
edHelper Subscribers: click here to build a printable worksheet.
Name _____________________________ Date ___________________
Arkansas
(Answer ID # 0849724)
Complete and show your work.
Sample
This is only a sample pre-made worksheet. Sign up now!
1 Mrs. Small's class in Danville had a reading contest. Amber read 427 pages for the month, while Samuel read only 398 pages. How many more pages did Amber read than Samuel?
2 At Hope and Joy Elementary, Emma lost her purse. In it, there were two dollar bills, three quarters, three dimes, six nickels, and seven pennies. How much money did Emma lose?
3 John and his family were traveling to Little Rock, Arkansas. He saw a sign that read, "Little Rock - 130 miles." His father drove for a while longer when John saw another sign that read, "Little Rock - 87 miles." How far had his father driven from the first sign to the second sign?
4 The Yell Fest will be held on May 12th. The White River Water Carnival in Batesville will be held on August 3rd. How many days are there between these two events?
5 sample printable - click above to make a printable with all questions
6 sample printable - click above to make a printable with all questions
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Search a number
149397 = 3192621
BaseRepresentation
bin100100011110010101
321120221020
4210132111
514240042
63111353
71161363
oct443625
9246836
10149397
11a2276
1272559
1353001
143c633
152e3ec
hex24795
149397 has 8 divisors (see below), whose sum is σ = 209760. Its totient is φ = 94320.
The previous prime is 149393. The next prime is 149399. The reversal of 149397 is 793941.
149397 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 149397 - 22 = 149393 is a prime.
149397 is a lucky number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (149393) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1254 + ... + 1367.
It is an arithmetic number, because the mean of its divisors is an integer number (26220).
2149397 is an apocalyptic number.
It is an amenable number.
149397 is a deficient number, since it is larger than the sum of its proper divisors (60363).
149397 is a wasteful number, since it uses less digits than its factorization.
149397 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2643.
The product of its digits is 6804, while the sum is 33.
The square root of 149397 is about 386.5190810296. The cubic root of 149397 is about 53.0616347159.
The spelling of 149397 in words is "one hundred forty-nine thousand, three hundred ninety-seven".
Divisors: 1 3 19 57 2621 7863 49799 149397
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https://www.machinedesign.com/home/article/21830368/todays-brain-teaser
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# Today's brain teaser
Sept. 22, 2011
OK, quick: What spot on the earth is the farthest away from the center of the earth? Before you say "the top of Mount Everest," consider this: The earth is not a perfect sphere. To figure out the right answer, you must approximate earth's diameter as ...
OK, quick: What spot on the earth is the farthest away from the center of the earth? Before you say "the top of Mount Everest," consider this: The earth is not a perfect sphere.
To figure out the right answer, you must approximate earth's diameter as an ellipsoid. You must also know the earth's polar diameter and its equatorial diameter. Then you can get down to business.
The process has been worked out in a blog post by John D. Cook, a research statistician at M. D. Anderson Cancer Center and associate faculty for the UT Graduate School of Biomedical Sciences. If you are interested in investing a few minutes to find out the answer, read his post:
### Lee Teschler | Editor
Leland was Editor-in-Chief of Machine Design. He has 34 years of Service and holds a B.S. Engineering from the University of Michigan, a B.S. Electrical Engineering from the University of Michigan;, and a MBA from Cleveland State University. Prior to joining Penton, Lee worked as a Communications design engineer for the U.S. Government.
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# How many cups in 12.5 ml ? 12.5 milliters milk, water, oil to cups.
How to covert 12.5ml to cups? A cup is a unit of volume that is equal to 0.0528344 liters, or approximately 12.5 milliliters. It is also equal to 1/18.927059 liters, or approximately 0.0528 cups. A milliliter is a unit of volume that is equal to 0.001 liters, or approximately 0.1250 milliliters. In the United States, a cup is also equal to 240 milliliters, or 1/4 liter.
How many cups is 12.5 ml
## Convert 12.5 milliliters water to cups
Ml to cups formula: 12.5 ml / 236.5883953392 = 0.05283441 Cups
ml
=
ml
### Ml liquid, water to Cups Conversion table / chart
Ml cups american Cups canada 11.1 0.046916925000002 0.048834139903212 11.2 0.047339600000002 0.049274087109547 11.3 0.047762275000002 0.049714034315882 11.4 0.048184950000002 0.050153981522217 11.5 0.048607625000002 0.050593928728553 11.6 0.049030300000002 0.051033875934888 11.7 0.049452975000002 0.051473823141223 11.8 0.049875650000002 0.051913770347558 11.9 0.050298325000002 0.052353717553894 12.5 0.050721000000002 0.052793664760229 12.1 0.051143675000002 0.053233611966564 12.2 0.051566350000002 0.053673559172899 12.3 0.051989025000002 0.054113506379234 12.4 0.052411700000002 0.05455345358557 12.5 0.052834375000002 0.054993400791905 12.6 0.053257050000002 0.05543334799824 12.7 0.053679725000002 0.055873295204575 12.8 0.054102400000002 0.056313242410911 12.9 0.054525075000002 0.056753189617246
See also Convert 25 ml to cups. How many cups is 25 ml?
## Convert 12.5 milliliters milk to cups
ml
=
ml
### Ml milk to Cups Conversion table / chart
Ml cups american Cups canada 11.1 0.046916925000002 0.048834139903212 11.2 0.047339600000002 0.049274087109547 11.3 0.047762275000002 0.049714034315882 11.4 0.048184950000002 0.050153981522217 11.5 0.048607625000002 0.050593928728553 11.6 0.049030300000002 0.051033875934888 11.7 0.049452975000002 0.051473823141223 11.8 0.049875650000002 0.051913770347558 11.9 0.050298325000002 0.052353717553894 12.5 0.050721000000002 0.052793664760229 12.1 0.051143675000002 0.053233611966564 12.2 0.051566350000002 0.053673559172899 12.3 0.051989025000002 0.054113506379234 12.4 0.052411700000002 0.05455345358557 12.5 0.052834375000002 0.054993400791905 12.6 0.053257050000002 0.05543334799824 12.7 0.053679725000002 0.055873295204575 12.8 0.054102400000002 0.056313242410911 12.9 0.054525075000002 0.056753189617246
## Convert 12.5 milliliters oil to cups
ml
=
ml
### Ml oil to Cups Conversion table / chart
Ml cups american Cups canada 11.1 0.046916925000002 0.048834139903212 11.2 0.047339600000002 0.049274087109547 11.3 0.047762275000002 0.049714034315882 11.4 0.048184950000002 0.050153981522217 11.5 0.048607625000002 0.050593928728553 11.6 0.049030300000002 0.051033875934888 11.7 0.049452975000002 0.051473823141223 11.8 0.049875650000002 0.051913770347558 11.9 0.050298325000002 0.052353717553894 12.5 0.050721000000002 0.052793664760229 12.1 0.051143675000002 0.053233611966564 12.2 0.051566350000002 0.053673559172899 12.3 0.051989025000002 0.054113506379234 12.4 0.052411700000002 0.05455345358557 12.5 0.052834375000002 0.054993400791905 12.6 0.053257050000002 0.05543334799824 12.7 0.053679725000002 0.055873295204575 12.8 0.054102400000002 0.056313242410911 12.9 0.054525075000002 0.056753189617246
To convert 12.5 milliliters to cups, divide the volume in milliliters by 236.588 (the number of milliliters in 1 cup). This gives 0.0527 cups, or approximately half a cup. Note that this is an approximation, since different measuring containers can vary slightly in size and therefore yield slightly different volumes for the same amount of liquid. If you need to get an exact conversion for your purposes, it’s best to use an online conversion tool or a kitchen scale instead of trying to estimate using this method.
## Exam:
12.5 mL equals how many cups?
• 12.5 ml milk to cups = 0.0527 cups
• 12.5 ml water to cups = 0.0527 cups
• 12.5 ml sugar to cups = 0.0527 cups
See also 284 ml is how many cups? 284 ml to cups conversion table
## F.A.Q 12.5 ml to cups
### How many cups is 12.5 ml?
To accurately convert 12.5 milliliters to cups, divide the volume by 236,588 (the number of milliliters in 1 cup). This results in approximately 0.0527 cups or half a cup!
### How much is 12.5 ml in cups?
Did you know that half a cup of liquid is equal to 12.5 milliliters? Converting between different units of measurement can be tricky, but the basic formula is easy: simply divide your volume in mL by 236.588 – the number representing 1 cup! Though it’s an approximation, this should get you close enough – just keep in mind that measuring containers come with varying amounts of tolerance when they are filled up to their brim.
### How many Cups are in 12.5 Milliliters?
To convert 12.5 milliliters to cups, divide the volume in milliliters by 236.588 (the number of milliliters in 1 cup).
### 12.5 Milliliters equals how many Cups?
To convert 12.5 milliliters to cups, divide the volume in milliliters by 236.588 (the number of milliliters in 1 cup). This gives 0.0527 cups, or approximately half a cup. Note that this is an approximation, since different measuring containers can vary slightly in size and therefore yield slightly different volumes for the same amount of liquid.
Now, let’s take a look at 12.5 milliliters to cups conversion table.
Source:
See more articles in the category: ML to cups
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# algebra
Sarina started depositing money in savings account. after 10 weeks she had \$592. after 18 weeks, she had saved \$968. what is a linear equation that represents the relationship between # of weeks and the amt Sarina saved. Let y= \$ and x= weeks
1. 👍 0
2. 👎 0
3. 👁 32
1. Let y= ax+b then
592=10a+b
968=18a+b
and we find a and b.
1. 👍 0
2. 👎 0
posted by Mgraph
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# MATLAB moving a point in the XY plane
In MATLAB, I have an XY plane that is represented by finite number of points. The possible values of x are stored in a vector X and the possible values of y are stored in another vector Y. I have a point, say A, such that A(1) belongs to X is the x-coordinate and A(2) belongs to Y is the y-coordinate.
This point A can move in one of 8 ways if it is in the middle:
`````` . . . . A . .
. A . OR . . OR . A
. . . . .
``````
Of course, the set of these points changes if the point A is on the edge (sometimes only 5, sometimes only 3 if it is a corner). How can I find the set of these "1-hop" neighboring points? What about the set of "k-hop" neighboring points? By set I mean two vectors one for x-coordinates and another for y-coordinates. Thanks!
-
is this homework? – slayton Jun 20 '12 at 20:24
Nope, actually I am trying to simulate something in networks and I need this to continue. – Alex Jun 20 '12 at 20:26
ok no problem, you description sounded homework-y to me – slayton Jun 20 '12 at 20:32
You haven't defined things well enough to answer. If X = [1 100] and Y = [7 7] then there are only two places the point can be, and they aren't next to each other, so the point can't move at all. If all values of X between some min and max are allowed, you need to tell us. If they are not you need to tell us what you mean by "1-hop" neigbhoring points, what the rules are to hop from one point to another. Amazingly, if you are able to tell us these things clearly, you will probably find you can code it. – mwengler Jun 20 '12 at 23:02
Consider the following code:
``````%# create grid of 2D coordinates
sz = [5 6];
[X,Y] = meshgrid(1:sz(2),1:sz(1));
%# point A
A = [1 2]
%# neighboring points
k = 2; %# hop size
[sx,sy] = meshgrid(-k:k,-k:k); %# steps to get to neighbors
xx = bsxfun(@plus, A(1), sx(:)); %# add shift in x-coords
xx = min(max(xx,1),sz(2)); %# clamp x-coordinates within range
yy = bsxfun(@plus, A(2), sy(:));
yy = min(max(yy,1),sz(1));
B = unique([xx yy],'rows'); %# remove duplicates
B(ismember(B,A,'rows'),:) = []; %# remove point itself
``````
The result for the point `A = (1,2)` with `k=2` hops:
``````B =
1 1
1 3
1 4
2 1
2 2
2 3
2 4
3 1
3 2
3 3
3 4
``````
and an illustration of the solution:
``````x A x x . .
x x x x . .
x x x x . .
. . . . . .
. . . . . .
``````
-
Thanks this works! – Alex Jun 21 '12 at 17:49
lets say `A = [Xcenter Ycenter]`
for K-hop, you can access points:
``````pointsX = [];
pointsY = [];
for i=-k:k
pointsX = [pointsX Xcenter+i];
pointsY = [pointsY Ycenter+i];
end
``````
Furthermore, you can filter these points by order coordinates and remove the outliers. e.g. consider
``````(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)
``````
Now you know that minimum allowed X and Y are 1, so just filter out points with any ordinate and/or abscissa lesser than that.
-
Tip: use the `{}` tag to format code nicely (or add 4 spaces before each line to do it by hand). – tmpearce Jun 21 '12 at 2:21
Thanks. This is not exactly complete for what I need though but it is a good start :) – Alex Jun 21 '12 at 17:52
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Question 99
# The number of solutions to the equation $$\mid x \mid (6x^2 + 1) = 5x^2$$ is
Solution
For x <0, -x($$6x^2+1$$) = $$5x^2$$
=> ($$6x^2+1$$) = -5x
=> ($$6x^2 + 5x+ 1$$) = 0
=>($$6x^2 + 3x+2x+ 1$$) = 0
=> (3x+1)(2x+1)=0 =>x=$$\ -\frac{\ 1}{3}$$ or x=$$\ -\frac{\ 1}{2}$$
For x=0, LHS=RHS=0 (Hence, 1 solution)
For x >0, x($$6x^2+1$$) = $$5x^2$$
=> ($$6x^2 - 5x+ 1$$) = 0
=>(3x-1)(2x-1)=0 =>x=$$\ \frac{\ 1}{3}$$ or x=$$\ \frac{\ 1}{2}$$
Hence, the total number of solutions = 5
Video Solution
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You are Here: Home >< Maths
# Another Angle Question.. Rhombus watch
Announcements
1. Do the same rules apply here as angles in a circle? I don't know.. how do you do this question:
Attached Images
2. (Original post by hearthethunder)
Do the same rules apply here as angles in a circle? I don't know.. how do you do this question:
From definitions of rhombus and Kite, get that system is symmetrical about BE. Draw BE
AED=25
ABD = 55 so BDA = 55 so ADE = 125
SO DAE = X = 180-(125+25) =30
Check with DAB = 180-(55+55) = 180-(55+25+30)
Aitch
3. (Original post by Aitch)
From definitions of rhombus and Kite, get that system is symmetrical about BE. Draw BE
ABD = 55 so BDA = 55 so ADE = 125
SO DAE = X = 30
Check with DAB = 180-(55+55) = 180-(55+25+30)
Aitch
Definitions:
http://mathworld.wolfram.com/Rhombus.html
http://mathworld.wolfram.com/Kite.html
Aitch
4. (Original post by Aitch)
From definitions of rhombus and Kite, get that system is symmetrical about BE. Draw BE
AED=25
ABD = 55 so BDA = 55 so ADE = 125
SO DAE = X = 180-(125+25) =30
Check with DAB = 180-(55+55) = 180-(55+25+30)
Aitch
Where on earth have you got 55 and 25 from.. I only see 110 and 50
5. (Original post by hearthethunder)
Where on earth have you got 55 and 25 from.. I only see 110 and 50
Draw the vertical BE. It bisects these two angles.
Aitch
6. Logical, X = 30,
think about all angles in rhombus must equal 360. (sames true for kite)
7. (Original post by Aitch)
Draw the vertical BE. It bisects these two angles.
Aitch
Ohh yeah so it does
8. (Original post by Vijay1)
Logical, X = 30,
think about all angles in rhombus must equal 360. (sames true for kite)
I don't get how you can apply that rule to this question..
9. (Original post by hearthethunder)
Where on earth have you got 55 and 25 from.. I only see 110 and 50
His first instruction was to construct the line BE, this splits the angle ABC = 110 into ABD = 55 = BCD. and AED = 25 = DEC.
10. ok... from where did you get 125?
11. (Original post by Neo1)
ok... from where did you get 125?
BDE is a straight line. Subtract 55 from it, get 125.
Aitch
12. Huh subtract 55 from where ? lol i dont get it...
13. (Original post by Neo1)
Huh subtract 55 from where ? lol i dont get it...
BDE is 180°
BDA is 55°
Therefore ADE is (180-55)° = 125°
Aitch
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## gaby236 3 years ago The average speed of one airplane is 200 mph faster than that of an automobile. If the automobile starts at 3 a.m. and the airplane starts from the same place at 11 a.m., they will have traveled the same number of miles by 1:00 p.m. If r represents the rate of the automobile, how many miles does the plane travel? 200r 2(200 + r) 10r
1. gaby236
i think my anwer is 2(200+r)
2. gaby236
show the work
3. robtobey
4. gaby236
but what is the answer b or c @robtobey
5. sauravshakya
Distance travelled by the plane = distance travelled by the automobile = 10r
6. ajit.shirsat
10r
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# Looping over initial values - guide/example code
#1
Dear All,
I need to ask how to run loop over initial values, I am working on (for me) relatively complicated model, and hence, I am unable to solve for steady-state analyticaly, and so, I need to boost power of the numeric steady-state solvers to search in larger space, because I am afraid, that my initial quess will be too far from true steady-state value to converge.
Is there any usefull guide/example code, or at least some function like set_param_value()?
I would be also happy for advice, with solver to use for such a task (personally, i found simple fsolve as most efficient)?
Any advice/insight would be wery welcome!
Best
Jan Žemlička
#2
In case you have such a challenging problem, you should still go for an analytic approach. Simplify the problem until just a few equations remain and then solve those numerically. That is typically easier and faster than going brute force.
#3
Dear professor Pfeifer,
thank you very much for your advice, my supervisor also suggested this approach, but I am not really sure, if this will work, so I plan to build steady state block with as much analytical symplifications as I would be able to derive, but I need to prepare such a “last resort” mechanism to be as sure, as possible, that I will be able to find steady state.
Is there some way, how to program this loop in dynare .mod file, or using looping over dynare file in Matlab? I saw some codes for looping over parameters, so I want to ask, if there is something like set_param_value() function for initial values. I will use this brute force option just once, to find best starting guess for further aplications, so runtime is not so much constraint for me, I just want to be really sure, that solver will converge.
Jan Žemlička
#4
Again, I cannot recommend this approach. It is a lot more promising to simplify the problem first and then loop over initial values for a lower-dimensional problem. If your model is big, you will be trying to fill a large-dimensional hypercube with your randomly drawn initial values. That is very unlikely to work.
If you want to loop over initial values, simply set the content of `oo_.steady_state` and call
``````[steady_state,params,info] = evaluate_steady_state(oo_.steady_state,M_,options_,oo_,~options_.steadystate.nocheck);
``````
The last output argument will tell you whether the search has been successful.
#5
Dear professor Pfeifer,
I plan to use this option, does syntax of this function change, when I will provide steady_state_model block? Should implementation of this loop looks something like this?
Best Regards
Jan Žemlička
#6
You should use something like
``````info=1;
iter=1;
while info && iter<100
iter=iter+1;
end
``````
#7
Thank you very much, now I think, that I understand, how to do it!
I just thinking about deterministic grid for this task, instead of random walk approach, create sequence of initial values (capital, hours worked, nominal rate,…) and create nested loops to step-by step search in reasonable area of state space.
Best Regards
Jan Žemlička
#8
That is fine as well. Simply set the entries of `oo_.steady_state` as you like in that loop. But remember: filling out a hypercube like that will be challenging.
#9
Dear professor Pfeifer,
thank you very much, I really appreciate your guidance!
Best Regards
Jan Žemlička
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https://math.answers.com/Q/What_is_six_twenty_fourths_in_percent_form
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0
# What is six twenty fourths in percent form?
Wiki User
2012-02-12 21:12:26
6/24 as a percent = 100*6/24 = 25%
Wiki User
2012-02-12 21:12:26
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Q: What is six twenty fourths in percent form?
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## 1528307000000
1,528,307,000,000 (one trillion five hundred twenty-eight billion three hundred seven million) is an even thirteen-digits composite number following 1528306999999 and preceding 1528307000001. In scientific notation, it is written as 1.528307 × 1012. The sum of its digits is 26. It has a total of 14 prime factors and 196 positive divisors. There are 555,744,000,000 positive integers (up to 1528307000000) that are relatively prime to 1528307000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 26
• Digital Root 8
## Name
Short name 1 trillion 528 billion 307 million one trillion five hundred twenty-eight billion three hundred seven million
## Notation
Scientific notation 1.528307 × 1012 1.528307 × 1012
## Prime Factorization of 1528307000000
Prime Factorization 26 × 56 × 11 × 138937
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 14 Total number of prime factors rad(n) 15283070 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,528,307,000,000 is 26 × 56 × 11 × 138937. Since it has a total of 14 prime factors, 1,528,307,000,000 is a composite number.
## Divisors of 1528307000000
196 divisors
Even divisors 168 28 14 14
Total Divisors Sum of Divisors Aliquot Sum τ(n) 196 Total number of the positive divisors of n σ(n) 4.13552e+12 Sum of all the positive divisors of n s(n) 2.60722e+12 Sum of the proper positive divisors of n A(n) 2.10996e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23625e+06 Returns the nth root of the product of n divisors H(n) 72.433 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,528,307,000,000 can be divided by 196 positive divisors (out of which 168 are even, and 28 are odd). The sum of these divisors (counting 1,528,307,000,000) is 4,135,523,470,872, the average is 210,996,095,45.,265.
## Other Arithmetic Functions (n = 1528307000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 555744000000 Total number of positive integers not greater than n that are coprime to n λ(n) 1736700000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56548669942 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 555,744,000,000 positive integers (less than 1,528,307,000,000) that are coprime with 1,528,307,000,000. And there are approximately 56,548,669,942 prime numbers less than or equal to 1,528,307,000,000.
## Divisibility of 1528307000000
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 4 0 8
The number 1,528,307,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1528307000000)
Base System Value
2 Binary 10110001111010110001100100010001011000000
3 Ternary 12102002211100020201200222
4 Quaternary 112033112030202023000
5 Quinary 200014433043000000
6 Senary 3130032231002212
8 Octal 26172614421300
10 Decimal 1528307000000
12 Duodecimal 208242b09368
20 Vigesimal 2jdjfif000
36 Base36 ji3ej0e8
## Basic calculations (n = 1528307000000)
### Multiplication
n×i
n×2 3056614000000 4584921000000 6113228000000 7641535000000
### Division
ni
n⁄2 7.64154e+11 5.09436e+11 3.82077e+11 3.05661e+11
### Exponentiation
ni
n2 2335722286249000000000000 3569700720130350443000000000000000000 5455598598480255494490001000000000000000000000000 8337829527247563834017529958307000000000000000000000000000000
### Nth Root
i√n
2√n 1.23625e+06 11518.7 1111.87 273.427
## 1528307000000 as geometric shapes
### Circle
Diameter 3.05661e+12 9.60264e+12 7.33789e+24
### Sphere
Volume 1.49527e+37 2.93516e+25 9.60264e+12
### Square
Length = n
Perimeter 6.11323e+12 2.33572e+24 2.16135e+12
### Cube
Length = n
Surface area 1.40143e+25 3.5697e+36 2.64711e+12
### Equilateral Triangle
Length = n
Perimeter 4.58492e+12 1.0114e+24 1.32355e+12
### Triangular Pyramid
Length = n
Surface area 4.04559e+24 4.20693e+35 1.24786e+12
## Cryptographic Hash Functions
md5 6cbb1341aa64af25ce7eb27a05619a06 094dc5803e29004d7714dfb62bfba8d21a54d87a 35a2431c0215c1473cc4448ae7279e045944cdd5f34f803b9005a503eb19abe0 54ec48d54ff93a362c9bf12c3b55f13f179f1b657a21907a40e2e8b37f8ad9c93606d81ee60902d6128dd90df0b6c0728d55e7e048cd426ed56a1c78a375a0a6 59201634eb92c6eb2f7d648642a7ff17933a2997
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# 993 Radian/Day in Revolution/Nanosecond
Angular Velocity
Revolution/Nanosecond
993 Radian/Day = 1.8291766028964e-12 Revolution/Nanosecond
## How many Revolution/Nanosecond are in 993 Radian/Day?
The answer is 993 Radian/Day is equal to 1.8291766028964e-12 Revolution/Nanosecond and that means we can also write it as 993 Radian/Day = 1.8291766028964e-12 Revolution/Nanosecond. Feel free to use our online unit conversion calculator to convert the unit from Radian/Day to Revolution/Nanosecond. Just simply enter value 993 in Radian/Day and see the result in Revolution/Nanosecond. You can also Convert 994 Radian/Day to Revolution/Nanosecond
## How to Convert 993 Radian/Day to Revolution/Nanosecond (993 rad/day to r/ns)
By using our Radian/Day to Revolution/Nanosecond conversion tool, you know that one Radian/Day is equivalent to 1.8420711006006e-15 Revolution/Nanosecond. Hence, to convert Radian/Day to Revolution/Nanosecond, we just need to multiply the number by 1.8420711006006e-15. We are going to use very simple Radian/Day to Revolution/Nanosecond conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Radian/Day} = \text{1.8420711006006e-15 Revolution/Nanosecond}$$
$$\text{993 Radian/Day} = 993 \times 1.8420711006006e-15 = \text{1.8291766028964e-12 Revolution/Nanosecond}$$
## What is Radian/Day Unit of Measure?
Radian per day is a unit of measurement for angular velocity. By definition, one radian per day represents change in the orientation of an object by one radian every day.
## What is the symbol of Radian/Day?
The symbol of Radian/Day is rad/day. This means you can also write one Radian/Day as 1 rad/day.
## What is Revolution/Nanosecond Unit of Measure?
Revolution per nanosecond is a unit of measurement for angular velocity. By definition, one revolution per nanosecond represents change in the orientation of an object by one revolution every nanosecond.
## What is the symbol of Revolution/Nanosecond?
The symbol of Revolution/Nanosecond is r/ns. This means you can also write one Revolution/Nanosecond as 1 r/ns.
## Radian/Day to Revolution/Nanosecond Conversion Table (993-1002)
9931.8291766028964e-12
9941.831018673997e-12
9951.8328607450976e-12
9961.8347028161982e-12
9971.8365448872988e-12
9981.8383869583994e-12
9991.8402290295e-12
10001.8420711006006e-12
10011.8439131717012e-12
10021.8457552428018e-12
## Radian/Day to Other Units Conversion Table
993 radian/day in degree/second is equal to0.65850357704272
993 radian/day in degree/millisecond is equal to0.00065850357704272
993 radian/day in degree/microsecond is equal to6.5850357704272e-7
993 radian/day in degree/nanosecond is equal to6.5850357704272e-10
993 radian/day in degree/minute is equal to39.51
993 radian/day in degree/hour is equal to2370.61
993 radian/day in degree/day is equal to56894.71
993 radian/day in degree/week is equal to398262.96
993 radian/day in degree/month is equal to1731732.71
993 radian/day in degree/year is equal to20780792.48
993 radian/day in radian/second is equal to0.011493055555556
993 radian/day in radian/millisecond is equal to0.000011493055555556
993 radian/day in radian/microsecond is equal to1.1493055555556e-8
993 radian/day in radian/nanosecond is equal to1.1493055555556e-11
993 radian/day in radian/minute is equal to0.68958333333333
993 radian/day in radian/hour is equal to41.38
993 radian/day in radian/week is equal to6951
993 radian/day in radian/month is equal to30224.44
993 radian/day in radian/year is equal to362693.25
993 radian/day in gradian/second is equal to0.73167064115857
993 radian/day in gradian/millisecond is equal to0.00073167064115857
993 radian/day in gradian/microsecond is equal to7.3167064115857e-7
993 radian/day in gradian/nanosecond is equal to7.3167064115857e-10
993 radian/day in gradian/minute is equal to43.9
993 radian/day in gradian/hour is equal to2634.01
993 radian/day in gradian/day is equal to63216.34
993 radian/day in gradian/week is equal to442514.4
993 radian/day in gradian/month is equal to1924147.45
993 radian/day in gradian/year is equal to23089769.43
993 radian/day in gon/second is equal to0.73167064115857
993 radian/day in gon/millisecond is equal to0.00073167064115857
993 radian/day in gon/microsecond is equal to7.3167064115857e-7
993 radian/day in gon/nanosecond is equal to7.3167064115857e-10
993 radian/day in gon/minute is equal to43.9
993 radian/day in gon/hour is equal to2634.01
993 radian/day in gon/day is equal to63216.34
993 radian/day in gon/week is equal to442514.4
993 radian/day in gon/month is equal to1924147.45
993 radian/day in gon/year is equal to23089769.43
993 radian/day in revolution/second is equal to0.0018291766028964
993 radian/day in revolution/millisecond is equal to0.0000018291766028964
993 radian/day in revolution/microsecond is equal to1.8291766028964e-9
993 radian/day in revolution/nanosecond is equal to1.8291766028964e-12
993 radian/day in revolution/minute is equal to0.10975059617379
993 radian/day in revolution/hour is equal to6.59
993 radian/day in revolution/day is equal to158.04
993 radian/day in revolution/week is equal to1106.29
993 radian/day in revolution/month is equal to4810.37
993 radian/day in revolution/year is equal to57724.42
993 radian/day in sign/second is equal to0.021950119234757
993 radian/day in sign/millisecond is equal to0.000021950119234757
993 radian/day in sign/microsecond is equal to2.1950119234757e-8
993 radian/day in sign/nanosecond is equal to2.1950119234757e-11
993 radian/day in sign/minute is equal to1.32
993 radian/day in sign/hour is equal to79.02
993 radian/day in sign/day is equal to1896.49
993 radian/day in sign/week is equal to13275.43
993 radian/day in sign/month is equal to57724.42
993 radian/day in sign/year is equal to692693.08
993 radian/day in mil/second is equal to11.71
993 radian/day in mil/millisecond is equal to0.011706730258537
993 radian/day in mil/microsecond is equal to0.000011706730258537
993 radian/day in mil/nanosecond is equal to1.1706730258537e-8
993 radian/day in mil/minute is equal to702.4
993 radian/day in mil/hour is equal to42144.23
993 radian/day in mil/day is equal to1011461.49
993 radian/day in mil/month is equal to30786359.23
993 radian/day in mil/year is equal to369436310.81
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Specific Heat and Phase Change
## How much energy one must put in per unit mass in order to raise the temperature 1 degree Celsius.
Levels are CK-12's student achievement levels.
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At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
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## Calorimetry 1: Heat Transfer and Temperature Change for One Object - Overview
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## Calorimetry 2: Heat Transfer and Temperature Change for Two Objects - Overview
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## Calorimetry 3: Heat Transfer and Phase Changes - Overview
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## Mixed Units - Example 1
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Converting with units in both the numerator and denominator
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## Calorimetry 1: Heat Transfer and Temperature Change for One Object - Example 1
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Solving problems involving heat transfer for one object using Q = mcÎT
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## Calorimetry 2: Heat Transfer and Temperature Change for Two Objects - Example 1
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Determining mass or specific heat capacity for problems involving heat transfer between two objects using Q = mcÎT
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## Calorimetry 3: Heat Transfer and Phase Changes - Example 1
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Solving problems involving phase changes using the equation Q = mL
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## Calorimetry 3: Heat Transfer and Phase Changes - Example 2
by CK-12 //basic
Determining the amount of heat necessary to heat and change the phase of an object using Q = mcÎT and Q = mL
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https://studydaddy.com/question/the-old-frothingslosh-pale-stale-ale-company-sells-type-z-cardboard-shipping-box
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Answered You can hire a professional tutor to get the answer.
QUESTION
# The Old frothingslosh pale stale ale company sells type z cardboard shipping boxes (their beer failed).
The Old frothingslosh pale stale ale company sells type z cardboard shipping boxes (their beer failed). The firm requires 500 boxes each week, and it meets this demand with a production capacity of 100 boxes per day (monday through friday) Set-up costs before each run are \$30.00. Holding costs are \$.03 per box per day. There are 310 production days per year. Determine the optimum number of production runs. What is the maximum size of the inventory? How long does it take for inventory to reach this maximum?
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https://leanprover-community.github.io/mathlib_docs/analysis/calculus/iterated_deriv.html
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# mathlib3documentation
analysis.calculus.iterated_deriv
# One-dimensional iterated derivatives #
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We define the n-th derivative of a function f : π β F as a function iterated_deriv n f : π β F, as well as a version on domains iterated_deriv_within n f s : π β F, and prove their basic properties.
## Main definitions and results #
Let π be a nontrivially normed field, and F a normed vector space over π. Let f : π β F.
• iterated_deriv n f is the n-th derivative of f, seen as a function from π to F. It is defined as the n-th FrΓ©chet derivative (which is a multilinear map) applied to the vector (1, ..., 1), to take advantage of all the existing framework, but we show that it coincides with the naive iterative definition.
• iterated_deriv_eq_iterate states that the n-th derivative of f is obtained by starting from f and differentiating it n times.
• iterated_deriv_within n f s is the n-th derivative of f within the domain s. It only behaves well when s has the unique derivative property.
• iterated_deriv_within_eq_iterate states that the n-th derivative of f in the domain s is obtained by starting from f and differentiating it n times within s. This only holds when s has the unique derivative property.
## Implementation details #
The results are deduced from the corresponding results for the more general (multilinear) iterated FrΓ©chet derivative. For this, we write iterated_deriv n f as the composition of iterated_fderiv π n f and a continuous linear equiv. As continuous linear equivs respect differentiability and commute with differentiation, this makes it possible to prove readily that the derivative of the n-th derivative is the n+1-th derivative in iterated_deriv_within_succ, by translating the corresponding result iterated_fderiv_within_succ_apply_left for the iterated FrΓ©chet derivative.
noncomputable def iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] (n : β) (f : π β F) (x : π) :
F
The n-th iterated derivative of a function from π to F, as a function from π to F.
Equations
noncomputable def iterated_deriv_within {π : Type u_1} {F : Type u_2} [normed_space π F] (n : β) (f : π β F) (s : set π) (x : π) :
F
The n-th iterated derivative of a function from π to F within a set s, as a function from π to F.
Equations
theorem iterated_deriv_within_univ {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
### Properties of the iterated derivative within a set #
theorem iterated_deriv_within_eq_iterated_fderiv_within {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} :
x = β n f s x) (Ξ» (i : fin n), 1)
theorem iterated_deriv_within_eq_equiv_comp {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} :
= β (fin n) F).symm) β n f s
Write the iterated derivative as the composition of a continuous linear equiv and the iterated FrΓ©chet derivative
theorem iterated_fderiv_within_eq_equiv_comp {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} :
n f s = β F) β
Write the iterated FrΓ©chet derivative as the composition of a continuous linear equiv and the iterated derivative.
theorem iterated_fderiv_within_apply_eq_iterated_deriv_within_mul_prod {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} {m : fin n β π} :
β n f s x) m = finset.univ.prod (Ξ» (i : fin n), m i) β’ x
The n-th FrΓ©chet derivative applied to a vector (m 0, ..., m (n-1)) is the derivative multiplied by the product of the m is.
theorem norm_iterated_fderiv_within_eq_norm_iterated_deriv_within {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} :
@[simp]
theorem iterated_deriv_within_zero {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} :
= f
@[simp]
theorem iterated_deriv_within_one {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {x : π} (h : s x) :
x = s x
theorem cont_diff_on_of_continuous_on_differentiable_on_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {n : ββ} (Hcont : β (m : β), βm β€ n β continuous_on (Ξ» (x : π), x) s) (Hdiff : β (m : β), βm < n β (Ξ» (x : π), x) s) :
cont_diff_on π n f s
If the first n derivatives within a set of a function are continuous, and its first n-1 derivatives are differentiable, then the function is C^n. This is not an equivalence in general, but this is an equivalence when the set has unique derivatives, see cont_diff_on_iff_continuous_on_differentiable_on_deriv.
theorem cont_diff_on_of_differentiable_on_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {n : ββ} (h : β (m : β), βm β€ n β s) s) :
cont_diff_on π n f s
To check that a function is n times continuously differentiable, it suffices to check that its first n derivatives are differentiable. This is slightly too strong as the condition we require on the n-th derivative is differentiability instead of continuity, but it has the advantage of avoiding the discussion of continuity in the proof (and for n = β this is optimal).
theorem cont_diff_on.continuous_on_iterated_deriv_within {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {n : ββ} {m : β} (h : cont_diff_on π n f s) (hmn : βm β€ n) (hs : unique_diff_on π s) :
s
On a set with unique derivatives, a C^n function has derivatives up to n which are continuous.
theorem cont_diff_within_at.differentiable_within_at_iterated_deriv_within {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {x : π} {n : ββ} {m : β} (h : n f s x) (hmn : βm < n) (hs : unique_diff_on π s)) :
s) s x
theorem cont_diff_on.differentiable_on_iterated_deriv_within {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {n : ββ} {m : β} (h : cont_diff_on π n f s) (hmn : βm < n) (hs : unique_diff_on π s) :
s) s
On a set with unique derivatives, a C^n function has derivatives less than n which are differentiable.
theorem cont_diff_on_iff_continuous_on_differentiable_on_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {s : set π} {n : ββ} (hs : unique_diff_on π s) :
cont_diff_on π n f s β (β (m : β), βm β€ n β s) β§ β (m : β), βm < n β s) s
The property of being C^n, initially defined in terms of the FrΓ©chet derivative, can be reformulated in terms of the one-dimensional derivative on sets with unique derivatives.
theorem iterated_deriv_within_succ {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} (hxs : s x) :
iterated_deriv_within (n + 1) f s x = s x
The n+1-th iterated derivative within a set with unique derivatives can be obtained by differentiating the n-th iterated derivative.
theorem iterated_deriv_within_eq_iterate {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} (hs : unique_diff_on π s) (hx : x β s) :
x = (Ξ» (g : π β F), s)^[n] f x
The n-th iterated derivative within a set with unique derivatives can be obtained by iterating n times the differentiation operation.
theorem iterated_deriv_within_succ' {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {s : set π} {x : π} (hxs : unique_diff_on π s) (hx : x β s) :
iterated_deriv_within (n + 1) f s x = s x
The n+1-th iterated derivative within a set with unique derivatives can be obtained by taking the n-th derivative of the derivative.
### Properties of the iterated derivative on the whole space #
theorem iterated_deriv_eq_iterated_fderiv {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {x : π} :
x = β(iterated_fderiv π n f x) (Ξ» (i : fin n), 1)
theorem iterated_deriv_eq_equiv_comp {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
= β (fin n) F).symm) β iterated_fderiv π n f
Write the iterated derivative as the composition of a continuous linear equiv and the iterated FrΓ©chet derivative
theorem iterated_fderiv_eq_equiv_comp {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
iterated_fderiv π n f = β F) β
Write the iterated FrΓ©chet derivative as the composition of a continuous linear equiv and the iterated derivative.
theorem iterated_fderiv_apply_eq_iterated_deriv_mul_prod {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {x : π} {m : fin n β π} :
β(iterated_fderiv π n f x) m = finset.univ.prod (Ξ» (i : fin n), m i) β’ x
The n-th FrΓ©chet derivative applied to a vector (m 0, ..., m (n-1)) is the derivative multiplied by the product of the m is.
theorem norm_iterated_fderiv_eq_norm_iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} {x : π} :
@[simp]
theorem iterated_deriv_zero {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} :
= f
@[simp]
theorem iterated_deriv_one {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} :
=
theorem cont_diff_iff_iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {n : ββ} :
The property of being C^n, initially defined in terms of the FrΓ©chet derivative, can be reformulated in terms of the one-dimensional derivative.
theorem cont_diff_of_differentiable_iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {n : ββ} (h : β (m : β), βm β€ n β differentiable π f)) :
cont_diff π n f
To check that a function is n times continuously differentiable, it suffices to check that its first n derivatives are differentiable. This is slightly too strong as the condition we require on the n-th derivative is differentiability instead of continuity, but it has the advantage of avoiding the discussion of continuity in the proof (and for n = β this is optimal).
theorem cont_diff.continuous_iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {n : ββ} (m : β) (h : cont_diff π n f) (hmn : βm β€ n) :
theorem cont_diff.differentiable_iterated_deriv {π : Type u_1} {F : Type u_2} [normed_space π F] {f : π β F} {n : ββ} (m : β) (h : cont_diff π n f) (hmn : βm < n) :
differentiable π f)
theorem iterated_deriv_succ {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
iterated_deriv (n + 1) f = deriv f)
The n+1-th iterated derivative can be obtained by differentiating the n-th iterated derivative.
theorem iterated_deriv_eq_iterate {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
The n-th iterated derivative can be obtained by iterating n times the differentiation operation.
theorem iterated_deriv_succ' {π : Type u_1} {F : Type u_2} [normed_space π F] {n : β} {f : π β F} :
iterated_deriv (n + 1) f = (deriv f)
The n+1-th iterated derivative can be obtained by taking the n-th derivative of the derivative.
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https://spinnaker8manchester.readthedocs.io/en/latest/_modules/spynnaker/pyNN/utilities/running_stats/
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# Source code for spynnaker.pyNN.utilities.running_stats
# Copyright (c) 2017-2019 The University of Manchester
#
# This program is free software: you can redistribute it and/or modify
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
import math
[docs]class RunningStats(object):
""" Keeps running statistics.
From: http://www.johndcook.com/blog/skewness_kurtosis/
"""
__slots__ = ["__mean", "__mean_2", "__n_items"]
def __init__(self):
self.__mean = 0.0
self.__mean_2 = 0.0
self.__n_items = 0
""" Adds an item to the running statistics.
:param x: The item to add
:type x: int or float
"""
old_n_items = self.__n_items
self.__n_items += 1
delta = x - self.__mean
delta_n = delta / self.__n_items
term_1 = delta * delta_n * old_n_items
self.__mean += delta_n
self.__mean_2 += term_1
[docs] def add_items(self, mean, variance, n_items):
""" Add a bunch of items (via their statistics).
:param float mean: The mean of the items to add.
:param float variance: The variance of the items to add.
:param int n_items: The number of items represented.
"""
if n_items > 0:
new_n_items = self.__n_items + n_items
mean_2 = variance * (n_items - 1.0)
delta = mean - self.__mean
delta_2 = delta * delta
new_mean = (((self.__n_items * self.__mean) + (n_items * mean)) /
new_n_items)
new_mean_2 = (self.__mean_2 + mean_2 +
(delta_2 * self.__n_items * n_items) / new_n_items)
self.__n_items = new_n_items
self.__mean = new_mean
self.__mean_2 = new_mean_2
@property
def n_items(self):
""" The number of items seen.
:rtype: int
"""
return self.__n_items
@property
def mean(self):
""" The mean of the items seen.
:rtype: float
"""
return self.__mean
@property
def variance(self):
""" The variance of the items seen.
:rtype: float
"""
if self.__n_items <= 1:
return 0.0
return self.__mean_2 / (self.__n_items - 1.0)
@property
def standard_deviation(self):
""" The population standard deviation of the items seen.
:rtype: float
"""
return math.sqrt(self.variance)
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http://www.chegg.com/homework-help/discovering-statistics-w-student-cd-tables-and-formula-card-2nd-edition-chapter-3.2-solutions-9781429227988
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View more editions
Solutions
Discovering Statistics
# TEXTBOOK SOLUTIONS FOR Discovering Statistics 0th Edition
• 2952 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
A deviation for a data value y is the difference between that value and the mean of the complete data which is with us. That is deviation is equal to.
The properties of the deviation are as follows:
If the value in the data is larger than the mean obtained, then the deviation will be greater than zero.
If the value given in the data is smaller than that of the mean, then the deviation will be less than zero.
If the value is equal to the mean of the data, then the deviation of the value will be zero.
We can also say that the deviation can be considered as a subtraction of mean of the data from the value of the data.
Corresponding Textbook
Discovering Statistics | 0th Edition
9781429227988ISBN-13: 1429227982ISBN: Authors:
Alternate ISBN: 9781429227537, 9781429227988, 9781429242141, 9781429295260
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https://www.engineeringtoolbox.com/amp/laminar-friction-coefficient-d_1032.html
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Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
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# Laminar Flow - Friction Coefficients
The friction coefficient - or factor - of a fluid flow at laminar conditions can be calculated as
λ = 64 / Re
= 64 μ / (d hu ρ)
= 64 ν / (d hu) (1)
where
λ = friction coefficient (non-dimensional)
Re = Reynolds Number (non-dimensional)
μ = absolute (dynamic) viscosity (Ns/m2, lb m/s ft)
d h= hydraulic diameter (mm, ft)
u = mean velocity in flow (m/s, ft/s)
ρ = density of fluid (kg/m3, lb m/ft3 )
ν = μ / ρ = kinematic viscosity (m2/s, ft2/s)
Equation (1) is only valid at laminar conditions where Reynolds Number is less than 2300 . For turbulent conditions where Reynolds Number exceeds 4000 the Colebrook equation should be used to calculate the friction coefficient.
In practice laminar flow is only actual for viscous fluids - like crude oil, fuel oil and other oils.
The friction coefficient for laminar flow indicated in the Moody diagram (SI based):
## Related Topics
### • Fluid Mechanics
The study of fluids - liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.
## Related Documents
### Colebrook Equation
Friction loss coefficients in pipes, tubes and ducts.
### Darcy-Weisbach Equation - Major Pressure and Head Loss due to Friction
The Darcy-Weisbach equation can be used to calculate the major pressure and head loss due to friction in ducts, pipes or tubes.
### Reynolds Number
Introduction and definition of the dimensionless Reynolds Number - online calculators.
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# Linear velocity of rigid body with vectors
1. Jan 21, 2012
### eximius
1. The problem statement, all variables and given/known data
A rigid body is rotating with angular velocity 2 rad/s about an axis in the
direction of the vector (i + j + k) and passing through the point Q = (0, 1, -1) on
the body. Find the linear velocity of the point P = (1, 0, 1) on the body.
(You may use the result v = ω x r .)
2. Relevant equations
v= ω x r
3. The attempt at a solution
Magnitude of vector joining P and Q = |P-Q| = r
.:. r = √6
v= ω x r
v= 2 x √6
v = 4.8990m/s
I'm assuming I'm completely wrong because the amount of work isn't enough for the amount of marks. I've seen similar questions online and I've noticed that x=1+t, y=1+2t and z=3t have been used in a number of the answers. Are these general rules or have they been derived from the question somehow? Any and all help would be greatly appreciated. Thanks.
2. Jan 21, 2012
### SammyS
Staff Emeritus
You haven't taken into account the vector nature of r and ω.
3. Jan 21, 2012
### eximius
So P-Q = (1,-1,2) and w = (2,2,2). Therefore v = (6,-2,-4)m/s ?
I understand your reply but it doesn't really tell me what to do.
4. Jan 21, 2012
### SammyS
Staff Emeritus
Well, you did take the cross product of r and what you thought was ω. So, maybe I helped point you in the right direction -- which is all I wanted to do.
By the way: ω ≠ (2,2,2). The magnitude of (2,2,2) is 2√(3) . The magnitude of ω is 2 .
5. Jan 22, 2012
### eximius
Erm... So what is the answer then? Am I even right in thinking that P-Q is equal to r? How do I get v? Sorry but you're being very cryptic and I just don't get it...
6. Jan 22, 2012
### eximius
Is it something to do with using P-Q as the normal to the plane and discovering the plane's equation with N.(x-Px) = 0? Or something like that?
7. Jan 22, 2012
### SammyS
Staff Emeritus
Yes P-Q = r .
Cryptic? I did mention that you had a problem with what you had for ω.
What is wrong with ω specifically? It does have the correct direction, but the wrong magnitude.
You multiplied i+j+k by 2 . However, the magnitude of the vector i+j+k is not 1 .
8. Jan 22, 2012
### eximius
Ahhh so the magnitude is √3. Therefore ω= 2*√3. So v = 2√3 * √6 = 6√2 = 8.49m/s ?
I'm sorry if I seemed annoyed or something. I was just confused and frustrated.
Edit: I have r in vector form, I need ω in vector form. The magnitude of ω is 2. So I need to get the vector form from this. But how?
Last edited: Jan 22, 2012
9. Jan 22, 2012
### SammyS
Staff Emeritus
First of all: 2*√3 = √(12) ≠ √6 .
The vector, (2, 2, 2) has a magnitude of 2*√3 , so multiply that by 1/√3 . What will that give you?
10. Jan 22, 2012
### eximius
The vector (2,2,2) doesn't have magnitude 2*√3, it has a magnitude of √(2^2 + 2^2 + 2^2) = √12, doesn't it? Is the intention to get the unit vector? I really don't understand at all. If you could please just tell me, I'm simply not getting it.
v and r are vectors, ω is a scalar
If P-Q = r, then r=(1,-1,2)
ω = 2rad/s in the direction of (i,j,k)
v=(6,-2,-4) from determinant method.
|v|=√(6^2 + (-2)^2 + (-4)^2) = 2*√14=7.48m/s
Am I even close?
11. Jan 22, 2012
### SammyS
Staff Emeritus
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## Topic 6.3: Transformations to Linear Regression (HOWTO)
Introduction Notes Theory HOWTO Examples Engineering Error Questions Matlab Maple
# Problem
Given data (xi, yi), for i = 1, 2, ..., n which is known to approximate an exponential curve, find the best fitting exponential function of the form y(x) = aebx.
# Assumptions
We will assume the model is correct and that the data is defined by two vectors x = (xi) and y = (yi).
# Tools
We will use algebra and linear regression.
# Process
Take the logarithm of the y values and define the vector φ = (φi) = (log(yi)).
Now, find the least-squares curve of the form c1 x + c2 which best fits the data points (xiφi). See the Topic 6.1 Linear Regression.
Having found the coefficient vector c, the best fitting curve is
y = ec2 ec1 x.
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Functional Analysis/C*-algebras
Functional Analysis Chapter 5: C^*-algebras
(May 27, 2008). Beware this chapter is less than a draft.
A Banach space ${\displaystyle {\mathcal {A}}}$ over ${\displaystyle \mathbf {C} }$ is called a Banach algebra if it is an algebra and satisfies
${\displaystyle \|xy\|\leq \|x\|\|y\|}$.
We shall assume that every Banach algebra has the unit ${\displaystyle 1}$ unless stated otherwise.
Since ${\displaystyle \|x_{n}y_{n}-xy\|\leq \|x_{n}\|\|y_{n}-y\|+\|x_{n}-x\|\|y\|\to 0}$ as ${\displaystyle x_{n},y_{n}\to 0}$, the map
${\displaystyle (x,y)\mapsto xy:{\mathcal {A}}\times {\mathcal {A}}\to {\mathcal {A}}}$
is continuous.
For ${\displaystyle x\in {\mathcal {A}}}$, let ${\displaystyle \sigma (x)}$ be the set of all complex numbers ${\displaystyle \lambda }$ such that ${\displaystyle x-\lambda 1}$ is not invertible.
5 Theorem For every ${\displaystyle x\in {\mathcal {A}}}$, ${\displaystyle \sigma (x)}$ is nonempty and closed and
${\displaystyle \sigma (x)\subset \{s\in \mathbf {C} ||s|\leq \|x\|\}}$.
Moreover,
${\displaystyle r(x){\overset {\mathrm {def} }{=}}\sup\{|z||z\in \sigma (x)\}=\lim _{n\to \infty }\|x^{n}\|^{1/n}}$
(${\displaystyle r(x)}$ is called the spectral radius of ${\displaystyle x}$)
Proof: Let ${\displaystyle G\subset {\mathcal {A}}}$ be the group of units. Define ${\displaystyle f:\mathbf {C} \to A}$ by ${\displaystyle f(\lambda )=\lambda 1-x}$. (Throughout the proof ${\displaystyle x}$ is fixed.) If ${\displaystyle \lambda \in \mathbf {C} \backslash \sigma (x)}$, then, by definition, ${\displaystyle f(\lambda )\in G}$ or ${\displaystyle \lambda \in f^{-1}(G)}$. Similarly, we have: ${\displaystyle G\subset f(\sigma (x))}$. Thus, ${\displaystyle x\in f^{-1}(G)\subset \sigma (x)}$. Since ${\displaystyle f}$ is clearly continuous, ${\displaystyle \mathbf {C} \backslash \sigma (x)}$ is open and so ${\displaystyle \sigma (x)}$ is closed. Suppose that ${\displaystyle |s|>\|x\|}$ for ${\displaystyle s\in \mathbf {C} }$. By the geometric series (which is valid by Theorem 2.something), we have:
${\displaystyle \left(1-{x \over s}\right)^{-1}=\sum _{n=0}^{\infty }\left({x \over s}\right)^{n}}$
Thus, ${\displaystyle 1-{x \over s}}$ is invertible, which is to say, ${\displaystyle s1-x}$ is invertible. Hence, ${\displaystyle s\not \in \sigma (x)}$. This complete the proof of the first assertion and gives:
${\displaystyle r(x)\leq \|x\|}$
Since ${\displaystyle \sigma (x)}$ is compact, there is a ${\displaystyle a\in \sigma (x)}$ such that ${\displaystyle r(x)=a}$. Since ${\displaystyle a^{n}\in \sigma (x^{n})}$ (use induction to see this),
${\displaystyle r(x)^{n}\leq \|x^{n}\|}$
Next, we claim that the sequence ${\displaystyle {x^{n} \over s^{n+1}}}$ is bounded for ${\displaystyle |s|>r(x)}$. In view of the uniform boundedness principle, it suffices to show that ${\displaystyle g\left({x^{n} \over s^{n+1}}\right)}$ is bounded for every ${\displaystyle g\in A^{*}}$. But since
${\displaystyle \lim _{n\to \infty }g\left({x^{n} \over s^{n+1}}\right)=0}$,
this is in fact the case. Hence, there is a constant ${\displaystyle c}$ such that ${\displaystyle \|x^{n}\|\leq c|s|^{n+1}}$ for every ${\displaystyle n}$. It follows:
${\displaystyle r(x)\leq \lim _{n\to \infty }\|x^{n}\|^{1/n}\leq \lim _{n\to \infty }c^{1/n}|s|=|s|}$.
Taking inf over ${\displaystyle |s|>r(x)}$ completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that ${\displaystyle \sigma (x)}$ is empty. Then for every ${\displaystyle g\in A^{*}}$, the map
${\displaystyle s\mapsto g((x-s)^{-1})}$
is analytic in ${\displaystyle \mathbf {C} }$. Since ${\displaystyle \lim _{s\to \infty }g((x-s)^{-1})=0}$, by Liouville's theorem, we must have: ${\displaystyle g((x-s)^{-1})=0}$. Hence, ${\displaystyle (x-s)^{-1}=0}$ for every ${\displaystyle s\in \mathbf {C} }$, a contradiction. ${\displaystyle \square }$
5 Corollary (Gelfand-Mazur theorem) If every nonzero element of ${\displaystyle {\mathcal {A}}}$ is invertible, then ${\displaystyle {\mathcal {A}}}$ is isomorphic to ${\displaystyle \mathbf {C} }$.
Proof: Let ${\displaystyle x\in {\mathcal {A}}}$ be a nonzero element. Since ${\displaystyle \sigma (x)}$ is non-empty, we can then find ${\displaystyle \lambda \in \mathbf {C} }$ such that ${\displaystyle \lambda 1-x}$ is not invertible. But, by hypothesis, ${\displaystyle \lambda 1-x}$ is invertible, unless ${\displaystyle \lambda 1=x}$.${\displaystyle \square }$
Let ${\displaystyle {\mathfrak {m}}}$ be a maximal ideal of a Banach algebra. (Such ${\displaystyle {\mathfrak {m}}}$ exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of ${\displaystyle {\mathfrak {m}}}$ consists of invertible elements, ${\displaystyle {\mathfrak {m}}}$ is closed. In particular, ${\displaystyle {\mathcal {A}}/{\mathfrak {m}}}$ is a Banach algebra with the usual quotient norm. By the above corollary, we thus have the isomorphism:
${\displaystyle {\mathcal {A}}/{\mathfrak {m}}\simeq \mathbf {C} }$
Much more is true, actually. Let ${\displaystyle \Delta ({\mathcal {A}})}$ be the set of all nonzero homeomorphism ${\displaystyle \omega :{\mathcal {A}}\to \mathbf {C} }$. (The members of ${\displaystyle \Delta ({\mathcal {A}})}$ are called characters.)
5 Theorem ${\displaystyle \Delta ({\mathcal {A}})}$ is bijective to the set of all maximal ideals of ${\displaystyle {\mathcal {A}}}$.
5 Lemma Let ${\displaystyle x\in {\mathcal {A}}}$. Then ${\displaystyle x}$ is invertible if and only if ${\displaystyle \omega (x)\neq 0}$ for every ${\displaystyle \omega \in \Delta ({\mathcal {A}})}$
5 Theorem ${\displaystyle \omega (x)=\omega ({\hat {x}})}$
An involution is an anti-linear map ${\displaystyle x\to x^{*}:A\to A}$ such that ${\displaystyle x^{**}=x}$. Prototypical examples are the complex conjugation of functions and the operation of taking the adjoint of a linear operator. These examples explain why we require an involution to be anti-linear.
Now, the interest of study in this chapter. A Banach algebra with an involution is called a C*-algebra if it satisfies
${\displaystyle \|xx^{*}\|=\|x\|^{2}}$ (C*-identity)
From the C*-identity follows
${\displaystyle \|x^{*}\|=\|x\|}$,
for ${\displaystyle \|x\|^{2}\leq \|x^{*}\|\|x\|}$ and the same for ${\displaystyle x^{*}}$ in place of ${\displaystyle x}$. In particular, ${\displaystyle \|1\|=1}$ (if ${\displaystyle 1}$ exists). Furthermore, the ${\displaystyle C^{*}}$-identity is equivalent to the condition: ${\displaystyle \|x\|^{2}\leq \|x^{*}x\|}$, for this and
${\displaystyle \|x^{*}x\|\leq \|x^{*}\|\|x\|}$ implies ${\displaystyle \|x\|=\|x^{*}\|}$ and so ${\displaystyle \|x\|^{2}\leq \|x^{*}x\|\leq \|x\|^{2}}$.
For each ${\displaystyle x\in {\mathcal {A}}}$, let ${\displaystyle C^{*}(x)}$ be the linear span of ${\displaystyle \{1,y_{1}y_{2}...y_{n}|y_{j}\in \{x,x^{*}\}\}}$. In other words, ${\displaystyle C^{*}(x)}$ is the smallest C*-algebra that contains ${\displaystyle x}$. The crucial fact is that ${\displaystyle C^{*}(x)}$ is commutative. Moreover,
Theorem Let ${\displaystyle x\in {\mathcal {A}}}$ be normal. Then ${\displaystyle \sigma _{A}(x)=\sigma _{C^{*}(x)}(x)}$
A state on ${\displaystyle C^{*}}$-algebra ${\displaystyle {\mathcal {A}}}$ is a positive linear functional f such that ${\displaystyle \|f\|=1}$ (or equivalently ${\displaystyle f(1)=1}$). Since ${\displaystyle S}$ is convex and closed, ${\displaystyle S}$ is weak-* closed. (This is Theorem 4.something.) Since ${\displaystyle S}$ is contained in the unit ball of the dual of ${\displaystyle {\mathcal {A}}}$, ${\displaystyle S}$ is weak-* compact.
5 Theorem Every C^*-algebra ${\displaystyle {\mathcal {A}}}$ is *-isomorphic to ${\displaystyle C_{0}(X)}$ where ${\displaystyle X}$ is the spectrum of ${\displaystyle {\mathcal {A}}}$.
5 Theorem If ${\displaystyle C_{0}(X)}$ is isomorphic to ${\displaystyle C_{0}(Y)}$, then it follows that ${\displaystyle X}$ and ${\displaystyle Y}$ are homeomorphic.
3 Lemma Let ${\displaystyle T}$ be a continuous linear operator on a Hilbert space ${\displaystyle {\mathcal {H}}}$. Then ${\displaystyle TT^{*}=T^{*}T}$ if and only if ${\displaystyle \|Tx\|=\|T^{*}x\|}$ for all ${\displaystyle x\in {\mathcal {H}}}$.
Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma.
3 Lemma Let ${\displaystyle N}$ be a normal operator. If ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are distinct eigenvalues of ${\displaystyle N}$, then the respective eigenspaces of ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are orthogonal to each other.
Proof: Let ${\displaystyle I}$ be the identity operator, and ${\displaystyle x,y}$ be arbitrary eigenvectors for ${\displaystyle \alpha ,\beta }$, respectively. Since the adjoint of ${\displaystyle \alpha I}$ is ${\displaystyle {\bar {\alpha }}I}$, we have:
${\displaystyle 0=\|(N-\alpha I)x\|=\|(N-\alpha I)^{*}x\|=\|N^{*}x-{\bar {\alpha }}x\|}$.
That is, ${\displaystyle N^{*}x={\bar {\alpha }}x}$, and we thus have:
${\displaystyle {\bar {\alpha }}\langle x,y\rangle =\langle N^{*}x,y\rangle =\langle x,Ny\rangle ={\bar {\beta }}\langle x,y\rangle }$
If ${\displaystyle \langle x,y\rangle }$ is nonzero, we must have ${\displaystyle \alpha =\beta }$. ${\displaystyle \square }$
5 Exercise Let ${\displaystyle {\mathcal {H}}}$ be a Hilbert space with orthogonal basis ${\displaystyle e_{1},e_{2},...}$, and ${\displaystyle x_{n}}$ be a sequence with ${\displaystyle \|x_{n}\|\leq K}$. Prove that there is a subsequence of ${\displaystyle x_{n}}$ that converges weakly to some ${\displaystyle x}$ and that ${\displaystyle \|x\|\leq K}$. (Hint: Since ${\displaystyle \langle x_{n},e_{k}\rangle }$ is bounded, by Cantor's diagonal argument, we can find a sequence ${\displaystyle x_{n_{k}}}$ such that ${\displaystyle \langle x_{n_{k}},e_{k}\rangle }$ is convergent for every ${\displaystyle k}$.)
5 Theorem (Von Neumann double commutant theorem) M is equal to its double commutant if and only if it is closed in either weak-operator topology or strong-operator topology.
Proof: (see w:Von Neumann bicommutant theorem)
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# Control Systems Questions and Answers – State Models for Linear Continuous-Time Systems
This set of Control Systems Questions and Answers focuses on “State Models for Linear Continuous-Time Systems”.
1. Consider the following servomotors:
1. AC-two phase servomotor
2. DC servomotor
3. Hydraulic servomotor
4. Pneumatic servomotor
The correct sequence of these servomotor in increasing order of power handling capacity is:
a) 2,4,3,1
b) 4,2,3,1
c) 2,4,1,3
d) 4,2,1,3
Explanation: Power handling capacity is the capacity of the system where the system can handle very high power and among the given is maximum DC servomotor.
2. Open loop transfer function of a system having one zero with a positive real value is called.
a) Zero phase function
b) Negative phase function
c) Positive phase function
d) Non-minimum phase function
Explanation: Non-minimum phase system has zero lying on the right half but not pole and combination of the all pass and minimum phase system can be non-minimum phase system.
3. Assertion (A): The stator winding of a control transformer has higher impedance per phase
Reason (R): The rotor of control transformer is cylindrical in shape.
a) Both A and R are true and R is correct explanation of A
b) Both A and R are true and R is not correct Explanation of A
c) A is True and R is false
d) A is False and R is true
Explanation: The control transformer has two parts as stator which is stationary having higher impedance and rotor which was rotatory was cylindrical in shape.
4. Assertion (A): In a shunt regulator, the control element is connected in shunt with the load to achieve constant output voltage.
Reason (R): The impedance of the control element varies to keep the total current flowing through the load and the control element constant.
a) Both A and R are true and R is correct explanation of A
b) Both A and R are true and R is not correct Explanation of A
c) A is True and R is false
d) A is False and R is true
Explanation: Shunt regulator refers to the winding connected in shunt or parallel and to achieve constant output voltage.
5. Assertion (A): In the error detector configuration using a synchro transmitter and synchro control transformer, the latter is connected to the error amplifier.
Reason (R): Synchro control transformer has almost a uniform reluctance path between the rotor and the stator.
a) Both A and R are true and R is correct explanation of A
b) Both A and R are true and R is not correct Explanation of A
c) A is True and R is false
d) A is False and R is true
Explanation: Synchro control transformer is amplifier that is used to amplify the error and synchro transmitter is used to detect error.
6. A tachometer is used as an inner loop in a position control servo-system. What is the effect of feedback on the gain of the sub-loop incorporating tachometer and in the effective time constant of the system?
a) Both are reduced
b) Gain is reduced but the time constant is increased
c) Gain is increased but the time constant is reduced
d) Both are increased
Explanation: Tachometer is the device that is used to control the speed of motor in the control system and Feedback and effective time constant of the system both are reduced on the gain of the sub-loop.
7. If the initial conditions are inherently zero, what does it physically mean?
a) The system is at rest but stores energy
b) The system is working but does not store energy
c) The system is at rests or no energy is stored in any of its parts
d) The system is working with zero reference input
Explanation: A system with zero initial condition is said to be at rest since there is no stored energy.
8. In case of DC servomotor the back-emf is equivalent to an “electric friction” which tends to :
a) Improve stability of the motor
b) Slowly decrease stability of the motor
c) Vary rapidly decrease stability of the motor
d) Have no effect on stability
Explanation: Back emf is the voltage that is generated in absence of the input and in case of DC servomotor tends to improve the stability of the motor.
9. Consider the following statements for the pneumatic and hydraulic systems:
1. The normal operating pressure of pneumatic control is very much higher than that of hydraulic control.
2. In, pneumatic control external leakage is permissible to a certain extent, but there should no leakage in a hydraulic control.
Which of the statements given above is/are correct?
a) 1 only
b) 2 only
c) Both 1 and 2
d) Neither 1 and 2
Explanation: Operating pressure of pneumatic as the control,action is mainly to control the flow of air is more and also no leakage is permitted in hydraulic systems.
10. The transfer function of a LTI system is given as 1/(s+1). What is the steady-state value of the unit-impulse response?
a) 0
b) 1
c) 2
d) Infinite
Explanation: Steady state value is the final value and it is calculated from the final vale theorem and final value theorem is applicable for the stable systems only.
Sanfoundry Global Education & Learning Series – Control Systems.
To practice all areas of Control Systems Assessment Questions, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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Free Essay
Research Theories
In: Other Topics
Submitted By fatima57242
Words 479
Pages 2
NUST COLLEGE OF ELECTRICAL AND MECHANICAL ENGINEERING
NUST COLLEGE OF ELECTRICAL AND MECHANICAL ENGINEERING
DIGITAL IMAGE PROCESSING
LAB REPORT 4 Submitted By:
NS Fatima Hassan
DE-34-CE-B-209
Submitted to:
Sir Saqib Ejaz
Code:
img=imread('fig01.jpg'); figure(1) imshow(img,[]); img=rgb2gray(img); rotatedim=imrotate(img,45,'bilinear'); threshold=graythresh(rotatedim);
BW = im2bw(rotatedim,threshold); figure(2) imshow(BW,[]) Output:
Figure [ 1 ]: Original gray scale image
Figure [ 2 ]: Rotated image after thresholding
Code:
img=imread('fig02.jpg'); figure(1) imshow(img,[]); img=rgb2gray(img) doubimg=im2double(img); %converting image to double figure(2) imshow(doubimg,[]); logtransform=log(doubimg); %log transform figure(3) imshow(logtransform,[]);
Output:
Figure 3: Original image
Figure 4: Image after log transformation
Figure 5: Image after Log Transform
Code: img=imread('fig03.jpg'); figure(1) imshow(img,[]); img=rgb2gray(img); doubimg=im2double(img); %power transformtaions powimg1=1*(doubimg.^5); figure(2) imshow(powimg1,[]); powimg2=1*(doubimg.^10); figure(3) imshow(powimg2,[]); powimg3=1*(doubimg.^2); figure(4) imshow(powimg3,[]); powimg4=1*(doubimg.^0.5); figure(5) imshow(powimg4,[]); powimg5=1*(doubimg.^0.1); figure(6) imshow(powimg5,[]); %log transformation logtransform=log10(doubimg); figure(7) imshow(logtransform,[]); %inverse log transformation inv=(10).^logtransform; figure(8) imshow(inv,[]);
Output:
Figure 5: Original image
Power Law Transformation
Figure 6a
Figure 7b
Figure 7c
Figure 7d
Figure 7e
Figure 7: Log Transformation
Figure 8: Inverse Log Transform
Code: img=imread('fig04b.jpg'); figure(1) imshow(img,[]); img=rgb2gray(img); doubimg=double(img); s=zeros(size(img,1),size(img,2)); for i=1:size(doubimg,1) for j=1:size(doubimg,2) if ((doubimg(i,j)<96 )||(doubimg(i,j)==96)) s(i,j)=doubimg(i,j)*(32/96); elseif (doubimg(i,j)>96 && (doubimg(i,j)<160 || doubimg(i,j)==160)) s(i,j)=((doubimg(i,j)-96)*((224-32)/(160-96)))+32; else s(i,j)=((doubimg(i,j)-160)*((225-254)/(255-160)))+224; end end end figure(2) imshow(s,[]); Output:
Figure [ 9 ]: Original Image
Figure [ 10 ]: High Contrast image
Code: img=imread('fig05.jpg'); figure(1) imshow(img,[]); img=rgb2gray(img); doubimg=double(img); for x=1:8 bp=bitget(img,x); figure(x+1) imshow(bp,[]); end Output:
Figure [ 11 ]: Bit planes 1 to 6
Figure [ 12 ]:Bit planes 7,8
Output:
Figure [ 13 ]a :Outputs Task 6
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Market Research Theory
...MARKET RESEARCH 1. Discuss the importance of attitude measurement, and describe tow different approaches to measuring people's attitudes toward a given object. There is a growing need among today's marketers to better understand their customer's attitudes and feelings toward the company's products, services, and delivery systems. Some researchers view "attitude" as a derived composite outcome of the interaction between a person's beliefs (i.e., cognitive thoughts) and expressed emotions (i.e., affective feelings) regarding those beliefs. Knowing these interactions can be helpful in predicting a person's behavior (i.e., conative action). Not all researchers accept this trilogy approach to measuring attitudes; some simply see attitudes as a global indicator of a person's feelings (i.e., affect = attitude) toward an object or behavior. No matter the approach, these is significant diagnostic value to both researchers and practitioners in understanding the different scale measurements used to capture people's belief structures versus emotional feelings versus behavior tendencies. Tell how to correctly design and text Likert, semantic differential, and behavior intention scales, and explain their strengths and weaknesses. Likert scale designs uniquely use a set of agreement/disagreement scale descriptors to capture a person's attitude toward a given object or behavior. Contrary to popular belief, a Likert scale format does not measure a person's complete attitude, only......
Words: 2923 - Pages: 12
For Rigour in Organizational Management Theory Research
Words: 1301 - Pages: 6
Discuss Theories and Research in Ell Development
...Week5-Final Paper (Classroom Scenario Analysis) To make curriculum accessible to English language learners (ELLs), teachers must be able to integrate strategies to help them develop social and academic language skills in English and provide support by using comprehensible input and scaffolding. Accommodations should be based on current theories and research in language and literacy development, and they should address the interrelationship between culture and language. For this assignment, you will read the "Classroom Scenario" from Pathways to teaching series: Practical strategies for teaching english language learners. In a five-to eight-page paper, complete the following: Part 1: Analysis Analysis the scenario by addressing the following: Identify the strategies used in the scenario to help ELLs develop social and academic language skills in English. Explain if these strategies are effective. Why or why not? Explain how instructional input and scaffolding are used to support ELLs. Discuss current theories and research in ELL development. Give specific examples of how the teacher used theory and research to support his practice. Provide specific examples to support your points. Make sure to discuss the teacher's or student's actions in the scenario, and align them to specific concepts learned during the course. Whenever appropriate, use course vocabulary to demonstrate your knowledge of how it is applied and activated......
Words: 325 - Pages: 2
Nursing Theory Research Paper
...A nursing theory is a structured framework of concepts and purposes intended to guide the nursing practice. Nursing theories are important in nursing practice as they allow the nurses to use their critical thinking and analytical skills to improve concept comprehension. The use of nursing theories in the practice can bring new knowledge and can influence the future nursing practice. Nursing theories developed to describe the nursing care, guide the nursing practice and provide a foundation for clinical decision making. The foundation of the nursing theory was placed by Florence Nightingale. The first nursing theory established in the late 1800s when there was a strong emphasis on the nursing education took place. There are several nursing...
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Outline and Evaluate the Contributions of Psychological Research (Theories and/or Studies) to Our Understanding of the Formation of Relationships
...evaluate the contributions of psychological research (theories and/or studies) to our understanding of the formation of relationships (24 marks) One theory of formation of relationships that has contributed to our understanding of the formation of relationships, is the reward/need satisfaction theory. Byrne and Clove suggest that this theory means mutual attraction occurs when each partner meets the other persons need through operant conditioning. This might be the need for financial satisfaction or love etc. The rewards and needs can come from various factors. One of these factors is proximity which describes the distance between you and the potential partner. If the proximity is close then the reward gained is less effort being put in in having to see them. Another factor similarity refers to how similar you are to the potential partner in regards to the interests you both share i.e. religion, beliefs, music etc. The more similar you are the high the reward of enjoying each other’s company. A last factor is physical attractiveness referring to how attractive you think the potential partner is. A supporting study was done by Cate et al where he asked 337 individuals to assess their existing relationships in terms of reward level and satisfaction. The results found that reward levels was the most superior out of all other factors in determining relationship satisfaction. This therefore increases the reliability of the ‘Need/Satisfaction Theory’ and does explain why we form......
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Theory and Research
...How are theory and research related to each other? Discuss in detail and refer to the major components of research design (the sample design, the data collection design, and the data analysis design). The relationship between theory and research: Theory is defined as a systematic explanation for a set of facts and laws. Theory in relationship to research provides the hypothesis; therefore it is the source of one's research project. Theory is a motivation for undertaking Social research; research test, modifies, and expands social theory. A researcher can use theory for direction in choosing a research design or work with in limited theoretical expectations. In either format, research is structured by ideas and expectations created through use of abstract, logical reasoning. Theory is one of the three main elements in the traditional model of scientific method. In this model, developing a theory begins with an interest in some aspect of the real world. one's theoretical analysis of the social factors believed to affect the research topic, clarifies the possible relationship among factors known as variables. Through the next element, operational, the theoretical considerations results in the formation of a hypothesis. The hypothesis defines general cause and effect concepts known as research variables which are illustrated in a model. The variables are denoted by the letters X, being the independent variable or cause, and the Y, the dependent variable......
Words: 478 - Pages: 2
Research Theory
...Week 5 Literature Review 1. Transparency is not a One-Way Street, (2003). Disaster Prevention and Management, 12(1), 71-72. Retrieved from http://search.proquest.com/docview/214378197?accountid=14872 This article reviews how the red Cross has not been transparent with its dealings and how this has caused a mistrust between the organization and the public. This article applies to how the Red Cross is not providing information to the public and this is causing many problems internally and externally for the organization 2. Red Cross, (2015). Mission, Vision and Value Statement of the Red Cross. Retrieved from www.redcross.org This site explains what the Red Cross is trying to accomplish and how it operates. This site states what the true mission is and what is expected of the members. This site gives an insight into what the Red Cross is doing wrong. 3. Foster, R. (1950). The American Red Cross: A History. NY: Harper and Brothers. This book contains the history of the Red Cross and what the organization was meant to be and it gives a look into where the mission began to fail. 4. Gilbo, P. (1981). The American Red Cross. New York: Harper and Row. This book gives information on the Charter between the US Government and the American red Cross. This information covers the Charter and how FEMA and the Red Cross are suppose to work together in a disaster. 5. Allen, J. (2005). The Scandalous History of the Red Cross. CounterPunch:......
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Theory and Research
...We all use Language to communicate, to express ourselves to get our ideas across and to connect with the person whom we are speaking. In school we listen to our teacher inorde3r for us to understand the lesson. We make some oral and activities to test us if we truly understand the message or lesson that the teacher sent because we have different opinion, instinct, ideas and understanding that’s why we have different answer. Also our teacher have a different ways on how they teach us, so that we can easily understand the message or lesson. Sometimes they use some gestured, loud voice, and different movement in order to emphasis the message the he/she will sent. Listening is very important in communicating because it is a combination of hearing what another person says and psychological involvement with the person who is talking. Listening requires more than hearing words. It requires a desired understand another human being, an attitude if respect and acceptance and a willingness to open one’s mind to try and see things from another point of view. Listening requires a high level of concentration and energy. It demands that we set aside our own thoughts and agendas, put ourselves in another’s shoes and try to see the world through that person’s eyes. True listening requires that we suspend judgment, evaluation, and approval in an attempt to understand another is frame of reference, emotions, and attitudes. When we listen effectively we gain information that is valuable to......
Words: 780 - Pages: 4
Gesalt Theory Research
...Family Counseling Approach Research Gestalt Family Therapy Gretchen Thomas Liberty University Abstract Not every client will be of the Christian faith. With that known, proper integration of psychology, spirituality and theology is imperative to have a successful and healthy therapeutic relationship. There are many theorists with distinctively diverse approaches. The Gestalt experiential family therapy process allows professionals the flexibility to individualize their procedures while maintaining an unyielding theoretical background. The way one views integration has a lot to do with their worldview, which determines their use or lack of use of spiritual methods in counseling as well as help the professional deal with on-Christian counselees. This manuscript provides a synopsis of the history of the Gestalt experiential family therapy framework, the leading figures, and in conclusion an overview of how a personal worldview interacts with the Gestalt theory. keywords: Gestalt, integration, experiential family therapy, worldview Family Counseling Approach Research Gestalt Family Therapy Introduction The experiential model or experiential approach to family therapy has many innovators. The symbolic experiential approach pioneered by Carl Whitaker highlights how the therapist provides counselees different ways to accept and deal with issues (Goldenberg and Goldenberg, 2013). Satir (1967) emphasizes the......
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Organizational Research and Theory
...Organizational Research and Theory Name: Institution: Date: Power is an essential factor in the running of an organization institution or even a state. A powerful institution will always ensure that things work out and at no instanced are things going wrong. However, one would wonder what exactly organizational power is. Power also has some benefits that come along with it. Organizational power, in this case, can be defined as the means by which conflict are resolved in an organization. Others define power as the ability of a group or a person to overcome the resistance of others so as to achieve the desired outcome (Jones, 2010). It is reasonable to state that some element of coercion exists in conflict resolution. Therefore, organizational power is specifically said to be the ability of A to make B to do a thing that B could not have done. Therefore, if a subunit with an organization has a lot of power, it can apply power to achieve the results that it desires. In conflict resolution element of power plays a crucial role. It influences the decisions to be made, for example, how resources shall be allocated to different subunits. Most decisions in an organization are made through bargaining, and when this is so then the side with more power significantly influences the outcome or the decision made. In other words, power determines the subunit that will suffer with an organization and the one that will benefit. As a matter of fact, conflicts arise in organizations...
Words: 669 - Pages: 3
Free Essay
Kinds of Research Theory
...Aesthetician" redirects here. For a cosmetologist who specializes in the study of skin care, see Esthetician. Aesthetics (/ɛsˈθɛtɪks/; also spelled æsthetics and esthetics) is a branch ofphilosophy dealing with the nature of art, beauty, and taste, with the creation and appreciation of beauty.[1][2] It is more scientifically defined as the study ofsensory or sensori-emotional values, sometimes called judgments of sentimentand taste.[3] More broadly, scholars in the field define aesthetics as "critical reflection on art, culture and nature."[4][5] More specific aesthetic theory, often with practical implications, relating to a particular branch of the arts is divided into areas of aesthetics such as art theory, literary theory, film theory and music theory. An example from art theory is aesthetic theory as a set of principles underlying the work of a particular artist or artistic movement: such as the Cubist aesthetic.[6] Contents Edit The word aesthetic is derived from the Greek αἰσθητικός (aisthetikos, meaning "esthetic, sensitive, sentient"), which in turn was derived from αἰσθάνομαι (aisthanomai, meaning "I perceive, feel, sense").[7] The term "aesthetics" was appropriated and coined with new meaning in the German form Æsthetik(modern spelling Ästhetik) by Alexander Baumgarten in 1735. Aesthetics and the philosophy of artEdit Aesthetics is for the artist as Ornithology is for the birds. — Barnett Newman[8][9] For some, aesthetics is considered a synonym for the philosophy......
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1. ### Texas Education on iTunes U
• Resource ID: ITUNES01
• Subject:
Stream or download podcasts, videos, and other multimedia lessons directly from iTunes U. OER resources available in iTunes U can be streamed on iOS devices, or they can be downloaded and delivered through other selected devices and/or platforms.
2. ### TXRCFP: Texas Response to Curriculum Focal Points for K-8 Mathematics Revised 2013
• Resource ID: TEA001
• Subject: Math
The Texas Response to Curriculum Focal Points Revised 2013 was created from the 2012 revision of the TEKS as a guide for implementation of effective mathematics instruction by identifying critical areas of content at each grade level.
3. ### Vertical Alignment Charts for Revised Mathematics TEKS
• Resource ID: Revised_Math_TEKS_VA
• Subject: Math
This resource provides vertical alignment charts for the revised mathematics TEKS.
4. ### Determining the Perimeter of a Polygon (Series and Activity 1)
• Resource ID: TEKS12_MATH_03_001
• Subject: Math
This activity provides an opportunity for students to investigate the perimeter of polygons.
5. ### Investigating and Comparing Life Cycles
• Resource ID: R4SCI0020
• Subject: Science
A Tier I life science instructional resource for grade 3
6. ### Texas Performance Standards Project
• Resource ID: GT001
• Subject:
These videos introduce the Texas Performance Standards Project (TPSP), a resource for providing differentiated instruction to gifted/talented (G/T) students.
7. ### Weathering and Decomposition
• Resource ID: R4SCI0025
• Subject: Science
A Tier 1 earth science instructional resource for grade 3
8. ### Interactive Math Glossary
• Resource ID: MATH_IMG_001
• Subject: Math
The Interactive Math Glossary is provided by the Texas Education Agency to help teachers explore and understand mathematics vocabulary.
9. ### Heating and Cooling
• Resource ID: R4SCI0040
• Subject: Science
A Tier 1 physical science instructional resource for grade 3.
10. ### Texas Online College and Career Readiness
• Resource ID: OCCRRC_001
• Subject:
The videos in this resource introduce the Online College and Career Readiness Resource Center. The OCCRRC is a comprehensive tool for college and career development in Texas.
11. ### Exploring Energy
• Resource ID: R4SCI0052
• Subject: Science
A Tier 1 energy instructional resource for grade 3.
12. ### Response to Intervention (RtI)
• Resource ID: RtI001
• Subject:
Response to Intervention (RtI) is a multi-tiered framework designed to recognize and assist struggling learners through a systematic approach that includes strategies for intervention across curriculum, instruction, assessment and implementation.
13. ### Forces at Work
• Resource ID: R4SCI0059
• Subject: Science
A Tier 1 force and motion instructional resource for grade 3
14. ### Mathematics TEKS: Supporting Information
• Resource ID: Revised_Math_TEKS_SI
• Subject: Math
This resource presents supporting information of the mathematics TEKS for Kindergarten through Grade 12.
15. ### Target 2% Lessons to Support Comprehension for Grades K through 5
• Resource ID: T2P0005
These comprehension activities are designed to enhance the instruction of all teachers of reading and to meet the learning needs of all students, as indicated by data.
16. ### Texas Lesson Study Briefing
• Resource ID: TXLS_Pilot
• Subject:
Texas Lesson Study (TXLS) is systematic, inquiry-based professional development that fosters a collaborative, professional-learning environment among teachers and can be implemented by all campuses.
17. ### Target 2% Lessons to Support Vocabulary for Grades K through 5
• Resource ID: T2P0003
Vocabulary knowledge is an essential component of comprehension, language proficiency, and reading growth for all students.These vocabulary activities are designed to enhance the instruction of all teachers of reading and to meet the learning needs of all students, as indicated by data.
18. ### Target 2% Lessons to Support Phonics for Grades K through 5
• Resource ID: T2P0002
Phonics is the relationship between the letters (graphemes) of written language and the sounds (phonemes) of spoken language. These phonics activities are designed to enhance the instruction of all teachers of reading and to meet the learning needs of all students, as indicated by data.
19. ### Target 2% Lessons to Support Fluency for Grades K through 5
• Resource ID: T2P0004
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Pinterest • The world’s catalogue of ideas
Phonics challenge based on Obb and Bob (phonics play). Was further differentiated with different coloured words for different groups of children.
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Teacher's Pet – Ideas & Inspiration for Early Years (EYFS), Key Stage 1 (KS1) and Key Stage 2 (KS2) | Dice Stamping
1
Golf Ball Ladybugs! Got some old golf balls at home? Then recycle them and make a cute decoration for your garden! Painting golf balls to look like ladybugs is easy so it's a great project to do with kids. Just don't forget to wear an appropriate mask when you're spray painting the golf balls. ;) Is this going to be your next family fun activity?
During this game, students will practice telling one more, one less, ten more, and ten less than a given number. This game can be used during Envision's Topic 9: Comparing and Ordering Numbers to 100.
Write numbers 1-10 on a beach ball. Make sure to repeat so kids can practice multiplying like numbers. When your child catches the ball, they multiply together the two numbers their hands land on before throwing it to the next person.
"One More" Activity; roll the dice ... add 1 more to cover up the total.... mmmm... 1 less.... change numbers on card and dice and do doubles! LH
One Less Math Game... could make a board for 1 more....... doubles.... half it.... I'll try it! LH
'Zap it!' One more or one less game - place sticks face down in the pot, children take turns to pull out a stick and say the number one more or one less than the number on the stick. If they pull out a stick with ' zap it!' written on, they must put all their sticks back in the pot. Work as a team to try to get all the sticks out of the pot apart from the ones with 'zap it'! written on! Very popular game in our class! LG☆
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http://betterlesson.com/lesson/resource/1903722/73873/beginning-and-ending-a-lesson
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## Beginning and Ending a Lesson - Section 1: Lesson Opener
Beginning and Ending a Lesson
# Preview and Goal-Setting
Unit 1: Models and Constructions
Lesson 2 of 14
## Big Idea: Students learn what will be expected of them in this course as they are encouraged to persevere in making sense of a pair of novel problems.
Print Lesson
10 teachers like this lesson
Standards:
Subject(s):
Math, Geometry, Geometric Modeling, Technology and Engineering, assessment, high expectations
54 minutes
### Tom Chandler
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https://robotics.shanghaitech.edu.cn/courses/ca/18s/hw/5/
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## Homework 5 Solving Travelling Salesman Problem Using Pthread
Computer Architecture I ShanghaiTech University
HW4 HW5 HW6
### Travelling Salesman Problem
#### Introduction of TSP
The travelling salesman problem (TSP) asks the following question:
"Given a list of `cities` and the `distances` between each pair of cities,
what is the `shortest` possible route that `visits each city and returns to the origin city`?"
(You want to "link" these cities to form a shortest route)
It is actually an NP-complete problem (there's NO polynomial algrithms to solve this problem), It is important in operations research and theoretical computer science and also has many kinds of applications such as planning, logistics, and the manufacture of microchips, etc...
Check this wikipedia link for more details.
#### Problem Setup
Welcome to the NP-complete world, in this homework, you will try to solve this really hard problem use some strategy and accelerate using `pthreads`. You are a travelling salesman, given a map and start point, and you need to find the shortest route which starts at the given point, goes over all these cities only once (except the start city) and finally goes back.
You should implement a travelling salesman problem(TSP) solver in C.
#### Input
The first line of input contains 3 integers `t`,`n` and `s`
• `t` means the number of threads (main thread is not included), ` 0< t <=20 `
• `n` stands for the total number of cities
• `s` stands for the start point, ` 0 <= s < n `
• `n` will be guaranteed to`<= 10` in Autolab testcases.
The next `n` lines represent a symmetric adjacency matrix `A` (because the edges are `undirected`).
`A[i][j]` represents the ` distance `between the cities `i` and `j`, for all valid `i, j`, `1000>=A[i][j]>=0` and they are all `integers`.
A distance of `0` means their is no road between city `i,j`
For example:
``` 2 4 0 0 100 100 100 100 0 35 25 100 35 0 30 100 25 30 0 ```
#### Output
One single integer tells the total length of the route if it's feasible, -1 otherwise. The following example won't necessary be the right answer.
``` 200\n ```
#### Requirement
• You only allowed to inclue: `stdio.h stdlib.h string.h pthread.h`
• We will check memory leak using valgrind.
• Comment Rule. You need to have MEANINGFUL comments no less than 25% of the total lines of code you wrote.
• You must create `t` number of threads and all of them should do `reasonable` things (Otherwilse we will deduct 50% of your points).
• Of course your program should output the `precise` value, so `Randomized Algorithms` and `k-Approximation Algorithms` will be very likely to fail the test cases.
• Your code must stay in 2 files `tsp.c` and `tsp.h`, which will be created by yourself. We will not offer a code template this time.
• We will use the following Makefile:
``````CC=gcc
CFLAGS=-Wpedantic -Wall -Wextra -Werror -std=c89 -pthread
tsp.o: tsp.c tsp.h
\${CC} \${CFLAGS} tsp.c -o tsp.o``````
• Memory limit will be 40960KB in Autolab.
• You can test your memory by type `ulimit -v 40960` in Terminal and then run `./tsp.o < input_file_name `
• .
• Submit to Autolab with a file named hw5.tar, containing exactly two files: `tsp.c tsp.h`, it should produce a tsp.o file using our Makefile and can be tested using this command `./tsp.o < input_file_name `
• Autolab will just test your code with some `small <= 10 ` testcases, to verify the correctness of your program and it will contribute `60%` of the whole points.
• After the autograding deadline, we will run your code on a larger server with higher performance with a bigger testcase to test your speed.
• AC all the testcases in Autolab `DOES NOT` guarentee your correctness in the larger testcases on the bigger server, if you have something wrong with our code and use some randomized Algorithm or approximation algrithms.
• For the other `40%` of points: if you fail on the bigger testcases (like find a sub-optimal sulution), you will got `0` points, for programs which output the right answer, the fastest 20 people will get full points, the top 21~50 will get 80%, top 51~80 get 60%, others get 40%
#### Hints
• Implement a Brute Force Algorithm (Generate and test all the `(n-1)!` permutations of routes) and optimize it using pthread might be a safe and reasonable choice.
• Consider a cleverer inplementation of Brute Force algorithm when you generate permutations. (Generate it online or use recursive search.)
• You can also try to model it as Linear Programming
• Dynamic Programming is fast but you may consider how to deal with data concurrency problem, it's very hard to parallelize.
• Approximation and Randomized Algorithm is not recommanded, but not restricted. Though it is more feasible and much faster in practice but it may find suboptimal sulutions.
• If you implement some advanced algorithms and parallelize it additionaly, you can come to jiangchy & nianzhl's office hour, we may give you some bonus if you really did a good job.
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TL;DR
On with TASK #2 from The Weekly Challenge #209. Enjoy!
# The challenge
You are given an array of accounts i.e. name with list of email addresses.
Write a script to merge the accounts where possible. The accounts can only be merged if they have at least one email address in common.
Example 1:
Input: @accounts = [ ["A", "a1@a.com", "a2@a.com"],
["B", "b1@b.com"],
["A", "a3@a.com", "a1@a.com"] ]
]
Output: [ ["A", "a1@a.com", "a2@a.com", "a3@a.com"],
["B", "b1@b.com"] ]
Example 2:
Input: @accounts = [ ["A", "a1@a.com", "a2@a.com"],
["B", "b1@b.com"],
["A", "a3@a.com"],
["B"m "b2@b.com", "b1@b.com"] ]
Output: [ ["A", "a1@a.com", "a2@a.com"],
["A", "a3@a.com"],
["B", "b1@b.com", "b2@b.com"] ]
# The questions
Should the merging be “stable”? I mean, should we preserve as much as possible the order of appearance of the different groups? It seems not, because in the second example the two “A” groups both appear before the “B” group, despite a “B” group appearing between them.
Which begs a related question: maybe it’s some kind of “stable”, but moving forward instead of keeping things backwards? I’m digressing.
Another question relates to the order of the addresses. The inputs are arranged in arrays, which seems to imply that order might be important. ON the other hand, these arrays contain semantically different data (a name, addresses), so maybe it’s more like a tuple and order does not matter. I’ll assume the latter.
# The solution
The solution in Perl is somehow intentionally long and complicated. I took the challenge of producing a stable result, i.e. try to preserve the order of appearance of addresses if possible. Additionally, I tried to minimize the copying and duplications and iterations and whatsnot, in pure evil spirit of premature optimization.
Addresses are iterated over and amassed in “groups” by name. Each group contains all disjoint addresses belonging to that name, trying to pack them as much as possible while we do the input’s sweep. If we can merge, we merge and move on to see if additional merging is possible (because previous addresses A and B might be disjoint, but both joined with later address C).
#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
use constant TRUE => (!0);
use constant FALSE => (!!0);
my @accounts = (
['A', 'a1@a.com', 'a2@a.com'],
['B', 'b1@b.com'],
['A', 'a3@a.com', 'a4@a.com'],
['B', 'b2@b.com', 'b1@b.com'],
['A', 'a8@a.com'],
['A', 'a3@a.com', 'a2@a.com'],
);
for my $merged (merge_accounts(\@accounts)->@*) { say '[', join(', ', map { +"'$_'"} $merged->@* ), ']'; } sub hashes_intersect ($h1, $h2) { my$n1 = scalar(keys($h1->%*)); my$n2 = scalar(keys($h2->%*)); ($h1, $h2) = ($h2, $h1) if$n1 > $n2; # now$h1 has *at most* as many elements as $h2, it's beneficial to # iterate over it for my$key (keys $h1->%*) { return TRUE if exists$h2->{$key}; } return FALSE; } sub merge_accounts ($aref) {
my %alternatives_for; # track each name separately
my %group_for; # track aggregated groups by order of appearance
for my $i (0 ..$aref->$#*) { my ($name, @addresses) = $aref->[$i]->@*;
$group_for{$i} = my $new = { i =>$i,
name => $name, addresses => { map {$_ => 1 } @addresses },
};
# Add this group like it's detached
my $all_groups =$alternatives_for{$name} //= []; push$all_groups->@*, $new; # sweep back to merge when necessary my$challenger = $all_groups->$#*;
my $resistant =$challenger - 1;
my $last_wiped; while ($resistant >= 0) {
my $cas =$all_groups->[$challenger]{addresses}; my$ras = $all_groups->[$resistant]{addresses};
if (hashes_intersect($cas,$ras)) {
$ras->%* = ($ras->%*, $cas->%*); # merge ($last_wiped, $challenger) = ($challenger, $resistant); delete$group_for{$all_groups->[$last_wiped]{i}};
$all_groups->[$last_wiped] = undef;
}
--$resistant; } # sweep ahead to remove wiped out stuff, if necessary if (defined($last_wiped)) {
my $marker = my$cursor = $last_wiped; while (++$cursor < $all_groups->$#*) {
$all_groups->[$marker++] = $all_groups->[$cursor]
if defined($all_groups->[$cursor]);
}
splice $all_groups->@*,$marker if $marker <$all_groups->@*;
}
}
my @accounts = map {
my $group =$group_for{$_}; [$group->{name}, sort { $a cmp$b } keys $group->{addresses}->%* ]; } sort {$a <=> $b } keys %group_for; return \@accounts; } Update: the code above was previously bugged in a subtle way in the sweep ahead to remove merged elements in$all_groups. Thanks to oldtechaa for telling me! This will teach me to keep things simpler the next time 🙄
For contrast, in the Raku implementation I chose to ditch the stability and opted for some copying of data around, which I think improves readability and maintainability. Otherwise, the approach is pretty much the same: sweep and merge, keeping disjoint addresses.
#!/usr/bin/env raku
use v6;
sub MAIN {
my @accounts =
['A', 'a1@a.com', 'a2@a.com'],
['B', 'b1@b.com'],
['A', 'a3@a.com', 'a4@a.com'],
['B', 'b2@b.com', 'b1@b.com'],
['A', 'a8@a.com'],
['A', 'a3@a.com', 'a2@a.com'],
;
for merge-accounts(@accounts) -> $merged { put '[',$merged.map({"'$_'"}).join(', '), ']'; } } sub merge-accounts (@accounts) { my %alternatives_for; for @accounts ->$account {
my ($name, @addresses) = @$account;
my $new = { name =>$name, addresses => @addresses.Set };
my @disjoint;
my $all = %alternatives_for{$name} //= [];
for @$all ->$candidate {
if ($new<addresses> ∩$candidate<addresses>) { # merge
$new<addresses> = ($new<addresses>.keys.Slip,
$candidate<addresses>.keys.Slip ).Set; } else { @disjoint.push:$candidate;
}
}
@disjoint.push: $new; %alternatives_for{$name} = @disjoint;
}
return %alternatives_for.values».Slip.flat
.map({[ $_<name>,$_<addresses>.keys.Slip ]})
.Array;
}
All in all, this challenge was a bit more… challenging than the average for me. All of this, of course, thanks to manwar!
Stay safe!
Comments? Octodon, , GitHub, Reddit, or drop me a line!
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# When is it easy to write down the Bhargava S-factorial?
In Manjul Bhargava's The Factorial Function and Generalizations he motivates a new type of factorial $n!_S$ using by generalizing three theorems:
• For $k, l \in \mathbb{Z}$, we have $k! \times l!$ divides $(k+l)!$.
• For any primitive polynomial $f(x) \in \mathbb{Z}[x]$ with $\deg f = k$ then $\mathrm{gcd}\{ f(a): a \in \mathbb{Z}\}$ divides $k!$
In the process of solving generalizing these two results, he invents a factorial for any set of integers $S \in \mathbb{Z}$. For any prime $p$, order the element of $S$ by:
• choose $a_0 \in S$
• find $a_1$ giving the smallest power of $(a_1 - a_0)$
• find $a_2$ giving the smallest power of $(a_2 - a_0)(a_2 - a_1)$
• ...
• find $a_k$ giving the smallest power of $\prod_{i< k} (a_k - a_i)$
Here is my question:
• Bhargava observes that the integers in order $1,2,3, \dots, n$ form a $p$-ordered sequence for all $p$.
• Later in the same paper he observes the squares do as well $1^2,2^2,3^2, \dots, n^2$
• He also observes the powers of $2$ form a simultaneous $p$-ordering $1,2,4,8,\dots$
Does there exist a characterization of $S$ for when such a simultaneous ordering exists for all $p$? How can we read off the primes for which this isn't the case?
To my knowledge, this question remains open. It is difficult to characterize such sets. I would point out that all Fermat numbers is one of those sets [1]. More generally, if $E = O_f (a)$ is an the orbit of an integer $a$ under the iterated action of a polynomial $f \in \mathbb{Z}[X]$, then the sequence is simultaneous ordering exists for all $p$ (for sequence $E$).
Y fares
• Sorry, I meant: 1) the set of Fermat's numbers 2) The sequence( f ^ n (a)) is a simultaneous ordering of E = O_f (a) Nov 2 '14 at 14:29
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# What does DC voltage reading from a DMM mean?
When a DMM is set to measure DC voltage, does a read-out value correspond to an output of the ADC inside the DMM, or do multiple ADC outputs are averaged over a time interval and then converted to the DMM read-out value?
An oscilloscope always reads fluctuations in voltage, but a DMM tends to give stable readings. I wonder where the source of the difference is.
• an oscilloscope will not show any fluctuations if you measure true DC voltage May 27, 2019 at 2:46
• @jsotola For the same voltage input, OSC readings tend to fluctuate more. May 27, 2019 at 3:16
A typical DMM will only give 1-3 readings per second, whereas a DSO is expected to make potentially billions of readings per second.
The primary consequence of this is that fewer readings per second lets you integrate the reading over time, which is how a DMM can give true RMS readings, whereas a DSO can only give instantaneous readings.
If you slow a DSO down to just a few readings per second and tell it to average the results, you will get something closer to what a DMM gives you. But realize that it is not designed to do this, so a DMM will still do a better job at the tasks it is made for.
Another aspect of the difference is that even high-end DSOs rarely exceed 8 bits per sample, meaning only 256 values per range. Even a cheap DMM will generally give at least 2000 counts per range, and many go more than an order of magnitude higher than that.
These two aspects — speed and precision — are inversely related: it takes time to do a 16-bit or so measurement with high accuracy, so 50000 count DMMs either can only give a few readings per second or they're really expensive. Flipping it around, it's cheap to make a DSO that can do 500 megasamples per second over a 100 MHz measurement bandwidth if you only ask for 8-bit sampling.
• So, both OSC and DMM average input voltage over time, but they use different scale of intervals? May 27, 2019 at 3:29
• @Nownuri: It's better to think of a DSO as measuring each reading as fast as it possibly can at the expense of accuracy, relative to a DMM. The compensating advantage is that you get a lot of readings, so you can build up a pretty good picture of what the signal is doing down at the microsecond scale and below. May 27, 2019 at 5:37
Having a display which 'jumps all over the place' is bad for user confidence and also bad for meeting the typical use-case of "Tell me: ABOUT what voltage is present?" in many cases.
"Once upon a time", and quite likely still, DMMs use(d) "dual slope" analog to digital conversion. This has the advantage of being able to achieve a large number of bits of accuracy (and/or resolution) at relatively low cost.
It has the disadvantage of tending to be slower to much slower per conversion cycle.
DMM's once (and probably still) used to often make the conversion period a multiple of the mains cycle time (60 or 50 Hz mains) to minimise the impact of mains AC noise.
AND if you have a somewhat suspect conversion system, not only averaging the result but doing some sort of sliding window averaging when the input signal is 'a little noisy' and/or the conversion process is pushing its limits somewhat will add apparent stability.
All these (or some subset) add up to a high resolution, OK accuracy stable display.
If you look in the spec sheet (that small sheet of close typed thin paper that often comes with meters) you'll find that on a 1999 display (which thus has 0.05% resolution), the accuracy is usually around 1% (ie merits a 199 display.)
On a digital display of actual voltage, the last digit "should" hunt between two adjacent digits, with the time on each proportional to the undisplayed remainder. eg a voltage of 1.9876 V on a perfect accuracy "three and a half digit" 1999 display should alternate between 1.987 V and 1.988 V. On avergae it should display the former 40% of the time and the latter 60% of the time as the undisplayed 6 is 6/10 of the way between 1.987 and 1.988. Many meters do NOT have the least significant digit "hunting" in this manner.
A DMM takes a certain amount of time to digitize each reading. There are trade offs to that: longer is less responsive in the case of “varying DC”, but averaged out AC noise better.
Precision instruments sometimes have this adjustable. Some call it “reading time”, for example. NI has a nice discussion in one of their manuals.
• Do you mean that each voltage reading corresponds to an ADC output, and the noise component is averaged out inside the ADC? May 27, 2019 at 3:26
• Different instruments work differently. Some average a bunch of ADC measurements, but there’s often some averaging in the analog front end. May 27, 2019 at 3:29
• Okay, so somewhere inside the DMM the reduction seems to happen. Does a DMM reading correspond to an ADC output? May 27, 2019 at 3:31
• Varies. Cheap ones, yes. Precision ones do quite a bit more. May 27, 2019 at 3:32
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# Edexcel Pure Maths Sample Paper 2
Questions and Worked Solutions for Edexcel Pure Maths Paper 2 Sample.
Related Pages
More A Levels Past Papers
Edexcel Pure Maths Paper 2 Sample Past Paper (page 59)
1. f (x) = 2x3 - 5 x2 + ax + a Given that (x + 2) is a factor of f(x), find the value of the constant a.
1. Some A level students were given the following question.
Solve, for –90° < θ < 90°, the equation
cos θ = 2 sin θ
The attempts of two of the students are shown below.
(a) Identify an error made by student A.
Student B gives θ = –26.6° as one of the answers to cos θ = 2 sin θ.
(b) (i) Explain why this answer is incorrect.
(ii) Explain how this incorrect answer arose.
1. Given y = x(2x + 1)4, show that
dy/dx = (2x + 1)n(Ax + B)
where n, A and B are constants to be found.
1. Given
f(x) = ex
g(x) = 3lnx, x > 0
(b) Show that there is only one real value of x for which gf (x) = fg(x)
1. The mass, m grams, of a radioactive substance, t years after first being observed, is modelled by the equation
m = 25e–0.05t
According to the model,
(a) find the mass of the radioactive substance six months after it was first observed,
(b) show that dm/dt = km, where k is a constant to be found.
1. Complete the table below. The first one has been done for you. For each statement you must state if it is always true, sometimes true or never true,
1. (a) Use the binomial expansion, in ascending powers of x, to show that
where k is a rational constant to be found.
A student attempts to substitute x = 1 into both sides of this equation to find an approximate value for √3.
(b) State, giving a reason, if the expansion is valid for this value of x.
1. Figure 1 shows a rectangle ABCD.
The point A lies on the y-axis and the points B and D lie on the x-axis as shown in Figure 1.
Given that the straight line through the points A and B has equation 5y + 2x = 10
(a) show that the straight line through the points A and D has equation 2y - 5x = 4
(b) find the area of the rectangle ABCD.
1. In a geometric series the common ratio is r and sum to n terms is Sn
1. Figure 2 shows a sketch of part of the graph y = f(x), where
f(x) = 2|3 - x| + 5, x ≥ 0
(a) State the range of f
(b) Solve the equation
f(x) = 1/2 x + 30
Given that the equation f(x) = k, where k is a constant, has two distinct roots,
(c) state the set of possible values for k.
1. (a) Solve, for –180° ≤ x < 180°, the equation
3 sin2 x + sin x + 8 = 9 cos2 x giving your answers to 2 decimal places.
(b) Hence find the smallest positive solution of the equation
3sin2(2θ – 30°) + sin (2θ – 30°) + 8 = 9 cos2(2θ – 30°)
1. (a) Express 10 cos θ - 3 sin θ in the form R cos (θ + α), where R > 0 and 0 < α < 90°
Give the exact value of R and give the value of α, in degrees, to 2 decimal places. The height above the ground, H metres, of a passenger on a Ferris wheel t minutes after the wheel starts turning, is modelled by the equation
H = a - 10 cos (80t)° + 3 sin (80t)°
where a is a constant.
Figure 3 shows the graph of H against t for two complete cycles of the wheel.
Given that the initial height of the passenger above the ground is 1 metre,
(b) (i) find a complete equation for the model,
(ii) hence find the maximum height of the passenger above the ground.
(c) Find the time taken, to the nearest second, for the passenger to reach the maximum height on the second cycle.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
It is decided that, to increase profits, the speed of the wheel is to be increased.
(d) How would you adapt the equation of the model to reflect this increase in speed?
1. A company decides to manufacture a soft drinks can with a capacity of 500 ml.
The company models the can in the shape of a right circular cylinder with radius r cm and height h cm.
In the model they assume that the can is made from a metal of negligible thickness.
(a) Prove that the total surface area, S cm2, of the can is given by
S = 2πr 2 + 1000/r
Given that r can vary,
(b) find the dimensions of a can that has minimum surface area.
(c) With reference to the shape of the can, suggest a reason why the company may choose not to manufacture a can with minimum surface area.
1. Figure 4 shows a sketch of the curve C with equation
y = 5x3/2 – 9x + 11, x ≥ 0
The point P with coordinates (4, 15) lies on C.
The line l is the tangent to C at the point P.
The region R, shown shaded in Figure 4, is bounded by the curve C, the line l and the y-axis. Show that the area of R is 24, making your method clear.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# SCM Repository
[matrix] Diff of /pkg/Matrix/DESCRIPTION
[matrix] / pkg / Matrix / DESCRIPTION
# Diff of /pkg/Matrix/DESCRIPTION
revision 3230, Mon Aug 21 16:02:51 2017 UTC revision 3231, Mon Sep 4 15:22:49 2017 UTC
# Line 1 Line 1
1 Package: Matrix Package: Matrix
2 Version: 1.2-12 Version: 1.2-12
3 Date: 2017-08-21 Date: 2017-09-04
4 Priority: recommended Priority: recommended
5 Title: Sparse and Dense Matrix Classes and Methods Title: Sparse and Dense Matrix Classes and Methods
6 Contact: Doug and Martin <Matrix-authors@R-project.org> Contact: Doug and Martin <Matrix-authors@R-project.org>
# Line 17 Line 17
17 comment = c("condest() and onenormest() for octave", comment = c("condest() and onenormest() for octave",
18 "Copyright: Regents of the University of California")) "Copyright: Regents of the University of California"))
19 , person("R Core Team", role = "ctb", comment="base R matrix implementation") , person("R Core Team", role = "ctb", comment="base R matrix implementation")
, person("R Foundation", role = "ctb", comment="some po/ translation files")
20 ) )
21 Maintainer: Martin Maechler <mmaechler+Matrix@gmail.com> Maintainer: Martin Maechler <mmaechler+Matrix@gmail.com>
22 Description: Classes and methods for dense and sparse matrices and Description: A rich hierarchy of matrix classes, including triangular,
23 operations on them using 'LAPACK' and 'SuiteSparse'. symmetric, and diagonal matrices, both dense and sparse and with
24 pattern, logical and numeric entries. Numerous methods for and
25 operations on these matrices, using 'LAPACK' and 'SuiteSparse' libraries.
26 Depends: R (>= 3.0.1) Depends: R (>= 3.0.1)
27 Imports: methods, graphics, grid, stats, utils, lattice Imports: methods, graphics, grid, stats, utils, lattice
28 Suggests: expm, MASS Suggests: expm, MASS
Legend:
Removed from v.3230 changed lines Added in v.3231
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BROWSE ALPHABETICALLY LEVEL: Elementary Advanced Both INCLUDE TOPICS: Basic Math Algebra Analysis Biography Calculus Comp Sci Discrete Economics Foundations Geometry Graph Thry History Number Thry Phys Sci Statistics Topology Trigonometry radian – scientific notation radian A dimensionless unit of measure of angles. An angle of one radian is given by the central angle of a circle subtending an arc of length equal to the radius of the circle. Consequently, 360 degrees is the same as 2p radians. See the related article for a more extensive exposition. Related article: Trig Functions and Identities range The set of elements to which a function maps the elements of its domain set. rational exponent An exponent of the form p/q, with p and q integers and q not zero. Evaluated as the qth root of the base, raised to the pth power, or equivalently, as the qth root of the pth power of the base. For a negative base, this operation is not defined except when q is odd. Irrational roots may be considered as limits of sequences of rational roots.Cf. laws of exponents. rational number An element of the set Q consisting of ordered pairs (p, q) of integers, with q not 0, and with the order relation (p, q) < (r, s) if and only if ps < rq as integers. (The ordered pairs are usually written p/q, i.e., as a fraction (ratio) with integer numerator and denominator.) The rational numbers are countably infinite.Cf. natural number, real number. Related MiniText: Number -- What Is How Many? ratio test ARTICLE A test for the convergence of a series. See the related article for a complete description. real number An element of the set R consisting of all of the rational numbers together with all of the irrational numbers. Sometimes called the continuum. Usually defined formally by a Dedekind cut of rational numbers. The real numbers form (uniquely) a complete ordered field, but are not algebraically complete.It is a famous theorem of Georg Cantor that the real numbers are not countable.Cf. complex number. Related MiniText: Number -- What Is How Many? real number line A geometrical line graphically representing the set of real numbers, in which every real number corresponds to a unique point on the line, and every point on the line corresponds to a unique real number. reflexive relation A relation “ ~ ” on a set X is reflexive if for every element x in X we have x ~ x. The relation “ ~ ” is called irreflexive if for every x we have that x ~ x is false. Note that a relation may be neither reflexive nor irreflexive.Cf. symmetric relation, transitive relation. regular polygon A polygon all of whose sides are equal in length and all of whose interior angles are equal. regular solid A polyhedron having congruent faces, which are themselves regular polygons. Also called Platonic solid. Related article: Platonic Solids relation An n-place relation is defined on a Cartesian product of n sets, and is represented by a set of ordered n-tuples. For example, the less-than (“<”) relation is a binary relation on numbers, and the membership relation (“e”) is a binary relation on sets. The property of forming a Pythagorean triple is a ternary relation on natural numbers, of which for example (3,4,5) is a member since 32 + 42 = 52.In a binary (two-place) relation, the set from which the abscissae are taken is called the domain, and the set providing the ordinates is called the range. Binary relations are classified according to whether they are reflexive, transitive, and/or symmetric.Cf. function, partial order, lattice. relatively prime Two natural numbers a and b are relatively prime if their greatest common divisor is 1. Riemann integral See integral. Riemann sum Let f be a real-valued function defined on the closed interval [a, b], and let D be a partition of [a, b], i.e., a = x0 < x1 < ... < xn = b, and where Dxi is the width of the i th subinterval. If c i is any point in the i th subinterval, then the sumis called the Riemann sum of f for the partition D. right angle An angle of 90 degrees (p/2 radians). Equivalently, it can be said that two right angles are supplemental angles, i.e., they add up to a straight line (180 degrees or p radians).Cf. complementary angles, acute, obtuse. root An nth root of a real or complex number x is a number which when multiplied by itself n times yields x.Of a polynomial p: A number x such that p(x) = 0. root test A test for the convergence of a series. See the related article for a complete description. Related article: Series scalar A quantity having only magnitude, not direction (typically an element of a field, such as the real numbers or complex numbers).Cf. vector. scalar product The scalar product, also called dot product, of two vectors is the sum of the products of the corresponding components of the two vectors. I.e., given two vectors x = (x1, x2, ..., xn) and y = (y1, y2, ..., yn), their scalar product is the scalar x1y1 + x2y2 + ... + xnyn.Cf. vector product. scalene A triangle is called scalene if all of its sides are unequal (equivalently, if all of its angles are unequal). scientific notation A number is written in scientific notation when it is written as the product of a real number between 1 and 10 and a power of 10. E.g., 320 is written in scientific notation as 3.2 × 102. On some calculators and in some textbooks, this may be written as 3.2E2. Scientific notation is a convenient way to represent very large and very small numbers. radian – scientific notation
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"Gluing" related terms, short phrases and links
Web keywen.com
Gluing Article History Tree Map
Encyclopedia of Keywords > Projective Plane > Gluing Michael Charnine
Keywords and Sections
Review of Short Phrases and Links
This Review contains major "Gluing"- related terms, short phrases and links grouped together in the form of Encyclopedia article.
### DefinitionsUpDw('Definitions','-Abz-');
1. The gluing is specified by a continuous function f from ∂ D n = S n -1 to X n −1. (Web site)
### Sheaf TheoryUpDw('SHEAF_THEORY','-Abz-');
1. The first major hurdle in sheaf theory is to see that this gluing or patching axiom is a correct abstraction from the usual idea in geometric situations.
### TrenchingUpDw('TRENCHING','-Abz-');
1. Lawn Belt DIY Irrigation Systems - Learn how easy it is to install a lawn irrigation system without any trenching or gluing. (Web site)
### Gluing MapsUpDw('GLUING_MAPS','-Abz-');
1. If these gluing maps are diffeomorphisms, then we obtain a smooth (or differentiable) manifold.
### SurfaceUpDw('SURFACE','-Abz-');
1. Her invariant is constructed by looking at how the gluing map acts on the first homology group of the surface.
### BoardUpDw('BOARD','-Abz-');
1. I thought of gluing the center of each board to keep from cupping.
### ExtendingUpDw('EXTENDING','-Abz-');
1. This process of extending the original branch Log by gluing compatible holomorphic functions is known as analytic continuation.
### EndUpDw('END','-Abz-');
1. Thus, informally, the mapping cylinder M f is defined constructed by gluing one end of to Y by f.
2. For example, consider the surface formed by gluing one end of a cylinder to a half sphere.
### Opposite FacesUpDw('OPPOSITE_FACES','-Abz-');
1. Equivalently, the n -torus is obtained from the n -dimensional hypercube by gluing the opposite faces together. (Web site)
### PairUpDw('PAIR','-Abz-');
1. Gluing each pair of opposite faces together using this identification yields a closed 3-manifold.
### Gluing AxiomUpDw('GLUING_AXIOM','-Abz-');
1. To check the gluing axiom, let { U i } i ∈ I be a collection of open sets, and let U be the union of the { U i }.
2. The gluing axiom of sheaf theory is rather general.
3. Sometimes the gluing axiom is split into two axioms, one for existence and one for uniqueness.
### SheavesUpDw('SHEAVES','-Abz-');
1. Notice that we did not use the gluing axiom in defining a morphism of sheaves.
### Simplest WayUpDw('SIMPLEST_WAY','-Abz-');
1. The simplest way to make sure it is properly lined up on the mount is to make some index marks on the mirror and the mount prior to gluing.
### QuotientUpDw('QUOTIENT','-Abz-');
1. The action of the gluing map on the first homology group is induced by its action on the fundamental group, filtered through this quotient.
### ClosedUpDw('CLOSED','-Abz-');
1. A cell is attached by gluing a closed n -dimensional ball D n to the (n −1)- skeleton X n −1, i.e., the union of all lower dimensional cells. (Web site)
### Single PointUpDw('SINGLE_POINT','-Abz-');
1. In mathematics, a rose (also known as a bouquet of circles) is a topological space obtained by gluing together a collection of circles along a single point.
### Cw ComplexUpDw('CW_COMPLEX','-Abz-');
1. By contrast, gluing a one-dimensional string to three dimensional ball makes an object called a CW complex, not a manifold. (Web site)
### PointsUpDw('POINTS','-Abz-');
1. This can be visualized as gluing these points together in a single point, forming a quotient space. (Web site)
### WayUpDw('WAY','-Abz-');
1. We compose these morphisms in the visually evident way, by gluing the loose ends at the top of one to the loose ends at the bottom of the other.
### TopologyUpDw('TOPOLOGY','-Abz-');
1. In topology, quotients of spaces are obtained by "gluing points together".
2. In mathematics, the idea of descent has come to stand for a very general idea, extending the intuitive idea of gluing in topology. (Web site)
### FacesUpDw('FACES','-Abz-');
1. However for nice equivalence relations (e.g., those given by gluing together polyhedra along faces), it is a metric. (Web site)
### PolygonUpDw('POLYGON','-Abz-');
1. This includes the tessellations associated to the process of gluing together the sides of a polygon. (Web site)
2. A polyhedron is the three-dimensional version of a polygon, a three-dimensional figure made by gluing polygons together.
### PresheafUpDw('PRESHEAF','-Abz-');
1. Then a presheaf on C can be thought of as a geometrical structure built by gluing together these shapes along their common pieces. (Web site)
### SheafUpDw('SHEAF','-Abz-');
1. A sheaf on a site, however, should allow gluing, just like sheaves in classical topology.
2. Then a presheaf on X is a contravariant functor from O(X) to the category of sets, and a sheaf is a presheaf which satisfies the gluing axiom. (Web site)
### EdgesUpDw('EDGES','-Abz-');
1. Koichi Hirata's web software for finding all ways of gluing the edges of a polygon so that it can fold into a convex polytope. (Web site)
### CuttingUpDw('CUTTING','-Abz-');
1. All that remains is the craft of scoring the edges, cutting out the pieces, and gluing them together. (Web site)
2. It was just a case of cutting down the edges and gluing on the new parts.
3. In this talk, I will study the behavior of the quantum Hilbert space under the cutting and gluing of the symplectic manifold.
### WoodUpDw('WOOD','-Abz-');
1. In carpentry, the process of fastening together two pieces of board by gluing blocks of wood in the interior angle.
2. He also discusses joining pieces of wood, gluing them, and finishing them.
### GlueUpDw('GLUE','-Abz-');
1. So the gluing axiom fails: It is not always possible to glue two sections which agree on overlaps.
### ContinuousUpDw('CONTINUOUS','-Abz-');
1. The logarithm can be made continuous by gluing together countably many copies, called sheets, of the complex plane along the branch cut. (Web site)
### BoundaryUpDw('BOUNDARY','-Abz-');
1. The genus of such a surface is defined to be the genus of the two-manifold, which is obtained by gluing the unit disk along the boundary.
2. Gluing the circles together will produce a new, closed manifold without boundary, the Klein bottle.
3. If N is a compact manifold with boundary, its double M can formed by gluing together a copy of N and its mirror image along their common boundary.
### BoundariesUpDw('BOUNDARIES','-Abz-');
1. Formally, the gluing is defined by a bijection between the two boundaries. (Web site)
2. The amusing fact that the Klein bottle can be obtained from gluing two Mobius strips along their boundaries is proven.
### DeformedUpDw('DEFORMED','-Abz-');
1. Less formally it means that the two objects can be deformed into each other without cutting or gluing. (Web site)
### TopologicallyUpDw('TOPOLOGICALLY','-Abz-');
1. A 3-sphere can be constructed topologically by "gluing" together the boundaries of a pair of 3-balls.
### ManifoldsUpDw('MANIFOLDS','-Abz-');
1. Gluing along boundaries Two manifolds with boundaries can be glued together along a boundary.
2. Exactly as in the case of manifolds, differentiability conditions can be imposed on the gluing maps to give a definition of a differentiable orbifold.
### ManifoldUpDw('MANIFOLD','-Abz-');
1. One method of identifying points (gluing them together) is through a right (or left) action of a group, which acts on the manifold.
2. One method of identifying points (gluing them together) is through a right (or left) action of a group (mathematics), which group action on the manifold.
3. A manifold is a mathematical space which is constructed, like a patchwork, by gluing and bending together copies of simple spaces.
### NailingUpDw('NAILING','-Abz-');
1. The scaffold hitch can also be used as a clamp knot to hold two objects together while nailing or gluing takes place.
### SpaceUpDw('SPACE','-Abz-');
1. By gluing together enough of these charts, we can cover the space, but the map from our union of spectra to the moduli space will in general be many to one. (Web site)
2. But the Thom complex is the quotient of the total space of the sphere bundle obtained by gluing all the points at infinity together. (Web site)
3. Quotient space In topology, a quotient space is (intuitively speaking) the result of identifying or "gluing together" certain points of some other space. (Web site)
### IntuitivelyUpDw('INTUITIVELY','-Abz-');
1. Intuitively two spaces are homeomorphic if one can be deformed into the other without cutting or gluing.
### Scripting LanguagesUpDw('SCRIPTING_LANGUAGES','-Abz-');
1. Because of the features described above, scripting languages allow very rapid development for applications that are gluing in nature.
2. Scripting languages provide a powerful tool for easily gluing together components of a system.
3. Scripting languages are designed for gluing: They assume the existence of a set of powerful components and are intended primarily for connecting components.
### TorusUpDw('TORUS','-Abz-');
1. A torus (or donut) is obtained by gluing opposite sides of a square together.
2. Gluing a Torus Click To Watch Gluing is a good method to construct new topological spaces from known ones. (Web site)
### Projective PlaneUpDw('PROJECTIVE_PLANE','-Abz-');
1. A projective plane may be obtained by gluing a sphere with a hole in it to a Möbius strip along their respective circular boundaries. (Web site)
2. A cross-cap that has been closed up by gluing a disc to its boundary is an immersion of the real projective plane. (Web site)
### GluingUpDw('GLUING','-Abz-');
1. Patchwork A manifold can be constructed by gluing together pieces in a consistent manner, making them into overlapping charts. (Web site)
2. A manifold is a space created by gluing together pieces of Euclidean space, called charts.
3. The doubling of a manifold M with boundary is the operation of gluing two copies of M along the boundaries.
### CategoriesUpDw('Categories','-Abz-');
1. Projective Plane
2. Information Technology > Computer Science > Programming Languages > Scripting Languages
3. Intuitively
4. Nailing
5. Science > Mathematics > Topology > Manifold
6. Books about "Gluing" in Amazon.com
Short phrases about "Gluing" Originally created: April 04, 2011. Links checked: April 06, 2013. Please send us comments and questions by this Online Form Please click on to move good phrases up.
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Kevork Keheian
Posted on
# Writing DAX Formulas
## A - Introduction
Every Dax formula, similar to programming variables should be defined/declared by a name and followed by the formula/value.
``````<Calculation name> = <DAX formula>
``````
#### For example:
``````Order Type = "Allocation"
``````
Here, the `Order Type` is the name of the DAX formula and `"Allocation"` is the value or `Order Type`.
## B - Reminder
As we mentioned in the introduction that each DAX formula, whether it is a calculated column, calculated table or a measure returns a value.
In other words "Allocation" is the returned value.
This DAX formula is considered as a measure. So, whenever we drag this measure into a values field, it will return the value "Allocated" to each field of the visual.
So the returned result of a DAX will be `"Allocation"` for each row.
## C - Referring a Table
With DAX you can duplicate a table or a column using the same method as the above, but instead of a text inside a quotation you use the name of the table.
#### For Example:
Let's suppose that you have a table called `Sales` in your model. And you want to duplicate the table using a Calculated Table you will use the below formula:
``````Sales Orders = Sales
``````
Here the value or Sales Order which is the `Sales` table that was already in the model does not need single quotes.
Single quotes are needed when the value is a reserved word such as `Date` or the table name does not include embedded spaces.
EX:
``````Sales Orders = 'Sales Table'
``````
or
``````Sales Dates = 'Date'
``````
## D - Referring a Column
Referred Columns in a formula should be enclosed within square brackets.
#### example:
``````Total Sales = SUM([Amount])
``````
In the above example the value in the `SUM` function is the column name.
The column names are unique in the table. Some references tell you that you can use the column without preceding the table name.
But, to improve the readability of your formulas, it is preferable that the column reference precedes with its table name.
#### example:
``````Total Sales = SUM(Sales[Amount])
``````
Another reason for preceding the table name in a column reference formula is that you never know when you will update your model. As we mentioned earlier, the column name is always unique in a table, but can be found multiple times in a model which includes multiple tables.
## E - Referring a Measure:
Finally, you can refer a measure in your formula.
Let's suppose that we have the below column referring formulas (they are called measures):
``````Total Sales = SUM(Sales[Amount])
``````
``````Total Purchases = SUM(Purchases[Amount])
``````
And you want to calculate the difference between `Sales amount` and `Purchase Amount`.
To do this there are multiple ways:
``````Revenue = SUM(Sales[Amount]) - SUM(Purchase[Amount])
``````
or
``````Revenue= [Total Sales] - [Total Purchases]
``````
Note: This is not the right way to calculate the profit, but just an example to show how to refer to a measure in DAX
Note: Measures are a model-level object.
Although it is not recommended, but you can also precede a measure reference in a DAX formula with its table name.
#### example:
``````Revenue = [Total Sales] - Sales[Total Purchases]
``````
As we mentioned earlier many reasons to precede the table name in column referring measure, another reason is to differentiate between column referring and measure referring formulas.
## Summary:
• You can refer a TABLE in a DAX Formula `Sales Orders = Sales`
• You can refer a COLUMN in a DAX Formula `Total Sales = SUM([Amount])`
• You can refer a MEASURE in a DAX Formula `Revenue = [Total Sales] - [Total Purchases]`
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Question 128
# 3 Girls and 4 boys are to be seated in a row on 7 chairs in such a way that all the three girls always sit together. In how many different ways can it be done ?
Solution
Total number of persons = 3 + 4 = 7
The girls always sit together.
Considering the three girls as 1 person, we have 5 persons which can be arranged in 5! ways
But corresponding to each way of these arrangements, the girls can be arranged together in 3! ways.
Hence, required number of words = $$5! \times 3!$$
= 120 * 6 = 720
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$x\gt-7$
$\bf{\text{Solution Outline:}}$ Use the properties of inequality to solve the given inequality, $8-x\lt15 .$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given is equivalent to \begin{array}{l}\require{cancel} 8-x\lt15 \\\\ -x\lt15-8 \\\\ -x\lt7 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} -x\lt7 \\\\ x\gt\dfrac{7}{-1} \\\\ x\gt-7 .\end{array}
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Differences of Mixed Numbers without Renaming
## Subtracting equivalent mixed/improper fractions
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Differences of Mixed Numbers without Renaming
Have you ever had to cut an extra piece off of a board? Well, Travis is doing exactly that. Take a look.
While working with his Uncle, Travis discovered that one of the boards selected was too long for the project. Travis had to take the board and cut it so that it would fit in the place allotted on the floor. First, he measured the board.
Travis discovered that the board was $6 \frac{10}{16}$ feet long.
Travis needs to cut $3 \frac{2}{16}$ feet from the board.
To figure this out, Travis knows that he needs to subtract. Here is the problem that he wrote in his notebook.
$6\frac{10}{16} - 3\frac{2}{16}=\underline{\;\;\;\;\;\;\;\;\;}$
Now Travis has to complete the subtract. Then he will know the length of the board.
To complete this task, you will need to know how to subtract mixed numbers. Pay attention and this Concept will teach you everything that you need to know.
### Guidance
Just as we can add mixed numbers, we can also subtract mixed numbers.
The same rule applies, always subtract the fraction parts first then the whole numbers.
$& \quad \ \ 6\frac{3}{8}\\& \underline{- \quad 4\frac{1}{8}\;}$
We start by subtracting the fractions first, and these fractions have the same denominator so we can simply subtract the numerators. Three-eighths take away one-eighth is two-eighths.
$\frac{3}{8}-\frac{1}{8}=\frac{2}{8}$
Next, we subtract the whole numbers. 6 - 4 is 2. Our answer is $2\frac{2}{8}$ . However, our work is not finished because we can simplify two-eighths.
$\frac{2}{8}=\frac{1}{4}$
Our final answer is $2\frac{1}{4}$ .
Solve a few of these on your own. Be sure that your final answer is in simplest form.
#### Example A
$4\frac{4}{5}-3\frac{1}{5}=\underline{\;\;\;\;\;\;\;\;\;}$
Solution: $1 \frac{3}{5}$
#### Example B
$6\frac{4}{6}-1\frac{2}{6}=\underline{\;\;\;\;\;\;\;\;\;}$
Solution: $5 \frac{2}{6} = 5 \frac{1}{3}$
#### Example C
$7\frac{8}{9}-4\frac{4}{9}=\underline{\;\;\;\;\;\;\;\;\;}$
Solution: $3 \frac{4}{9}$
Have you figured out how to help Travis with the boards? Here is the original problem once again.
While working with his Uncle, Travis discovered that one of the boards selected was too long for the project. Travis had to take the board and cut it so that it would fit in the place allotted on the floor. First, he measured the board.
Travis discovered that the board was $6 \frac{10}{16}$ feet long.
Travis needs to cut $3 \frac{2}{16}$ feet from the board.
To figure this out, Travis knows that he needs to subtract. Here is the problem that he wrote in his notebook.
$6\frac{10}{16} - 3\frac{2}{16}=\underline{\;\;\;\;\;\;\;\;\;}$
Now Travis has to complete the subtract. Then he will know the length of the board.
To solve this problem, we can subtract the wholes and the parts separately.
$3 \frac{8}{16}$
This is the answer to the subtraction problem.
But wait, our work is not done yet! You can simplify this answer.
Our final answer is $3 \frac{1}{2}$ feet.
### Vocabulary
Mixed Number
a number that has a whole number and a fraction.
### Guided Practice
Here is one for you to try on your own.
$12\frac{46}{49} - 10\frac{39}{39}=\underline{\;\;\;\;\;\;\;\;\;}$
To find the difference, we have to subtract the wholes and the parts separately.
$2 \frac{7}{49}$
But our work is not done because the fraction part of this mixed number can be simplified.
Our answer is $2 \frac{1}{7}$ .
### Practice
Directions: Subtract the following mixed numbers. Be sure that your answer is in simplest form.
1. $6\frac{2}{9}-4\frac{1}{9}=\underline{\;\;\;\;\;\;\;\;\;}$
2. $5\frac{6}{10}-2\frac{1}{10}=\underline{\;\;\;\;\;\;\;\;\;}$
3. $8\frac{2}{8}-4\frac{1}{8}=\underline{\;\;\;\;\;\;\;\;\;}$
4. $12\frac{4}{8}-4\frac{2}{8}=\underline{\;\;\;\;\;\;\;\;\;}$
5. $6\frac{9}{10}-4\frac{2}{10}=\underline{\;\;\;\;\;\;\;\;\;}$
6. $15\frac{6}{15}-5\frac{3}{15}=\underline{\;\;\;\;\;\;\;\;\;}$
7. $18\frac{4}{12}-7\frac{2}{12}=\underline{\;\;\;\;\;\;\;\;\;}$
8. $20\frac{5}{20}-19\frac{1}{20}=\underline{\;\;\;\;\;\;\;\;\;}$
9. $5\frac{2}{5}-1\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}$
10. $8\frac{1}{2}-4\frac{1}{4}=\underline{\;\;\;\;\;\;\;\;\;}$
11. $6\frac{1}{3}-2\frac{1}{6}=\underline{\;\;\;\;\;\;\;\;\;}$
12. $5\frac{1}{4}-3\frac{2}{10}=\underline{\;\;\;\;\;\;\;\;\;}$
13. $8\frac{1}{3}-2\frac{1}{4}=\underline{\;\;\;\;\;\;\;\;\;}$
14. $12\frac{3}{4}-2\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}$
15. $18\frac{6}{9}-12\frac{1}{4}=\underline{\;\;\;\;\;\;\;\;\;}$
### Vocabulary Language: English
Mixed Number
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
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### Recent Question/Assignment
ENB360 ASSIGNMENT 1
Consider two incompressible Newtonian fluids of different viscosities flowing down an inclined plane and subject to gravity. Assume that the flow of both fluids is steady state and purely along the inclined plane so that the velocity only depends on the direction normal to the plane.
Assume that the flow of both fluids is steady state and purely along the inclined plane so that the velocity depends only on the direction normal to the plane.
The plane is set at an angle ? from the horizontal. The depth of the fluid closest to the inclined plane is h(1), the depth of the topmost fluid only is h(2). The total height of both fluids is H = h(1) + h(2).
We thus have to solve the Navier-Stokes equations for two different fluids.
Parameters
Let
? = 45?, h(1) = 0.005m h(2) = 0.01m, g = 9.8ms-2,
?(1) =?(2) = 6kgm-3.
Boundary conditions
Along the inclined plane, a ‘no-slip’ condition is applied so that the velocity of the fluid at the plane is zero.
At the interface between the two fluids, the velocities of the two fluids must match.
Also at the fluid-fluid interface, the viscous stress tensor for the two fluids must match.
As the top-most fluid has a fluid-air interface, we assume the viscous stress at this interface is zero.
Problems
1. Draw a schematic of the configuration including the 2 layers of fluid flowing down the inclined plan, all the parameters as well as the coordinate system. (2 Points)
2. List all the assumptions in mathematical notation and write the reduced form of the viscous stress tensor. (3 Points)
3. From the text, write explicitly the four boundary conditions required to solve the problem. (2 Points)
4. Based on the assumptions you gave in Problem 2, and the assumption of thin film flow, write the momentum equations to solve in Cartesian coordinates. (3 Points)
5. Using the boundary conditions, solve the equations to determine the velocity profiles of both fluids. (5 Points)
6. (a) Assuming µ(1) constant, equal to 0.01Pa s, plot the velocity profiles as a function of µ(2). (1 Point)
(b) Assuming µ(2) constant, equal to 0.02Pa s, plot the velocity profiles as a function of µ(1). (1 Point) (c) Discuss. (2 Points)
7. Assuming µ(2) constant, equal to 0.02Pa s, calculate the velocity at the top of the external fluid layer. (1 Point)
CRICOS No. 00213J ENB360 1
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Monte Carlo Simulation
Discover the intriguing world of Monte Carlo Simulation in Corporate Finance. This crucial tool, originally conceived in the realm of physics, now provides profound insights in the business sphere. Unpack its definition and grasp its importance in the finance industry. March through detailed steps of the Monte Carlo Simulation process, and comprehend the concept and role of convergence in this method. From theoretical frameworks to practical applications, this exploration of Monte Carlo Simulation will augment your Business Studies knowledge beautifully.
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# Monte Carlo Simulation
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Discover the intriguing world of Monte Carlo Simulation in Corporate Finance. This crucial tool, originally conceived in the realm of physics, now provides profound insights in the business sphere. Unpack its definition and grasp its importance in the finance industry. March through detailed steps of the Monte Carlo Simulation process, and comprehend the concept and role of convergence in this method. From theoretical frameworks to practical applications, this exploration of Monte Carlo Simulation will augment your Business Studies knowledge beautifully.
## Understanding Monte Carlo Simulation in Corporate Finance
In Corporate Finance, Risk Management and decision making are paramount. Various mathematical and statistical techniques aid in achieving these goals, one of the most renowned being the Monte Carlo Simulation.
### Definition of Monte Carlo Simulation
The Monte Carlo Simulation is a computational algorithm that relies on repeated random sampling to obtain numerical results. Essentially, it's a technique used to understand the impact of Risk and uncertainty in prediction and forecasting models.
This algorithm is called Monte Carlo Simulation due to its basis in chance operations, mirroring the random processes at play in the Monte Carlo Casino in Monaco.
For instance, to calculate the value of a corporate project with uncertain variables like fluctuating interest rates, unpredictable market conditions, and volatile costs, a Monte Carlo simulation would be run multiple times (even thousands or millions) with different random inputs for these variables. It would then output a range of potential outcomes, which helps the stakeholders to assess the Risk involved in the project.
### The Importance of Monte Carlo Simulation in Finance
Monte Carlo simulation is an indispensable tool in financial analysis. This is due, in part, to its ability to factor in a myriad of variables and their possible combinations.
• It provides a comprehensive view of what may happen in the future and allows for better strategic planning.
• It helps analysts and investors calculate the risk and quantify the impact of adverse situations on investment plans.
• It allows for Scenario Analysis by representing different limitations and possibilities of financial models.
Interestingly, Monte Carlo Simulations grew in popularity with the advent of computers. The computational power of modern machines allows simulations to run millions of times in a short span, providing a high-resolution view of possible outcomes.
#### Monte Carlo Finance Simulation and Investment Strategies
Monte Carlo simulation is also a critical companion for devising investment strategies. It helps investors and portfolio managers to understand the likelihood of getting different outcomes from their Investment Decisions. For instance, it may provide the probability distribution of certain ROI (Return on Investment) levels.
High Returns, High risk High Returns, Low risk Low returns, High risk Low returns, Low risk
The previous table represents possible outcomes of investment strategies. The Monte Carlo Simulation would provide a probability distribution for these outcomes, helping investors make well-informed decisions.
Also, it helps in creating robust Financial Planning by showing the most likely outcomes and yielding a greater level of confidence.
## Walking Through the Monte Carlo Simulation Process
Monte Carlo simulation is a method that allows for the modelling of complex scenarios involving uncertainty or randomness. It plays a huge role in many sectors including business, finance, project management, energy, research and so on. This walkthrough will provide an in-depth look at the Monte Carlo Simulation process to aid your grasp of risk and uncertainty in real-world scenarios.
### Basic Monte Carlo Simulation Steps
Executing a Monte Carlo Simulation involves a number of key steps. Here's a simple breakdown:
1. Identify a problem: Every simulation begins with a problem that needs solving. This could be the risks involved in pursuing a corporate project, figuring out the best investment mix for a portfolio, or determining price elasticity for a product.
2. Define a model: The problem is then converted into a mathematical model. This can be a simple formula or a complex system of equations, depending on the scenario at hand.
3. Specify the inputs: Identify the uncertain parameters or variables in the model and specify their probability distributions. This can be a normal distribution, lognormal distribution, uniform distribution, etc.
4. Generate random variables: Use a random number generator (often built into the software you are using) to produce values for the uncertain parameters.
5. Calculate the output: The random values are input into the model to calculate the output. This is repeated countless times to achieve a spectrum of results or outputs.
6. Analyse the result: After running the simulation numerous times (can be thousands or millions), analyse the distribution of the results to understand the risk or uncertainty facing the task.
For a very simple illustration, consider a dice game where you want to know your chances of getting a total of 7 when you throw two dice. You could do a Monte Carlo Simulation to model this scenario with a simple model like $Output = Dice 1 + Dice 2$
### An Illustrative Monte Carlo Simulation Example
Imagine an investment scenario where a fund manager aims to understand the possible 20-year returns of a $100,000 investment in a portfolio. This portfolio is comprised of Bonds with an expected annual return of 4%, and stocks with an expected return of 8%. Here are the steps to follow: 1. Firstly, the problem is identified - determining the possible 20-year returns of a$100,000 investment in a portfolio (with a 4% expected return for Bonds and an 8% expected return for stocks).
2. Secondly, a model would be defined to represent the portfolio returns. Usually, the returns would be compounded annually to calculate the total portfolio value. The details of the model would depend on how the portfolio is balanced and any other factors considered.
3. Next, the uncertain parameters – the annual returns for bonds and stocks – are determined and their probability distributions are specified. Often, these returns in the financial world are assumed to follow a normal (GAussian) distribution based on historical data.
4. Then, the Monte Carlo method works by generating random annual returns for bonds and stocks according to their respective distributions, for 20 years.
5. These random returns are inserted into the model to calculate the portfolio value after 20 years. This process is then repeated a large number (like a million) of times.
6. Finally, the output - the different possible portfolio values after 20 years, are analysed. The average could be used as an estimate of the expected return, and the distribution of the results can show the level of risk involved.
### Understanding the Monte Carlo Simulation Formula
Whilst Monte Carlo Simulation utilises complex algorithms, underpinning it all is a relatively simple concept represented by the formula:
$X= \sum_{i=1}^{N} \frac {f(X_i)} {Pr(X_i)}$
Here, $$X$$ is the expected outcome, $$f(X_i)$$ is the value of the output for the $$i$$th scenario, and $$Pr(X_i)$$ is the probability of the $$i$$th scenario. $$N$$ is the total number of scenarios.
The formula essentially represents a weighted sum where each outcome's contribution to the total is weighted by its probability. Rest assured, the computational aspect of this formula is taken care of by the simulation software, so the user only needs to focus on defining a sound model and accurately representing uncertainty in the inputs.
Ensuring the accurate representation of uncertainty is one of the most challenging aspects of Monte Carlo Simulation. However, once this is achieved properly, users are rewarded with a powerful tool for understanding and managing all kinds of risk and uncertainty.
## Broad Applications of Monte Carlo Simulation
Monte Carlo Simulation isn't just locked into finance, its power, flexibility, and utility have seen it applied to a diverse range of endeavours. It provides value in helping model complex systems and evaluate the impact of risk and uncertainty, making it a valuable instrument in a variety of fields including business, energy, logistics, environment, and many more.
### Diverse Monte Carlo Simulation Applications in Business Studies
In the field of business studies, Monte Carlo simulation is an invaluable tool for analysing complex and unpredictable systems. Its unique approach allows for extrapolating valuable insights that inform business decisions, strategic planning, cost estimation, Risk Management, and Scenario Analysis.
Here are some significant applications:
• Project Management: The simulation can aid in formulating budgets and scheduling timelines for projects. By running a series of simulations of possible costs and timeframes, a project manager can better manage risks and contingencies.
• Marketing Research: Cultivating strategies to effectively reach consumers involves dealing with various uncertainties like market size, competition, and consumer behaviour. Monte Carlo Simulation can run through various scenarios helping companies decide on the best course of action.
• Operational Risk: For many businesses, cross-functionalities can induce a level of complexity and unpredictability. Monte Carlo Simulation enables organisations to conduct a thorough operational risk analysis to ensure smooth business operations.
• Investment Decisions: Financial investments are fraught with risk. The analysis via Monte Carlo simulation can reveal the range of possible outcomes for investments and thus help businesses make well-informed risk-return trade-offs.
Take the case of a logistics company. The company faces uncertainties in the form of fluctuating fuel prices, varying demands, and varying delivery times, among others. The Monte Carlo simulation can handle all these random parameters simultaneously and can thus provide the company with a distribution of potential profits. Such profound insights can significantly drive operational performance and strategic growth for the company.
### Convergence of Monte Carlo Simulation Demystified
In the Monte Carlo simulation process, convergence is a key concept. It refers to the point at which the result of the simulation (the output) stabilises, giving the user greater certainty about the validity of the results and the robustness of the model. The essence of convergence lies in the Law of Large Numbers, a principle that supports the reliability of the Monte Carlo method.
The Law of Large Numbers, in basic terms, says that as the number of experiments increases, the average of the results gets closer and closer to the expected value. So, if you draw a diagram where the x-axis represents the number of simulations (or iterations), and the y-axis represents the average result, as x increases, the fluctuation in y decreases. Eventually, y tends to settle down to a constant value; this is what is referred to as convergence in Monte Carlo Simulations.
Consider a simple Monte Carlo Simulation in which you are estimating the mean of a normal distribution from a sample. Initially, as you take more samples the mean may change dramatically. However, as you keep increasing the number of samples, the mean will stabilise and converge to the actual mean of the distribution. This is a good example of convergence in Monte Carlo Simulations.
#### The Role of Convergence in the Monte Carlo Process
It's crucial to appreciate that good convergence is an indication of robust simulation. The output gives the user confidence in the reliability of the estimates provided by the Monte Carlo method. However, it's essential to bear in mind that reaching convergence doesn't necessarily imply getting more precise estimates. It simply means that running the simulations more times won't result in drastic changes in the expected outcome.
To check the convergence, some prefer to run a series of trials and make statistical tests on the results of the trials. Others prefer to visualise the iterations and observe the stability of the results. Whatever the approach, understanding and checking for convergence is a significant step in the Monte Carlo simulation process.
Also, it's important to note that the number of iterations required to reach convergence may vary from case to case. It depends quite a bit on the complexity of the simulation, the setup of the model and the nature of the uncertainties being simulated. Therefore, it's vital to understand the drivers of convergence to ensure a reliable and informative outcome.
With the ability to analyse a wide spectrum of outcomes and asses probabilities for each, Monte Carlo Simulations provide a rich perspective on risk management, facilitating informed decision making across different contexts and applications.
## Monte Carlo Simulation - Key takeaways
• The Monte Carlo simulation is a computational algorithm that uses repeated random sampling to obtain numerical results and is primarily used to understand the impact of risk and uncertainty in prediction and forecasting models.
• In corporate finance, the Monte Carlo simulation allows analysts and investors to calculate the risk and quantify the impact of adverse situations on investment plans and helps in formulating better strategic plans.
• The Monte Carlo simulation process involves identifying a problem, defining a mathematical model for the problem, specifying the inputs or uncertain parameters in the model, running the simulation using a random number generator, and then analysing the distribution of results to understand the risk or uncertainty.
• The Monte Carlo Simulation formula can be represented as a weighted sum where each outcome's contribution to the total is weighted by its probability, symbolized as [ X= Σ_{i=1}^{N} f(X_i) / Pr(X_i) ] where, X is the expected outcome, f(X_i) is the output value for the i-th scenario, Pr(X_i) is the probability of the i-th scenario, and N is the total number of scenarios.
• Convergence in Monte Carlo Simulation refers to the point at which the result of the simulation stabilises, giving greater certainty about the robustness of the model and the validity of the results. The concept of convergence is driven by the Law of Large Numbers, signifying that as the number of experiments increases, the average of results gets closer to the expected value.
## Frequently Asked Questions about Monte Carlo Simulation
A Monte Carlo simulation is a computerised mathematical technique, used in business studies, which allows people to account for risk in quantitative analysis and decision making. It provides a range of possible outcomes and the probabilities they will occur for any choice of action.
Monte Carlo simulation works by defining a mathematical model of a given problem, applying a sequence of random inputs to the model, and then analysing the distribution of results to determine probabilistic estimates and assess risk, uncertainty, or variability.
Monte Carlo simulation is used in business studies for risk assessment, decision-making, and forecasting. It utilises computational algorithms to simulate the impacts of risk in quantitative analysis and decision-making, based on probability distributions.
The accuracy of the Monte Carlo simulation largely depends on the number of iterations or runs. The more iterations, the higher the accuracy as the model is better able to capture variations and uncertainties. However, even with high iterations, the accuracy of results is contingent on the quality of input data.
A simulation is a computational method that imitates a real-life process or situation. On the other hand, a Monte Carlo simulation is a type of simulation that uses random sampling and statistical analysis to predict the outcomes of uncertain scenarios or decisions.
## Monte Carlo Simulation Quiz - Teste dein Wissen
Question
What is the Monte Carlo Simulation?
Show answer
Answer
The Monte Carlo Simulation is a procedure used to understand the impact of risk and uncertainty in forecast models. It allows decision-makers to assess the possible outcomes of their decisions and manage risk effectively.
Show question
Question
What does a Monte Carlo Simulation rely on and why is technology crucial for it?
Show answer
Answer
A Monte Carlo Simulation relies on the theory of probability and probability distributions associated with variables. Technology is crucial because powerful computers generate vast numbers of random sampling distributions and analyze millions of outcomes quickly.
Show question
Question
What are the primary applications of Monte Carlo Simulations in business studies?
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Answer
Monte Carlo Simulations are used in various sectors like finance for asset pricing and investments, supply chain for demand forecasting, project management for risk analysis, and marketing for market research and strategic planning.
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Question
What are the necessary steps to prepare for a Monte Carlo Finance Simulation?
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Answer
First, identify the financial problem. Second, understand and quantify all affecting variables. Third, assign each variable a probability distribution. Fourth, run the simulation using a specialized software tool. Fifth, analyze and interpret the results. Don't forget to document each step for transparency and traceability.
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Question
What does a typical Monte Carlo Finance Simulation involve?
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Answer
It often involves modelling uncertain parameters over time, such as the price of a stock. It includes variables like initial stock price, rate of return, and volatility, each assigned a probability distribution. The simulation runs numerous iterations, each calculating a probable outcome, which helps in understanding future behaviour.
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Question
How are Monte Carlo Simulations deployed in financial scenarios?
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Answer
They can be used in portfolio management to model behaviour under various conditions, in financial auditing for risk assessment, and in investment banks for pricing derivatives and strategic decisions. They need realistic and reliable estimates, preferably based on historical data, for successful deployments.
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Question
What are the four fundamental steps in the Monte Carlo Simulation process?
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Answer
The four fundamental steps in the Monte Carlo Simulation process are: Model development, Input distribution selection, Model computation, and Output analysis.
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What is the 'convergence' in the context of Monte Carlo Simulations?
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Answer
In Monte Carlo Simulations, 'convergence' refers to the point at which the addition of more trials doesn't significantly alter the simulation's outcomes, indicating the model is approaching a 'true value'.
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Question
What does the mathematical formula for Monte Carlo Simulation represent?
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Answer
The formula for Monte Carlo Simulation represents the expected value of a function of the random variable, simplified to make calculating complex, multidimensional models possible.
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Question
What is the significance of Monte Carlo Investment Simulations in risk management?
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Answer
Monte Carlo Simulations provide a comprehensive view of risk by offering a probabilistic range of potential outcomes, helping to understand the impact of constituents' risk on overall portfolio risk, and improving quality of decision-making. They provide insights into risk and a visual representation of possible outcomes.
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Question
What role does Monte Carlo Simulation play in portfolio diversification?
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Answer
In portfolio diversification, Monte Carlo Simulation generates thousands of scenarios for each asset considering various economic conditions. This offers a robust understanding of how an asset might perform, leading to a more optimised portfolio, and enhanced understanding of investment risk and return.
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Question
What are some Do's and Don'ts when using Monte Carlo Simulations?
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Answer
Do's include using realistic parameters, running sufficient simulations, and acknowledging the method's limitations. Don'ts involve assuming the model as reality, making excessive assumptions, and overlooking unquantifiable risks.
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Question
What are some advanced applications of Monte Carlo Simulation?
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Answer
Monte Carlo Simulation finds advanced application in quantum physics for solving Schroedinger's equation, in computational biology for DNA sequence alignment and phylogenetic analysis, and in artificial intelligence for game strategy development such as in AlphaGo.
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Question
What is Convergence in the context of Monte Carlo Simulation and what are the vital principles that explain it?
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Answer
Convergence in Monte Carlo Simulation refers to the statistics of results reaching a stable value as the number of trials increase. The Law of Large Numbers (LLN) and the Central Limit Theorem (CLT) are vital principles that explain it.
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Question
What measures can be taken to attain proficiency in Monte Carlo Simulation?
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Answer
Proficiency in Monte Carlo Simulation can be achieved by mastering the basics like probability distributions, stochastic systems, LLN and CLT, developing coding skills in a language that supports Monte Carlo Simulation such as Python or R and understanding real-world applications of the method.
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Question
What is the Monte Carlo Simulation in the context of corporate finance?
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Answer
The Monte Carlo Simulation is a computational algorithm that uses repeated random sampling to obtain numerical results. It's used to understand the impact of risk and uncertainty in prediction and forecasting models.
Show question
Question
Why is Monte Carlo Simulation important in finance?
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Answer
It provides a comprehensive future view and better strategic planning, helps calculate risk and quantify impacts of adverse situations, and allows for financial models scenario analysis.
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How does Monte Carlo Simulation aid in devising investment strategies?
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Answer
Monte Carlo simulation helps investors understand the likelihood of different outcomes from their investment decisions by providing the probability distribution of certain ROI levels.
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Question
Why is the algorithm called Monte Carlo Simulation?
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Answer
The term 'Monte Carlo' is due to its basis in chance operations, mirroring the random processes at play in the Monte Carlo Casino in Monaco.
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Question
What are the key steps in executing a Monte Carlo Simulation?
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Answer
The steps include identifying a problem, defining a model, specifying the inputs, generating random variables, calculating the output, and analysing the result.
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Question
What does the formula of the Monte Carlo Simulation represent?
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Answer
The formula represents a weighted sum where each outcome's contribution to the total is weighted by its probability.
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Question
What is Monte Carlo Simulation used for?
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Answer
Monte Carlo Simulation is used for modelling complex scenarios involving uncertainty or randomness, to understand risk in various sectors such as business, finance, energy, and others.
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Question
What is the role of random variables in Monte Carlo Simulations?
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Answer
Random variables are generated to produce values for the uncertain parameters. They are then input into the model to calculate the output repeatedly to achieve a spectrum of results.
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Question
What are some of the key uses of Monte Carlo simulation in business studies?
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Answer
Monte Carlo simulation is used for project management, marketing research, operational risk analysis, and investment decisions.
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Question
What is convergence in the context of Monte Carlo Simulation?
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Answer
Convergence in Monte Carlo Simulation is the point at which the result of the simulation stabilises, providing certainty about the validity of the results and the robustness of the model.
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Question
How is Monte Carlo Simulation beneficial beyond the field of finance?
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Answer
Monte Carlo Simulation has broad applications including modelling complex systems, evaluating risk and uncertainty, and optimising logistical and operational decisions in industries like business, energy, logistics, and environment.
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Question
How can one ensure good convergence in the Monte Carlo Simulation process?
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Answer
To ensure good convergence, one can run a series of trials and make statistical tests on the results or visualise the iterations and observe the stability of the results.
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## Test your knowledge with multiple choice flashcards
What is the Monte Carlo Simulation?
What does a Monte Carlo Simulation rely on and why is technology crucial for it?
What are the primary applications of Monte Carlo Simulations in business studies?
Flashcards in Monte Carlo Simulation27
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What is the Monte Carlo Simulation?
The Monte Carlo Simulation is a procedure used to understand the impact of risk and uncertainty in forecast models. It allows decision-makers to assess the possible outcomes of their decisions and manage risk effectively.
What does a Monte Carlo Simulation rely on and why is technology crucial for it?
A Monte Carlo Simulation relies on the theory of probability and probability distributions associated with variables. Technology is crucial because powerful computers generate vast numbers of random sampling distributions and analyze millions of outcomes quickly.
What are the primary applications of Monte Carlo Simulations in business studies?
Monte Carlo Simulations are used in various sectors like finance for asset pricing and investments, supply chain for demand forecasting, project management for risk analysis, and marketing for market research and strategic planning.
What are the necessary steps to prepare for a Monte Carlo Finance Simulation?
First, identify the financial problem. Second, understand and quantify all affecting variables. Third, assign each variable a probability distribution. Fourth, run the simulation using a specialized software tool. Fifth, analyze and interpret the results. Don't forget to document each step for transparency and traceability.
What does a typical Monte Carlo Finance Simulation involve?
It often involves modelling uncertain parameters over time, such as the price of a stock. It includes variables like initial stock price, rate of return, and volatility, each assigned a probability distribution. The simulation runs numerous iterations, each calculating a probable outcome, which helps in understanding future behaviour.
How are Monte Carlo Simulations deployed in financial scenarios?
They can be used in portfolio management to model behaviour under various conditions, in financial auditing for risk assessment, and in investment banks for pricing derivatives and strategic decisions. They need realistic and reliable estimates, preferably based on historical data, for successful deployments.
More about Monte Carlo Simulation
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# Difference between revisions of "2011 AMC 8 Problems/Problem 5"
## Problem
What time was it $2011$ minutes after midnight on January 1, 2011?
$\textbf{(A)}\ \text{January 1 at 9:31PM}$
$\textbf{(B)}\ \text{January 1 at 11:51PM}$
$\textbf{(C)}\ \text{January 2 at 3:11AM}$
$\textbf{(D)}\ \text{January 2 at 9:31AM}$
$\textbf{(E)}\ \text{January 2 at 6:01PM}$
## Solution 1
There are $60$ minutes in an hour. $2011/60=33\text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$
## Solution 2
Like the previous solution, $2011$ minutes is $33$ hours and $31$ minutes. That means it has to be January 1 at 9:31PM or January 2 at 9:31AM because they are the only ones that end with 31. But, the answer can’t be in January 1 because there are 24 hours in a day and 33 is greater than 24. So, the answer is $\Rightarrow\boxed{\textbf{(D)}\text{ January 2 at 9:31AM}}$. -RealCXY
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# What are the most fundamental classes of mathematical algorithms? [closed]
Mathematical means either useful for Mathematics or based on Mathematics.
I guess one has to include the following items:
(1) Euclid and LLL
(2) Newton and variations based on the existence of attracting fixpoints for maps.
(3) Algorithms for linear algebra and, in particular, FFT
(4) Simplex algorithm and other algorithms using convexity properties
(5) Quadrature formula (integration, differential equations etc.)
(6) Factorization (integers, polynomials etc.) and primality proving
(7) Algorithms for computations with Groebner bases.
(8) Sorting and searching
(9) WZ and similar algorithms yielding automated proofs
This list is surely a strict subset of a satisfying answer. Which important (classes of) algorithms are missing? (I guess there must be also be something in probability and perhaps in geometry.)
Let me also specify that I would like the list to contain general-purpose algorithms, not algorithms for specific tasks, like encryption, error-correction, computations of class numbers for number fields etc. I guess, algorithms for elliptic curves are on the boundary and perhaps already slightly outside the list (many of them are however used for integer-factorization and are thus implicitely contained in item (6) of the previous list).
A last question: Are we missing important fundamental algorithms? (This is of course tricky, since we are probably not aware of their potential usefulness). By this I do not mean a algorithm for factorizing integers in polynomial time or an (impossible) algorithm deciding the existence of solutions for arbitrary Diophantine equations but algorithms useful for a large class of problems for which we have presently only case by case tricks.ge
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## closed as too broad by Steven Landsburg, Chris Godsil, Ricardo Andrade, Lucia, S. Carnahan♦Oct 5 '14 at 3:53
There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.
Community wiki at the very least. – Loop Space Jun 9 '10 at 8:24
Roland, but why do you wish to collect those important algorithms? (You do not indicate your motivation! :-) ) – Wadim Zudilin Jun 9 '10 at 10:56
Sorry, I'm voting to close. The question as stated seems hopelessly broad, like asking for a list of important theorems. I strongly believe that the answer to your last question is "yes". – S. Carnahan Jun 9 '10 at 12:43
I think the intent of the question is a good one, but this is a bit of a Celestial Emporium of Benevolent Knowledge approach to categorisation. "Euclid" covers a pretty diverse bunch of things, if it covers any technique found in the Elements. – Charles Stewart Jun 9 '10 at 13:00
In 100 Digits Challenge: an Extended Review (March 2005) Jon Borwein lists "the 20th century's Top Ten" algorithms:
(1) 1946: The Metropolis Algorithm for Monte Carlo. Through the use of random processes, this algorithm offers an effcient way to stumble toward answers to problems that are too complicated to solve exactly.
(2) 1947: Simplex Method for Linear Programming. An elegant solution to a common problem in planning and decision-making.
(3) 1950: Krylov Subspace Iteration Method. A technique for rapidly solving the linear equations that abound in scientific computation.
(4) 1951: The Decompositional Approach to Matrix Computations. A suite of techniques for numerical linear algebra.
(5) 1957: The Fortran Optimizing Compiler. Turns high-level code into efficient computer-readable code.
(6) 1959: QR Algorithm for Computing Eigenvalues. Another crucial matrix operation made swift and practical.
(7) 1962: Quicksort Algorithms for Sorting. For the efficient handling of large databases.
(8) 1965: Fast Fourier Transform. Perhaps the most ubiquitous algorithm in use today, it breaks down waveforms (like sound) into periodic components.
(9) 1977: Integer Relation Detection. A fast method for spotting simple equations satisfied by collections of seemingly unrelated numbers.
(10) 1987: Fast Multipole Method. A breakthrough in dealing with the complexity of $n$-body calculations, applied in problems ranging from celestial mechanics to protein folding.
Please refer to the original paper for many examples illustrating Jon's point.
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Interesting. (1), (9) and (10) should probably be added to my list. (3), (4), (6) and (8) are contained in point (3) and (7) is contained in (8). I am not sure about (5), it should at least be an arbitrary compiler and not necessary Fortran. – Roland Bacher Jun 9 '10 at 9:37
Strange to ignore totally the area of cryptography. The emphasis is obviously on "scientific computing". – Charles Matthews Jun 9 '10 at 10:05
Integer Relation Detection (1977) precedes LLL (1982, not included). How queer! :-) – Robin Chapman Jun 9 '10 at 10:13
One argument for excluding cryptography: it should not exist in a perfect world! More convincingly, it is not really general purpose but adresses a concrete practical spaect of life. Including it, one has also to accept algorithms used at Wall-Street, algorithms useful for polls etc. – Roland Bacher Jun 9 '10 at 12:13
Wadim: Greedy algorithm! – Victor Protsak Jun 9 '10 at 18:44
All algorithms are based on mathematics, so if you want the word "mathematical" in your subject to be meaningful you need to be more specific.
Borwein's list in another answer is very heavily numerical. Here's a more combinatorial list, drawn from the topics I would typically cover in an undergraduate introductory algorithms class:
• integer arithmetic (including the elementary algorithms for addition, multiplication, etc but also faster divide-and-conquer multiplication, the equivalence in complexity between multiplication and division, modular exponentiation, and gcds; maybe also matrix multiplication and RSA cryptography)
• sorting (both comparison-based and integer sorting), searching sorted lists, and median finding
• pattern matching in strings
• dynamic programming (longest common subsequence, knapsack and subset sum problems, etc)
• graph algorithms (breadth first search, depth first search, topological ordering, minimum spanning trees, and shortest paths)
• computational geometry (nearest neighbors and convex hulls)
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"All algorithms are based on mathematics..." This might be too narrow and poses the question how to define 'algorithm'. Algorithms might also be phenomenons which can be found in some "collective behavior of decentralized, self-organized systems, natural or artificial", for example in communication (see Wikipedia 'Swarm intelligence' [1]). Algorithms in this sense are not based on mathematics, only described by formal systems and sometimes also mimicked by mathematicians (bees algorithm). [1] en.wikipedia.org/wiki/Swarm_intelligence – Bruce Arnold Jun 16 '10 at 16:13
An important class of algorithms which both make use of many fundamental algorithms already appearing in your list and contain significant ideas not reflected yet in your list are algorithms for computation with semialgebraic sets (i.e., algorithms in real algebraic geometry). These algorithms have a very different* flavour than their counterparts' over algebraically closed fields such as Buchberger's algorithm and related techniques used for Groebner basis computation, Wu's method and other techniques in elimination theory, etc.
Quantifier elimination (effective semialgebraic projection) has always been a central concern in algorithmic real algebraic geometry, so there is perhaps some feeling that these algorithms should be placed under your bullet point (9). But, the wealth of techniques developed in the context of real quantifier elimination are independently interesting and useful outside of the goal of automated proof.
Important algorithms include those which compute:
• Cylindrical Algebraic Decompositions (Collins et al),
• Signed subresultants (Collins et al),
• Betti numbers and Euler-Poincar\'e Characteristic (Basu, Basu-Pollack-Roy),
• Connected component sampling (Basu-Pollack-Roy),
• Roadmaps and Connectedness (Canny, Grigor'ev-Vorobjov, Heintz-Roy-Solerno, Gournay-Risler),
• Positivstellensatz witnesses via Semidefinite programming (Parrilo, Choi, Harrison, Lam, Powers, Woermann et al) [perhaps this is partially covered by your bullet point (4)].
In a certain sense, the properties Collins exploited in defining cylindrical algebraic decompositions have been generalised to what are now called o-minimal structures,' which has led to a rich and very active research area at the intersection of model theory and semialgebraic and subanalytic geometry (see L. van den Dries' Tame Topology and O-minimal Structures'').
(* Though it should be mentioned that fundamental algorithms in real algebraic geometry often make use of fundamental algorithms in classical algebraic geometry. )
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Fixed (should have been Vorobjov, and updated Poincar\'e with correct accent). Thank you! – Grant Olney Passmore Jun 9 '10 at 13:45
No problems! Is there a simple place where the collection of these algorithms is described? – Wadim Zudilin Jun 9 '10 at 13:49
Thankfully, yes! The monumental tome Algorithms in Real Algebraic Geometry' by Basu-Pollack-Roy is invaluable. A review is here ( math.tamu.edu/~rojas/bpr.pdf ) and with the 2nd edition, they've made it freely available online ( perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted1.html ). I refer to it daily! – Grant Olney Passmore Jun 9 '10 at 14:05
What about algorithms coming from graph theory? This seems to be a rich source with lots of real world applications too. To mention just a few
• Minimum weight spanning tree
• Maximum weight matchings
• Colouring algorithms
• Minor-testing
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The top 10 data mining algorithms identified by the IEEE International Conference on Data Mining (ICDM) in December 2006, which are presented in this article and this subsequent book are the following:
See here for further discussions and viewpoints on the importance of more recent algorithms in data mining.
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I'm surprised this list doesn't include interior point methods, which (unlike the simplex method) demonstrate that a large class of nonlinear optimization problems (namely, convex SDPs) can be solved in polynomial time.
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Though I agree it'd be nice for IPMs to have a separate bullet, in his defense, bullet point 4 does say and other algorithms using convexity properties.' – Grant Olney Passmore Jun 9 '10 at 14:37
Two different revolutionary data compressing Ziv and Lempel algorithms: LZ77 and LZ78 (or ZL77 and ZL78, perhaps depending on the selection of a natural language?).
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Online Math Lesson – P6W26Q13
At first, John had some money. He spent 1⁄4 of it on a T-shirt and 2⁄5 of the remainder on a pair of shoes. After that, his parents gave him \$120. The ratio of the total amount of money he had at [...]
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# What are stem and leaf diagrams in maths?
## What are stem and leaf diagrams in maths?
A stem and leaf diagram shows numbers in a table format. It can be a useful way to organise data to find the median, mode and range of a set of data.
## How do you do a stem and leaf plot in math?
To make a stem and leaf plot, each observed value must first be separated into its two parts:
1. The stem is the first digit or digits;
2. The leaf is the final digit of a value;
3. Each stem can consist of any number of digits; but.
4. Each leaf can have only a single digit.
What is a stem and leaf diagram?
A stem-and-leaf diagram, also called a stem-and-leaf plot, is a diagram that quickly summarizes data while maintaining the individual data points. In such a diagram, the “stem” is a column of the unique elements of data after removing the last digit. This diagram was invented by John Tukey.
### How do you answer a stem and leaf question?
Correct answer: Each of the numbers in the leftmost column, or “stems” represents the tens digits of one or more scores; each digit in one of the rows, or “leaf”, represents the units digit of a score. highest (or lowest) score. Counting up 27 scores, we get to the “4” in the “7” row, which represents a score of 74.
### What are the 7 steps for constructing a stem and leaf plot?
How to Make a Stem-and-Leaf Plot
1. Step 1: Determine the smallest and largest number in the data. The game stats:
2. Step 2: Identify the stems.
3. Step 3: Draw a vertical line and list the stem numbers to the left of the line.
4. Step 4: Fill in the leaves.
5. Step 5: Sort the leaf data.
What is the key for a stem and leaf plot?
When reading a stem and leaf plot, you will want to start with the key. It will guide you on how to read the other values. The key on this plot shows that the stem is the tens place and the leaf is the ones place. Stem and leaf plots are similar to horizontal bar graph, but the actual numbers are used instead of bars.
#### What makes up a stem and leaf diagram?
A stem and leaf diagram is a way of displaying a collection of numbers. The ‘stem’ consists of the first part of every number, usually the first digit (s) and the ‘leaf’ consists of the latter part of every number, usually the last digit.
#### How do you find the number of leaves in a tree?
The numbers in brackets at the end of each leaf tells us the number of leaves belonging to each stem. Note the similarity to a bar chart (just rotate 90 0 anti-clockwise). Once we have completed our diagram we can then analyse the data.
What is the split between stem and leaf?
To put the number 78.9 into a stem and leaf diagram, the ‘stem’ would be 78 and the ‘leaf’ would be 9. In this case, the key would indicate that the split between stem and leaf is a decimal.
## How many values are in the 7 stem diagram?
In the French side of the stem and leaf diagram, there are 5 values listed in the 7 stem group. The modal group is therefore the 70 s or the 70 – 79 group, since values that belong to this group must begin with a 7 .
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# How to find the center of a circle given two points
So I will choose two points and a radius and show that it is possible to solve the equations to find the Centre of the circle. Then we will have its equation! Let the two pints be (5, 4) and (3, 6) and
Solve Now
## Finding the Center-Radius Form of a Circle Given
How to find the equation of a circle given two pointsMore videos:Finding the equation of a circle given a midpoint and a radius:http://youtu.be/HBLoKLaDL4sFi
Get Started
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## Finding the Center of a circle given two points and a radius
Recommended: Please try your approach on {IDE} first, before moving on to the solution. Midpoint Formula: The midpoint of two points, (x1, y2) and (x2, y2) is : M = ( (x 1 + x 2) / 2, (y 1 + y 2) / 2) The center of the circle is the
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To determine what the math problem is, you will need to take a close look at the information given and use your problem-solving skills. Once you have determined what the problem is, you can begin to work on finding the solution.
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# How would you find the number, 3/7 of which is equal to t? (let's say it was any number, just need the equation)
2 months ago
## Solution 1
Guest #3617
2 months ago
Let the number be x
(3/7) x = t
I can't think of anything useful to add to this. How about an example?
Let x = 49
(3/7) x = t
(3/7) * 49 = t seven will go into 49
3 * 49 = t
t = 147
## 📚 Related Questions
Question
Amy has for gerbils Which Way 20 G 22 G 24 and 26 G what is the total weight of the gerbils in milligrams
Solution 1
First, I'll translate the problem into English:
Amy has four gerbils whose masses are 20 g, 22 g, 24 g, and 26 g.
What is the total mass of the gerbils in milligrams?
First add the masses in grams:
20 g + 22 g + 24 g + 26 g = 92 g
Now convert grams into milligrams.
1 gram = 1000 milligrams
92 g * 1000 = 92,000 mg
Answer: The total mass is 92,000 mg.
Question
How do i simplify 40 15×40 2
Solution 1
I am going to assume that you meant 40(15 x 40)2,, but if not then please let me know and I will fix my answer for you.
The first step for solving this is to multiply the numbers in the parenthesis.
40 x 600 x 2
Now multiply the first two numbers together.
24000 x 2
48000
Let me know if you have any further questions.
:)
Question
Emilio borrows $1200 from a bank with 8% simple interest per year. How much interest will he have to pay back in 2 years? Solution 1 The interest per year on the amount of money that Emillio borrowed from the bank at the rate of 8% interest is$192
• Interest can be regarded as the charges that is added to the principal, usually some percentage of the principal.
• Principal is the amount of money that is been lent out.
• Rate can be regarded as the percentage increase of the principal.
• Time (t) can be regarded as the period that is needed to pay the money that was lent out back.
• were were given rate (r) as%
• withas ?
From the the formula for expressing
• Then substitute into the formula
Therefore, the interest on the money is \$192
Solution 2
Simple Interest equation is I=p×r×t
P= is the principal amount (1200)
R= is the interest rate per year (8% or 0.08)
T= time in years (2)
I= 1200 × 0.08 × 2= 192.00
Interest is 192.00
Question
Fernando has a bucket that holds 3 gallons of water.he is filling the bucket using a 1-pint container.how many times will he have to fill the pint container in order to fill the bucket
Solution 1
48 times he will need to fill the pint container.
Question
solve the equation. choose the answer in simplest form. 4x= 8/7 (A) 1/28 (B) 1/7 (C) 2/7 (D) 7/2
Solution 1
4x=8/7
x=(8/7)÷4
x=2/7
Solution 2
4x = 8/7
/4 /4
x = (8/7)/4 (Keep, change, flip)
x = 8/7 * 1/4
x = 8/28
x = 2/7
Question
The width of a rectangle is twice it’s height if the perimeter is 54 inches find the area
Solution 1
The area is 162.
You can get this by solving for the sides. One side will be 9 and the other will be 18. This is because the perimeter must be 6x since one side is x and the other would be 2x.
Solution 2
Let the height be x.
⇒ Height = x
The width of a rectangle is twice it’s height.
⇒ Width = 2x
Perimeter = 54
⇒2 (x + 2x) = 54
Solve x:
2 (x + 2x) = 54
Combine like terms:
2 (3x) = 54
Distribute 2:
6x = 54
Divide both sides by 6:
x = 9
Height = x = 9
Width = 2x = 2(9) = 18
Area = 9 x 18 = 162 in²
Question
What is the length of the diagonal of a square with 8 foot sides?
Solution 1
Using the Pythagorean Theorem,
a² + b² = c²
c² = 8² + 8²
c² = 64 + 64
c² = 128
c = √128
c =8√2 (radical form) or 11.31 feet (nearest hundredth)
Question
A library is open til' 5:00 P.M. On Wednesdays, the library closes 2 hours early. How many hours for a month is the library open.
Solution 1
So
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7
ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add
25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
437.143+64.2856=501.429
so aprox 501.429
the real equation is where n represents the number of days in the month
apros 501.429
Question
How do i solve 3 (2y-4)=8y+9-9y
Solution 1
3(2y - 4) = 8y + 9 - 9y
Distribute 3:
6y - 12 = 8y + 9 - 9y
Combine like terms:
6y - 12 = -y + 9
7y - 12 = 9
7y = 21
Divide both sides by 3:
y = 3
Solution 2
The first step for solving this equation is to distribute 3 through the parenthesis.
6y - 12 = 8y + 9 - 9y
Collect the like terms with y.
6y - 12 = -y + 9
Move the variable to the left side and then change its sign.
6y + y - 12 = 9
Now move the constant to the right side and change its sign.
6y + y = 9 + 12
Add the terms on the left side of the equation.
7y = 9 + 12
Add the numbers on the right side.
7y = 21
Lastly,, divide both sides of the equation by 7 to find your final answer.
y = 3
This means that y = 3 is the correct answer to your question.
Let me know if you have any further questions.
:)
Question
the temperature last night was 10 degrees below 0. The next morning by 9 a.m. the temperature had risen 12 degrees. what was the temperature at 9 a.m.?
Solution 1
It was 2 degrees.................................................................................................................................................................................
Solution 2
It was two degrees at 9am
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The Scientific Method
She also found dead bugs on the bathroom floor, gross stains in the bedside tables and a canister of nitrous oxide, which some people inhale. But nothing was more shocking than what INSIDE EDITION.
The devastated mother of a newborn boy who died when he was mistakenly given the wrong gas at a Sydney hospital has described how she held her baby’s lifeless body and urged him to "Wake up, wake up".
Nitrous oxide: such a simple little molecule. when British chemist Joseph Priestly—the same dude who discovered oxygen (sounds weird that anyone can discover oxygen, doesn’t it?)—first identified.
A biological mechanism has been discovered that helps convert nitrogen-based fertilizer into nitrous oxide, an ozone-depleting. Cornell University. "Bacterial mechanism converts nitrogen to.
It made me dance about the laboratory like a madman,” said Humphry Davy in 1799, after having inhaled from an oiled green air bag his first lungful of nitrous oxide, later known. So he might: aged.
Ostrom was part of a team that discovered an ancient thriving colony. sulfur and supersaturated nitrous oxide—the highest ever measured in a natural aquatic environment. "It’s an extreme.
Software Atomic Operation Bank The bank has deployed software robots in retail banking operations, agri-business, trade and forex, treasury and human resources management, among other areas. The deployment has reduced the response. Bank of America. and co-head of global software. Before Citigoup, he worked at Deutsche Bank AG for 13 years focused on areas of technology banking including internet
Thermodynamics Closed System Equation First Law of Thermodynamic: Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms. ∆(Energy of the system) + ∆(Energy of surroundings) = 0 ∆Ut = Q + W → ∆(nU) = Q + W dUt = dQ + dW
29, near evidence of a party involving the recreational use of nitrous oxide, or NOS. Julio Cesar Munoz. They say evidence of a “NOS party” was in the vicinity of where his body was discovered.
Once they finally gained access they discovered a grisly grotto full of handcuffs, chains and several canisters of nitrous oxide. Nitrous oxide, also known as laughing gas, is sometimes used by.
Cornell researchers have discovered a biological mechanism that helps convert nitrogen-based fertilizer into nitrous oxide, an ozone-depleting greenhouse gas. The paper was published online Nov. 17 in.
Michigan State University scientists have pinpointed a new source of nitrous oxide, a greenhouse gas that’s more potent. hot spots has mystified soil microbiologists since it was discovered several.
The devastated mother of a newborn boy who died when he was mistakenly given the wrong gas at a Sydney hospital has described how she held her baby’s lifeless body and urged him to "Wake up, wake up".
"Nitrous oxide has been used for literally hundreds of years in this country without any problems." It was discovered in 1772 by British scientist Joseph Priestley and within 30 years the chemist.
nitrous oxide and fancy dress. The party was fantastically shot, not least because it managed to suggest that Delaney was a wounded bear out of his depth among the upper crust. It also brought several.
Georganism Cool Math 3 Georganism. Georganisms are sweet little creatures on a mysterious island with varied skills and limitations. Your mission? Explore the world by working together! Controls: Arrows = Move/Jump/Eat/Swim, Spacebar = Change character, Enter = Combine characters Games Similar. Adam and Eve: Zombies. Feed Us 3. Play Georganism 3 at Y3Games.org. Play Georganism 3 for free, play
Quantum 4front Order Form A comprehensive list of options, accessories and configurations is provided on the product order form. Your local MMS Medical product specialist will be happy to advise you. QUANTUM 4FRONT PERMOBIL F3 PERMOBIL F5 We cover the whole of Ireland. With bases in Cork, Dublin and Galway MMS Medical provide All-Ireland coverage. request call back. Quantum
Scientists have pinpointed a new source of nitrous. oxide quickly," said Sasha Kravchenko, an MSU plant, soil and microbial scientist and lead author of the study. "But the reason for these.
Nurse midwife Karli-Rae Kerrschneider wanted the same supportive birth experience she promises her own patients — and that included the use of nitrous oxide, or laughing. She protested further when.
It is highly reactive and toxic; it is used as a signaling molecule; it depletes the ozone layer in our planet’s atmosphere; and it is the precursor of the greenhouse gas nitrous oxide (N 2 O. the.
These nitrous oxide canisters were found in one of Newport’s parks last. in larger amounts each time we did a litter pick "As a concerned local councillor I discovered this was a growing problem.
Authorities say 41-year-old Evangelos R. Scullion was arrested after 939 pounds of Nitrous Oxide was discovered during a traffic stop earlier this week along the Turnpike, Scullion was stopped for a.
When Organisms Fight For Limited Resources When tiny marine organisms grow uncontrolled. and the receptor they bind (pharmacological target), in which a radiolabeled toxin competes for a limited number of receptor binding sites with the. said that the agency has the best genetically-modified organisms detection laboratory in Africa. According to him, “If anybody is saying that the NBMA does not have
Nurse midwife Karli-Rae Kerrschneider wanted the same supportive birth experience she promises her own patients — and that included the use of nitrous oxide, or laughing gas. She protested further.
Ncert Class 6 Social Science Book Download This chapter is all about the magnetic field, field lines, field due to a current carrying conductor, AC and DC generator, Advantages of AC over DC and domestic electric circuits. NCERT Solutions for. Jan 07, 2019 · Download NCERT textbooks and NCERT solutions of CBSE Class 10th Maths, Science, Hindi, English and Social Science subjects in
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5
# Four charges are placed at the corners of a square of side a 1m, with G1 = Gz =-1C, Q=q4 =+1C. Initially there Is no charge at the center of the square: The Coulomb...
## Question
###### Four charges are placed at the corners of a square of side a 1m, with G1 = Gz =-1C, Q=q4 =+1C. Initially there Is no charge at the center of the square: The Coulomb constant is 8.98755 x 10"N.m'/c' .BEFOREAFTERSa) Find the wark required to bring the charge (rom infinity and place it at the center of the squareSb) What is the magnitude of the total electrostatic energy of the final 5 charge system? Use Scientific Notation for answverl
Four charges are placed at the corners of a square of side a 1m, with G1 = Gz =-1C, Q=q4 =+1C. Initially there Is no charge at the center of the square: The Coulomb constant is 8.98755 x 10"N.m'/c' . BEFORE AFTER Sa) Find the wark required to bring the charge (rom infinity and place it at the center of the square Sb) What is the magnitude of the total electrostatic energy of the final 5 charge system? Use Scientific Notation for answverl
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# How do you analyze a truss?
## How do you analyze a truss?
There are two major methods of analysis for finding the internal forces in members of a truss; the Method of Joints, which is typically used for the case of creating a truss to handle external loads, and the Method of Sections, which is normally used when dealing modifying the internal members of an existing truss.
## What is truss in structural analysis?
A truss is an assembly of straight or curved bars biarticulated at their ends, which forms a stable structure. For their lightness and strength, trusses are widely used to solve the problems of range, resistance and aesthetics.
How do you calculate truss members?
The simplest form of this equation is to take the length of your roof and divide it by 2. For example, if your roof is 40-feet long, it will need a total of 20 trusses.
### What is truss analysis used for?
The objective of truss analysis is to determine the reactions and member forces. The methods used for carrying out the analysis with the equations of equilibrium and by considering only parts of the structure through analyzing its free body diagram to solve the unknowns.
### What is the purpose of analyzing the truss in sections?
The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. The method of sections is usually the fastest and easiest way to determine the unknown forces acting in a specific member of the truss.
What are truss members?
Trusses are structures that are composed entirely of two-force members. Each member of a truss is assumed to be a straight member that can only have forces applied on the ends of that member. The ends are pinned together so that they allow rotation.
#### What are the two methods of truss analysis?
Methods of analysis of trusses: The two common methods of analysis of trusses are the method of joint and the method of section (or moment).
#### What is truss equation?
Note: For a 3D determinate truss: 3n = m+r. If the truss is “determinate” then this condition is satisfied.
What is zero member in a truss?
Definitions. Zero-Force Members: structural members that support No loading but aid in the stability of the truss. Two-Force Members: structural members that are: a) subject to no applied or reaction moments, and b) are loaded only at 2 pin joints along the member.
## What is the purpose of a truss?
A truss gives a stable form capable of supporting considerable external load over a large span with the component parts stressed primarily in axial tension or compression. The individual pieces intersect at truss joints, or panel points.
What are the two analysis of trusses?
### What are pin joints?
A revolute joint (also called pin joint or hinge joint) is a one-degree-of-freedom kinematic pair used frequently in mechanisms and machines. The joint constrains the motion of two bodies to pure rotation along a common axis.
What is the truss analysis?
Truss Analysis – Method of joints: In method of joints, we look at the equilibrium of the pin at the joints. Since the forces are concurrent at the pin, there is no moment equation and only two equations for equilibrium viz.. Therefore we start our analysis at a point where one known load and at most two unknown forces are there.
#### How to determine the forces in the members of a truss?
There are two methods of determining forces in the members of a truss – Method of joints and method of sections. We start with the method of joints: In method of joints, we look at the equilibrium of the pin at the joints.
#### What happens if the number of members in a truss is less?
On the other hand, less number of members will make the truss unstable and it will collapse when loaded. This will happen because the truss will not be able to provide the required number of forces for all equilibrium conditions to be satisfied. Statically determinate trusses are known as simple trusses.
Is a truss determinate or determinate?
For this truss example, there are 11 members, 3 reaction components (hinge at A and roller at E), and 7 joints: Since D is zero, the truss is determinate and can be fully analysed using the three equations of static equilibrium. Now, how do we know if a truss is stable or not?
Simple Steps
1. Always Start by calculating reactions at supports.
2. Pick a point with a known force and look at in isolation.
3. Use vector geometry and the sum of forces = 0 to solve the other member forces.
4. Repeat the process until all members are solved.
5. Remember to look out for Zero Members.
## What is tension in a truss?
On truss bridges, a tension member is subject to forces that pull outward at its ends. Even on a “wooden” truss bridge, these members are often individual metal pieces such as bars or rods. Compressive forces push or compress together and are heavier. The individual members form a triangular pattern.
## What is the difference between compression and tension?
The main difference between tension and compression is that tension refers to forces that attempt to elongate a body, whereas compression refers to forces that attempt to shorten the body.
How do you identify compression and tension in a truss?
If the magnitudes of a calculated force is positive and it is pointing away from the joint, you have tension. If it is negative and pointing away from the joint you have compression.
### Is truss compression or tension?
Trusses are, normally, designed to carry axial forces in its members, which are either tension or compression or reversible tension/compression depending on the worst cases of loading and load combinations. Truss members are connected at joints using welds or bolts.
### What is tension and compression?
Back to definition. Tension is a force that stretches something. Compression is a force that squeezes something together. Materials are only useful if they can withstand forces. Force flows through a material like water flows through a pipe.
What are the main basic principles of solving the truss problems?
Simple Steps
• Always Start by calculating reactions at supports.
• Make a slice through the members you wish to solve.
• Treat the half structure as its own static truss.
• Solve the truss by taking the sum of forces = 0.
• Take the moment about a node of more than one unknown member.
#### Which member are in tension in a truss?
Tension members in trusses are called ties and these are members which are being stretched. It is an industry convention that the arrows are shown pulling in on themselves. This is in contrast to the tension in a beam in which the tension forces pull outwards from the beam as shown in the bottom diagram.
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