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# Fun, Hard Math Questions....
+1
346
3
+4096
Mods, if you want to post this, at the Stcky Topics discussion, you are welcome to!
Let's see if anyone here, can solve this, fun, challenging question...
Three players \(A,B\) and \(C\) play the following game: On each of three cards an integer is written. These three numbers \(p,q,r\) satisfy \(0 < p < q < r\). The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.
This process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, \(A\) has 20 counters in all, \(B\) has 10 and \(C\) has 9. At the last round \(B\) received \(r\) counters. Who received \(q\) counters on the first round?
This is from the 1974 IMO...Please do not search this problem up, and copy the solution here...
If you guys like this, I can post a problem from IMO every day, and you guys can solve it.
Have fun!
Feb 19, 2018
#1
+7354
+6
After the last round, A has 20 counters, B has 10, and C has 9 . So...
(p + q + r) * the num of rounds = 20 + 10 + 9 = 39
p + q + r = 39 / the num of rounds
Since p + q + r must be an integer, the num of rounds must be 3, 13, or 39 .
If the num of rounds = 13 , p + q + r = 3 , but no 3 different positive integers add to 3 .
If the num of rounds = 39 , p + q + r = 1 , but no 3 different positive integers add to 1 .
So the num of rounds must be 3 , and p + q + r = 13
By trial and error, I think I found 3 different integers that work: 1, 4, and 8
sum of counters earned in 3 rounds for player A = 8 + 8 + 4 = 20
sum of counters earned in 3 rounds for player B = 1 + 1 + 8 = 10
sum of counters earned in 3 rounds for player C = 4 + 4 + 1 = 9
q = the middle number = 4
On the first round, player C earned 4 counters.
I just got lucky by guessing those numbers for p, q, and r.....Maybe there is a way to do it without guessing!
Feb 20, 2018
#2
+99592
+2
Impressive, hectictar !!!
CPhill Feb 20, 2018
#3
+4096
+2
That is correct! Good job, hectictar! Amazing!
tertre Feb 20, 2018 | 734 | 2,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-18 | latest | en | 0.873998 |
https://forum.math.toronto.edu/index.php?PHPSESSID=a822j273a85q6l7745l44ouus5&topic=1536.0 | 1,675,867,481,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00553.warc.gz | 268,199,929 | 6,599 | ### Author Topic: TT2A-P4 (Read 4585 times)
#### Victor Ivrii
• Elder Member
• Posts: 2602
• Karma: 0
##### TT2A-P4
« on: November 20, 2018, 05:52:56 AM »
(a) Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} \hphantom{-}1 & \hphantom{-}2\\ -5 &-1\end{pmatrix}\mathbf{x}.$$
(b) Sketch trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).
#### Samarth Agarwal
• Jr. Member
• Posts: 6
• Karma: 11
##### Re: TT2A-P4
« Reply #1 on: November 20, 2018, 08:59:17 AM »
First, try to find the eigenvalues with respect to the parameter
$$A=\begin{bmatrix} 1&2\\ -5&-1\\ \end{bmatrix}$$
$$det(A-rI)=(1-r)(-1-r)+10=0$$
$$r^2 + 9 = 0$$
$$r = \pm 3i$$
The eigenvector is \\ \begin{bmatrix} -2\\ 1-3i \end{bmatrix}
Therefore x_1 =
$$\begin{bmatrix} -2\\ 1-3i \end{bmatrix} (\cos3t + i\sin3t)$$
$$= \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + i \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix}$$
Therefore the general solution
$$x(t) = c_1 \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + c_2 \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix}$$
« Last Edit: November 20, 2018, 09:14:25 AM by Samarth Agarwal »
#### Mengfan Zhu
• Jr. Member
• Posts: 10
• Karma: 5
##### Re: TT2A-P4
« Reply #2 on: November 22, 2018, 02:51:44 PM »
To be clear, I did it step by step to get the general real solution
If there are any mistakes, please tell me below ^_^
#### Jingze Wang
• Full Member
• Posts: 30
• Karma: 25
##### Re: TT2A-P4
« Reply #3 on: November 22, 2018, 03:46:14 PM »
Hello Samarth, I think your graph is not right, since the eigenvalues have no real parts, then graph should be center instead of spiral.
#### Michael Poon
• Full Member
• Posts: 23
• Karma: 10
• Physics and Astronomy Specialist '21
##### Re: TT2A-P4
« Reply #4 on: November 22, 2018, 05:06:53 PM »
Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically.
#### Victor Ivrii
Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically. | 810 | 2,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-06 | latest | en | 0.489347 |
https://eevibes.com/mathematics/linear-algebra/what-are-the-block-matrices/ | 1,679,444,422,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00574.warc.gz | 276,636,699 | 15,563 | # Introduction to Block Matrices
What are the Block Matrices? In matrix multiplication, whenever the order is high and we have some repeated entries in the matrix we prefer to use the concept of block matrices multiplication.
In this approach a submatrix of a matrix A is obtained by deleting certain number of rows and/or columns. It is also called partitioning of matrix and these small portions of the matrix are called the Blocks of A.
## Example of Block Matrices
Consider the following matrix A
Here it can be seen how the matrix A is divided into sub-matrices called block and finally A has been written in terms of the Block matrices.
Now if we want to multiply two matrices, we can use the concept of Block matrix multiplication given that the order of multiplication should be satisfied.
### Compute AB using block multiplication for the given matrices.
#### Solution:
In the above example the resultant matrix is of order 3×3.
This type of multiplication is helpful when we have repeated sequence of Identity matrices. | 204 | 1,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-14 | longest | en | 0.935693 |
https://lbs-to-kg.appspot.com/5/af/49.3-pond-in-kilogram.html | 1,722,865,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00481.warc.gz | 293,943,003 | 6,159 | Pond In Kilogram
# 49.3 lbs te kg49.3 Pond te Kilogram
lbs
=
kg
## Hoe om 49.3 pond skakel na kilogram?
49.3 lbs * 0.45359237 kg = 22.362103841 kg 1 lbs
## Skakel 49.3 lbs gemeenskaplike gewigte
Eenhede van metingGewig
Mikrogram22362103841.0 µg
Milligram22362103.841 mg
Gram22362.103841 g
ons788.8 oz
Pond49.3 lbs
Kilogram22.362103841 kg
Stone3.5214285714 st
Amerikaanse ton0.02465 ton
Ton0.0223621038 t
Imperial ton0.0220089286 Long tons
## Alternatiewe spelling
49.3 lbs te Kilogram, 49.3 lbs na Kilogram, 49.3 lb te kg, 49.3 lb na kg, 49.3 lbs te kg, 49.3 lbs na kg, 49.3 Pond te kg, 49.3 Pond na kg, 49.3 lb te Kilogram, 49.3 lb na Kilogram | 279 | 653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.106419 |
https://www.answerswave.com/ExpertAnswers/find-all-points-on-the-circle-x2-y2-100-where-the-slope-is-3-x-y-3-4-x-y-1-3-4-smaller-y-value-x-lar-aw587 | 1,701,850,267,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00794.warc.gz | 713,447,206 | 6,222 | Home / Answered Questions / Other / find-all-points-on-the-circle-x2-y2-100-where-the-slope-is-3-x-y-3-4-x-y-1-3-4-smaller-y-value-x-lar-aw587
# (Solved): Find All Points On The Circle X2 + Y2 = 100 Where The Slope Is 3. (x, Y) = ( 3, -4 (x, Y) = (1 -3,4 ...
Find all points on the circle x2 + y2 = 100 where the slope is 3. (x, y) = ( 3, -4 (x, y) = (1 -3,4 * ) (smaller y-value) x ) (larger y-value)
We have an Answer from Expert | 173 | 434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-50 | latest | en | 0.728748 |
https://docs.salome-platform.org/latest/gui/ADAO/en/ref_assimilation_keywords.html | 1,679,754,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945333.53/warc/CC-MAIN-20230325130029-20230325160029-00296.warc.gz | 268,104,248 | 6,794 | # 13.17. List of commands and keywords for data assimilation or optimisation case¶
We summarize here all the commands available to describe a calculation case by avoiding the particularities of each algorithm. It is therefore a common inventory of commands.
The set of commands for an data assimilation or optimisation case is related to the description of a calculation case, that is a Data Assimilation procedure or an Optimization procedure.
The first term describes the choice between calculation or checking. In the graphical interface, each of the three types of calculation, individually more oriented to data assimilation, optimization methods or methods with reduction (some algorithms are simultaneously in various categories), is imperatively indicated by one of these commands:
ASSIMILATION_STUDY
Required command. This is the general command describing one case using data assimilation methods. In the graphical interface, it hierarchically contains all the other commands.
OPTIMIZATION_STUDY
Required command. This is the general command describing one case using optimization methods. In the graphical interface, it hierarchically contains all the other commands.
REDUCTION_STUDY
Required command. This is the general command describing one case using methods with reduction. In the graphical interface, it hierarchically contains all the other commands.
The nested terms are sorted in alphabetical order. They are not necessarily required for all algorithms. The various commands are the following:
AlgorithmParameters
Dictionary. This variable indicates the data assimilation or optimization algorithm chosen by the keyword “Algorithm”, and its potential optional parameters. The potential choices by this keyword “Algorithm” are available through the graphical interface or in the reference documentation containing a specific sub-section for each of them. There are for example the “3DVAR”, the “Blue”, etc. Optionally, the command also allows to add parameters to control the chosen algorithm. Their values are defined either explicitly or in a “Dict” type object. See the Description of options of an algorithm by “AlgorithmParameters” for the detailed use of this command part.
Background
Vector. The variable indicates the background or initial vector used, previously noted as . Its value is defined as a “Vector” or “VectorSerie” type object. Its availability in output is conditioned by the boolean “Stored” associated with input.
BackgroundError
Matrix. This indicates the background error covariance matrix, previously noted as . Its value is defined as a “Matrix” type object, a “ScalarSparseMatrix” type object, or a “DiagonalSparseMatrix” type object, as described in detail in the section Requirements to describe covariance matrices. Its availability in output is conditioned by the boolean “Stored” associated with input.
ControlInput
Vector. This variable indicates the control vector used to force the evolution model at each step, usually noted as . Its value is defined as a “Vector” or “VectorSerie” type object. When there is no control, it has to be a void string ‘’.
Debug
Boolean value. This variable define the level of trace and intermediary debug information. The choices are limited between 0 (for False) and 1 (for True).
EvolutionError
Matrix. The variable indicates the evolution error covariance matrix, usually noted as . It is defined as a “Matrix” type object, a “ScalarSparseMatrix” type object, or a “DiagonalSparseMatrix” type object, as described in detail in the section Requirements to describe covariance matrices. Its availability in output is conditioned by the boolean “Stored” associated with input.
EvolutionModel
Operator. The variable indicates the evolution model operator, usually noted , which describes an elementary step of evolution. Its value is defined as a “Function” type object or a “Matrix” type one. In the case of “Function” type, different functional forms can be used, as described in the section Requirements for functions describing an operator. If there is some control included in the evolution model, the operator has to be applied to a pair .
ExecuteInContainer
String. This variable allows to choose the execution mode in YACS in a specific container. In its absence or if its value is “No”, no separate container is used for execution and it runs in the main YACS process. If its value is “Mono”, a specific YACS container is created and it is used to host the execution of all nodes in the same process. If its value is “Multi”, a specific YACS container is created and it is used to host the execution of each node in a specific process. The default value is “No”, and the possible choices are “No”, “Mono” and “Multi”.
Warning
In its present version, this command is experimental, and therefore remains subject to changes in future versions.
InputVariables
Optional command. This variable allows to indicates the name and size of physical variables that are bundled together in the state vector. This information is dedicated to data processed inside an algorithm.
Observation
List of vectors. The variable indicates the observation vector used for data assimilation or optimization, and usually noted . Its value is defined as an object of type “Vector” if it is a single observation (temporal or not) or “VectorSeries” if it is a succession of observations. Its availability in output is conditioned by the boolean “Stored” associated in input.
ObservationError
Matrix. The variable indicates the observation error covariance matrix, usually noted as . It is defined as a “Matrix” type object, a “ScalarSparseMatrix” type object, or a “DiagonalSparseMatrix” type object, as described in detail in the section Requirements to describe covariance matrices. Its availability in output is conditioned by the boolean “Stored” associated with input.
ObservationOperator
Operator. The variable indicates the observation operator, usually noted as , which transforms the input parameters to results to be compared to observations . Its value is defined as a “Function” type object or a “Matrix” type one. In the case of “Function” type, different functional forms can be used, as described in the section Requirements for functions describing an operator. If there is some control included in the observation, the operator has to be applied to a pair .
Observers
List of functions linked to variables. This command allows to set internal observers, that are functions linked with a particular variable, which will be executed each time this variable is modified. It is a convenient way to monitor variables of interest during the data assimilation or optimization process, by printing or plotting it, etc. Common templates are provided to help the user to start or to quickly make his case.
OutputVariables
Optional command. This variable allows to indicates the name and size of physical variables that are bundled together in the output observation vector. This information is dedicated to data processed inside an algorithm.
StudyName
String. This variable is an open string to identify or describe the ADAO study by a name or a sentence.
StudyRepertory
String. If available, the directory specified by this string is used as base name for calculation, and also used to find all the script files given by name without path, that can define some other commands by scripts.
UserDataInit
Script name. This variable allows to initialize some parameters or data automatically before algorithm input processing. It indicates a script file name to be executed before entering in initialization phase of chosen variables.
UserPostAnalysis
Multiline string. This variable allows to process some parameters or data automatically after data assimilation or optimization algorithm processing. Its value is defined either as a predefined pattern name, or as a script file name, or as a string, allowing to put post-processing code directly inside the ADAO case. Common templates are provided to help the user to start or to quickly make his case. We refer to the description of Requirements to describe a post-processing after an ADAO calculation for the list of templates and their format. Important note: this processing is only performed when the case is executed in TUI or exported to YACS. | 1,614 | 8,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | longest | en | 0.875544 |
https://engineering.stackexchange.com/questions/38525/2-complement-code-and-carry-zero-overflow-and-negative-flags | 1,621,316,169,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00417.warc.gz | 248,716,492 | 36,415 | # 2 complement code and Carry, Zero, Overflow and Negative Flags
Assuming I have a negative 8-bit number represented in 2 complement code, let's say -12, what number can I add to it to set all flags mentioned in the title (Zero, Negative, Overflow and Carry), or at least 3 of them?
My understanding now is that 8 bit 2 complement code can only represent numbers from -128 to 127, so when I have -12 I can add +12 and I will set the carry and zero flag. The other way would be to add a negative number to -12 that will be equal to a positive 8 bit number(e.g. -12-120), which means I will set my overflow and carry flags.
So is there a way to set all of the flags or at least three of them ?
• Can 0 be a negative number? The flags are based upon the result of caculation. So no! – StainlessSteelRat Nov 6 '20 at 1:19
• It seems obvious that the answer can't be zero and negative at the same time. – user253751 Nov 11 '20 at 18:21 | 250 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-21 | longest | en | 0.922969 |
https://trizenx.blogspot.com/ | 1,716,288,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00410.warc.gz | 505,780,694 | 37,493 | ## Posts
### Computational number theory in Sidef
In this tutorial we're going to look how we can use Sidef for doing various computations in number theory.
### Lossless data compression
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### Digital steganography: hiding data in images
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### Pseudoprimes: construction methods
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### Primality testing algorithms
In this post we're going to look at some algorithms for checking if a given number is either prime or composite. Édouard Lucas , Leonhard Euler , Carl Pomerance and Pierre de Fermat Prime numbers play an important role in public-key cryptography (e.g. RSA algorithm ) which require fast generation of large random prime numbers that have around 600 digits (or more). The simplest way to generate a prime number, is to pick a random odd integer and check its primality, repeating this process until a prime is found. This approach requires a fast algorithm for checking the primality of a number. In 2002, Manindra Agrawal , Neeraj Kayal , and Nitin Saxena published the AKS primality test , unconditionally proving that the primality of a number can be checked in polynomial time, with an asymptotic time complexity of Õ(log(n)^12). Later on, in 2005, Carl Pomerance and Hendrik Lenstra improved the complexity of the AKS algorithm to run in just Õ(log(n)^6) steps.
### Primorial in algorithms
In this post we present several practical algorithms based on primorials .
### Special-purpose factorization algorithms
In this post, we will take a brief look at a nice collection of special-purpose factorization algorithms. Most of them old and well-known. Some of them new.
### Partial sums of arithmetical functions
In this post we're going to look at some very interesting generalized formulas for computing the partial sums of some arithmetical functions .
### Continued fraction factorization method
The factorization method that we'll discuss in this post, it's called the continued fraction factorization method (CFRAC), and is quite an old method, but still pretty interesting, sharing many concepts and ideas with other modern factorization methods.
### Curiosities in number theory
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### Investigating the Fibonacci numbers modulo m
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### Representing integers as the sum of two squares
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### Representing integers as the difference of two squares
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### Various representations for famous mathematical constants
In this unusual post, much like in the older post, The beauty of Infinity , we're listing the most famous mathematical constants as representations of infinite series , infinite products and limits .
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The Mandelbrot set and its complex beauty.
### RSA algorithm
RSA is a practical public-key cryptographic algorithm, which is widely used on modern computers to communicate securely over large distances. The acronym of the algorithm stands for Ron Rivest , Adi Shamir and Leonard Adleman , which first published the algorithm in 1978. # Algorithm overview Choose p and q as distinct prime numbers Compute n as n = p*q Compute \phi(n) as \phi(n) = (p-1) * (q-1) Choose e such that 1 < e < \phi(n) and e and \phi(n) are coprime Compute the value of d as d ≡ e^(-1) mod \phi(n) Public key is (e, n) Private key is (d, n) The encryption of m as c, is c ≡ m^e mod n The decryption of c as m, is m ≡ c^d mod n # Generating p and q In order to generate a public and a private key, the algorithm requires two distinct prime numbers p and q, which are randomly chosen and should have, roughly, the same number of bits. By today standards, it is recommended that each prime number to have a | 1,217 | 5,675 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-22 | latest | en | 0.889698 |
https://zbmath.org/?q=an:1120.14011 | 1,685,299,090,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00792.warc.gz | 1,220,978,099 | 13,044 | ## Rationally connected foliations after Bogomolov and McQuillan.(English)Zbl 1120.14011
The main strategy of Mori theory is to relate the existence and properties of rational curves on a projective manifold $$X$$ to the positivity properties of the anticanonical bundle $$-K_X$$. For example, it is a well-known theorem that a projective manifold whose anticanonical bundle is ample, is rationally connected [cf. J. Kollár, Y. Miyaoka and S. Mori, J. Algebr. Geom. 1, No. 3, 429–448 (1992; Zbl 0780.14026)]. Initiated by Y. Miyaoka’s landmark paper [in: Algebraic geometry, Proc. Summer Res. Inst., Brunswick/Maine 1985, part 1, Proc. Symp. Pure Math. 46, No. 1, 245–268 (1987; Zbl 0659.14008)], algebraic geometers want to refine this analysis by shifting the focus from the anticanonical bundle to subsheaves of the tangent bundle.
More precisely one can ask: given a saturated subsheaf $$F \subset T_X$$ that satisfies a certain positivity property, what can we say about the existence of rational curves $$C$$ on $$X$$ that point in the directions of $$F$$, i.e. such that the tangent sheaf $$T_C$$ is included in $$F| _C$$?
In the article under review a fundamental result concerning this problem is established. The main theorem states the following:
Let $$X$$ be a normal complex projective variety, and let $$C$$ be a projective curve that is contained in the smooth locus of $$X$$. Let $$F \subset T_X$$ be an integrable subsheaf which is locally free along $$C$$. Suppose now that $$F| _C$$ is an ample vector bundle, then a $$F$$-leaf that meets $$C$$ in a general point is algebraic and its closure is rationally connected.
This result can be seen as a foliated version of the classical theorem that a complex projective manifold is rationally connected if and only if it contains a very free rational curve. Note furthermore that the main theorem (as the title of the paper says) has been stated in a slightly more general form in a preprint by F. Bogomolov and M. McQuillan [Rational curves on foliated varieties, IHES preprint, 2001]. The authors of the article under review follow the basic approach of the Bogomolov-McQuillan paper, but they give simplified and much more accessible proofs of the technical key points. Using classical techniques from foliation theory, the following corollary is shown:
Let $$X$$ be a normal complex projective variety, and let $$T_{X/Q}$$ be the relative tangent sheaf of the rationally connected quotient, that is the unique almost holomorphic fibration $$X \rightarrow Q$$ such that the general fibre is rationally connected and the base $$Q$$ is not uniruled. Let now $$C$$ be a general complete intersection curve on $$X$$ and suppose that the restriction $$T_X| _C$$ contains an ample subsheaf $$F_C$$. Then $$F_C$$ is contained in the relative tangent sheaf $$T_{X/Q}$$ at a general point of the curve $$C$$.
This corollary is a refined version of Miyaoka’s famous characterisation of non-uniruled varieties as the varieties whose cotangent bundle is nef on general complete intersection curves.
### MSC:
14E30 Minimal model program (Mori theory, extremal rays) 14J45 Fano varieties 14J40 $$n$$-folds ($$n>4$$) 14F17 Vanishing theorems in algebraic geometry 57R30 Foliations in differential topology; geometric theory
### Citations:
Zbl 0780.14026; Zbl 0659.14008
Full Text:
### References:
[1] Feodor A. Bogomolov and Michael L. McQuillan. Rational curves on foliated varieties. IHES, Preprint, February 2001. [2] Jean-Benoît Bost, Algebraic leaves of algebraic foliations over number fields, Publ. Math. Inst. Hautes Études Sci. 93 (2001), 161 – 221 (English, with English and French summaries). · Zbl 1034.14010 [3] César Camacho and Alcides Lins Neto, Geometric theory of foliations, Birkhäuser Boston, Inc., Boston, MA, 1985. Translated from the Portuguese by Sue E. Goodman. · Zbl 0568.57002 [4] Olivier Debarre, Higher-dimensional algebraic geometry, Universitext, Springer-Verlag, New York, 2001. · Zbl 0978.14001 [5] Hubert Flenner, Restrictions of semistable bundles on projective varieties, Comment. Math. Helv. 59 (1984), no. 4, 635 – 650. · Zbl 0599.14015 [6] Tom Graber, Joe Harris, and Jason Starr, Families of rationally connected varieties, J. Amer. Math. Soc. 16 (2003), no. 1, 57 – 67. · Zbl 1092.14063 [7] A. Grothendieck, Éléments de géométrie algébrique. IV. Étude locale des schémas et des morphismes de schémas. III, Inst. Hautes Études Sci. Publ. Math. 28 (1966), 255. · Zbl 0144.19904 [8] A. Grothendieck, Éléments de géométrie algébrique. IV. Étude locale des schémas et des morphismes de schémas. III, Inst. Hautes Études Sci. Publ. Math. 28 (1966), 255. · Zbl 0144.19904 [9] Robin Hartshorne, Cohomological dimension of algebraic varieties, Ann. of Math. (2) 88 (1968), 403 – 450. · Zbl 0169.23302 [10] Robin Hartshorne, Ample vector bundles on curves, Nagoya Math. J. 43 (1971), 73 – 89. · Zbl 0218.14018 [11] Andreas Höring. Uniruled varieties with splitting tangent bundle. preprint math.AG/0505327, 2005. [12] János Kollár, Extremal rays on smooth threefolds, Ann. Sci. École Norm. Sup. (4) 24 (1991), no. 3, 339 – 361. · Zbl 0753.14036 [13] Flips and abundance for algebraic threefolds, Société Mathématique de France, Paris, 1992. Papers from the Second Summer Seminar on Algebraic Geometry held at the University of Utah, Salt Lake City, Utah, August 1991; Astérisque No. 211 (1992) (1992). [14] János Kollár, Rational curves on algebraic varieties, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics], vol. 32, Springer-Verlag, Berlin, 1996. · Zbl 0877.14012 [15] János Kollár, Yoichi Miyaoka, and Shigefumi Mori, Rationally connected varieties, J. Algebraic Geom. 1 (1992), no. 3, 429 – 448. · Zbl 0780.14026 [16] Adrian Langer, Semistable sheaves in positive characteristic, Ann. of Math. (2) 159 (2004), no. 1, 251 – 276. · Zbl 1080.14014 [17] Robert Lazarsfeld, Positivity in algebraic geometry. I, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics], vol. 48, Springer-Verlag, Berlin, 2004. Classical setting: line bundles and linear series. Robert Lazarsfeld, Positivity in algebraic geometry. II, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics [Results in Mathematics and Related Areas. 3rd Series. A Series of Modern Surveys in Mathematics], vol. 49, Springer-Verlag, Berlin, 2004. Positivity for vector bundles, and multiplier ideals. [18] Yoichi Miyaoka, Deformations of a morphism along a foliation and applications, Algebraic geometry, Bowdoin, 1985 (Brunswick, Maine, 1985) Proc. Sympos. Pure Math., vol. 46, Amer. Math. Soc., Providence, RI, 1987, pp. 245 – 268. [19] Shigefumi Mori, Projective manifolds with ample tangent bundles, Ann. of Math. (2) 110 (1979), no. 3, 593 – 606. · Zbl 0423.14006 [20] C. S. Seshadri, Fibrés vectoriels sur les courbes algébriques, Astérisque, vol. 96, Société Mathématique de France, Paris, 1982 (French). Notes written by J.-M. Drezet from a course at the École Normale Supérieure, June 1980. [21] Frank W. Warner, Foundations of differentiable manifolds and Lie groups, Scott, Foresman and Co., Glenview, Ill.-London, 1971. · Zbl 0241.58001
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2,289 | 7,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | longest | en | 0.9123 |
https://math.answers.com/Q/Why_don%27t_prime_numbers_greater_than_10_end_in_5_or_0 | 1,652,940,764,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00532.warc.gz | 466,699,262 | 37,997 | 0
# Why don't prime numbers greater than 10 end in 5 or 0?
Wiki User
2016-02-18 23:33:12
Any number ending in 5 that isn't 5 is a multiple of 5. Any number ending in zero is even. The only even Prime number is 2.
Wiki User
2016-02-18 23:33:12
Study guides
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12 Reviews | 108 | 305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-21 | latest | en | 0.872056 |
http://www.computer-timeline.com/timeline/leonard-stowe/ | 1,716,690,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00430.warc.gz | 38,027,502 | 25,704 | # Leonard Stowe
In the late 1870s the government clerk from Wellington, New Zealand—Leonard Stowe, devised a simple adding machine, which he patented in 1880 in Great Britain and Germany (patent Nr. DE11907). The English patent for Stowe’s Patent Calculating Machine (see the lower patent drawing) was assigned to Stowe brothers, 32, Essex Street Strand, London.
A working model of the machine was exhibited at the International Exhibition of 1880-1881 in Melbourne, Australia, and received an honorable mention (It was advertised as: This machine combines simplicity with accuracy, has no complicated machinery in it, and is guaranteed to last for years without repair. Any sums in Addition, either in plain figures or in money, can be performed with it.) Unfortunately, it seems there is no example of the device, known to survive to the present, so we have only the patent application at our disposal.
The machine of Stowe consists of a number of cylinders, or rollers, with figures printed thereon, zero or naught being represented by a red square. To the left of the cylinders are a number of indicators, each of which records every perfect revolution of its corresponding cylinder.
To place the machine ready for work the indicators must be turned until zero or naught is visible; and the cylinders must be turned so that the zero on the extreme left i.e., nearest to the indicators is visible.
Let’s see an example of adding operation, using the patent drawing (see the image below).
4
5
7
3
4
Turn the cylinder towards you till 4 appears in the first column at the left end—stop, start again from the Zero or Red Square, now visible till you come to 5 in that column; start again from the Zero or Red Square, now visible, till you come to 7 in that column; start again from the Zero or Red Square, now visible, till you come to 3 in that column; start again from, the Zero or Red Square, now visible, till you come to 4 in that column.
On looking at the indicator you will find figure 2 recorded, and figure 3 will be the figure now visible at the left end of the cylinder, the two together making 23 which is the required total.
Note: The result of any addition will always be found by reading the figures recorded on the indicators together with the first figure on the left end of the cylinders.
#### Biography of Leonard Stowe
Leonard Stowe was born on 11 March 1837, in Trolly Hall, a large townhouse, still preserved in Buckingham, Aylesbury Vale District, South East England. He was the son of William Stowe (1791-1860), a surgeon and natural history enthusiast at Buckingham, and his wife, Mary Stowe (Rogers). His eldest brother, William Henry Stowe (1825–1855), was an English scholar and journalist.
Leonard Stowe attended a school at Iffley, near Oxford, and later (1853-1856) studied at the celebrated Rugby School in Warwick-shire (one of the oldest independent schools in Britain), when Dr. Meyrick Goulburn was headmaster.
On 2 September 1858, Leonard Stowe went aboard the ship (barque) Lady Alice, traveling from Gravesend, near London, to Nelson (a town on the eastern shores of Tasman Bay, in the South Island), New Zealand. In the 19th century, the voyage from England to New Zealand takes a lot of time (more than four months), so the ship arrived in Nelson on 14 January 1859.
In Nelson, Stowe had some years’ experience of station life under Arthur Penrose Seymour, a run-holder from the Awatere district. In 1863 Stowe was appointed secretary to Thomas Carter, third Superintendent of the province of Marlborough, and from 1864 Stowe acted under Seymour, when he was raised to the fourth Superintendent of the province of Marlborough.
In 1865 Stowe became Clerk of the Legislative Council, a position which he occupied for more than 20 years. In 1889 he was appointed as a Clerk of the Parliament. Stowe has filled many other offices during his career (see the lower photo from 1902), and he was a notable figure in the Parliamentary life of New Zealand for over 30 years. He was also a headmaster of Nelson College for several years.
Leonard Stowe’s first wife, Mary Jane Iles-Stowe, died on 26 March 1868, aged only 30. Three years later, on 31 May 1871, he married in Nelson, New Zealand, to Jane Greenwood (1838-1931), an artist.
Jane was born on 18 April 1838, in Charenton-le-Pont, Île-de-France, France, the 3rd daughter of Dr. John Danforth Greenwood (1803-1890), a successful physician from Mitcham, Surrey, England, and Sarah Greenwood (nee Field) (1809-1889). She arrived in Nelson, NZ, with her parents and siblings (7 brothers and sisters, as her youngest brother was born on board the ship, en-route to New Zealand) aboard the Phoebe ship on 29 March 1843 (they left England from Gravesend on 16 Nov 1842, as Dr. Greenwood had secured the position of Surgeon Superintendent and Justice of the Peace, receiving free passage for himself and his family in return.)
In Nelson Danforth was working as a doctor, farmer, magistrate, Captain (of the Nelson Militia), Clergyman, and Flax Agent. Later in his life, he took a few public positions, like a member of the Legislative Council, Inspector of Schools, Principal of Nelson College, and Sergeant of Arms to the House of Representatives in Wellington. Sarah Danforth was a keen artist and letter writer and took to the new job of housekeeping with gusto, and later between 1865 and 1868, she ran a successful school with six of their daughters (they had 12 children).
Leonard and Jane Stowe had five children: three sons— William Reginald (9 Mar 1872–9 Feb 1949 ), Harry, and Leonard Acland (11 Aug 1876–7 Nov 1876), and two daughters—Emily Muriel (10 Apr 1875–28 May 1971), and Mary Sylvia (28 Jul 1873–20 May 1927).
Jane Stowe was a delightful watercolorist and used to exhibit her paintings at the New Zealand Academy of Fine Arts from 1883 until 1931. She received many prizes and honorable mentions.
Leonard and Jane Stowe lived in a nice wooden house called Te Moana (later known as Tiakiwai) at 2 Tinakori Road, Wellington (see the nearby photo from the 1880s).
Leonard Stowe died 83 years of age on 25 April 1920 and was buried in Bolton Street Cemetery, Wellington. His wife Jane died of bronchitis in Wellington on 5 November 1931, aged 93. | 1,497 | 6,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.926654 |
http://delta.cs.cinvestav.mx/~mcintosh/comun/contours/node6.html | 1,513,326,344,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00468.warc.gz | 67,577,768 | 3,014 | Next: Four-point two-dimensional approximation Up: Contours for <PLOT> Previous: Specialized data points
# Three-point two-dimensional approximation
The fundamental hypothesis of the <PLOT> contour programs is that linear interpolation will suffice. Since the monomials 1, x, and y form a basis or first degree functions of two variables, the Vandermonde matrix for the combination will have dimension 3. Usually planar data is taken from a square or rectangular grid; if the corners of the grid squares are used as base points, there will be one too many. Rather than worry about all the ways that contour lines can cross a square, <PLOT> divides each square into two triangles, which are then treated separately. The three vertices of each half square fit nicely into the interpolation formula. When the approximation of linearity is grossly erroneous, the existence of a preferred diagonal becomes quite visible in the erratic wanderings of the contour.
For this case the basic equation has the form
The Vandermonde matrix is
The inverse matrix is
with Vandermonde determinant
V = x2y3+x3y1+x1y2-y2x3-y3x1-y1x2.
However it is arrived at, the equation can be exhibited in the explicit form
For the purposes of contouring, it is fortunate that the basic equation is an implicit equation of considerable symmetry. For purposes of interpolation, z can be expressed as a function of x and y, but it is equally possible to assign z a value and solve for y as a function of x.
Even so, the equations appear rather formidable. If a specific contour value z0 has been chosen, the term in z can be combined with the constant term obtain an equation for a line in a plane:
For uniformity of appearance, z0 could be subtracted from the last rows of the two left hand determinants. Further consolidation results from selecting a specific value for one of the remaining variables, say x0. Then the formula for y would be
still, this and similar results follow directly from original determinant.
Next: Four-point two-dimensional approximation Up: Contours for <PLOT> Previous: Specialized data points | 450 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-51 | latest | en | 0.920152 |
https://datalab.marine.rutgers.edu/ooi-lab-exercises/lab-2-the-display-of-oceanographic-data/lab-2-4/ | 1,721,638,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00273.warc.gz | 163,755,025 | 28,822 | Lab 2.4 – Station profiles, how to read a standard oceanography graph
Fundamental concept: Understand how to read station profiles to identify variability and trends in the data
Preparation for: Lab 5 – Density
Estimated time to complete: 30 minutes
Materials needed (none)
Is there structure to the water in the ocean?
Although ocean water looks much the same wherever you go, once you start to measure properties such as salinity and temperature you would see that there are considerable differences. A standard type of graph used to examine some of these differences is called a station profile. Typically these are graphs showing a data type, such as temperature, salinity, or something else, plotted against the depth of the water. Oceanographers make these graphs because many of the properties of seawater change with depth.
Temperature is one property that does just this. Temperature often decreases as depth increases since the source of the heat is the sun. Sunlight is quickly absorbed in the upper layers of water resulting in higher temperatures there than in deeper water. To make a station profile for temperature you take all the temperature data collected at one oceanographic station. This would be a place where the research vessel stopped and lowered a temperature measuring device (one of the sensors on a CTD) into the water, and collected data as the instrument was lowered to the sea floor. A station profile plot shows the way that temperature varies with increasing depth. Typically station profiles are plotted with depth increasing down on the y-axis and the property of interest is plotted on the x-axis. This convention, plotting with depth increasing down, is used because it makes it easier to picture the distribution of the property in the ocean, where depth does increase in the downward direction.
Figure 2.4.1. Sea surface temperature anomaly (the difference between normal and observed temperatures) in September 2014 (Courtesy: NOAA Fisheries)
What can oceanographers learn from temperature data? Just as there are heat waves on the surface of the Earth, there are heat waves in the ocean. One such oceanic heat wave occurred from 2013 to 2015. During this time there was a huge pool of warm water in the Pacific, nicknamed The Blob by the office of the Washington State Climatologist. Figure 2.4.1 shows the difference in sea surface temperature between “normal” times and the time when The Blob was present in the North Pacific. The huge area covered in red shows the location of the anomalously warm water. This feature was not just interesting to those who study ocean temperature. The high temperatures of this water impacted the weather along the west coast of the U.S. and adversely impacted marine life because it was missing some of the chemicals necessary for the marine algae called phytoplankton. Those climatologists seem to have a sense of humor! The cause of The Blob is linked to a persistent high pressure ridge in the atmosphere, nicknamed The Ridiculously Resilient Ridge. In the activity below you will work with a type of graph called a station profile. This kind of data in The Blob helped oceanographers figure out the depth of the warm water and how much heat it contained.
Station Profiles Exploration
The map in Figure 2.4.1 shows the surface area of The Blob. But how deep is this anomalously warm water? A plot showing the temperature of the water with depth, called a station profile, can be used to answer this question, especially if we compare a station profile from a “normal” time to one from The Blob.
The station profile below (Figure 2.4.2) shows the temperature in the North Pacific at a time before The Blob developed. It shows a few things that are often present in station profiles, a surface mixed layer, a depth range where the temperature undergoes significant change (called the thermocline) and a deeper layer with either uniformly cold water or very small changes in temperature with increasing depth. The surface mixed layer often is mixed due to the action of currents and waves, although in some places and seasons mixing may occur due to heat being removed from the surface water. The result would be that the water would become colder, and therefore more dense so it would sink. But regardless of the cause of the mixing, the surface mixed layer is easy to identify because of its uniform temperature.
Quick Check Questions:
We can look at a profile from 2014, when the warm surface water of The Blob was present, and answer the same questions.
To see the effect of The Blob on the waters in the North Pacific it is useful to examine both profiles on the same set of axes, as shown below (Figure 2.4.4).
In this introduction to station profiles we have examined temperature. But also useful are profiles of other water properties, such as salinity and density. These profiles also often show a surface mixed layer, a “cline” (halocline in the case of salinity and pycnocline in the case of density) and a deeper layer of uniform, or slowly changing, properties.
Station Profiles Application
Now that you have practiced reading profile graphs, we will put your skills to use to examine temperature profiles at different locations in the North Atlantic Ocean. Below is a station profile showing temperature at the Coastal Pioneer array, off the mid-Atlantic U.S. coast.
Figure 2.4.5.
Interpretation Questions:
1. Why is it useful to make a station profile graph with this orientation (depth increasing downward)?
2. Identify the maximum and minimum temperature values in this station profile graph from the Pioneer array (Figure 2.4.5).
3. How does the temperature of the water change as you go deeper in the water?
4. The depth range where the temperature changes the most rapidly is called the thermocline. What is the depth of the bottom of the thermocline in this profile?
5. North Atlantic fin whales migrate through the area of the Pioneer array. These whales breathe air at the surface and dive to feed on krill, squid and other prey. If a fin whale dove from the surface to 100 meters deep at the time and location that this profile was collected, how much change in temperature would it experience? (Assume that surface temperatures are equivalent to the temperature at a depth of 35 meters)
Now let’s also look at a temperature profile graph from the Irminger Sea array in the deep North Atlantic basin. (Note: While most data points can be read by hovering your cursor over it, You must read the surface data point manually from the graph axis as practice).
Figure 2.4.6
Application Questions:
1. Identify the maximum and minimum temperature values in this station profile graph from the Irminger Sea (Figure 2.4.6). How does the temperature of the water change as you go deeper in the water?
2. Now compare the Pioneer and Irminger temperature profiles. How similar or different are these two station profiles?
3. Click the buttons in Figure 2.4.6 to match the depth and temperature scales. Did your answer to the previous question change when you did this?
4. Why do you think these two temperature profiles are so different? [Hint: Click below to view the location of the two arrays]
Reflection question:
1. When comparing two or more data sets why is it important to compare the scales? Use an example from this station profile activity to support your answer. | 1,533 | 7,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.938756 |
https://goprep.co/ex-7-q2-how-many-words-can-be-formed-using-the-letter-of-i-1nkxuq | 1,603,827,191,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894426.63/warc/CC-MAIN-20201027170516-20201027200516-00042.warc.gz | 334,151,601 | 30,985 | Q. 25.0( 1 Vote )
# How many words can be formed using the letter of UNIVERSITY using each letter at most once?i) If each letter must be usedii) If some or all the letters may be omitted.
Letters in the word UNIVERSITY: E, I, N, R, S, T, U, V, Y
There are 9 distinct letters (’I’ is repeated once).
Let us say that each letter is used only once in formation of words of any length.
Let P(m, n) denote the number of permutations of selecting n items from m, ie., m! / (m-n)!
(i) If all 9 letters are to be used = 9!
(ii) If words contain 0 to 9 letters: Sum P(9, n) with n denoting the number of letters to be present in the word.
P(9,0) + P(9,1)+ P(9,2) + P(9,3) +.... + P(9,8) + P(9,9)
= 986410
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A309254 Shifts left by 2 places under series reversion. 3
1, 1, 1, -1, 1, 1, -11, 41, -83, -77, 1621, -8503, 25689, -11283, -436307, 3380827, -15378675, 37938353, 87204085, -1755462619, 12722279385, -61250294425, 164643256457, 465385122833, -10366591960879, 88472237039417, -527994960273655, 2118908595206935, -1118896610239543 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,7 COMMENTS Series reversion of the g.f. gives the same sequence with two initial terms (1, 1) dropped. LINKS Table of n, a(n) for n=1..29. FORMULA G.f. satisfies A((A(x) - x - x^2)/x^2) = x. MATHEMATICA Nest[InverseSeries[#] x^2 + x^2 + x &, x + O[x]^2, 14][[3]] CROSSREFS Cf. A067145, A309564. Sequence in context: A282321 A031389 A178495 * A360154 A217624 A097991 Adjacent sequences: A309251 A309252 A309253 * A309255 A309256 A309257 KEYWORD sign AUTHOR Vladimir Reshetnikov, Aug 07 2019 STATUS approved
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Last modified July 18 06:54 EDT 2024. Contains 374377 sequences. (Running on oeis4.) | 473 | 1,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.592482 |
https://classroomsecrets.co.uk/free-consolidation-of-steps-1-2-year-4-area-learning-video-clip/ | 1,638,737,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363216.90/warc/CC-MAIN-20211205191620-20211205221620-00139.warc.gz | 233,588,952 | 66,079 | Free Consolidation of Steps 1-2 Year 4 Area Learning Video Clip | Classroom Secrets
MathsYear 4Spring Block 2 (Area)01 What is Area? › Free Consolidation of Steps 1-2 Year 4 Area Learning Video Clip
Free Consolidation of Steps 1-2 Year 4 Area Learning Video Clip
Step 1 and 2: Free Consolidation of Steps 1-2 Year 4 Area Learning Video Clip
Meet Hadriana, Cicero and Vitus who live in the town of Pompeii. Help them to explore the villa that Hadriana’s father is building and work out the area of different rooms and mosaics. Pupils will help them to count the squares in mosaics and shapes on the wall and work out the sizes of different rooms, all before disaster strikes….
This Learning Video Clip has been designed as a consolidation tool for steps 1 and 2. It contains content relevant to all these steps and can be used in parts to recap a particular step or in on its own at the end of a teaching sequence for the included steps.
More resources for Spring Block 2
(2 votes, average: 5.00 out of 5)
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[/s2If]
Discussion points for teachers
1. Estimate how many mosaic tiles will be needed to cover the area.
Discuss the reference mosaic square. Discuss how many would be needed to work out the area of the mosaic. Discuss what it means to estimate.
25 squares
2. The cubiculum covers a bigger area than the tablinium! Do you agree with Cicero? Use the reference square to help you.
Discuss the surface of both rooms. Discuss the number of squares and how the reference square should be used to estimate on both rooms.
Cicero is correct. The cubiculum covers 19 squares, the tablinium covers 18 squares.
3. Which shape is made up of the most squares? What is the area of each shape?
Discuss the number of squares in each shape. Discuss an estimate before counting. Discuss how different shapes can appear to cover bigger areas.
The gardens are made up of the most squares. Gardens = 21 squares; Kitchen = 16 squares; Reception = 19 squares.
4. There are 6 squares in one row and 4 rows altogether. What is the total area of the bath house?
Discuss how this could be shown as a multiplication question. Discuss the relationships and count to prove the answer.
24 squares.
5. How many squares are there of each colour?
Discuss the different colours and count the squares to work out each one.
Blue = 4 squares; Green = 5 squares; Red = 5 squares; Purple = 4 squares; Yellow = 3 squares; Black = 3 squares
6. Count the shape on the wall. Do you agree with Hadriana?
Discuss the shape of the green snake. Discuss ways of working out the answer, can a multiplication be used?
Hadriana is incorrect. There are 22 squares.
National Curriculum Objectives
Mathematics Year 4: (4M7b) Find the area of rectilinear shapes by counting squares
1. This resource is available to play with a Taster subscription. | 681 | 2,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.910825 |
https://socratic.org/questions/55872919581e2a7bb787746e | 1,576,388,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00370.warc.gz | 533,890,526 | 6,379 | # Question 7746e
Jun 21, 2015
You have 13.6 g of beryllium fluoride in that solution.
#### Explanation:
You're dealing with a solution of known molarity of beryllium fluoride, $B e {F}_{2}$.
Molarity is defined as moles of solute, in your case beryllium fluoride, per liters of solution. This means that if you know the molarity of the solution and its volume, you can determine how many moles of solute you have present.
$C = \frac{n}{V} \implies n = C \cdot V$
${n}_{B e {F}_{2}} = \text{0.442 M" * 655 * 10^(-3)"L" = "0.2895 moles}$ $B e {F}_{2}$
To get the mass of beryllium fluoride that would contain this many moles, use the compound's molar mass
0.2895cancel("moles") * "47.01 g"/(1cancel("mole")) = "13.609 g"# $B e {F}_{2}$
Rounded to three sig figs, the answer will be
${m}_{B e {F}_{2}} = \textcolor{g r e e n}{\text{13.6 g}}$ | 293 | 849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-51 | longest | en | 0.867033 |
https://goprep.co/q3c-draw-the-circumcircle-of-pmt-in-which-pm-5-6-cm-p-60-m-i-1nkcvr | 1,618,616,043,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00140.warc.gz | 381,371,317 | 24,274 | Q. 3 C4.8( 4 Votes )
# Solve any the thr
Step-1: construct the segment PM having length 5.6 cm and using protractor draw lines at the angles P and M which measures 60° and 70° respectively mark the intersection point as T thus ∆PMT is ready
Step-3: construct perpendicular bisector of line PM by keeping the needle of compass at point P and taking approximately more than half of PM distance in compass draw arc above and below PM
Step-4: keeping the same measurement in compass keep the needle at point M and draw intersecting arcs above and below segment PM
Step-5: join the intersections of arcs to get a line ‘a’ which is perpendicular bisector of segment PM.
Step-6: Similarly by repeating steps 3,4,5 construct a perpendicular bisector for line TM so instead of P substitute T and repeat steps 3,4,5 we will get a line ‘b’ perpendicular bisector of segment TM
Step-7: keep the needle of compass at point of intersection of line a and b and from there take distance till any vertex of triangle PTM and construct the circle
The circle is required circumcircle to ∆PTM
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<span lang="EN-USMHB - Mathematics Part-2 | 421 | 1,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-17 | latest | en | 0.855569 |
https://quantumcomputing.stackexchange.com/questions/2149/grovers-algorithm-what-to-input-to-oracle?noredirect=1 | 1,726,302,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00482.warc.gz | 430,500,018 | 44,005 | Grover's algorithm: what to input to Oracle?
I am confused about what to input to Oracle in Grover's algorithm.
Don't we need to input what we are looking for and where to find what we are looking for to Oracle, in addition to the superpositioned quantum states?
For example, assume we have a list of people's names {"Alice", "Bob", "Corey", "Dio"}, and we want to find if "Dio" is on the list. Then, Oracle should take $1/2(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$ as an input and output $1/2(|00\rangle + |01\rangle + |10\rangle - |11\rangle)$. I kind of understand that.
But don't we also need to input the word "Dio" and the list {"Alice", "Bob", "Corey", "Dio"} to Oracle? Otherwise, how can Oracle return output? Is it not explicitly mentioned since Oracle is a black box and we do not have to think about how to implement it?
• Oracle has the ability to recognize if the word "Dio" is in the list.
• To do so, Oracle takes the superpositioned quantum states as an input, where each quantum state represents the index of the list.
• So, input $|00\rangle$ to Oracle means, check if the word "Dio" is in the index 0 of the list and return $-|00\rangle$ if yes and return $|00\rangle$ otherwise.
• In our case, Oracle returns $1/2(|00\rangle + |01\rangle + |10\rangle - |11\rangle)$.
• But what about the list and the word?
• While not phrased in the same way, I believe your question is more or less the same as this one: Grover's algorithm: where is the list? Commented May 26, 2018 at 3:54
Although popular explanations of Grover's algorithm talk about searching over a list, in actuality you use it to search over possible inputs 0..N-1 to a function. The cost of the algorithm is $O(\sqrt{N} \cdot F)$ where $N$ is the number of inputs you want to search over and $F$ is the cost of evaluating the function. If you want that function to search over a list, you must hardcode the list into the function.
Hard coding the function to use a list of $N$ items is usually a very bad idea, because it tends to cause $F$ to equal $O(N)$. Which would make the total cost of Grover's algorithm $O(\sqrt{N} \cdot F) = O(\sqrt{N} \cdot N) = O(N^{1.5})$. Which sort of defeats the whole purpose, since $N^{1.5} > N$.
• Would you not input an ordered list, making the lookup much quicker? Of course, you might want to then include the cost of ordering the list, but I guess that still comes out as $O(\sqrt{N}\log(N))$ overall. Commented May 26, 2018 at 4:09
• @DaftWullie The issue is that Grover must do a lookup under superposition, and this requires a multiplexer circuit with N AND gates (or other non-Clifford operations). A quantum AND gate (i.e. a Toffoli) has non-negligible cost when performing error correction. This cost is technically also present in the classical machine (i.e. RAM has O(N) AND gates), it just happens to be negligible and even avoidable in that context. Commented May 28, 2018 at 1:35
• I don't understand what you're saying. Would you be able to express a question, and answer it, to show the details? (I don't think I can phrase a good enough question at this point) Commented May 28, 2018 at 14:06
• @DaftWullie I think the question would be something like "how do I give a quantum computer read access to a classical database and how expensive is it". Commented May 28, 2018 at 18:37
If you have a list that has the input names {"Alice", "Bob", "Corey", "Dio"} and you want to recognize the if the word Dio is on the list, then you need a way to encode the list's contents as input for the oracle. So the question is not what is on this list, the question is how many different inputs do you want to be able to encode? What is the total vocabulary?
If there are 4 possible names (Alice, Bob, Corey and Dio), and you want to be able to encode any list with 4 of these names, then your input is four elements of two bits each: abcdefgh and the encoding is Alice=00, Bob=01, Corey=10 and Dio=11. You then would have an oracle that returns 1 if ab=11, cd=11, ef=11 or gh=11.
If the inputs can be any 5-character string, encoded in 8-bit ASCII, then the input to the oracle is going to be $$5\times8\times N$$ where $$N$$ is the maximum number of items that might be in your list. So the input could be quite large. However, the Oracle doesn't need to know all possible inputs; it merely needs to return 1 if any of the inputs contain "Dio" in any allowable position. | 1,193 | 4,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-38 | latest | en | 0.892285 |
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### Course: Class 9 (Assamese)>Unit 1
Lesson 4: Laws of exponents for real numbers
# Intro to rational exponents
What does it mean to take a number by a power which is a unit fraction? For example, what is the result of 3 raised to ½? Created by Sal Khan.
## Want to join the conversation?
• At , Sal asks, "But what is the square root of 4, especially the principle root mean?" and then goes on to ask, "what is a number that if I were to multiply it by itself...I'm going to get 4." I understand that +2 is apparently "the principle root", but what about -2, which if multiplied by itself is also equal to 4? What kind of root is -2 and why is it that only the "principle root" is given as the answer to "what is the square root of 4?"
• The principle root and square root are two different things. The square root is asking that question, which number squared equals that number, say, 4. This does leave two answers, positive and negative, so you were correct. However, the principle root is basically the absolute of the square root, or √|x|, which means that it is only positive. This was created, I think, for geometry, because you can't have a triangle with side lengths of -3, -4, and -5. Also, if this is a comfort to you, I didn't know about principle roots until recently. We'd say that the square root of 4 is ±2, for that same reason you mentioned.
• Should I memorize some of the basic exponents?
Example:
4^3
• I recommend memorizing the perfect squares up to 30 especially if u want to compete in math.
• In fractional exponents, i'm curious on what to do if there is a fraction such as 5/7 or 9/17 as an exponent. Do you take the square root and then multiply, or do something else?
• Take the root equivalent to the denominator (bottom), and raise to the power of the numerator (top).
• What do you do when you have for example 2/3 to the power of 2?
• so if 8^1/3 = ³√8 then would 8^2/3 = 2³√8?
• No, unfortunately this would be wrong. You propably have not learned this yet, but you can rewrite any exponential expression of the form x^(n*m) as (x^n)^m.
So when you look at your example of 8^2/3, you could rewrite it as 8^(2*1/3).
By matching the corresponding parts to x^(n*m), this could then be expressed in the form of (x^n)^m:
(8^2)^(1/3)
= 64^(1/3)
= 4
Alternatively you could swap the 2 and the 1/3, which might make the problem easier. You can do this because of the the commutative property of multiplication, which allows you to "choose" wether you see 2 as the m or the n (the same thing goes for the 1/3). This would give you:
(8^(1/3))^2
= (2)^2
= 4
• okay what are the 3 cube roots of 8? (cube numbers have 3 roots, square numbers have 2 roots)
by cube number I mean x^3=y
by square number I mean x^2=y
now I already know 2 is a cube root but it is not the Only cube root. there are two others. what are those other cube roots?
• The other cube roots are 2 and 2. 8 is just 2*2*2. Or 2^3. That is it's prime factorization, nothing else. Hope this helped!
• At no part of the video did Sal explain what to do if a number is raised to, for example, the 2/5th power? What if the fraction has an integer larger than 1 for its numerator?
• What a good question!
If you have ANY fractional power, the denominator tells what root to take and the numerator tells what power to raise that number to.
For example, 16^3/2 means take the square root of 16, then raise that to the 3rd power
(getting 64 as the answer).
Another example, 32^(2/5) means take the fifth root of 32, then raise that to the 2nd power (getting 4 as the answer).
• I know that 7^0 is one, and is so for all numbers other than zero. But what I want to know is why it isn't zero. Or, in other words, why isn't 7^0 equal to zero?
• Sal writes all the rational exponents as fractions. But can exponents be in decimal form?
For example, x^(-2.5) and x^(-5/2), are both of them correct?
• The two versions are equivalent. However, the fraction form is easier to understand. The denominator of the fraction tells you the radical index. You have a denominator of 2, so it indicates a square root. If the denominato is 3, then the problem is working with a cube root. It it is 4, then the problem is working with a 4throot.
If you have a problem like: (-8)^(2/3) you can see that you need to do a cube root (the 3 in the denominator) and then square the result (the 2 in the numerator.
(-8)^(2/3) = cubert(-8)^2 = (-2)^2 = 4
If the exponent is in decimal form, that information is not visible. You would have to convert to a fraction to make the info visible. There is also the risk that you convert a fraction to decimal, find it repeats and you then round the decimal value. If you happen to do this, then you have changed the exponent. For example: An exponent of 1/3 = Do a cube root. If you convert it to decimal form: 1/3 = 0.33333... with the 3 repeating. If it gets rounded to 0.3, the exponent would then be 3/10 which means do the 10th root, then cube the result.
Hope this helps.
• Why is anything raised to the zero power 1? It doesn't make sense to me and seems like a made-up answer. It seems to me like anything raised to the zero power should equal zero. That kind of answer seems more logical.
(1 vote)
• The concept of a number raised to the zero power equals one can be explained in several ways and is based on basic multiplicative concepts. Looking at the pattern established when a number is raised to different powers, each one less than the next, helps explain the concept.
When a number such as 2 is raised to different powers, a particular pattern is seen as the exponent changes:
2^6 = 2*2*2*2*2*2 = 64 2^5 = 2*2*2*2*2 = 32 2^4 = 2*2*2*2 = 16 2^3 = 2*2*2 = 8 2^2 = 2*2 = 4 2^1 = 2
As the exponent value moves from 6 to 1, we see that the resulting values are reduced, consecutively, dividing by 2: 64/2 = 32, 32/2 = 16, 16/2 = 8, 8/2 = 4 and 4/2 = 2. Extrapolating from this pattern, an exponent of 0 will result in an answer of 2/2 = 1, proving 2^0 = 1.
The number 2 was used to provide an example; however, this concept applies to all nonzero numbers. | 1,765 | 6,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.962007 |
https://example-a.com/answer/71532968 | 1,709,253,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00160.warc.gz | 236,927,011 | 7,373 | databasefunctional-dependenciesdecomposition
# Finding the key for R
``````R = {A, B, C, D, E, F, G, H, I, J}
F =
{{A,B} -> {C},
{A}-> {D,E},
{B} -> {F},
{C}-> {B},
{F}->{G,H},
{D}->{I,J}
``````
The question is: What is the key for R?
I assume based on how the question has been formulated that there is one single candidate key which they want me to find.
If i have AB+ i can determine all the attributes in the relation meaning AB is a superkey. The proper subsets of the superkey AB which are {A} and {B} are not superkeys, hence why AB then is a candidate key. But from what i can tell there is another candidate key aswell which we can find if we have AC+. Is this correct or am i making a mistake somewhere?
Solution
• Yes, you are correct: assuming that `F` is a cover of the dependencies of `R`, the relation has two candidate keys: `{A, B}` and `{A, C}`.
This can be easily shown by computing both `{A,B}+` and `{A,C}+`. | 271 | 937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.922349 |
https://stats.stackexchange.com/questions/274135/two-dice-rolls-same-number-in-sequence | 1,580,147,330,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251705142.94/warc/CC-MAIN-20200127174507-20200127204507-00361.warc.gz | 666,342,826 | 34,162 | # Two dice rolls - same number in sequence
I am currently studying Statistical Inference class on Coursera. In one of the assignments, the following question comes up.
| Suppose you rolled the fair die twice.
What is the probability of rolling the same number two times in a row?
1: 2/6
2: 1/36
3: 0
4: 1/6
Selection: 2
| You're close...I can feel it! Try it again.
| Since we don't care what the outcome of the first roll is, its probability is 1.
The second roll of the dice has to match the outcome of the first,
so that has a probability of 1/6. The probability of both events occurring is 1 * 1/6.
I do not understand this bit. I understand that the two die rolls are independent events and their probabilities can be multiplied, so the outcome should be 1/36.
Can you please explain, why I am wrong?
• 1/36 is the probability of getting a 5 and then a 3, for example (both of which are events with probability 1/6), but that's not what the question is about. – user253751 Apr 18 '17 at 5:44
• @immibis I didn't follow your comment at first! But, of course, you are absolutely right! – The Monk Apr 20 '17 at 16:35
The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6).
The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36). Another way to think about it is that you don't care what the first number is, you just need the second number to match it (with probability 1/6 ).
• More clearly, the probability for the first die to match your criteria is 100%, it will always have a number. The second die has a 1/6 of matching the criteria. – Mooing Duck Apr 17 '17 at 22:56
To make it perfectly clear, consider the sample space for rolling a die twice.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
There are 36 equally likely possible outcomes, 6 of which define the event "rolling the same number two times in a row". Then, the probability of this event occurring is $\frac{6}{36}$, which is equal to $\frac{1}{6}$.
• This is the best way of answering this kind of questions. +1 :) – pratyay Apr 18 '17 at 14:35
• I like the way the answer was edited. – Michael R. Chernick Apr 19 '17 at 3:25
Conceptually, this is just asking "what are the chances a second die matches the result of the first". Suppose I rolled a die, secretly, and asked you to match the outcome with your own roll.
No matter which number I rolled, there is a 1/6 chance that your die matches my roll, as there is a 1/6 chance any die roll comes up any specific number.
If you roll a 1 then on the second roll (for a fair 6 sided die) the probability that the second roll is a 1 is 1/6 (assuming independence. This would be true for any other possible first roll.
Hope this helps :
Probability for the first roll to turn up as 1 : 1/6 Probability for the second roll also to turn up as 1 : 1/6
Therefore , probability that the first two rolls turn up as 1 is (1/6*1/6) = 1/36
Now the probability that the first two rolls turn up as 2 is (1/6*1/6) = 1/36 . . . . Same applies for 3,4,5,6
So the probability for any number to turn up consecutively twice is (1/36+1/36+1/36+1/36+1/36+1/36) = (6/36) = 1/6
• I think you are just repeating answers that already have been asked. – Michael R. Chernick Apr 18 '17 at 13:23
i would look at it as a combination problem . where you are asked what possible combinations are thre that have same numbers on first and second roll. combinations are 6 (11,22,33,44,55,66) from a total possibilities 6*6=36 so probability is 6/36
Since I didn't see this exact way of framing it above:
For your first roll there are 6 possible answers, and 6 acceptable answers (as any number 1-6 is acceptable).
6/6
For the second roll there are 6 possible answers, but now only 1 will match the first roll.
1/6
6/6 * 1/6 = 1/6 | 1,307 | 4,148 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-05 | latest | en | 0.934448 |
www.hirehop.jp | 1,722,872,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00118.warc.gz | 632,354,127 | 18,587 | HireHop is probably the best and most fearure rich cloud rental Equipment software, however it also has a resource for users to market their products for free to potential customers on HireHop‘s equipment rental portal. As it has been a popular source for customers to source their hired rigging equipment, we decided to give some advice for when hiring motors and truss, as you need to know the rigging weight loading calculations involved.
Lets take an imaginary scenario that a rigger has suspended a truss on it’s two end points, each motor is rated at 500kg, however the evenly distributed load on the truss is 1,300kg, so the rigger attaches a third 500kg rated motor to pickup the centre of the truss. Thinking with 3 motors rated at 500kg each there would be no problem, to his dismay, the truss comes crashing down, centre point first.
The reason this didn’t work in our imaginary scenario is due to multi-point beam load calculations that he failed to account for. The rigger in this case incorrectly assumed that the three motors would share the load between them equally and didn’t take into account the Three Moment Theorem. This is a very complex formula, however to simplify matters, we have illustrated simplified various rigs below and shown the various loads as percentages of the entire load at each point.
As you can see from the top image with two points, each point carries 50% of the load. If you look at the one below with 3 points you will see that the centre point supports 62% of load and outer points only support 19% of the load. Applying that to what happened to our rigger:
Load = 1,300Kg
Each Outer Point = 0.19 x 1300 = 247Kg per point (19% of the load)
Centre Point = 0.62 x 1300 = 806Kg on centre point (62% of the load)
As you can see, the 500Kg centre point was overloaded at 806Kg and thus collapsed. What the rigger should have done is put up 2 extra motors:
Load = 1,300Kg
Each Outer Point = 0.13 x 1300 = 169Kg per point
Centre Points = 0.37 x 1300 = 481Kg per point
For our scenario, ideally the rigger should use 5 points to give himself a larger margin of error, as with 4 points there is only a 19Kg margin on the centre points.
IMPORTANT
• It is very important to note that the above is for an evenly distributed load and that the motors are all moving at the same speed. If for example on a three point pickup the centre motor moved faster than the outer two motors, the entire weight would be taken up on the centre motor with the outer two going slack.
• It is not advised to rely on the above when using manual chain blocks as these never climb at the same speed. For a manual chain block, each block should be able to support the entire load.
• Please remember to include the weight of the truss and all fixtures when calculating the load.
• If you are calculating the load on the points above the motor, remember to add the weight of the motor to each point.
• The mathematics is theoretical in an ideal world and factors can be different in the real world, so always include an error margin and never get too close to the maximum loading.
For more in depth rigging calculations, see the Prolyte Black Book | 752 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.945548 |
https://stats.stackexchange.com/questions/51718/assessing-approximate-distribution-of-data-based-on-a-histogram/51721 | 1,568,602,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572471.35/warc/CC-MAIN-20190916015552-20190916041552-00270.warc.gz | 691,301,521 | 39,805 | # Assessing approximate distribution of data based on a histogram
Suppose I want to see whether my data is exponential based on a histogram (i.e. skewed to the right).
Depending on how I group or bin the data, I can get wildly different histograms.
One set of histograms will make is seem that the data is exponential. Another set will make it seem that data are not exponential. How do I make determining distributions from histograms well defined?
• Why not forget about the histograms, because the problems you describe are well established, and consider alternative tools such as qq plots and goodness of fit tests? – whuber Mar 8 '13 at 18:28
### The difficulty with using histograms to infer shape
While histograms are often handy and sometimes useful, they can be misleading. Their appearance can alter quite a lot with changes in the locations of the bin boundaries.
This problem has long been known*, though perhaps not as widely as it should be -- you rarely see it mentioned in elementary-level discussions (though there are exceptions).
* for example, Paul Rubin[1] put it this way: "it's well known that changing the endpoints in a histogram can significantly alter its appearance". .
I think it's an issue that should be more widely discussed when introducing histograms. I'll give some examples and discussion.
Why you should be wary of relying on a single histogram of a data set
Take a look at these four histograms:
That's four very different looking histograms.
If you paste the following data in (I'm using R here):
Annie <- c(3.15,5.46,3.28,4.2,1.98,2.28,3.12,4.1,3.42,3.91,2.06,5.53,
5.19,2.39,1.88,3.43,5.51,2.54,3.64,4.33,4.85,5.56,1.89,4.84,5.74,3.22,
5.52,1.84,4.31,2.01,4.01,5.31,2.56,5.11,2.58,4.43,4.96,1.9,5.6,1.92)
Brian <- c(2.9, 5.21, 3.03, 3.95, 1.73, 2.03, 2.87, 3.85, 3.17, 3.66,
1.81, 5.28, 4.94, 2.14, 1.63, 3.18, 5.26, 2.29, 3.39, 4.08, 4.6,
5.31, 1.64, 4.59, 5.49, 2.97, 5.27, 1.59, 4.06, 1.76, 3.76, 5.06,
2.31, 4.86, 2.33, 4.18, 4.71, 1.65, 5.35, 1.67)
Chris <- c(2.65, 4.96, 2.78, 3.7, 1.48, 1.78, 2.62, 3.6, 2.92, 3.41, 1.56,
5.03, 4.69, 1.89, 1.38, 2.93, 5.01, 2.04, 3.14, 3.83, 4.35, 5.06,
1.39, 4.34, 5.24, 2.72, 5.02, 1.34, 3.81, 1.51, 3.51, 4.81, 2.06,
4.61, 2.08, 3.93, 4.46, 1.4, 5.1, 1.42)
Zoe <- c(2.4, 4.71, 2.53, 3.45, 1.23, 1.53, 2.37, 3.35, 2.67, 3.16,
1.31, 4.78, 4.44, 1.64, 1.13, 2.68, 4.76, 1.79, 2.89, 3.58, 4.1,
4.81, 1.14, 4.09, 4.99, 2.47, 4.77, 1.09, 3.56, 1.26, 3.26, 4.56,
1.81, 4.36, 1.83, 3.68, 4.21, 1.15, 4.85, 1.17)
Then you can generate them yourself:
opar<-par()
par(mfrow=c(2,2))
hist(Annie,breaks=1:6,main="Annie",xlab="V1",col="lightblue")
hist(Brian,breaks=1:6,main="Brian",xlab="V2",col="lightblue")
hist(Chris,breaks=1:6,main="Chris",xlab="V3",col="lightblue")
hist(Zoe,breaks=1:6,main="Zoe",xlab="V4",col="lightblue")
par(opar)
Now look at this strip chart:
x<-c(Annie,Brian,Chris,Zoe)
g<-rep(c('A','B','C','Z'),each=40)
stripchart(x~g,pch='|')
abline(v=(5:23)/4,col=8,lty=3)
abline(v=(2:5),col=6,lty=3)
(If it's still not obvious, see what happens when you subtract Annie's data from each set: head(matrix(x-Annie,nrow=40)))
The data has simply been shifted left each time by 0.25.
Yet the impressions we get from the histograms - right skew, uniform, left skew and bimodal - were utterly different. Our impression was entirely governed by the location of the first bin-origin relative to the minimum.
So not just 'exponential' vs 'not-really-exponential' but 'right skew' vs 'left skew' or 'bimodal' vs 'uniform' just by moving where your bins start.
Edit: If you vary the binwidth, you can get stuff like this happen:
That's the same 34 observations in both cases, just different breakpoints, one with binwidth $1$ and the other with binwidth $0.8$.
x <- c(1.03, 1.24, 1.47, 1.52, 1.92, 1.93, 1.94, 1.95, 1.96, 1.97, 1.98,
1.99, 2.72, 2.75, 2.78, 2.81, 2.84, 2.87, 2.9, 2.93, 2.96, 2.99, 3.6,
3.64, 3.66, 3.72, 3.77, 3.88, 3.91, 4.14, 4.54, 4.77, 4.81, 5.62)
hist(x,breaks=seq(0.3,6.7,by=0.8),xlim=c(0,6.7),col="green3",freq=FALSE)
hist(x,breaks=0:8,col="aquamarine",freq=FALSE)
Nifty, eh?
Yes, those data were deliberately generated to do that... but the lesson is clear - what you think you see in a histogram may not be a particularly accurate impression of the data.
### What can we do?
Histograms are widely used, frequently convenient to obtain and sometimes expected. What can we do to avoid or mitigate such problems?
As Nick Cox points out in a comment to a related question: The rule of thumb always should be that details robust to variations in bin width and bin origin are likely to be genuine; details fragile to such are likely to be spurious or trivial.
At the least, you should always do histograms at several different binwidths or bin-origins, or preferably both.
Alternatively, check a kernel density estimate at not-too-wide a bandwidth.
One other approach that reduces the arbitrariness of histograms is averaged shifted histograms,
(that's one on that most recent set of data) but if you go to that effort, I think you might as well use a kernel density estimate.
If I am doing a histogram (I use them in spite of being acutely aware of the issue), I almost always prefer to use considerably more bins than typical program defaults tend to give and very often I like to do several histograms with varying bin width (and, occasionally, origin). If they're reasonably consistent in impression, you're not likely to have this problem, and if they're not consistent, you know to look more carefully, perhaps try a kernel density estimate, an empirical CDF, a Q-Q plot or something similar.
While histograms may sometimes be misleading, boxplots are even more prone to such problems; with a boxplot you don't even have the ability to say "use more bins". See the four very different data sets in this post, all with identical, symmetric boxplots, even though one of the data sets is quite skew.
[1]: Rubin, Paul (2014) "Histogram Abuse!",
Blog post, OR in an OB world, Jan 23 2014
• Practically every graph of necessity bins data like this. The bins are just small enough (the width of one pixel along the axis) that it doesn't matter? – AJMansfield Jul 11 '13 at 19:13
• @AJMansfield This is a bit like saying "every distribution is discrete" - while literally true, it obscures the relevant issue. A typical number of bins in a binned estimator is vastly smaller than a typical number of pixels... and with any graphics that make use of anti-aliasing, the 'effective' number of pixels is larger (in that it's potentially possible to distinguish differences of positions between pixels) – Glen_b Jul 12 '13 at 0:45
• The fundamental issue is that histograms heavily rely on the bin size. It is difficult to determine this a priori. – user46925 Mar 20 '16 at 14:51
A kernel density or logspline plot may be a better option compared to a histogram. There are still some options that can be set with these methods, but they are less fickle than histograms. There are qqplots as well. A nice tool for seeing if data is close enough to a theoretical distribution is detailed in:
Buja, A., Cook, D. Hofmann, H., Lawrence, M. Lee, E.-K., Swayne,
D.F and Wickham, H. (2009) Statistical Inference for exploratory
data analysis and model diagnostics Phil. Trans. R. Soc. A 2009
367, 4361-4383 doi: 10.1098/rsta.2009.0120
The short version of the idea (still read the paper for details) is that you generate data from the null distribution and create several plots one of which is the original/real data and the rest are simulated from the theoretical distribution. You then present the plots to someone (possibly yourself) that has not seen the original data and see if they can pick out the real data. If they cannot identify the real data then you don't have evidence against the null.
The vis.test function in the TeachingDemos package for R help implement a form of this test.
Here is a quick example. One of the plots below is 25 points generated from a t distribution with 10 degrees of freedom, the other 8 are generated from a normal distribution with the same mean and variance.
The vis.test function created this plot and then prompts the user to choose which of the plots they think is different, then repeats the process 2 more times (3 total).
• @ScottStafford, I added a copy of the plot above. This one uses qqplots but the function will also generate histograms or density plots could be programmed. – Greg Snow Apr 18 '13 at 19:13
Cumulative distribution plots [MATLAB, R] – where you plot the fraction of data values less than or equal to a range of values – are by far the best way to look at distributions of empirical data. Here, for example, are the ECDFs of this data, produced in R:
This can be generated with the following R input (with the above data):
plot(ecdf(Annie),xlim=c(min(Zoe),max(Annie)),col="red",main="ECDFs")
lines(ecdf(Brian),col="blue")
lines(ecdf(Chris),col="green")
lines(ecdf(Zoe),col="orange")
As you can see, it's visually obvious that these four distributions are simply translations of each other. In general, the benefits of ECDFs for visualizing empirical distributions of data are:
1. They simply present the data as it actually occurs with no transformation other than accumulation, so there's no possibility of accidentally deceiving yourself, as there is with histograms and kernel density estimates, because of how you're processing the data.
2. They give a clear visual sense of the distribution of the data since each point is buffered by all the data before and after it. Compare this with non-cumulative density visualizations, where the accuracy of each density is naturally unbuffered, and thus must be estimated either by binning (histograms) or smoothing (KDEs).
3. They work equally well regardless of whether the data follows a nice parametric distribution, some mixture, or a messy non-parametric distribution.
The only trick is learning how to read ECDFs properly: shallow sloped areas mean sparse distribution, steep sloped areas mean dense distribution. Once you get the hang of reading them, however, they're a wonderful tool for looking at distributions of empirical data.
• Is there any documentation available to read CDFs? eg what if my cdf distribution like you have showed above then how can we classify \ guesstimate it into chisquare, normal or other distribution based on looks – stats101 Aug 12 '16 at 15:22
Suggestion: Histograms usually only assign the x-axis data to have occurred at the midpoint of the bin and omit x-axis measures of location of greater accuracy. The effect this has on the derivatives of fit can be quite large. Let us take a trivial example. Suppose we take the classical derivation of a Dirac delta but modify it so that we start with a Cauchy distribution at some arbitrary median location with a finite scale (full width half-maximum). Then we take the limit as the scale goes to zero. If we use the classical definition of a histogram and do not change bin sizes we will capture neither the location or the scale. If however, we use a median location within bins of even of fixed width, we will always capture the location, if not the scale when the scale is small relative to the bin width.
For fitting values where the data is skewed, using fixed bin midpoints will x-axis shift the entire curve segment in that region, which I believe relates to the question above.
STEP 1 Here is an almost solution. I used $n=8$ in each histogram category, and just displayed these as the mean x-axis value from each bin. Since each histogram bin has a value of 8, the distributions all look uniform, and I had to offset them vertically to show them. The display is not the correct answer, but it is not without information. It correctly tells us that there is an x-axis offset between groups. It also tells us that the actual distribution appears to be slightly U shaped. Why? Note that the distance between mean values is further apart in the centers, and closer at the edges. So, to make this a better representation, we should borrow whole samples and fractional amounts of each bin boundary sample to make all the mean bin values on the x-axis equidistant. Fixing this and displaying it properly would require a bit of programming. But, it may just be a way to make histograms so that they actually display the underlying data in some logical format. The shape will still change if we change the total number of bins covering the range of the data, but the idea is to resolve some of the problems created by binning arbitrarily.
STEP 2 So let's start borrowing between bins to try to make the means more evenly spaced.
Now, we can see the shape of the histograms beginning to emerge. But the difference between means is not perfect as we only have whole numbers of samples to swap between bins. To remove the restriction of integer values on the y-axis and complete the process of making equidistant x-axis mean values, we have to start sharing fractions of a sample between bins.
Step 3 The sharing of values and parts of values.
As one can see, the sharing of parts of a value at a bin boundry can improve the uniformity of distance between mean values. I managed to do this to three decimal places with the data given. However, one cannot, I do not think, make the distance between mean values exactly equal in general, as the coarseness of the data will not permit that.
One can, however, do other things like use kernel density estimation.
Here we see Annie's data as a bounded kernel density using Gaussian smoothings of 0.1, 0.2, and 0.4. The other subjects will have shifted functions of the same type, provided one does the same thing as I did, namely use the lower and upper bounds of each data set. So, this is no longer a histogram, but a PDF, and it serves the same role as a histogram without some of the warts. | 3,868 | 13,888 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-39 | latest | en | 0.901479 |
https://english.stackexchange.com/questions/253167/little-vs-few-when-talking-about-amounts-of-people | 1,702,043,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00295.warc.gz | 262,291,140 | 41,931 | # Little vs few when talking about amounts of people
The amount of people who surrounded him was little.
vs
The amount of people who surrounded him was few.
Are they both grammatically correct? In the first one, the "little" is talking about the amount of people, and not people itself, in which "few" would be used.
Also, which one would generally be used in normal situations?
• The people were few and the amount was little. Jun 17, 2015 at 19:21
• Countable nouns use few, uncountable ones use little. (English grammar rule when learning English). For example: a little chocolate (you can say 2 chocolates, or a few chocolates, but then you mean the piece of chocolate wrapped up in a paper), but a few people (2 people 3 people...) Jun 17, 2015 at 19:29
• Jun 17, 2015 at 19:30
• First off, it's not "amount" of people, but "number" of people -- and that should solve the problem, too. HTH.
– Kris
Jun 18, 2015 at 7:22
People are countable. Therefore, talking about a generalized "amount" of people is not a great idea. Your sentence would be better formed as "The number of people who surrounded him was small." This in turn suggests that you could replace "the number of people who surrounded him" with "the crowd surrounding him".
• Or better yet, "A few people surrounded him". Why discuss the number if you're not giving it? Jun 17, 2015 at 19:18
• Because the exact number is less important (to the speaker in this hypothetical context) than the fact that it is small. Jun 17, 2015 at 19:24
• That's what a few means -- 'some integer less than some `minimal integer`". Jun 17, 2015 at 19:35
Couple of alternate ways of saying it:
1. Few people surrounded him.
2. The number of people surrounding him was small.
Your sentences sound quite queer due to your construction:
The amount of people who surrounded him was xy.
At first I read the sentence as
The amount of people who surrounded him was xy.
and was expecting a which instead of the who. That is of course my problem, but why not use a sentence like
Only a few people surrounded him. (while his fellow had by comparison gathered a rather large crowd) **
There is another aspect to your question also.
• few people
• a little amount
Using few with amount should not be correct, since amount is singular while few must be used with a plural.
The adjective is referring to the subject - amount, so only
The amount of people who surrounded him was little.
can be correct, even if one would prefer another construction.
** While I was writing my answer John Lawler commented and R Sahu posted similar sentences.
• That's not the actual answer. It is not about if the noun is singular or plural, it's about if the noun is countable or not. Uncountable nouns use little, a little, and much, whereas countable nouns use numbers (1, 2, 3), few, a few, many. Jun 17, 2015 at 19:32
• @JuanRocamonde I thought about that, too, but amount itself is not countable. There can be several/few/five amounts. But one amount cannot be few/many. It just can be any/some/little/big, or is that wrong? Money behaves unlike amount. You can have some money, much money, but not moneys or many money, not even few money. Jun 17, 2015 at 19:49 | 803 | 3,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.984698 |
http://doc.raqsoft.com/esproc/func/delete.html | 1,680,171,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949181.44/warc/CC-MAIN-20230330101355-20230330131355-00599.warc.gz | 14,036,076 | 20,951 | • esProc
Tutorial
Function Reference
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Code Reference
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JSON-style Parameter Guide
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# delete()
Here’s how to use delete() function.
## T .delete()
Description:
Delete specified records from a table sequence.
Syntax:
T.delete(k) Delete the kth record T.delete(p) Delete records whose sequence numbers exist in p T.delete(A) Delete records that exist in sequence A
Note:
The function deletes the specified records from table sequence T, and automatically updates the index, if any, and checks distinctness. The deleted records will be always saved in the delete buffer.
Parameter:
T A table sequence k A positive integer, which specifies the position of a record to be deleted p An integer sequence with the length of n, which specifies the positions of the records to be deleted A A sequence, which specifies the records to be deleted
Option:
@n Return the deleted record or a record sequence of the deleted records
Return value:
The table sequence T, some records of which have been deleted
Example:
A 1 =demo.query("select * from EMPLOYEE") 2 =A1.delete(1) Delete the first record 3 =A1.delete([2,4,6]) Delete the second, the forth and the sixth records. 4 =A1.select(EID>5) 5 =A1.delete(A4) Delete the records whose EID is more than 5. 6 =A1.delete@n(1) Return the deleted record
Note:
We use the store address, instead of the field names or field values, to judge whether the records specified by A exist in T. So to delete the specified records, we generally use the function T.delete(T.select(…)) to locate them.
Related function:
## A. delete()
Description:
Delete specified members from a sequence.
Syntax:
A.delete(k) Delete the kth member A.delete(p) Delete members whose sequence numbers exist in p; count backward when a member of sequence p is less than 0
Note:
The function deletes the kth member or the members whose sequence numbers exist in p from sequence A, and automatically updates the index, if any, and checks distinctness.
Parameter:
A A sequence k A positive integer that indicates the position of a member to be deleted in the sequence p An integer sequence with the length of n that specifies the positions of the members to be deleted
Option:
@n Return the deleted record or a record sequence of the deleted records
Return value:
A sequence
Example:
A 1 =["a","c","d","e","f"] 2 =A1.delete([2,4,5]) [a,d] 3 =demo.query("select * from STUDENTS") 4 =A3.delete@n(2) Return the deleted record 5 =["a","c","d","e","f"].delete([3,-5]) Count backward to get the value [c,e,f] since one member of the sequence parameter is less than 0
Related function:
## T.delete( P )
Description:
Delete specified record(s) from an entity table.
Syntax:
T.delete(P)
Note:
The function deletes from entity table T the records that match key values of record sequence P. The primary key should be in ascending order. If T is the base table having an attached table, delete records having same dimensions from the attached table.
Parameter:
T An entity table P A record sequence having same structure as T
Option:
@n Return deleted record(s) from the record sequence
Example:
A 1 =file("emp.ctx") An existing composite table file 2 =A1.open() Open the composite table’s base table 3 =A2.cursor().fetch() Get data in the base table: 4 =A2.attach(table3) Return the base table’s attached table table3 5 =A4.cursor().fetch() Get data of the attached table 6 =create(EID,NAME).record([1,"aaa",21,"bbb"]) Return a table sequence 7 =A2.delete@n(A6) Delete records from both the table sequence and the base table where the former’s key values are contained in the latter’s corresponding key values; @n option is used to return the deleted record 8 =A2.cursor().fetch() Check the base table to see that the record where key EID value is 1 is deleted 9 =A4.cursor().fetch() Check the attached table to see that the record with same dimension value is deleted
## T.delete( P )
Description:
Delete one or more specified records from a pseudo table.
Syntax:
T.delete(P)
Note:
The function deletes records of record sequence P from pseudo table T by comparing their primary keys.
Parameter:
T A pseudo table P A record sequence having the same structure with T
Option:
@n Return the deleted records
Example:
A 1 =create(file).record(["D:/file/pseudo/Employee.ctx"]) 2 =pseudo(A1) Generate a composite table object 3 =create(Dept,AvgSalary).record(["HR",7000,"CSD",6018.04]) Return a table sequence 4 =A2.delete@n(A3) Delete records from A2’s pseudo table according to table sequence A3’s primary key values, and return the deleted records 5 =A4.import() | 1,140 | 4,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | longest | en | 0.870284 |
https://chrisvaughnmusic.com/music/dj-dinda.html | 1,669,551,129,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00655.warc.gz | 200,882,023 | 5,383 | # Free help with math problems
Apps can be a great way to help learners with their math. Let's try the best Free help with math problems. We will also look at some example problems and how to approach them.
## The Best Free help with math problems
This Free help with math problems provides step-by-step instructions for solving all math problems. A math tutor can be an invaluable resource for this. By definition, a word problem is a mathematical problem that involves words rather than numbers or symbols. You might see words like "if it rains tomorrow, how many inches of rain will there be?" Word problems usually involve numbers or quantities, but they also include words that represent concepts such as length, time, area and volume. However, they often look different from standard mathematical problems because they rely more on language than mathematics. For example, you might be given the word "lose" and asked how many pounds of weight you would have to lose to reach a certain weight goal.
Intermediate algebra involves solving a variety of problems involving addition, subtraction, multiplication, and division. This type of problem typically involves the application of previous knowledge to solve new problems. Examples of intermediate algebra include finding the area of a rectangular plot using a right triangle with unknown lengths and finding the value of an unknown quantity in a series by summing certain terms. Intermediate algebra problems tend to be more complex than elementary algebra problems because they require more advanced math skills such as addition and subtraction. Consequently, students may need additional time to practice these skills before they are ready to tackle these types of problems. In order to effectively solve an intermediate algebra problem, students must understand the process of adding and subtracting integers. In addition, they need to understand the concept of multiplication and division. To practice intermediate algebra, students can use online resources such as Khan Academy or their math textbooks. By practicing intermediate algebra problems at home, students will be able to develop better math skills that will help them tackle any future algebra problem that comes their way.
Word math problems can be written, oral, or mathematically based. There are two main types of word math: word scramble and word patterning. Scrambled words are scrambled letters that must be rearranged in order to form a word. Word patterning tasks are more complex, requiring you to identify the parts of a word that match up with each other (such as letter, number, or symbol). Word math problems can help improve your vocabulary and sentence structure. In addition, they can help keep you sharp as you age by keeping your mind active and engaged.
When you’re given a non-linear equation like: (3x^2+4x+1)^(1/3) =3x^4 +4x^2 – 1 You need to identify the roots of the equation so that you can work out how to solve it. Once you’ve identified the roots, you can find the solution by plugging them into the equation and solving for x. There are several different ways you can approach solving exponential equations. You can check whether or not you’ve solved for one root and if so, check whether or not you’ve solved for all of the roots by working backwards from the solution back to the original equation. You can also use a graphing calculator and try to plot the function on a chart so that you can see at a glance whether or not you have found all of the roots.
A best app with Full Support If anyone thinks math is difficult, he should try it out but only if he knows basic of mathematics (Maybe I am wrong but everyone should try it Thank You the app App Developer Team)
### Angelina Thompson
This app is the best. This app will make you understand math better, and one of the things I really love about this app is that they have this setting where it will show you all the steps of the way and every time, I use it makes me get to learn the strategies that they are showing me. I 100% recommend this app😄
### Gemma Barnes
Solve the triangle App that can solve any math problem Answer my math word problem Equiation solver Double equation solver Geometry photo calculator | 839 | 4,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-49 | latest | en | 0.955386 |
https://nrich.maths.org/public/leg.php?code=53&cl=4&cldcmpid=543 | 1,550,598,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490806.45/warc/CC-MAIN-20190219162843-20190219184843-00192.warc.gz | 637,172,976 | 8,508 | # Search by Topic
#### Resources tagged with Simultaneous equations similar to System Speak:
Filter by: Content type:
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### There are 41 results
Broad Topics > Algebra > Simultaneous equations
### System Speak
##### Age 16 to 18 Challenge Level:
Five equations... five unknowns... can you solve the system?
### Always Two
##### Age 14 to 18 Challenge Level:
Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.
### Matchless
##### Age 14 to 16 Challenge Level:
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### Overturning Fracsum
##### Age 14 to 16 Challenge Level:
Solve the system of equations to find the values of x, y and z: xy/(x+y)=1/2, yz/(y+z)=1/3, zx/(z+x)=1/7
### All-variables Sudoku
##### Age 11 to 18 Challenge Level:
The challenge is to find the values of the variables if you are to solve this Sudoku.
### Leonardo's Problem
##### Age 14 to 18 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Pair Squares
##### Age 16 to 18 Challenge Level:
The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.
##### Age 16 to 18 Challenge Level:
Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.
### Negatively Triangular
##### Age 14 to 16 Challenge Level:
How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ?
### Pareq Calc
##### Age 14 to 16 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Intersections
##### Age 14 to 18 Challenge Level:
Change one equation in this pair of simultaneous equations very slightly and there is a big change in the solution. Why?
##### Age 14 to 16 Challenge Level:
Four jewellers share their stock. Can you work out the relative values of their gems?
### CD Heaven
##### Age 14 to 16 Challenge Level:
All CD Heaven stores were given the same number of a popular CD to sell for £24. In their two week sale each store reduces the price of the CD by 25% ... How many CDs did the store sell at. . . .
### Graphs of Changing Areas
##### Age 16 to 18 Challenge Level:
Use graphs to gain insights into an area and perimeter problem, or use your knowledge of area and perimeter to gain insights into the graphs...
### Surds
##### Age 14 to 16 Challenge Level:
Find the exact values of x, y and a satisfying the following system of equations: 1/(a+1) = a - 1 x + y = 2a x = ay
##### Age 11 to 16 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
##### Age 16 to 18 Challenge Level:
Find a quadratic formula which generalises Pick's Theorem.
### Simultaneous Equations Sudoku
##### Age 11 to 16 Challenge Level:
Solve the equations to identify the clue numbers in this Sudoku problem.
### Real(ly) Numbers
##### Age 16 to 18 Challenge Level:
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
### Fruity Totals
##### Age 7 to 16 Challenge Level:
In this interactivity each fruit has a hidden value. Can you deduce what each one is worth?
### Roots and Coefficients
##### Age 16 to 18 Challenge Level:
If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?
### Which Is Bigger?
##### Age 14 to 16 Challenge Level:
Which is bigger, n+10 or 2n+3? Can you find a good method of answering similar questions?
### Which Is Cheaper?
##### Age 14 to 16 Challenge Level:
When I park my car in Mathstown, there are two car parks to choose from. Can you help me to decide which one to use?
### What's it Worth?
##### Age 11 to 16 Challenge Level:
There are lots of different methods to find out what the shapes are worth - how many can you find?
### Coffee
##### Age 14 to 16 Challenge Level:
To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
##### Age 11 to 16 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### LCM Sudoku II
##### Age 11 to 18 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Rudolff's Problem
##### Age 14 to 16 Challenge Level:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
### Intersection Sudoku 1
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
### How Many Balls?
##### Age 16 to 18 Challenge Level:
A bag contains red and blue balls. You are told the probabilities of drawing certain combinations of balls. Find how many red and how many blue balls there are in the bag.
### Polycircles
##### Age 14 to 16 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Double Time
##### Age 16 to 18 Challenge Level:
Crack this code which depends on taking pairs of letters and using two simultaneous relations and modulus arithmetic to encode the message.
### Building Tetrahedra
##### Age 14 to 16 Challenge Level:
Can you make a tetrahedron whose faces all have the same perimeter?
### Intersection Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
### Walls
##### Age 16 to 18 Challenge Level:
Plane 1 contains points A, B and C and plane 2 contains points A and B. Find all the points on plane 2 such that the two planes are perpendicular.
### Escriptions
##### Age 16 to 18 Challenge Level:
For any right-angled triangle find the radii of the three escribed circles touching the sides of the triangle externally.
### Battery Modelling
##### Age 16 to 18 Challenge Level:
Find out how to model a battery mathematically
### A Method of Defining Coefficients in the Equations of Chemical Reactions
##### Age 14 to 18
A simple method of defining the coefficients in the equations of chemical reactions with the help of a system of linear algebraic equations.
### Cobalt Decay
##### Age 16 to 18 Challenge Level:
Investigate the effects of the half-lifes of the isotopes of cobalt on the mass of a mystery lump of the element. | 1,837 | 7,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-09 | longest | en | 0.848222 |
http://math.stackexchange.com/questions/195327/proving-that-m-leq-n-using-finite-spanning-sets-and-linear-independence | 1,469,316,242,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00274-ip-10-185-27-174.ec2.internal.warc.gz | 153,331,861 | 18,288 | Proving that $m \leq n$, using finite spanning sets and linear independence.
Let $V$ be a vector space in $F$, and suppose that $V$ has a finite spanning set $S=\{v_1,\ldots,v_n\}$. Show that if $T=\{u_1,\ldots,u_m\}$ is a linearly independent subset of V, then $m\leq n$. (We are not assuming $T$ is a subset of $S$)
-
Just use the definition of dimension. Recall: when you want to find the (finite) dimension of a vector space, the "problem" is that spanning sets might be too big and linearly independent sets might be too small: more precisely, when you have a spanning set $S$ with $n$ elements, then $n\geq \dim V$. And, given any subset $T$ of $m$ linearly independent vectors, you have $m\leq \dim V$. The dimension is the number of vectors you collect when you do "your best" in any of these two cases: you can find a minimal set of generators, or a maximal independent system. – Brenin Sep 13 '12 at 18:13
I think the title of this question can be improved... – ivan Sep 13 '12 at 18:57
Hint: Since $S$ is a spanning set, you can write $u_i=\sum\limits_{j=1}^n\lambda_{ij}v_j$ for some $\lambda_{ij}\in F$. If $m>n$, can you find a linear dependence among these elements?
Hint 2: We have $u_1=\sum\limits_{j=1}^n\lambda_{1j}v_j$ and $u_2=\sum\limits_{j=1}^n\lambda_{2j}v_j$. If $\lambda_{21}\neq 0$, then $$u_1-\frac{\lambda_{11}}{\lambda_{21}}u_2=0v_1+x_2v2+\cdots + x_nv_n$$ while otherwise $u_2=0v_1+x_2v_2+\cdots + x_nv_n$ for some $x_2,\ldots,x_n\in F$. Either way, we've found a linear combination of the first $2$ elements of $T$ which eliminates $v_1$. Proceeding in this manner, what do we get using the first $n+1$ elements of $T$?
So if you let $u_m=\lambda_1 v_1 + ... + \lambda_n v_n$ then I'm unclear, but I'll continue with my line of thought. Assume $\lambda_n \neq 0$ then we can say $v_n=\frac{u_m-\lambda_1 v_1 - ... }{\lambda_n}$? I'm confused, sorry. – tk2 Sep 13 '12 at 17:56
You can select a basis from $S$ and you can extend $T$ a basis. These basises have the same length. | 667 | 2,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.772846 |
https://calculatorshub.net/telecom-calculators/convolution-calculator/ | 1,726,620,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00008.warc.gz | 127,477,764 | 28,286 | Home » Simplify your calculations with ease. » Telecom Calculators » Convolution Calculator Online
# Convolution Calculator Online
## Introduction
Digital signal processing is a cornerstone of modern technology, powering everything from image rendering software to audio compression algorithms. At the heart of many of these processes is a concept known as convolution. The Convolution Calculator, an essential tool for many engineers and mathematicians, makes it easy to apply this concept. This article will unravel the workings of this calculator, its underlying formula, and offer a practical example of its usage.
## Understanding Convolution
At its core, convolution is a mathematical operation that takes two functions and produces a third function that portrays how the shape of one is modified by the other. It's commonly used in signal processing and statistics to analyze and manipulate data.
The Convolution Calculator uses this concept, providing a platform to perform convolution operations between two functions with relative ease and accuracy.
## Mechanics of the Convolution Calculator
The Convolution Calculator is built to accept two functions as inputs. The user is required to input these functions, often in the form of arrays or sets of numbers, into the provided fields.
Upon clicking the 'Calculate' button, the calculator retrieves these values and checks for their validity. If the inputs pass the validity check, the calculator proceeds to perform the convolution operation, following the prescribed formula. The result is then displayed in the designated output field.
## The Convolution Formula
The mathematical formula for convolution of two functions, say f(t) and g(t), is given as:
``````(f * g)(t) = ∫ f(τ)g(t - τ) dτ
``````
This formula signifies the integral over the entire domain of the functions, 'f(τ)' and 'g(t - τ)' represent the input functions, 't' is the variable of integration, and the symbol '∫' is indicative of integration.
## Example of Convolution Calculation
For instance, consider two functions: f(t) = t and g(t) = t^2.
In this case, the convolution of f and g, computed through the formula, would be:
``````(f * g)(t) = ∫ f(τ)g(t - τ) dτ
= ∫ τ(t - τ)^2 dτ
``````
The Convolution Calculator, given these inputs, would compute and present this result.
## Real-World Application of Convolution
Convolution is not just a theoretical mathematical operation; it has practical applications in several fields.
1. Signal Processing: Convolution helps describe the relationship between a system's input and output, where the system is defined by a function h(t).
2. Image Processing: It plays a crucial role in filtering operations, such as smoothing, sharpening, and edge detection.
3. Data Analysis: Convolution aids in understanding the probability density functions in statistical analysis and data science.
## Common FAQs
What is Convolution?
Convolution is a mathematical operation that combines two functions to generate a third, illustrating how the shape of one is altered by the other.
What is the formula used in the Convolution Calculator?
The convolution of two functions f and g is calculated using the formula: (f * g)(t) = ∫ f(τ)g(t - τ) dτ.
How is Convolution used in real-world applications?
Convolution has a wide range of applications, especially in digital signal processing, image processing, and data analysis. It is a fundamental operation in these fields.
## Conclusion
To conclude, the Convolution Calculator is a powerful tool, engineered with precision and finesse. With the ability to perform complex calculations swiftly and accurately, it offers invaluable assistance to professionals and students in the realm of digital signal processing and beyond. As technology continues to evolve, tools like these will continue to play a pivotal role in shaping the future. | 794 | 3,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.888682 |
https://documen.tv/question/write-the-epression-as-a-decimal-number-5-1-1-1-10-8-1-100-6-1-1000-17786840-64/ | 1,638,781,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00548.warc.gz | 285,762,555 | 15,530 | ## Write the expression as a decimal number. 5×1+1× 1 10 +8× 1 100 +6× 1 1000 =
Question
Write the expression as a decimal number.
5×1+1×
1
10
+8×
1
100
+6×
1
1000
=
in progress 0
4 months 2021-07-31T19:00:15+00:00 1 Answers 1 views 0 | 108 | 237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-49 | latest | en | 0.752992 |
https://expert-tutor.net/qnt561-phoenix-consumer-food-concepts-descriptive-measures-paper/ | 1,680,265,157,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00517.warc.gz | 301,552,045 | 12,142 | # QNT561 Phoenix Consumer Food Concepts & Descriptive Measures Paper
Purpose of Assignment
The purpose of this assignment to orient students to the key concepts in statistics. This assignment will introduce students to the language of statistics. Students will also get a chance to warm-up on evaluating some basic descriptive statistics using Excel® prior to the course start.
Assignment Steps
This assignment has an Excel® dataset spreadsheet attached. You will be required to only do one of the three datasets.
Resource: Microsoft Excel®, Statistics Concepts and Descriptive Measures Data Set
Choose one of the following datasets to complete this assignment:
• Consumer Food
• Financial
• Hospital
Answer each of the following in a total of 90 words:
• For each column, identify whether the data is qualitative or quantitative.
• Identify the level of measurement for the data in each column.
• For each column containing quantitative data:
• Evaluate the mean and median
• Interpret the mean and median in plain non-technical terms
• Use the Excel =AVERAGE function to find the mean
• Use the Excel =MEDIAN function to find the median
• For each column containing quantitative data:
• Evaluate the standard deviation and range
• Interpret the standard deviation and range in plain non-technical terms
• Use the Excel =STDEV.S function to find the standard deviation
• For range (maximum value minus the minimum value), find the maximum value using the Excel =MAX function and find the minimum value using the Excel’s =MIN function
Click the Assignment Files tab to submit your assignment. | 313 | 1,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.786177 |
http://www.physicsforums.com/showthread.php?p=4150851 | 1,369,286,244,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702849682/warc/CC-MAIN-20130516111409-00044-ip-10-60-113-184.ec2.internal.warc.gz | 659,459,557 | 8,360 | ## Local Existence and Global Existence of differential equations
Hi everyone! I'm newbie in this forum, please help me for my question.
In differential equation we know that the differential equation has a solution and uniqueness. which is usually called the existence and uniqueness theorem. my question, what is the difference of local existence and global existence existence from the point of view of functional analysis?
thanks before...
Your question is hard to answer without more information. What kind of differential equations are we talking about? ODE, parabolic/hyprebolic PDE? For problems with a distinguished time variable, the solution is a mapping from R into a space of functions (for PDE). You can think of it as a one parameter family of functions. Local existence means that that this mapping is defined near 0. Global existence means it extends for all time.
I forget about it, I mean Ordinary differential Equation in C([a,b],Rn). C([a,b],Rn) is notation for mapping of continuous functions in [a,b] into Rn, I can write f element of C([a,b],Rn) then f=(f1,f2,f3,...,fn). what do you mean about "the solution is mapping from R into a space of functions" is which have notation C([a,b],Rn)? for your last sentences, I can understand. thanks :)
## Local Existence and Global Existence of differential equations
If you have a partial differential equation like the heat equation then your unknown function is something like u(t,x), where t is the time variable and x is the spatial variable. For a fixed t, u(t,x) is the heat distribution over the domain. So a solution of the PDE is viewed as a mapping from the time domain into the domain of heat distributions (which is a space of functions). The reason I thought you might be talking about PDE is because you referred to functional analysis.
I'm still not clear what type of equation you are talking about, but it sounds like an ODE:
y'=F(t,y),
where y is a vector valued function y(t)=(y1(t),.... , yn(t)).
In that case, local/global existence refers to whether the solutions are defined for all values of t. Does this answer your original question?
I think about space of functions in my mind is similar to what you mean. but we have difference on notations. Yes, you are right like your example. yes, yes, you are right, that was I mean. thak you very much for your help. I would learn more.
Tags differential eq, existence, global, local | 540 | 2,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2013-20 | longest | en | 0.92863 |
http://fab.cba.mit.edu/classes/864.05/people/rocha/week06/annealing.py.txt | 1,553,656,850,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-13/segments/1552912207618.95/warc/CC-MAIN-20190327020750-20190327042750-00529.warc.gz | 77,610,806 | 1,074 | from numarray import * from MLab import * from pylab import * def energy(spins, bonds): n = size(spins) e = bonds[:n-1] * spins[:n-1] * spins[1:] return - (sum(e) + bonds[n-1] * spins[n-1] * spins[0]) def anneal(spins, bonds, alpha, steps = 100): spins = spins.copy() n = size(spins) e = zeros((steps,1), Float64); for t in range(steps): # Energy E = energy(spins,bonds) # Flip a random spin r = int(n*rand()) spins[r] = - spins[r] # New Energy and Delta En = energy(spins,bonds) dE = En - E e[t] = E # Reject Move? p = exp(-alpha * t * dE) if ((dE > 0) & (p < rand())): spins[r] = - spins[r] # Reject return e def main() : steps = 6000 alphas = array([0.1, 0.01, 0.001]) N = 100 spins = 2 * (rand(N,1) > 0.5) - 1 bonds = randn(N,1) for i in range(size(alphas)): e = anneal(spins, bonds, alphas[i], steps) plot(e) title("Simulated Annealing"); show() if __name__ == "__main__" : main() | 317 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | longest | en | 0.539741 |
https://projectbink.com/how-to-get-the-absolute-value-in-google-sheets/ | 1,653,161,263,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00470.warc.gz | 531,246,923 | 14,678 | # How to Get the Absolute Value in Google Sheets
The magnitude of a number is the absolute value of that number, regardless of whether it is positive or negative. In other terms, it is the distance between it and zero.
The absolute value of -10 and 10 are both 10, because they are both at the same distance from zero.
And sometimes while working on sheets, we may need to find absolute value of a number and we have ways to find it. Let’s see.
ABS Function
The best way to return the absolute value for values in your spreadsheet is to use the ABS function. The syntax for this function is –
=ABS(value) here,
value – the number for which the absolute value should be returned. It can be a cell reference or a value entered directly into the formula. To obtain the absolute value of the returned value, you can use math operations inside the formula.
The above example shows you the best results of using this function.
Convert Negative Numbers To Positive Numbers
When converting negative numbers to positive ones, the ABS function is unquestionably the best option, especially if you have a huge dataset to work with.
However, there are plenty of additional ways to convert negative numbers to positive in Google Sheets.
Adding a minus sign (-) before the cell reference of the cell containing your negative number is one simple technique.
As you can see in the example above, I have a formula in column B that uses cell references from column A. Each cell reference has a minus sign (-) preceding it. This changes the value of the number to its inverse. As a result, negative numbers become positive.
However, this method has the disadvantage of being more prone to errors than the prior method. If you use this option on a positive number, it will convert it to a negative, therefore use caution when using it. | 372 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-21 | longest | en | 0.875001 |
https://kr.mathworks.com/matlabcentral/cody/problems?term=tag%3A%22recursion%22 | 1,643,128,788,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304859.70/warc/CC-MAIN-20220125160159-20220125190159-00491.warc.gz | 399,287,106 | 21,146 | Cody
Problems
1 – 29 of 29
Problem Title Likes Solvers Difficulty
Problem 1827. Negation the hard way
Created by: Andrew Newell
Tags recursion
1
62
Problem 43086. Recursion at variable input
Created by: Jamil Kasan
9
52
Problem 42647. Recursion - Fun
Created by: Manish Joshi
0
32
Problem 44543. Normie Function
Created by: Naimul Hassan
2
26
Problem 48990. Solve the recursion
Created by: Alisio
0
14
Problem 47385. Find Logic 28
Created by: Dev Gupta
1
24
Problem 2040. Additive persistence
Created by: Jean-Marie Sainthillier
6
140
Problem 292. Infernal Recursion
Created by: Matt Fig
Tags recursion
0
74
Problem 44073. Fractal: area and perimeter of Koch snowflake
Created by: Jihye Sofia Seo
1
22
Problem 46591. Ackerman Function
Created by: Andrew Mohebbi
Tags recursion
2
21
Problem 734. Ackermann's Function
Created by: Richard Zapor
4
60
Problem 2838. Optimum Egyptian Fractions
Created by: Jan Orwat
0
12
Problem 52355. ICFP2021 Hole-In-Wall: Solve Problem 47, Score=0, Figure Vertices 11, Hole Vertices 10
Created by: Richard Zapor
0
1
Problem 1836. Negation and new variables
Created by: Jean-Marie Sainthillier
2
18
Problem 1978. Sokoban: Puzzle 10.45
Created by: Richard Zapor
0
6
Problem 2005. BattleShip - Seaman (1) thru Admiral(6) : CPU Time Scoring(msec)
Created by: Richard Zapor
1
3
Problem 2026. Skyscrapers - Puzzle
Created by: Richard Zapor
1
5
Problem 43963. Finding operators in a MATLAB function in a string.
Created by: Adrien Rapicault
1
7
Problem 579. Spiral In
Created by: Nicholas Howe
3
73
Problem 2085. Sudoku Solver - Standard 9x9
Created by: Richard Zapor
7
32
Problem 44955. Spell the number
Created by: Graham
0
4
Problem 1301. RISK Calculator - Large Armies, High Accuracy, Fast
Created by: Richard Zapor
3
16
Problem 47453. Slitherlink I: Trivial
Created by: Richard Zapor
0
1
Problem 47478. Slitherlink V: Assert/Evolve/Check (large)
Created by: Richard Zapor
1
1
Problem 633. Create Circular Perfect Square Sequence
Created by: Richard Zapor
Tags recursion
1
9
Problem 1173. Binpack Contest: Retro - - Best Packing
Created by: Richard Zapor
0
6
Problem 47463. Slitherlink II: Gimmes
Created by: Richard Zapor
2
0
Problem 47468. Slitherlink III: Evolve
Created by: Richard Zapor
0
0
Problem 47473. Slitherlink IV: Recursive (medium size)
Created by: Richard Zapor
0
0
1 – 29 of 29 | 759 | 2,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | longest | en | 0.671075 |
https://www.whizwriters.com/mathematics-1599/ | 1,603,643,161,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00015.warc.gz | 943,336,972 | 11,579 | # Mathematics
Insert <, >, or = between each pair of numbers to form a true statement.
31. 0.463 _____ 0.46300
a. >
b.<
c. =
Arrange in order from smallest to largest. 32. 6.09, 6.9, 6.099, 6.999
a. 6.999, 6.9, 6.099, 6.09
b. 6.09, 6.9, 6.099, 6.999
c. 6.099, 6.09, 6.999, 6.9
d. 6.09, 6.099, 6.9, 6.999
Solve.
33. According to his ultra-precise scale, Jeremy gained 3.249 pounds in a three-month period. Round this amount off to the nearest hundredth.
a. 3.2 pounds
b. 3.25 pounds
c. 0.25 pounds
d. 3.26 pounds
Divide.
34. (2.24) divided by 7
a. 1.32
b. 13.2
c. 3.2
d. 0.32
35. (3.9732) divided by 516
a. 0.0077
b. 0.0087
c. 0.78
d. 0.77
36. (163.2186) divided by 2.8 Round to the nearest hundredth.
a. 58.33
b. 58.332
c. 58.29
d. 5.83
Solve.
37. 0.028 × n = 0.7
a. 24.8
b. 26
c. 25
d. 0.0196
Write as an equivalent decimal. If a repeating decimal is obtained, use notation such as 0.4, 0.23, or 0.567
38. 317/20
a. 3.85
b. 3.855
c. 3.85
d. 3.85
Simplify by using the order of operations. Round your answer to the nearest hundredth. 39. 2.9 + (13.9 – 3.8) ∙ 7.2
a. -38.36
b. 99.18
c. 75.62
d. 93.60
First round each number to one non-zero digit. Then perform the calculation using the rounded numbers to obtain an estimate.
40. 51.7 – 5.17
a. estimate: 45
b. estimate: 55
c. estimate: 54
d. estimate: 46
Round as indicated.
41. Round 25.4564 to the nearest hundredth.
a. 25.56
b. 25.36
c. 25.46
d. 25.44
Divide.
42. (0.00786) divided by 0.04
a. 0.201
b. 0.1965
c. 0.1977
d. 0.1974
Write the ratio in lowest terms. 43. 20:14
a. (7/10)
b. (20/14)
c. (14/20)
d. (10/7)
Write the following as a rate in lowest terms. 44. \$685 for 20 rolls of film
a. (\$137/20 rolls of film)
b. (\$685/4 rolls of film)
c. (\$5/20 rolls of film)
d. (\$137/4 rolls of film)
Write the rate as a unit rate. 45. 8 cents for 4 marbles
a. 2 cents/marble
b. 0.5 cents/marble
c. 12 cents/marbles
d. 32 cents/marble
46. 420 people in 20 buses
a. 210 people/bus
b. 400 people/bus
c. 21 people/bus
d. 0.048 person/bus
Solve using a proportion. Round your answer to the nearest hundredth when necessary.
47. Find which is the better buy (lower cost per ounce) by finding each unit price rounded to three decimal places if necessary.
Jelly:
\$1.41 for 6 ounces. \$2.35 for 10 ounces
a. \$2.35 for 10 ounces
b. \$1.41 for 6 ounces
c. Both cost the same per ounce.
d. None of the above.
Write as a proportion.
48. If 60 pounds of potatoes is enough to feed 150 children, then 68 pounds should be the right amount for 170 children.
a. (60 pounds/150 children) = (170 children/68 pounds)
b. (150 children/60 pounds) = (68 pounds/170 children)
c. (60 pounds/150 children) = (68 pounds/170 children)
d. (60 pounds/170 children) = (68 pounds/150 children)
Solve each proportion for the given variable.
49. (n/22) = (6/11)
a. 12
b. 40c. 3
d. 24
50. (x/3.9) = (0.08/9) Round to the nearest hundredth.
a. 0.03
b. 438.75 c. 2.81
d. 28.85
Find the value of n. Round to the nearest hundredth if necessary.
51. (1200 revolutions/20 minutes) = (n revolutions/21 minutes)
a. 1134
b. 1260
c. 2.86
d. 25,200
52. (n pounds/4 kilograms) = (11.35 pounds/5 kilograms)
a. 8.63
b. 9.08
c. 2.84
d. 9.53
Solve using a proportion. Round your answer to the nearest hundredth when necessary.
53. On an architect’s blueprint, 1 inch corresponds to 12 feet. If an exterior wall is 68 feet long, find how long the blueprint measurement should be. Write answer as a mixed number if necessary.
a. 113/17 inches
b. 51 inches
c. 5inches
d. 68 inches
Solve.
54. In a survey of 100 people, 8 preferred relish on their hot dogs. What percent preferred relish?
a. 0.8%
b. 0.08%
c. 2/25%
d. 8%
Write as a decimal. 55. 0.079%
a. 0.00079
b. 7.9
c. 0.0079
d. 0.000079
Write the percent as a decimal.
56. In the last election Mr. Thomas received 0.2% of the vote.
a. 0.02
b. 2
c. 0.2
d. 0.002
Write as a fraction or as a mixed number
57. 50%
a. 5
b. 1/4
c. 1/2
d. 1
58. 171 3/7%
a. 171/7
b. 33/7
c. 6/7
d. 15/7
Solve. Round decimals to the nearest thousandth and percents to the nearest tenth of a percent. 59. Write the equivalent decimal and percent for (7/12).
a. 0.703;70.3%
b. 0.583;58.3%
c. 0.703;703%
d. 0.583;5.83%
Solve.
60. 37% of what is 74?
a. 0.5
b. 50
c. 2000
d. 200
61. A \$210 camera is on sale at 5% off. How much will the camera cost?
a. \$2089.50
b. \$1.05
c. \$199.50
d. \$10.50
62. A stock broker is paid \$600 per month plus 2% of the total sales of stocks that she sells. Last month Cora sold \$68,000 worth of stock. What was her total income for the month?
a. \$13,600
b. \$1360
c. \$14,200
d. \$1960
63. Raya borrowed \$10,000 to finish college at an interest rate of 6.5% per year. How much interest will Raya need to pay next year?
a. \$10,650
b. \$6500
c. \$16,500
d. \$650
Convert. When necessary, round the answer to two decimal places. 64. 2pounds = ____ ounces
a. 52
b. 44
c. 32
d. 45
Solve the problem.
65. Dennis is hosting a dinner party. His recipe calls for 4 ounces of chicken per person. If he is expecting 24 guests, how many pounds of chicken should he buy? Round your answer to two decimal places if necessary.
a. 1536 pounds
b. 96 pounds
c. 6 pounds
d. 12 pounds
66. A rectangular window is 3 feet 10 inches wide and 6 feet 9 inches tall. If the window needs to be sealed with insulation that costs \$1.22 per inch, how much will it cost to insulate the perimeter?
a. \$265.96
b. \$309.88
c. \$26.60
d. \$319.15
Change to a convenient unit of measure and add. 67. 53 cm + 9 dm + 19 mm
a. 1449 cm
b. 144.9 cm
c. 243.9 cm
d. 81 cm
Write true or false for the statement.
68. A grown man might weigh 101 kilograms.
a. True
b. False
Perform the conversion. Round to the nearest hundredth when necessary. 69. 33 L to qt
a. 14.98 qt
b. 39.60 qt
c. 31.22 qt
d. 34.88qt
Convert as indicated. When necessary, round to the nearest tenth of a degree. 70. 230° F to degrees Celsius
a. 446° C
b. 145.6° C
c. 110° C
d. 95.8° C
Solve.
71. The flow rate of a safety discharge pipe at a dam is rated for a maximum of 204,000 gallons per hour. Convert this rate to pints per second. Round your answer to the nearest tenth. During a flood, the flow rate at the dam is measured as 428 pints per second. Could the safety discharge pipe safely handle this flow rate?
a. 226.7 pints per second; no
b. 214.8 pints per second; no
c. 566.7 pints per second; yes
d. 453.3 pints per second; yes
72. Brian and Heather drove 426 miles in 6 hours. What was their average speed in kilometers per hour? (Round your answer to the nearest whole number, if necessary.)
a. 686 km/hr
b. 44 km/hr
c. 114 km/hr
d. 71 km/hr
Perform the conversion. Round to the nearest hundredth when necessary. 73. 898 yd = ____ m
a. 821.13
b. 2945.44
c. 978.82
d. 273.89
Find the area of the rectangle or square. 75. Length = 0.53 m, width = 0.24 m
a. 0.2544 m to power of ((2))
b. 0.77 m to power of ((2))
c. 1.54 m to power of ((2))
d. 0.1272 m to power of ((2))
Find the perimeter and area of the rhombus. 76. The height is 8 m and the base is 17 m
a. P = 34 m, A = 68 m to power of ((2))
b. P = 17 m, A = 25 m to power of ((2))
c. P = 68 m, A = 136 m to power of ((2))
d. P = 136 m, A = 68 m to power of ((2))
Find the area of the trapezoid.
The height is 5 m and the bases are 7 m and 8 m.
a. 140 m to power of ((2))
b. 75 m to power of ((2))
c. 10 m to power of ((2))
d. 37.5 m to power of ((2))
Find the perimeter of the triangle.
An equilateral triangle whose side measures 8 mi
a. 23 mi
b. 32 mi
c. 24 mi
d. 16 mi
Find the diameter of a circle if the radius has the value given. 79. r = 22 ft
a. 44 ft
b. 1519.76 ft c. 69.08 ft
d. 11 ft
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How do you write 453 percent as a decimal?
Wiki User
2012-07-09 17:43:51
453% = 4.53 (divide a percent by 100 to get the decimal form).
Wiki User
2012-07-09 17:43:51
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https://pballew.blogspot.com/2024/06/on-this-day-in-math-june-29.html | 1,722,888,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00631.warc.gz | 373,994,892 | 31,136 | ## Saturday 29 June 2024
### On This Day in Math - June 29
Jeannie, Happy birthday.
The 180th day of the year; 180 can be formed with the only the first two primes... 180 = 22 x 32 x (2+3) *Prime Curios
180 is the sum of two square numbers: $12^2 + 6^2$. It can also be expressed as either the sum of six consecutive primes: 19 + 23 + 29 + 31 + 37 + 41, or the sum of eight consecutive primes: 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37.
180 is a digitally balanced number, its eight binary digits contain four ones and four zeros, 10110100, they match up into two sets of balanced zero-one pairs, the first four digits, 1011, aligning perfectly with their digital opposite in the last four 0100. Creating the binary digits for 0, 1, 2, and 3
Beautiful trigonometry, arctan1 + arctan2 + arctan3 = pi/2 =180o
The digits 1, 8, 0 are the only digits in a 15,601 digit prime that is both a palindrome and strobogrammitic. *@fermatslibrary
EVENTS
In 3123 BC, a Sumerian astronomer saw a devastating asteroid, perhaps a half-mile wide, according to an interpretation of a clay tablet, made by researchers from Bristol University, reported in The Times on 31 Mar 2008. The ancient date was indicated by a computer recreation of the night sky using symbols on the tablet recording the positions of constellations The Planiform tablet found by Henry Layard at Nineveh, likely a 700 BC copy of the astronomer's notes, described in cuneiform a "white stone bowl approaching" that "vigorously swept along." The asteroid probably crashed into the Austrian Alps, leaving a swath of cataclysmic damage such as, for example, the Genesis destruction of Sodom and Gomorrah.*TIS
Planisphere tablet, British Museum
1456 According to one story that first appeared in a 1475 posthumous biography and was subsequently embellished and popularized by Pierre-Simon Laplace, Callixtus III excommunicated the 1456 apparition of Halley's Comet, believing it to be an ill omen for the Christian defenders of Belgrade from the besieging armies of the Ottoman Empire. No known primary source supports the authenticity of this account. The 29 June 1456 papal bull of Callixtus III calling for a public prayer for the success of the crusade, makes no mention of the comet. By 6 August, when the Turkish siege was broken the comet had not been visible in either Europe or Turkey for several weeks. *Wik
(John Francis Rigaud, 1785)*Wik
1785 Letitia Ann Sage became the first British woman to fly. From St George's Fields on the south side of the Thames, Vincenzo Lunardi and his partner Biggin, with two invitees, Mrs. Sage and a Colonel Hastings were supposed to make the flight, but the Hydrogen balloon wouldn't take off because of the weight. (Mrs Sage, a actress and model was also a somewhat large woman, rumored to weigh appx 200 pounds.) Lunardi and Hastings stepped down, and the balloon took off with Biggin and Mrs. Sage. It landed 90 minutes later, near Harrow, where the two aeronauts had to be rescued by a group of boys from Harrow School from the angry farmer whose crops were damaged. *Wik (There were even suggestions that rather more amorous events had occurred in the flight.)
1799 The Royal Charter for the Royal Institute is promised. Ever since its founding year the Royal Institution has maintained close links with the Royal Family. On 29 June 1799, George Finch, Earl of Winchilsea (1752-1826), the President of what had until then had been called simply the “Institution” reported to a meeting of its committee of Managers ‘that he had had the Honour of mentioning this Institution to his Majesty [George III], and that his Majesty was graciously pleased to honour it with His Patronage and to allow it to be called the Royal Institution’. The actual charter was presented on January 13 in 1800. *Royal Institute web page
1803 An open letter to the public, and the Congress of the United States on the topic "Of The Construction of Iron Bridges" is posted by Thomas Paine. Paine had discussed this work with President Jefferson in a letter while he was in England. *The National Intelligencer and Washington Advertiser, (Washington, DC) Wednesday, June 29, 1803; Issue CCCCXIX;
The letter is on line here
. Applying principles advocated by Paine, the designers of the first iron arch bridge in the United States created a structure that is still in service. Historic American Engineering Record, National Park Service, delineated by Christopher H. Marston, 1992. Library of Congress.
1877 After proving that the points in a square can be put in one-to-one correspondence with the points on a line segment Cantor wrote his friend Dedekind “Je le vois, mais je ne le crois pas.” (I see it, but I don’t believe it.) [Dauben, Georg Cantor, p. 55]*VFR
1927 Gellivara 1073: Minor planet discovered September 14, 1923 by Johann Palisa at Vienna. Named for the small town Gällivare in Swedish Lapland where in the year 1927 astronomers from several countries observed the Total Solar Eclipse of 1927 Named by the astronomer J. Rheden and endorsed by Anna Palisa.*NSEC
A Poster advertising viewing of Solar Eclipse from London, Midland, and Scotland Railway *GreatAmericanEclipse @AmericanEclipse
In 1954, the Atomic Energy Commission, by a vote of 4 to 1 decided against reinstating Dr. J. Robert Oppenheimer's access to classified information. The Atomic Energy Act of 1946 required consideration of "the character, associations, and loyalty" of the individuals engaged in the work of the Commission. Substantial defects of character and imprudent and dangerous associations, particularly with known subversives who place the interests of foreign powers above those of the United States, were considered reasons for disqualification. The Commission regarded his associations with persons known to him to be Communists exceeded tolerable limits of prudence and self-restraint, and lasted too long to be justified as merely the intermittent and accidental revival of earlier friendships.*TIS
1956 The interstate highway system was signed into law by President Eisenhower. Even (odd) numbered roads run East–West (North–South) with the numbers increasing from South to North (West to East). Roads with three digit numbers are loops around cities (when the first digit is even) or spurs (first digit odd); In either case the last two digits are the main road number. *VFR
Eisenhower had seen the speed and efficiency in moving troops and equipment on the four-lane autobahns in Germany during WW II. The idea of federal support of interstate limited-access routes in the U.S. had begun with a study under the Federal-Aid Highway Act of 1938. Little progress was made on building these roads while federal funding was low. When the Federal-Aid Highway Act of 1956 committed federal funds to the States for 90% of the cost, construction began in earnest for the System of Interstate and Defense Highways having at least four lanes with no at-grade railroad crossings. *TIS
In the summer of 1919, a young Lieutenant Colonel named Dwight D. Eisenhower participated in the first Army transcontinental motor convoy. The expedition consisted of 81 motorized Army vehicles that crossed the United States from Washington, DC, to San Francisco, a venture covering a distance of 3,251 miles in 62 days. The expedition was manned by 24 officers and 258 enlisted men. The convoy was to test the mobility of the military during wartime conditions. As an observer for the War Department, Lt. Col. Eisenhower learned first-hand of the difficulties faced in traveling great distances on roads that were impassable and resulted in frequent breakdowns of the military vehicles. These early experiences influenced his later decisions concerning the building of the interstate highway system during his presidential administration. *Eisenhower Library
2023 - My Jeannie is celebrating her birthday today, and I'm celebrating having her in my life... all the good I ever do is a reflection of a single sun.
BIRTHS
1716 Joseph Stepling, (29 June 1716 in Regensburg; 11 July 1778 in Prague) His fields included astronomy, physics and mathematics. At the age of 17 he documented with great accuracy the 1733 lunar eclipse. Later Euler was among his long list of correspondents. He transposed Aristotelian logic into formulas, thus becoming an early precursor of modern logic. already adopted the atomistic conception of matter he radically refused to accept Aristotelian metaphysics and natural philosophy. In 1748, at the request of the Berlin Academy, he carried out an exact observation of a solar and lunar eclipse in order to determine the precise location of Prague. During Stepling's long tenure at Prague, he set up a laboratory for experimental physics and in 1751 built an observatory, the instruments and fittings of which he brought up to the latest scientific standard.
Even though he passed up a professorship in philosophy in favor of a chair in mathematics, Empress Maria Theresa appointed him director of the faculty of philosophy at Prague as part of the reform of higher education. He was very interested in cultivating the exact sciences and founded a society for the study of science modeled on the Royal Society of London. In their monthly sessions. over which he presided until his death, the group carried out research work and investigations in the field of pure mathematics and its appiication to physics and astronomy. A great number of treatises of this academy were published.
Stepling corresponded with the outstanding contemporary mathematicians and astronomers: Christian Wolf. Leonhard Euler. Christopher Maire, Nicolas-Louis de Lacaille, Maximilian Heli, Joseph Franz, Rudjer Boskovic, Heinrich Hiss, and others. Also, Stepling was particularly successful in educating many outstanding scientists, including Johann Wendlingen, Jakob Heinisch, Johannes von Herberstein, Kaspar Sagner, Stephan Schmidt, Johann Korber, and Joseph Bergmann. After his death, Maria Theresia ordered a monument erected in the library of the University of Prague *Joseph MacDonnell, Fairfield Univ web page
1818 Pietro Angelo Secchi (29 June 1818 – 26 February 1878) Italian Jesuit priest and astrophysicist, who made the first survey of the spectra of over 4000 stars and suggested that stars be classified according to their spectral type. He studied the planets, especially Jupiter, which he discovered was composed of gasses. Secchi studied the dark lines which join the two hemispheres of Mars; he called them canals as if they where the works of living beings. (These studies were later continued by Schiaparelli.) Beyond astronomy, his interests ranged from archaeology to geodesy, from geophysics to meteorology. He also invented a meteorograph, an automated device for recording barometric pressure, temperature, wind direction and velocity, and rainfall. *TIS
1868 George Ellery Hale (June 29, 1868 – February 21, 1938) born. American astronomer known for his development of important astronomical instruments. To expand solar observations and promote astrophysical studies he founded Mt. Wilson Observatory (Dec 1904). He discovered that sunspots were regions of relatively low temperatures and high magnetic fields. Hale hired Harlow Shapley and Edwin Hubble as soon as they finished their doctorates, and he encouraged research in galactic and extragalactic astronomy as well as solar and stellar astrophysics. Hale planned and tirelessly raised funds for the 200" reflecting telescope at the Palomar Mountain Observatory completed in 1948, after his death, and named for him - the Hale telescope.*TIS
1893 Prasanta Chandra Mahalanobis FRS (29 June 1893 – 28 June 1972) was an Indian scientist and applied statistician. He is best remembered for the Mahalanobis distance, (a statistical measure of the distance between a point P and a distribution D, - a multi-dimensional generalization of the idea of measuring how many standard deviations away P is from the mean of D. ) and for being one of the members of the first Planning commission of free india. He made pioneering studies in anthropometry in India. He founded the Indian Statistical Institute, and contributed to the design of large-scale sample surveys *Wik
1893 Eduard Cech, (June 29, 1893 – March 15, 1960) Czech topologist born in Stračov, Bohemia (then Austria-Hungary, now Czech Republic). His research interests included projective differential geometry and topology. In 1921–1922 he collaborated with Guido Fubini in Turin. He died in Prague. *Wik
1904 Witold Hurewicz (June 29, 1904 - September 6, 1956) born. Hurewicz is best remembered for two remarkable contributions to mathematics, his discovery of the higher homotopy groups in 1935-36, and his discovery of exact sequences in 1941. His work led to homological algebra. It was during Hurewicz's time as Brouwer's assistant in Amsterdam that he did the work on the higher homotopy groups; "...the idea was not new, but until Hurewicz nobody had pursued it as it should have been. Investigators did not expect much new information from groups, which were obviously commutative...". *Wik He died in 1956 when he fell off a pyramid while attending a conference in Mexico.
1942 K. Jon Barwise (June 29, 1942 – March 5, 2000) an American mathematician, philosopher and logician who proposed some fundamental revisions to the way that logic is understood and used.*Wik
1972 Edray Herber Goins (born June 29, 1972, Los Angeles) is an American mathematician. He specializes in number theory and algebraic geometry. His interests include Selmer groups for elliptic curves using class groups of number fields, Belyi maps and Dessin d'enfants.
Goins was born in Los Angeles in 1972. His mother, Eddi Beatrice Brown, was a teacher. He attended public schools in South Los Angeles and got his BSc in mathematics and physics in 1994 from California Institute of Technology, where he also received two prizes for mathematics. He completed his PhD in 1999 on “Elliptic Curves and Icosahedral Galois Representations” from Stanford University, under Daniel Bump and Karl Rubin.
He served for many years on the faculty of Purdue University.[8] He has also served as visiting scholar at both the Institute for Advanced Study in Princeton, and Harvard.[6][9] Goins took a position at Pomona College in 2018.
His summers have focused on engaging underrepresented students in research in the mathematical sciences. He currently runs the NSF-funded Research Experience for Undergraduates (REU) "Pomona Research in Mathematics Experience (PRiME)", a program that Goins started in 2016 at Purdue University under the title "Purdue Research in Mathematics Experience (PRiME)". He is noted for his 2018 essay, "Three Questions: The Journey of One Black Mathematician". He was elected to the 2019 Class of Fellows of the Association for Women in Mathematics.
From 2015 to 2020, Goins served as president of the National Association of Mathematicians (NAM).
1979 Artur Avila Cordeiro de Melo (born 29 June 1979) is a Brazilian mathematician working primarily in the fields of dynamical systems and spectral theory. He is one of the winners of the 2014 Fields Medal, being the first Latin American and lusophone to win such award. He has been a researcher at both the IMPA and the CNRS (working a half-year in each one). He has been a professor at the University of Zurich since September 2018.
At the age of 16, Avila won a gold medal at the 1995 International Mathematical Olympiad and received a scholarship for the Instituto Nacional de Matemática Pura e Aplicada (IMPA) to start a M.S. degree while still attending high school in Colégio de São Bento and Colégio Santo Agostinho in Rio de Janeiro. He completed his M.S. degree in 1997. Later he enrolled in the Federal University of Rio de Janeiro (UFRJ), earning his B.S in mathematics.
At the age of 19, Avila began writing his doctoral thesis on the theory of dynamical systems. In 2001 he finished it and received his PhD from IMPA.
Much of Artur Avila's work has been in the field of dynamical systems. In March 2005, at age 26, Avila and Svetlana Jitomirskaya proved the "conjecture of the ten martinis," a problem proposed by the American mathematical physicist Barry Simon. Mark Kac promised a reward of ten martinis to whoever solved the problem: whether or not the spectrum of a particular type of operator is a Cantor set, given certain conditions on its parameters. The problem had been unsolved for 25 years when Avila and Jitomirskaya answered it affirmatively. Later that year, Avila and Marcelo Viana proved the Zorich–Kontsevich conjecture that the non-trivial Lyapunov exponents of the Teichmüller flow on the moduli space of Abelian differentials on compact Riemann surfaces are all distinct.
DEATHS
1895 T(homas) H(enry) Huxley (4 May 1825 – 29 June 1895) was an English biologist , known as "Darwin's Bulldog" for his promotion of Darwinism which led him to an advocacy of agnosticism (a word he coined). At the age of 12 he was reading advanced works on geology, and by early adolescence he recorded the results of simple self-conducted experiments. As a ship's assistant surgeon on HMS Rattlesnake he studied marine specimens by microscope. During the 1850's he published papers on animal individuality, the cephalous mollusks (ex. squids), the methods of paleontology, and the methods and principles of science and science education. *TIS
1924 Robert Simpson Woodward (July 21, 1849–June 29, 1924) was an American physicist and mathematician, born at Rochester, Michigan. He graduated C.E. at the University of Michigan in 1872 and was appointed assistant engineer on the United States Lake Survey. In 1882 he became assistant astronomer for the United States Transit of Venus Commission. In 1884 he became astronomer to the United States Geological Survey, serving until 1890, when he became assistant in the United States Coast and Geodetic Survey. In 1893 he was called to Columbia as professor of mechanics and subsequently became professor of mathematical physics as well. He was dean of the faculty of pure science at Columbia from 1895 to 1905, when he became president of the Carnegie Institution of Washington, whose reputation and usefulness as a means of furthering scientific research was widely extended under his direction. He was elected to the National Academy of Sciences in 1896. In 1898-1900 he was president of the American Mathematical Society, and in 1900 president of the American Association for the Advancement of Science. In 1915 he was appointed to the Naval Consulting Board. He died in 1924 in Washington, D.C.*Wik
An illustration of the transit of Venus of 1882. Ceiling mural in the Paris Observatory. *Wik
1966 Damodar Dharmananda Kosambi (31 July 1907 – 29 June 1966) was an Indian polymath with interests in mathematics, statistics, philology, history, and genetics. He contributed to genetics by introducing the Kosambi map function. In statistics, he was the first person to develop orthogonal infinite series expressions for stochastic processes via the Kosambi–Karhunen–Loève theorem. He is also well known for his work in numismatics and for compiling critical editions of ancient Sanskrit texts. His father, Dharmananda Damodar Kosambi, had studied ancient Indian texts with a particular emphasis on Buddhism and its literature in the Pali language. Damodar Kosambi emulated him by developing a keen interest in his country's ancient history. He was also a Marxist historian specialising in ancient India who employed the historical materialist approach in his work. He is particularly known for his classic work An Introduction to the Study of Indian History. *Wik
2013 Margherita Hack, Knight Grand Cross OMRI ( 12 June 1922 – 29 June 2013) was an Italian astrophysicist and scientific discriminator. The asteroid 8558 Hack, discovered in 1995, was named in her honor.
An athlete in her youth, Hack played basketball and competed in track and field during the National University Contests, called the Littoriali under Mussolini's fascist regime, where she won the long jump and the high jump events.
She was full professor of astronomy at the University of Trieste from 1964 to the 1st of November 1992, when Hack was placed "out of role" for seniority. She has been the first Italian woman to administrate the Trieste Astronomical Observatory from 1964 to 1987, bringing it to international fame.
Member of the most physics and astronomy associations, Margherita Hack was also director of the Astronomy Department at the University of Trieste from 1985 to 1991 and from 1994 to 1997. She was a member of the Accademia Nazionale dei Lincei (national member in the class of mathematical physics and natural sciences; second category: astronomy, geodesic, geophysics and applications; section A: astronomy and applications). She worked at many American and European observatories and was for long time member of working groups of ESA and NASA. In Italy, with an intensive promotion work, she obtained the growth of activity of the astronomical community with access to several satellites, reaching a notoriety of international level.
Hack has published several original papers in international journals and several books both of popular science and university level. In 1994 she was awarded with the Targa Giuseppe Piazzi for the scientific research, and in 1995 with the Cortina Ulisse Prize for scientific dissemination.
In 1978, Margherita Hack founded the bimonthly magazine L'Astronomia, whose first issue came out in November 1979;[20] later, together with Corrado Lamberti, she directed the magazine of popular science and astronomy culture Le Stelle.
Credits
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts
*NSEC= NASA Solar Eclipse Calendar
*RMAT= The Renaissance Mathematicus, Thony Christie
*SAU=St Andrews Univ. Math History
*TIA = Today in Astronomy
*TIS= Today in Science History
*VFR = V Frederick Rickey, USMA
*Wik = Wikipedia
*WM = Women of Mathematics, Grinstein & Campbell | 5,074 | 22,047 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.887372 |
http://personal.bgsu.edu/~carother/pi/geometry.html | 1,516,757,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892892.86/warc/CC-MAIN-20180124010853-20180124030853-00073.warc.gz | 263,484,344 | 1,092 | # Geometry Review
We have used several simple facts here:
1. A triangle inscribed in a semicircle, as shown below, is a right triangle.
2. Given triangle inscribed in semicircle , as shown below, the central angle is twice the angle .
[Main Index] | [Pi Index]
Neal Carothers - carother@bgnet.bgsu.edu | 79 | 306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.883676 |
http://woodworkingplans2013.com/tedswoodworking-review-buy-woodworking-plans.html | 1,550,398,281,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481832.13/warc/CC-MAIN-20190217091542-20190217113542-00294.warc.gz | 297,880,755 | 8,393 | With the advances in modern technology and the demands of industry, woodwork as a field has changed. The development of Computer Numeric Controlled (CNC) Machines, for example, has made us able to mass-produce and reproduce products faster, with less waste, and often more complex in design than ever before. CNC Routers can carve complicated and highly detailed shapes into flat stock, to create signs or art. Rechargeable power tools speed up creation of many projects and require much less body strength than in the past, for example when boring multiple holes. Skilled fine woodworking, however, remains a craft pursued by many. There remains demand for hand crafted work such as furniture and arts, however with rate and cost of production, the cost for consumers is much higher.
There are a few things particleboard is NOT. It’s not medium density fiberboard (MDF)—a material with greater density and weight composed of more uniform particles. It’s NOT oriented strand board (OSB), a material composed of large wood chips and strands that’s structurally equivalent to plywood. It does NOT have great nail or screw holding ability, nor is it all that water resistant; water can quickly cause the material to swell and lose structural integrity. But if you need something flat and cheap for use in a dry place, particleboard will do you proud. Learn how to make a plastic laminate tabletop with a particleboard substrate.
This time I surprised one of my favorite dice games and took it outside. I made a set of wooden dice in just a few hours, and instead of sitting in the room and doing nothing, we are taking our dice game out into the yard. With this set of wooden dice, dice games are becoming our favorite backyard game. Check out the step by step tutorial below so you can make your own.
How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
Working on one side at a time, glue and nail the side to the back. Apply glue and drive three 1-5/8-in. nails into each shelf, attach the other side and nail those shelves into place to secure them. Clamps are helpful to hold the unit together while you’re driving nails. Center the top piece, leaving a 2-in. overhang on both sides, and glue and nail it into place. Paint or stain the unit and then drill pilot holes into the top face of each side of the unit and screw in the hooks to hold your ironing board. Mount the shelf on drywall using screw-in wall anchors.
## When it comes to woodworking for beginners, I think it’s important to just learn how to use a few of the most essential woodworking tools for beginners. There are so many awesome tools available on the market today, it can be quite overwhelming as well as expensive to try to buy them all and know how to use them. Once you learn the basics of the most essential tools you will be able to start building in no time and feel comfortable learning any other new tools in the future.
Finding a toolbox for a mechanic, for his hand tools, is not a big challenge at all - there are dozens of the tool boxes available on the market, from huge roll-around shop cases to small metal boxes. Plumbers, electricians, and farmers are well served, too, with everything from pickup-truck storage to toolboxes and belts. But, if you are a shop-bound woodworker then the case changes. You get to need a tool box that suits the range and variety of hand tools that most woodworkers like to have. For those who deny making do with second best, there's only one solution, you’ve to build a wooden toolbox that should be designed expressly for a woodworking shop.
Build this handy stool in one hour and park it in your closet. You can also use it as a step to reach the high shelf. All you need is a 4 x 4-ft. sheet of 3/4-in. plywood, wood glue and a handful of 8d finish nails. Cut the plywood pieces according to the illustration. Spread wood glue on the joints, then nail them together with 8d finish nails. First nail through the sides into the back. Then nail through the top into the sides and back. Finally, mark the location of the two shelves and nail through the sides into the shelves. Don’t have floor space to spare? Build these super simple wall-mounted shoe organizers instead!
This is probably one the easiest woodworking projects you will find here. Although easy, a doormat is an equally important and useful item for households. As you can see in the image below, you will only need some 2X2 wooden boards and rope to build a simple doormat. This doormat is mostly useful for outdoor and porch. It will easily remove all the mud from your shoes with just one wipe. It is also very easy to clean and looks fabulous even if it is dirty.
Sanding small items is tricky, as they’re hard to clamp in a vise to work on them. So instead of bringing the sandpaper to the workpiece, I bring the workpiece to the sandpaper. I glue sheets of sandpaper to a piece of plywood; 60 and 100-grit on one side and 150 and 220-grit on the other. Spray adhesive works well for this. Since there’s sandpaper on both sides, my sanding board doesn’t slide around on the bench. Check out these small projects!
By video tutorial, you will get step by step process instructions of making a nice wooden folding sling chair from scratch. However, my first wooden chair was not the best one, but it was good enough to motivate me to make some more folding chairs like this one. If I can make this, you too can make one yourself. You can browse the internet for more folding sling chairs ideas and start making one now.
“I do a lot of finish-sanding freehand, without a sandpaper block, so I can smooth edges and get into nooks and crannies. But the finer grits are usually bonded to thinner paper and, at least for me, the paper is too thin and ends up tearing long before the grit wears out. So I apply duct tape to the back of the sandpaper. The sandpaper is still flexible enough to sand a tight radius and it’s far more durable. You can use this super-strong sandpaper like a shoeshine rag.” — Chuck Merchant
This rack can be built from old unused wood pallets you can find around the house. So it is also a great way to recycle those old pallets. You can also find a step by step tutorial at instructables.com for which I have included the source link below. This tutorial helps you to make a wood wine rack from the scratch. So what are waiting for? Just grab the items you need and start building a cool wooden rack for those nice wine bottles of yours.
Have you got an old whiskey barrel at home that you haven’t used for ages? If yes, this project is for you. You can make a really beautiful coffee table from that old whiskey barrel in a few easy steps. Apart from a coffee table, whiskey barrels can also be used to build several other furniture items. But that is a talk for later. Here, we will discuss how to make a coffee table from a whiskey barrel.
```I know this seems really simple and you may already know how to read a tape measure so just hear me out on this one! Often times with woodworking you need to make exact measurements and cuts and it’s rarely pretty even numbers like 15 inches or 15 1/2 inches. It’s usually like 15 5/8 inches or 15 9/16 inches. So, really knowing how to read a tape measure in its entirety is important. And I’ve created a quick post on how to read a tape measure the easy way along with a helpful free printable.
```
```Commonly used woodworking tools included axes, adzes, chisels, pull saws, and bow drills. Mortise and tenon joints are attested from the earliest Predynastic period. These joints were strengthened using pegs, dowels and leather or cord lashings. Animal glue came to be used only in the New Kingdom period.[3] Ancient Egyptians invented the art of veneering and used varnishes for finishing, though the composition of these varnishes is unknown. Although different native acacias were used, as was the wood from the local sycamore and tamarisk trees, deforestation in the Nile valley resulted in the need for the importation of wood, notably cedar, but also Aleppo pine, boxwood and oak, starting from the Second Dynasty.[4]
```
This was not actually a tutorial post to the woodworking plan ideas but the aim of the post was to give some easy and free woodworking ideas to the readers. If you have some time to entertain yourself and also willing to add some new stuff to your furniture you can take any idea from the list and start working on it. Be sure to see both post tutorial and video tutorials of the plan you have selected, it will make you understand everything nicely.
When cutting full sheets with my circular saw, I use plastic shelving units as sawhorses. The height is just right and by using three of them, I can make cuts in any direction and the plywood is fully supported. And because the shelving units are made of plastic, I can cut right into them without worrying that they’ll damage my saw blade. Plastic shelves are available for \$20 at home centers. — John Tinger. Check out these tips for making long cuts with a circular saw.
```Finally, as I shared in my post last week on how I learned woodworking, I learned how to use power tools by watching YouTube videos and then just trying them out for myself. I highly recommend this method to learn how to use your power tools. There are lots of videos on specific models of tools too. So, watch a few how-to videos and very importantly, review the tool manual and safety guide for your own tools. Then, go ahead and try the tool out yourself and start using it!
```
Sanding curves is tricky. Sometimes you need a sanding pad that’s both firm and flexible. A small notepad works great. Just wrap sandpaper around the pad and bend the pad to whatever arc you need. Slip the one end of the sandpaper between the pages to help hold it in place on the pad. Give this a try the next time you’re working on a project that has curves and tough to reach spots.
It is one of the easiest woodwork projects we are going to discuss today. Although it looks very easy to make, I still could not find any good tutorial on the internet that explains how to build this one. So I am here sharing an article link that gets the closest. The article explains how to make different kinds of DIY candle holders and what items you may need for the project.
Sanding concave molding doesn’t have to be difficult. Find a deep socket that fits the contour of your molding. Wrap a piece of sand- paper around the socket and hold it in place with your fingers. Your sanding will be uniform and the delicate edges of the molding won’t round over. — Eric and Cheryl Weltlich. In this video, Travis talks about his favorite sanding tips.
Harbor Freight carries a wide range of high quality woodworking tools that are perfect for the home garage and professional carpentry shop alike. If you have a love of woodcrafting, Harbor Freight has the essential tools and accessories you need to tackle woodworking projects efficiently and safely, no matter your skill level. Our huge selection of drill presses, saws, sanders and planers will see you through countless jobs while our dust collection options keep your workspace clear of sawdust. Whatever you’re looking for, from basic woodworking tools to more specialty items, Harbor Freight has what you need to get the job done.
The source above is not exactly a tutorial, but it gives you a basic idea of how the author built a Quirky Pallet Art to enhance the look of their old house. You can also find another tutorial at the link below. It shares a step by step procedure for making a wooden pallet sign. The final product is not exactly the same as the one above, but the basic idea is the same. | 2,623 | 12,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-09 | latest | en | 0.958648 |
https://www.stata.com/statalist/archive/2010-07/msg00227.html | 1,656,695,068,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00068.warc.gz | 1,064,247,543 | 4,984 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: RE: RE: estimation with a time trend.
From natasha agarwal To statalist@hsphsun2.harvard.edu Subject Re: st: RE: RE: estimation with a time trend. Date Mon, 5 Jul 2010 17:03:13 +0100
```On Mon, Jul 5, 2010 at 4:49 PM, Maarten buis <maartenbuis@yahoo.co.uk> wrote:
> --- On Mon, 5/7/10, natasha agarwal wrote:
>> I wanted to introduce a time trend in my estimation where
>> my years go from 2001-2005.
>>
>> With time trend I mean:
>>
>> panel identifier year time trend
>> 1 2001 1
>> 1 2002 2
>> 1 2003 3
>> 1 2004 4
>> 1 2005 5
>> 2 2001 1
>> 2 2002 2
>>
>> The only way I knew how to do it was to use the egen
>> -group- command.That is why I wanted to do the egen
>> -group- command.
>
> I am still confused: What is "time trend" supposed to
> measure that cannot be measured with "year"?
The results with the time dummies showed that the estimated
coefficients on the time dummies was increasing over time. I am
pasting the results of the same.
Fixed-effects (within) regression Number of obs = 46959
Group variable: number Number of groups = 12911
R-sq: within = 0.2300 Obs per group: min = 1
between = 0.4727 avg = 3.6
overall = 0.4656 max = 5
F(7,12910) = 626.65
corr(u_i, Xb) = 0.1507 Prob > F = 0.0000
(Std. Err. adjusted for 12911 clusters in number)
Robust
lnrval Coef. Std. Err. t P>t [95% Conf. Interval]
lnk .2036338 .0107336 18.97 0.000 .1825942 .2246733
lnw .5677307 .0168696 33.65 0.000 .5346637 .6007977
lnvfdi .0217376 .0198568 1.09 0.274 -.0171848 .0606599
y14 -.3255905 .0171191 -19.02 0.000 -.3591466 -.2920345
y15 -.2313728 .0130272 -17.76 0.000 -.256908 -.2058375
y16 -.1464979 .0097877 -14.97 0.000 -.1656832 -.1273126
y17 -.1293191 .0081628 -15.84 0.000 -.1453195 -.1133188
y18 (dropped)
_cons .3633778 .2559695 1.42 0.156 -.1383602 .8651158
sigma_u .95102062
sigma_e .55962159
rho .74279561 (fraction of variance due to u_i)
So I decided to introduce one a time variable like a trend which is
shown as before.
Given your
> earlier remarks you did not think that "the 0 outside the
> range of your data" was a problem in your fixed effects
> model. So what problem did you want to solve by creating
> that "time trend" variable?
>
> Mind you, I don't think it is a bad idea to ensure that
> 0 happens at a meaningful point in time, but it does not
> seem to be the problem that you wanted to solve. We need
> to make sure we understand what the problem is you want
> to solve before we can help.
>
> Again, there is no need to use -egen group()-, and it is
> dangerous as your year variable may be unequally spaced
> (even if you think that it is not, real data has the
> nasty property of always deviating from what you think
> should be true in your data...). You could easily check
> it before hand, but there is an easier and much saver
> solution: just subtract a constant:
>
> *--------- begin example --------
> drop _all
> input id year
> 1 2001
> 1 2002
> 1 2003
> 1 2004
> 1 2005
> 2 2001
> 2 2002
> end
> gen trend = year - 2000
> list
> *--------- end example ----------
> (For more on examples I sent to the Statalist see:
> http://www.maartenbuis.nl/example_faq )
>
> To repeat: do not use -egen group()- for this purpose,
> it is dangerous. It will work in this example, but it
> can just too easily backfire in real data.
>
> Hope this helps,
> Maarten
>
> --------------------------
> Maarten L. Buis
> Institut fuer Soziologie
> Universitaet Tuebingen
> Wilhelmstrasse 36
> 72074 Tuebingen
> Germany
>
> http://www.maartenbuis.nl
> --------------------------
>
>
>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 1,461 | 4,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.726678 |
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United States (204,421)
CIS 1166 (4)
Chapter
# QuantLogic2.pdf
6 Pages
59 Views
School
Temple University
Department
Computer & Information Science
Course
CIS 1166
Professor
David Shulman
Semester
Spring
Description
Quanti▯cational Logic Part 2 1 Nested Quanti▯ers Part 2 of this note on quanti▯cational logic discusses nested quanti▯ers. Here we might be quantifying over a statement that itself contains quanti▯ers. In terms of computer implementation, we have a loop within a loop. There is nothing essentially new with nested quanti▯ers. We have covered the basic principles, but it helps to show more examples of how quanti▯ers work and there are also special abbreviations that are often used when we have nesting and that are helpful to know. 2 Nesting the Same Quanti▯er You might nest several universal quanti▯ers to express some mathematical laws such as 8x8y(x + y = y + x) 8x8y8z((x + y) + z = x + (y + z)) 8x8y8z(x < y ^ y < z ! x < z) Here we might let our domain for all variables be the reals and we are expressing commutative and associative laws of addition and transitivity of less than. If S means SPEAKS as in part 1 of this note but this time our domain for the ▯rst input is all people in our class and for the second input is all Romance languages spoken by at least ▯ve million people, then we might write 8x8yS(x;y) to means that everyone in our class speaks all the languages we care about (Spanish, Italian, French, Romanian, Catalan etc.) One point ot note is that often when we write mathematical or other laws we do not explicitly write down all our quanti▯ers. We might write the formula x + y = y + x and mean that the formula holds for all (relevant) x 1 and y. Thus there is a convention that variables that are not quanti▯ed over by some explicitly written quanti▯er and that do not get speci▯c values by other means are assumed to be quanti▯ed over by a universal quanti▯er. We also have use for nested existential quanti▯cation. If we want to say that someone speaks some lang2age,2we m2ght write 9x9yS(x;y). We would write 9x9y9z(x + y = z ) to say that there are x, y, z in our chosen domain that provide a solution for the equation x + y = z . 2 n If we want to state Fermat’s last theorem, we would say :(9x9y9z9n(x + y = z ^n > 2)). Here our domain of interest for all variables might as well be the integers. If we want to say that our chosen domain has at least two elements we could write 9x9y(x 6= y). 2.1 Commutativity If you are dealing with the same kind of quanti▯er, it does not matter how you order the quanti▯ers. 8x8yP ▯ 8y8xP. 9x9yP ▯ 9y9xP. (We shall see below that order de▯nitely does matter when the quanti▯ers are di▯erent in kind.) 2.2 An Abbreviatory Convention Since ordering of the same kind of quanti▯er does not a▯ect truth value, there is a shortcut that is often used. Instead of writing 8x8yP, we might write 8s;yP Instead of writing 9x9yP, we might write 9x;yP. There is a similar abbreviation if we had three rather than two quanti▯ers so we might write 8x;y;z(x < y^y < z ! x < z). DO NOT USE THIS ABBREVIATION if on a quiz or midterm I tell you only to use basic quanti▯ers and do not use this abbreviation on the ▯nal unless you are speci▯cally told that it is o.k. to use this abbreivation. There is nothing wrong with the abbreviation; it is just that we might be testing whether you understand what is really happening at the most fundamental level. When I write something like 9x;yP, I am really saying that there is a pair x, y that makes P true and when I write something like 8x;yP, I am really saying that all pairs x, y are such that P is true. 2 3 Heterogeneous Nested Quanti▯cation Sometimes we need both forall and there exists. Let my domain be real numbers for all variables, then we have 9x8y(x + y = y), there is an x such that for all y, x + y = y. This statement is true (x = 0). Notice here we are using the same x for all y. We also have with the same domain, the real numbers, 8x9y(x + y = 0). Here y is just the negative of x and we have to use a di▯erent y for each x. So there is a di▯erence in meaning depending on which quanti▯er is ▯rst. Another examples using our predicate S whose ▯rst input is a person and whose second input is a language: If I write 8x9y(S(x;y)) that says everyone speaks some language (but di▯erent people might not be able to communicate because they have no language in common). 9y8x(S(x;y)) says there is a universal language; there is a language y that everyone speaks so everyone can communicate with everyone. If I say 8y9x(S(x;y)), I am saying that every language is spoken by someone; there are no languages without speakers. If I say 9x8y(S(x;y)) I am saying there is someone who speaks every language. Because the 9x comes before the 8y I have to use the same x for all y and thus I am saying that there is some one (the same person) who knows every single language; there is no language this person does not know. We see from these examples that with di▯erent kinds of quanti▯ers order very much matters. Another example with di▯erent quanti▯ers: To say that every two people have a language in common you would say 8x8y9z(S(x;z)^S(y;z)) but here the language z that x and y have in common could be di▯erent for di▯erent pairs x, y. 4 Nested QUanti▯ers and Negation I shall illustrate how one might apply De Morgan rules when there are many quanti▯ers to rewrite expressions. :8x9y8z:(:S(x;y) _ :S(y;z))
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So we can recommend you notes for your school. | 1,617 | 5,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-13 | latest | en | 0.916745 |
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## Presentation on theme: "How Leopards get Their Spots Will Brennan How Zebras Get Their Stripes."— Presentation transcript:
How Leopards get Their Spots Will Brennan How Zebras Get Their Stripes
Spots and Stripes Skin coloration is caused by melanin What is the cause of specific patterns such as spots and stripes?
Modeling animals Spot and stripe arrangement are random and distinct, yet share a definite pattern… This led mathematicians to realize that there must be some way to model the phenomena
Models Emerge Two models emerge: –One was developed by James Murray, and the other was developed by David Young Both models both incorporate the same elements however: reaction- diffusion
Young Model Young’s model is based off of cellular automata modeling Young has 4 assumptions on his model: –There are two types of cells; colored(D) and uncolored(U) –The colored cells secrete 2 morphagens; an inhibitor(I) and an activator(A)
Diffusion of Morphagens As the D cells release the A and I morphagens, they diffuse throughout the environment Near the D cells, there is a higher concentration of A, but this is inverse with its distance from D
What’s it all mean? Cell type is determined by the concentration of the morphagens in its area If over a U cell, A>I then the cell will switch to a D and start producing morphagens Conversely, if over a D cell, I>A, then it will change to a U
How does this explain the patterns? if AD - w*ID > 0 set the central cell to D, if AD - w*ID < 0 set the central cell to U if AD - w*ID = 0 leave the central cell unchanged.
Model in Action Young's model Young's model The shape of the patterns can then be changed by inputting different variables for the area of the model
Murray’s model Also based on reaction-diffusion, but concentrated on the rate of diffusion as opposed to concentration Also had two chemicals; an inhibitor and an activator working on the cells
Speed is of the essence The chemicals work at different speeds; –The activator is slower –While the inhibitor is faster This disparity in speed allows the inhibitor to surround the activator during diffusion, causing a spot.
An Analogy Forest Fires mimic this same dynamic Fires burn first, but diffuse slowly. Firefighters respond, spraying untouched trees surrounding the fire with anti- inflammatory chemicals, containing the fire in Spots
Theory in action By changing the rate of diffusion then, the pattern will be different. Murray also found as Young did that shape plays a role in the development of the pattern
Spots vs Stripes Since spotted leopards and striped tigers are about the same adult size, he concluded that it must happen during development i.e. the zebra resembles a long thin pencil like shape during development, resulting in its stripes
Harmony Although Murray and Young used slightly different methods to model the formation of spots and stripes, they both agreed that it can be solved through mathematical modeling
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```Fundamentals of
Electric Circuits
Chapter 14
the prior written consent of McGraw-Hill Education.
Overview
• This chapter will introduce the idea of the
transfer function: a means of describing the
relationship between the input and output of
a circuit.
• Bode plots and their utility in describing the
frequency response of a circuit will also be
introduced.
• The concept of resonance as applied to LRC
circuits will be covered as well
• Finally, frequency filters will be discussed.
2
Frequency Response
• Frequency response is the variation in a
circuit’s behavior with change in signal
frequency.
• This is significant for applications involving
filters.
• Filters play critical roles in blocking or
passing specific frequencies or ranges of
frequencies.
• Without them, it would be impossible to have
multiple channels of data in radio
communications.
3
Transfer Function
• One useful way to analyze the
frequency response of a
circuit is the concept of the
transfer function H(ω).
• It is the frequency dependent
ratio of a forced function Y(ω)
to the forcing function X(ω).
H
Y
X
4
Transfer Function
• There are four possible input/output
combinations:
H Voltage gain
H Current gain
Vo
Vi
I o
I i
H Transfer impedance
H Transfer admittance
Vo
I i
I o
Vi
5
Zeros and Poles
• To obtain H(ω), we first convert to frequency
domain equivalent components in the circuit.
• H(ω) can be expressed as the ratio of
numerator N(ω) and denominator D(ω)
polynomials.
N
H
D
• Zeros are where the transfer function goes to
zero.
• Poles are where it goes to infinity.
• They can be related to the roots of N(ω) and
D(ω)
6
Decibel Scale
• We will soon discuss Bode plots.
• These plots are based on logarithmic scales.
• The transfer function can be seen as an
expression of gain.
• Gain expressed in log form is typically
expressed in bels, or more commonly
decibels (1/10 of a bel)
GdB 10 log10
P2
P1
7
Bode Plots
• One problem with the transfer function is
that it needs to cover a large range in
frequency.
• Plotting the frequency response on a
semilog plot (where the x axis is plotted in
log form) makes the task easier.
• These plots are referred to as Bode plots.
• Bode plots either show magnitude (in
decibels) or phase (in degrees) as a function
of frequency.
8
Standard Form
• The transfer function may be written in terms
of factors with real and imaginary parts. For
example:
K j 1 j / z 1 j 2 / j /
1
H
2
1
1
k
1 j / p1 1 j 2 2 / n j / n
k
2
• This standard form may include the following
seven factors in various combinations:
– A gain K
– A pole (jω)-1 or a zero (jω)
– A simple pole 1/(1+jω/p1) or a simple zero
(1+jω/z1)
– A quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] or zero
1/[1+j21ω/ ωn+ (jω/ ωk)2]
9
Bode Plots
• In a bode plot, each of these factors is
graphically.
• Gain, K: the magnitude is 20log10K and the
phase is 0°. Both are constant with
frequency.
• Pole/zero at the origin: For the zero (jω), the
slope in magnitude is 20 dB/decade and the
phase is 90°. For the pole (jω)-1 the slope in
magnitude is -20 dB/decade and the phase is
-90°
10
Bode Plots
• Simple pole/zero: For the simple zero, the
magnitude is 20log10|1+jω/z1| and the phase
is tan-1 ω/z1.
• Where:
H dB 20 log10 1
j
z1
20 log10
as
z1
• This can be approximated as a flat line and
sloped line that intersect at ω=z1.
• This is called the corner or break frequency
11
Bode Plots
• The phase can be plotted as a series straight
lines
• From ω=0 to ω≤z1/10, we let =0
• At ω=z1 we let =45°
• For ω≥10z1, we let = 90°
• The pole is similar, except the corner
frequency is at ω=p1, the magnitude has a
negative slope
12
Bode Plots
• Quadratic pole/zero: The magnitude of the
quadratic pole 1/[1+j22ω/ ωn+ (jω/ ωn)2] is 20log10 [1+j22ω/ ωn+ (jω/ ωn)2]
• This can be approximated as:
H dB
0
as 0
40 log10
as
n
• Thus the magnitude plot will be two lines,
one with slope zero for ω<ωn and the other
with slope -40dB/decade, with ωn as the
corner frequency
13
Bode Plots
• The phase can be expressed as:
0 0
2 2 / n
tan 1
90 n
2
2
1 / n
180
• This will be a straight line with slope of 90°/decade starting at ωn/10 and ending at 10
ωn.
• For the quadratic zero, the plots are inverted.
14
Bode Plots
15
Bode Plots
16
Resonance
• The most prominent feature of the frequency
response of a circuit may be the sharp peak
in the amplitude characteristics.
• Resonance occurs in any system that has a
complex conjugate pair of poles.
• It enables energy storage in the firm of
oscillations
• It allows frequency discrimination.
• It requires at least one capacitor and
inductor.
17
Series Resonance
• A series resonant circuit
consists of an inductor and
capacitor in series.
• Consider the circuit shown.
• Resonance occurs when the
imaginary part of Z is zero.
• The value of ω that satisfies
this is called the resonant
frequency
0
1
LC
18
Series Resonance
• At resonance:
–
–
–
–
The impedance is purely resistive
The voltage Vs and the current I are in phase
The magnitude of the transfer function is minimum.
The inductor and capacitor voltages can be much more than
the source.
• There are two frequencies above and below
resonance where the dissipated power is half the
max:
2
R
1
R
1
2L
2 L LC
2
R
1
R
2
2L
2 L LC
19
Quality Factor
• The “sharpness” of the resonance is
measured quantitatively by the quality factor,
Q.
• It is a measure of the peak energy stored
divided by the energy dissipated in one
period at resonance.
Q
0 L
R
1
0CR
• It is also a measure of the ratio of the
resonant frequency to its bandwidth, B
B
R 0
L Q
20
Parallel Resonance
• The parallel RLC circuit
shown here is the dual of the
series circuit shown
previously.
• Resonance here occurs when
the imaginary part of the
• This results in the same
resonant frequency as in the
series circuit.
21
Series Resonance
• The relevant equations for the parallel
resonant circuit are:
2
1
1
1
1
2 RC
2 RC LC
1
B
RC
2
1
1
1
2
2 RC
2 RC LC
R
Q 0 RC
0 L
22
Passive Filters
• A filter is a circuit that is designed to pass
signals with desired frequencies and reject
or attenuate others.
• A filter is passive if it consists only of
passive elements, R, L, and C.
• They are very important circuits in that many
technological advances would not have been
possible without the development of filters.
23
Passive Filters
• There are four types of filters:
– Lowpass passes only low
frequencies and blocks high
frequencies.
– Highpass does the opposite of
lowpass
– Bandpass only allows a range of
frequencies to pass through.
– Bandstop does the opposite of
bandpass
24
Lowpass Filter
• A typical lowpass filter is formed
when the output of a RC circuit
is taken off the capacitor.
• The half power frequency is:
c
1
RC
• This is also referred to as the
cutoff frequency.
• The filter is designed to pass
from DC up to ωc
25
Highpass Filter
• A highpass filter is also
made of a RC circuit, with
the output taken off the
resistor.
• The cutoff frequency will be
the same as the lowpass
filter.
• The difference being that the
frequencies passed go from
ωc to infinity.
26
Bandpass Filter
• The RLC series resonant
circuit provides a bandpass
filter when the output is taken
off the resistor.
• The center frequency is:
0
1
LC
• The filter will pass frequencies
from ω1 to ω2.
• It can also be made by feeding
the output from a lowpass to a
highpass filter.
27
Bandstop Filter
• A bandstop filter can be
created from a RLC circuit by
taking the output from the LC
series combination.
• The range of blocked
frequencies will be the same
as the range of passed
frequencies for the bandpass
filter.
28
Active Filters
• Passive filters have a few drawbacks.
– They cannot create gain greater than 1.
– They do not work well for frequencies below the
audio range.
– They require inductors, which tend to be bulky
and more expensive than other components.
• It is possible, using op-amps, to create all the
common filters.
• Their ability to isolate input and output also
makes them very desirable.
29
First Order Lowpass
• If the input and feedback
elements in an inverting
amplifier are selectively
replaced with capacitors, the
amplifier can act as a filter.
• If the feedback resistor is
replaced with a parallel RL
element, the amplifier
becomes a lowpass filter.
• The corner frequency will be:
c
1
Rf C f
30
First Order Highpass
• Placing a series RL
combination in place of the
input resistor yields a
highpass filter.
• The corner frequency of the
filter will be:
c
1
Ri Ci
31
Bandpass
• To avoid the use of an inductor, it is possible
to use a cascaded series of lowpass active
filter into a highpass active filter.
• To prevent unwanted signals passing, their
gains are set to unity, with a final stage for
amplification.
32
Bandreject
• Creating a bandstop filter requires using a
lowpass and highpass filter in parallel.
• Both output are fed into a summing amplifier.
• It will function by amplifying the desired
signals compared to the signal to be
rejected.
33
```
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Spectrum analyzer wikipedia, lookup | 3,479 | 10,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-43 | latest | en | 0.867333 |
http://mathoverflow.net/questions/40834/is-there-a-group-scheme-equivalent-of-the-theorem-that-any-lie-group-is-diff-to | 1,469,787,701,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00276-ip-10-185-27-174.ec2.internal.warc.gz | 157,650,065 | 15,526 | # Is there a group-scheme equivalent of the theorem that any Lie group is diff. to a compact one cross R^n?
I'm rather ignorant in both fields, but I would still like to endeavor asking this question. I've just learned that any Lie group is diffeomorphic to a compact Lie group cross $\mathbb{R}^n$. While there is no (to my knowledge) equivalent to diffeomorphic in the group-schemes language, this does have obvious implications about the cohomology of Lie groups (which has an analog in the group-scheme language.)
So: Is there an analog of this theorem for group-schemes? What is it? What can we say?
-
For algebraic groups over a perfect field $k$, one has Chevalley's Theorem. It says that every algebraic group $G$ over $k$ contains a unique closed normal linear subgroup $H$ such that $G/H$ is an abelian variety. The abelian variety is the analogue of the compact Lie group, and the linear group $H$ is the analogue of affine space.
You mean to say that every (connected) linear group over $\mathbb{R}$ is diffeomorphic to $\mathbb{R}^n$? This seems like a stronger version of the theorem that every linear group is rational, but I've never heard of it before. Also, in the theorem you state, is it implied that $H$ is connected? – James D. Taylor Oct 2 '10 at 16:21
@James D. Taylor---what is a "(connected) linear group over $\mathbf{R}$" and what does it mean for such a thing to be diffeomorphic to $\mathbf{R}^n$? For me, GL_1 is a connected linear algebraic group over the reals. It's not diffeomorphic to anything though, because it's not a sensible topological space. Its real points can be given the structure of a sensible topological space---but this space is not connected. So there are issues here. Another example is the restriction of scalars from $\mathbf{C}$ to $\mathbf{R}$ of $GL(1)$; this has real points isomorphic to $\mathbf{C}^\times$. – Kevin Buzzard Oct 2 '10 at 20:54
...[ran out of space]...and $\mathbf{C}^\times$ isn't diffeo to $\mathbf{R}^n$ because it isn't simply connected. Perhaps these examples above may help to clarify what your question is. – Kevin Buzzard Oct 2 '10 at 20:57
Yes, thanks Kevin. My question is: in what way does Chevalley's theorem answer the question? It's portrayed as a more general theorem than the one I stated in the question. I'm trying to understand why. In order to relate the two, I assumed that cfranc had it in his mind that $H$ is diff. to $\mathbb{R}^n$. Indeed you're right that there are problems with the definition of "diff." here, and even with "connected". Still -- for his answer to make sense he must have had an idea of how these two things are related in the classical case. – James D. Taylor Oct 2 '10 at 23:02
Dear James D. Taylor: It doesn't really answer your question. If $G$ is a conn'd lin. alg. group over field $k$ of char. 0 then qt map $G \rightarrow G/U$ for unipotent radical $U$ admits splitting: $G = L \rtimes U$ over $k$ with conn'd reductive $L$ ("Levi factor"). Now $U$ is $k$-isomorphic to affine space as a variety, and if $G$ does not contain ${\rm{GL}}_1$ as $k$-subgroup and $k = \mathbf{R}$ then $L(\mathbf{R})$ is maximal compact subgp of $G(\mathbf{R})$. A pseudo-answer to your question, perhaps. For conn'd reductive $G$ it says nothing. But what would we want for $G = {\rm{SL}}_2$? – BCnrd Oct 2 '10 at 23:23 | 924 | 3,332 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.936015 |
https://tw.knowledge.search.yahoo.com/search?p=right+on&ei=UTF-8&fr2=sortBy&b=1&context=%7C%7Cgsmcontext%3A%3Aage%3A%3A1y%7Cgsmcontext%3A%3Amarket%3A%3Ahk | 1,611,552,890,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00155.warc.gz | 618,517,748 | 17,109 | # Yahoo奇摩 網頁搜尋
1. ### A slave to the rhythm尾斷 Michael Jackson 開頭讀a "sleve"之後又讀a "sthat" 點解個字會唔同左?
... M.J. sings pop songs, on love songs, the blurring become unclear in ...shorten rhythm= a linking in idea in such a singing, right . eg:-something like that---->sthat' or she is a slave...
分類：社會及文化 > 語言 2020年06月14日
2. ### Maths problem: how to do, thanks?
...s ) ∠APC = (∠APB + ∠APC)/2 = 180°/2 = 90° ...... ( adj. ∠s on st. li. ) ∴ ΔACP is a right -angled triangle. (bii) AC = √(AP² + CP²) ...... ( ...
分類：科學及數學 > 數學 2020年08月08日 | 222 | 536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.36172 |
https://philoid.com/question/35748-in-each-of-the-question-show-that-the-given-differential-equation-is-homogeneous-and-solve-each-of-them-x-ydy-x-ydx-0 | 1,695,650,212,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00787.warc.gz | 497,853,381 | 11,373 | ## Book: Mathematics Part-II
### Chapter: 9. Differential Equations
#### Subject: Maths - Class 12th
##### Q. No. 3 of Exercise 9.5
Listen NCERT Audio Books - Kitabein Ab Bolengi
3
##### In each of the question, show that the given differential equation is homogeneous and solve each of them.(x – y)dy – (x + y)dx = 0
(x - y)dy = (x + y)dx
Here, putting x = kx and y = ky
= k0.f(x,y)
Therefore, the given differential equation is homogeneous.
(x - y)dy – (x + y)dx = 0
To make it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
Integrating both sides we get,
2vdv = dt
The required solution of the differential equation.
3
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# CYB 130
\$9.00 This Tutorial was purchased 6 times & rated B+ by student like you. CYB 130 Week 1 Discussion zyBooks Reflection CYB 130 Week 2 Discussion Code Structure CYB 130 Week 3 Discussion Loop Statements CYB 130 Week 4 Discussion Containers CYB 130 Week 5 Discussion Handling Errors .....
\$75.00 This Tutorial was purchased 8 times & rated A by student like you. CYB 130 Week 1 Python LAB 2.30 Driving costs CYB 130 Week 1 Python LAB 2.16 Input Welcome message CYB 130 Week 1 Python LAB 2.32 Using math functions CYB 130 Week 1 Python LAB 2.31 Expression for calories burned during workout CYB 130 Week 1 Python LAB 2.29 Divide by x .....
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\$7.00 This Tutorial was purchased 4 times & rated A by student like you. CYB 130 Week 1 Conditional and Variables Write a Python script that determines the highest day temp and highest night temp of the following variables: day1Temp = 78 day2Temp = 84 day3Temp = 98 night1Temp = 55 night2Tem.....
\$7.00 This Tutorial was purchased 4 times & rated A+ by student like you. CYB 130 Week 1 Data Types and Variables Write a Python script that accepts the radius of a circle and computes the area of that circle. Use the formula: Area of a circle = 3.14 * radius ^2. Assign the initial radius variable the value: 1.5 Your out.....
\$3.00 This Tutorial was purchased 5 times & rated B+ by student like you. Respond to the following in a minimum of 175 words: Now that you have spent one week navigating your zyBook, what observations have you made about its design? Which elements were most helpful to you? Which features are still troublesome? What strategies have you applied that could b.....
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\$3.00 This Tutorial was purchased 9 times & rated A by student like you. CYB 130 Week 1 Python LAB 2.29 Divide by x Write a program using integers user_num and x as input, and output user_num divided by x three times. Ex If the input is 2000 2 Then the output is 1000 500 250 Note In Python 3, integer division discards fractions. Ex 6 // 4.....
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\$3.00 This Tutorial was purchased 12 times & rated A by student like you. CYB 130 Week 1 Python LAB 2.31 Expression for calories burned during workout The following equations estimate the calories burned when exercising (source) Women Calories = ( (Age x 0.074) — (Weight x 0.05741) + (Heart Rate x 0.4472) — 20.4022 ) x Time / 4.184 .....
\$3.00 This Tutorial was purchased 5 times & rated No rating by student like you. CYB 130 Week 1 Python LAB 2.32 Using math functions Given three floating-point numbers x, y, and z, output x to the power of z, x to the power of (y to the power of z), the absolute value of (x minus y), and the square root of (x to the power of z). Output each floating-point .....
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\$3.00 This Tutorial was purchased 3 times & rated No rating by student like you. Respond to the following in a minimum of 175 words: This week you will learn about basic code structure. The term structure, as it relates to programming, refers to the decisions you make to design your program to best meet its objective. Python provides features to create clean, ef.....
\$7.00 This Tutorial was purchased 3 times & rated No rating by student like you. CYB 130 Week 2 Loops and Conditional Statements User input can be obtained using the following statement and storing it into a variable for future use: num = input(“Enter a number: “) The Python shell will wait for the user to input a number and then continue for.....
\$3.00 This Tutorial was purchased 9 times & rated No rating by student like you. CYB 130 Week 2 Python LAB 3.11: Input and formatted output: Right-facing arrow Given input characters for an arrowhead and arrow body, print a right-facing arrow. Ex: If the input is: * # Then the output is: # ******## ******### ***.....
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\$3.00 This Tutorial was purchased 12 times & rated A by student like you. CYB 130 Week 2 Python LAB 3.13: Input and formatted output: House real estate summary Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summ.....
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\$3.00 This Tutorial was purchased 9 times & rated B+ by student like you. CYB 130 Week 2 Python LAB 3.28: Leap year A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra d.....
\$8.00 This Tutorial was purchased 3 times & rated A+ by student like you. CYB 130 Week 2 Using Loops Write a Python script that asks the user to input an integer and then a character. Use those values to print the character in the following pattern: r r r r r r r r r r r r r r r &n.....
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\$3.00 This Tutorial was purchased 4 times & rated No rating by student like you. Respond to the following in a minimum of 175 words: Most programming languages provide loop statements that help users iteratively process code. In Python you can write loops that handle many situations. What is the intuition behind using a loop statement? What do you gain from usin.....
\$8.00 This Tutorial was purchased 3 times & rated A+ by student like you. CYB 130 Week 3 Python and HTML Write a Python script that builds an HTML file of your own design. Add an image tag into your HTML and link it to an image on your hard drive. Make sure you store the image in the same directory as the html file, otherwise it might not find the file......
\$3.00 This Tutorial was purchased 15 times & rated A+ by student like you. CYB 130 Week 3 Python LAB 4.14 LAB: Count input length without spaces, periods, or commas Given a line of text as input, output the number of characters excluding spaces, periods, or commas. Ex: If the input is: Listen, Mr. Jones, calm down. the output is: .....
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\$3.00 This Tutorial was purchased 18 times & rated A by student like you. CYB 130 Week 3 Python LAB 4.17: Print string in reverse Write a program that takes in a line of text as input, and outputs that line of text in reverse. The program repeats, ending when the user enters "Quit", "quit", or "q" for the line of text. Ex: If the inp.....
\$3.00 This Tutorial was purchased 9 times & rated No rating by student like you. CYB 130 Week 3 Python LAB 4.18: Smallest and largest numbers in a list Write a program that reads a list of integers into a list as long as the integers are greater than zero, then outputs the smallest and largest integers in the list. Ex: If the input is: 10.....
\$3.00 This Tutorial was purchased 21 times & rated A by student like you. CYB 130 Week 3 Python LAB 4.19: Output values in a list below a user defined amount Write a program that first gets a list of integers from input. The input begins with an integer indicating the number of integers that follow. Then, get the last value from the input, which indic.....
\$3.00 This Tutorial was purchased 4 times & rated B+ by student like you. Respond to the following in a minimum of 175 words: Python lists are commonly used to store data types. Lists are a collection of information typically called a container. Think of a physical container that can hold all kinds of objects, not just one object of the same typ.....
\$8.00 This Tutorial was purchased 5 times & rated No rating by student like you. CYB 130 Week 4 Individual: Functional Programming Write a 2-part program as follows: Part 1: Write a function to convert Celsius to Fahrenheit. Part 2: Write a function to convert Fahrenheit to Celsius. Both of the functions (Celsius to Fahrenheit and Fahrenheit to Cels.....
\$7.00 This Tutorial was purchased 5 times & rated No rating by student like you. CYB 130 Week 4 Individual: Object Oriented Programming (OOP) As defined on pp. 449 to 468 in Ch. 17, “Object-Oriented Programming,” of Introduction to Computing and Programming in Python, a Turtle is an object from the class Turtle with the following features: Every turtle unde.....
\$3.00 This Tutorial was purchased 8 times & rated A+ by student like you. CYB 130 Week 4 Python LAB 5.18: Miles to track laps One lap around a standard high-school running track is exactly 0.25 miles. Write the function miles_to_laps() that takes a number of miles as an argument and returns the number of laps. Complete the program to output the number of laps. .....
\$3.00 This Tutorial was purchased 10 times & rated A+ by student like you. CYB 130 Week 4 Python LAB 5.19: Driving costs - functions Driving is expensive. Write a program with a car's miles/gallon and gas dollars/gallon (both floats) as input, and output the gas cost for 10 miles, 50 miles, and 400 miles. Output each floating-point value with two dig.....
\$3.00 This Tutorial was purchased 7 times & rated No rating by student like you. CYB 130 Week 4 Python LAB 5.20: Step counter A pedometer treats walking 2,000 steps as walking 1 mile. Write a program whose input is the number of steps, and whose output is the miles walked. Output each floating-point value with two digits after the decimal point, which can .....
\$3.00 This Tutorial was purchased 10 times & rated A+ by student like you. CYB 130 Week 4 Python LAB 5.21: Leap year - functions A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days.....
\$3.00 This Tutorial was purchased 9 times & rated B+ by student like you. CYB 130 Week 4 Python LAB 5.22: Swapping variables Write a program whose input is two integers and whose output is the two integers swapped. Ex: If the input is: 3 8 the output is: 8 3 Your program must define and call the following function. swap_val.....
\$3.00 This Tutorial was purchased 10 times & rated A+ by student like you. CYB 130 Week 4 Python LAB 5.23: Exact change - functions Write a program with total change amount as an integer input that outputs the change using the fewest coins, one coin type per line. The coin types are dollars, quarters, dimes, nickels, and pennies. Use singular and plural coin names as .....
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\$3.00 This Tutorial was purchased 7 times & rated A by student like you. CYB 130 Week 5 Python LAB 6.24: Varied amount of input data Statistics are often calculated with varying amounts of input data. Write a program that takes any number of integers as input, and outputs the average and max. Ex: If the input is: 15 20 0 5 the output is: 10 2.....
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\$3.00 This Tutorial was purchased 8 times & rated A+ by student like you. CYB 130 Week 5 Python LAB 6.26: Elements in a range Write a program that first gets a list of integers from input. That list is followed by two more integers representing lower and upper bounds of a range. Your program should output all integers from the list that are within that range (inclusi.....
\$3.00 This Tutorial was purchased 5 times & rated No rating by student like you. CYB 130 Week 5 Python LAB 6.5: Checker for integer string Forms often allow a user to enter an integer. Write a program that takes in a string representing an integer as input, and outputs yes if every character is a digit 0-9. Ex: If the input is: 1995 the output is: ye..... | 4,413 | 17,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.904367 |
https://tbc-python.fossee.in/convert-notebook/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_10.ipynb | 1,702,264,033,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00523.warc.gz | 618,779,617 | 37,124 | Chapter 10:Theory of Failures¶
Example No.10.10.1,Page No.401¶
In [1]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
P_e=300 #N/mm**2 #Elastic Limit in tension
FOS=3 #Factor of safety
mu=0.3 #Poissoin's ratio
P=12*10**3 #N Pull
Q=6*10**3 #N #Shear force
#Calculations
#Let d be the diameter of the shaft
#Direct stress
#P_x=P*(pi*4**-1*d**3)**-1
#After substituting values and further simplifying we get
#P_x=48*10**3
#Now shear stress at the centre of bolt
#q=4*3**-1*q_av
#After substituting values and further simplifying we get
#q=32*10**3*(pi*d**2)**-1
#Principal stresses are
#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5
#After substituting values and further simplifying we get
#p1=20371.833*(d**2)**-1
#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5
#After substituting values and further simplifying we get
#P2=-5092.984*(d**2)**-1
#q_max=((P_x*2**-1)**2+q**2)**0.5
#From Max Principal stress theory
#Permissible stress in Tension
P1=100 #N/mm**2
d=(20371.833*P1**-1)**0.5
#Max strain theory
#e_max=P1*E**-1-mu*P2*E**-1
#After substituting values and further simplifying we get
#e_max=21899.728*(d**2*E)**-1
#According to this theory,the design condition is
#e_max=P_e*(E*FOS)**-1
#After substituting values and further simplifying we get
d2=(21899.728*3*300**-1)**0.5 #mm
#Max shear stress theory
#e_max=shear stress at elastic*(FOS)**-1
#After substituting values and further simplifying we get
d3=(12732.421*6*300**-1)**0.5 #mm
#Result
print"Diameter of Bolt by:Max Principal stress theory",round(d,2),"mm"
print" :Max strain theory",round(d2,2),"mm"
print" :Max shear stress theory",round(d3,2),"mm"
Diameter of Bolt by:Max Principal stress theory 14.27 mm
:Max strain theory 14.8 mm
:Max shear stress theory 15.96 mm
Example No.10.10.2.Page No.402¶
In [2]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
M=40*10**6 #N-mm #Bending moment
T=10*10**6 #N-mm #TOrque
mu=0.25 #Poissoin's ratio
P_e=200 #N/mm**2 #Stress at Elastic Limit
FOS=2
#Calculations
#Let d be the diameter of the shaft
#Principal stresses are given by
#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)
#After substituting values and further simplifying we get
#P1=4.13706*10**8*(d**3)**-1 ............................(1)
#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)
#After substituting values and further simplifying we get
#P2=-6269718*(pi*d**3)**-1 ..............................(2)
#q_max=(P1-P2)*2**-1
#After substituting values and further simplifying we get
#q_max=2.09988*10**8*(d**3)**-1
#Max Principal stress theory
#P1=P_e*(FOS)**-1
#After substituting values and further simplifying we get
d=(4.13706*10**8*2*200**-1)**0.33333 #mm
#Max shear stress theory
#q_max=shear stress at elastic limit*(FOS)**-1
#After substituting values and further simplifying we get
d2=(2.09988*10**8*4*200**-1)**0.33333
#Max strain energy theory
#P_3=0
#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1
#After substituting values and further simplifying we get
d3=(8.62444*10**12)**0.166666
#Result
print"Diameter of shaft according to:MAx Principal stress theory",round(d,2),"mm"
print" :Max shear stress theory",round(d2,2),"mm"
print" :Max strain energy theory",round(d3,2),"mm"
Diameter of shaft according to:MAx Principal stress theory 160.52 mm
:Max shear stress theory 161.33 mm
:Max strain energy theory 143.2 mm
Example No.10.10.3,Page No.403¶
In [3]:
import math
from math import sin, cos, tan, pi, radians
#Initilization of Variables
f_x=40 #N/mm**2 #Internal Fliud Pressure
d1=200 #mm #Internal Diameter
q=300 #N/mm**2 #Tensile stress
#Calculations
#From Lame's Equation we have,
#Hoop Stress
#f_x=b*(x**2)**-1+a ..........................(1)
#p_x=b*(x**2)**-1-a .........................(2)
#the boundary conditions are
x=d1*2**-1 #mm
#After sub values in equation 1 and further simplifying we get
#40=b*100**-1-a ..........................(3)
#Max Principal stress theory
#q*(FOS)**-1=b*100**2+a ..................(4)
#After sub values in above equation and further simplifying we get
#From Equation 3 and 4 we get
a=80*2**-1
#Sub value of a in equation 3 we get
b=(f_x+a)*100**2
#At outer edge where x=r_0 pressure is zero
r_0=(b*a**-1)**0.5 #mm
#thickness
t=r_0-r1 #mm
#Max shear stress theory
P1=b*(100**2)**-1+a #Max hoop stress
P2=-40 #pressure at int radius (since P2 is compressive)
#Max shear stress
q_max=(P1-P2)*2**-1
#According max shear theory the design condition is
#q_max=P_e*2**-1*(FOS)**-1
#After sub values in equation we get and further simplifying we get
#80=b*(100**2)**-1+a
#After sub values in equation 1 and 3 and further simplifying we get
b2=120*100**2*2**-1
#from equation(3)
a2=120*2**-1-a
Thickness of metal by:Max Principal stress theory 41.42 mm | 1,739 | 4,892 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-50 | longest | en | 0.580056 |
https://123deta.com/document/ynxx5gjq-riemann-function-positivity-function-representation-functions-probabilistic-aspects.html | 1,627,529,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00261.warc.gz | 90,792,165 | 43,347 | # Riemann's zeta function and T-positivity (3): Kummer function and inner product representation (Functions in Number Theory and Their Probabilistic Aspects)
42
## 全文
(1)
(2)
By
### Okabe
*
Abstract
We considerRiemanns zetafunction from theviewpoint of the
### theory
ofstationaryGaus‐
sian processes. In the previous two papers
12 we
### proved
that Riemanns zeta function
satisfies an
### ordinary
differential equation with time
### delay
and then obtained a new represen‐
tation of the
### KMO‐Langevin
system which is the characteristics for the
### non‐negative
definite
function associated with Riemanns zeta function. As a continuation of the previous papers,
### first,
we introduce in this paper a derived Kummer function and prove a new representation
theorem for an
### analytic
continuation for Riemanns zeta
an
### analytic
continuation of the derived Kummer function.
### Second,
we prove an inner
### product
representa‐
tiontheorem for the
### analytic
continuationof Riemanns zetafunction and the derived Kummer
### function, by constructing
aHamiltonianoperatorassociated withastationaryGaussianprocess
with \mathrm{T}‐positivity.
Introduction
Riemanns
### hypothesis
for the zeta function
### $\zeta$= $\zeta$(s)=\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{s}} ({\rm Re}(s)>1)
has remained unsolved for 151 years
thegamma
Riemann
obtained the
for an
### analytic
continuation of the zeta function
Recieved May 30, 2011. Revised July 28, 2011. Accepted August 2, 2011.
2000 Mathematics Subject
### Classification(s):
Primary 11\mathrm{M}38; Secondary 60\mathrm{G}25, 60\mathrm{G}12, 82\mathrm{C}05.
Key Words: Riemanns zeta function, stationary Gaussian process with T‐positivity, Kummer
function, Hamiltonian operator, inner product representation
This work is partially supported by Grand‐in‐Aid for Scientific Research
### ((\mathrm{B})
No.23340026, Chal‐
lenging ExploratoryResearch
### No.23654017),
Global Center of ExcellenceNo.17340024,theMinistry
ofEducation, Sicence, Sports and Culture, Japan.
*
Kawasaki, 214‐8571, Japan.
(3)
:
where
and
### $\theta$= $\theta$(t)
are the gamma function and the theta
respec‐
defined
,
.
We note that
### \displaystyle \frac{ $\theta$(t)-1}{2}=\sum_{n=1}^{\infty}e^{- $\pi$ n^{2}t} (t>0)
.
Since the second term of the
side in
is
with
### respect
to
s\in \mathrm{C}, we see from the
### properties
of the gamma function that Riemanns zeta function
can be
### analytically
continued so that it is
at the
where it has a
### pole
of order 1 with residue 1 and vanishes on the set
### \{-2n;n\in \mathrm{N}\}
, to
be called the set of the trivial zero
Riemanns
that real
### parts
of all non‐trivial zero
### points
of the zeta function
### $\zeta$= $\zeta$(s)
lie on the vertical line
### ([16],[5]).
The purpose of this paper consists of the
### following
two: one is to introduce a
derived Kummer function associated with the Kummer function and the theta function
and to prove a new
theorem for the
### analytic
continuation of Riemanns
zeta
an
### analytic
continuation of the derived Kummer
### function;
second is to prove an another
### representation
theorem in terms of the innner
for
the
### analytic
continuation of Riemanns zeta
a Hamiltonian
### operator
associated with a
### stationary
Gaussian process with \mathrm{T}
The main
### point
is that when we rewrite the second term in the
side of
into
### \displaystyle \int_{1}^{\infty}\frac{ $\theta$(t)-1}{2}(t^{-\frac{1+\mathrm{s}}{2}}+t^{-\frac{2-\mathrm{s}}{2}})dt=\int_{0}^{\infty}\frac{ $\theta$(t+1)-1}{2}((t+1)^{-\frac{1+\mathrm{s}}{2}}+(t+1)^{-\frac{2-\mathrm{s}}{2}})dt,
we note that for each
, the terms
and
on
the
### right‐hand
side of the above
can
as the
transform of
bounded
### complex
valued Borel measures. In
we have
,
(4)
where for each
,
is a bounded
### complex
valued Borel measure
on
defined
### $\Gamma$_{s}(d $\lambda$)\displaystyle \equiv\frac{1}{ $\Gamma$(s)}e^{- $\lambda$}$\lambda$^{s-1}d $\lambda$.
We note that if s is a
real
then
### $\Gamma$_{s}
is the gamma distribution with
mean s and variance s.
The detailed content of this paper is as follows.
In Section
we recall the
the
### analytic
continuation for Riemanns zeta func‐
tion
due to
### Riemann,
because the method used there is used in the
in this
paper
In Section
### 3,
we introduce a derived Kummer function defined
### by combining
the
Kummerfunction and the theta
and obtain its
that the
of the
in the
### analytic
continuation of Riemanns zeta
function can be
as the
### Laplace
transform of a bounded
### complex
valued Borel
measure defined on
we prove a new
theorem of the
### analytic
cointinuation for Riemanns zeta
the
### analytic
continuation
for the derived Kummer function.
In Section
### 4,
we introduce other two kinds of functions derived from the Kummer
function and the theta function and
their
### analytic
continuation and prove a
recurrence formula among
### according
to the recurrence formula with
to
### parameters
of the Kummer function.
In order to
### clarify
a mathematical structure of the notion of \mathrm{T}
### ‐positivity coming
from the axiomatic field
from the
of the
### theory
of stochas‐
tic processes, we constructed in
the Hamiltonian
on the real
### splitting
space associated with a
### stationary
Gaussian process with \mathrm{T}
### ‐positivity
and derived an
infinite‐dimensional
the time
### evolutiojn
of the above pro‐
cess. In Section
the same
as in
### [7],
we construct a Hamiltonian
on the
### splitting
space associated with a
### stationary
Gaussian
process with \mathrm{T}
we
a note
the
### Shwinger
function
of order 2 and the
### Wightmann
function of order 2 in the axiomatic field
In Section
weprove aninner
theoremfor the
### analytic
con‐
tinuationof Riemannszeta function and the derived Kummer
the bounded
### complex
valued Borel measure used in Section 3 to the gamma distribution
and
the Hamiltonian
### operator
associated with the
Gaussian process
with \mathrm{T}
### ‐positivity
whose covarinace function is
the
### Laplace
transform of the
gamma distribution
We shall
Riemanns
the inner
represen‐
(5)
Section 6 in the
### forthcoming
paper. We would like to dedicate our
thanks for
Prof.
### University,
and the referee for
The
### analytic
continuation for Riemanns zeta function due to
Riemann
we
the fundamental
### properties concerning
the gamma function
which are used in this paper,
### together
with the beta function
defined
Theorem 2.1.
The gamma
can be
### analytically
continued
so that it has no zero
and is
at
the set
### poles
with
order 1 with residue 1.
.
.
.
.
### (1.4),
we note that the theta function
satisfies the
functional
,
which is
Theorem 2.2.
For any
### s\in\{s\in \mathrm{C};{\rm Re}(s)>1\},
(6)
Proof. Fix any n\in \mathrm{N} and
.
the
of variables
in
we have
and so
the above with
### respect
to n\in \mathrm{N}, we see from
that
the
of variables
and
### (2.2)
to the first term of the
we have
On the other
direct
we have
and
to
### (2.3),
we see that Theorem 2.2 holds.
Lemma 2.3.
The theta
the
ties:
### e^{- $\pi$ t}\displaystyle \leq\frac{ $\theta$(t)-1}{2}\leq e^{- $\pi$ t}(1-e^{- $\pi$})^{-1} (t>1)
.
Proof. It follows from
that the
### e^{- $\pi$ t}\displaystyle \leq\frac{ $\theta$(t)-1}{2}
holds. Onthe other
the
(7)
to
we see that
### =e^{- $\pi$ t}(1-e^{- $\pi$})^{-1},
which proves Lemma 2.3.
Lemma
we prove
Lemma 2.4.
The
is
with
### respect
tos\in \mathrm{C}.
Proof. Fix any
such that
### |s_{0}|<N
with a natural number N. Since
### \displaystyle \frac{\partial}{\partial s}t^{-s}=-(\log t)t^{-s}
, we see from Lemma 2.3 that for any s\in \mathrm{C} such that
!
.
convergence
### theorem,
we see that Lemma 2.4 holds.
### applying
Lemma 2.4 to Theorem
we have
Theorem 2.5.
The
can be
### analytically
continued on \mathrm{C} so that it is
at two
s=0,
where it has
### of
order 1 with residue 1.
### using
Theorem 2.1 and Theorem
we see that
Theorem 2.6.
Riemanns zeta
can be
### analytically
con‐
tinued on \mathrm{C}
the
and it is
at the
### point
s=1, where it has a
### of
order 1 with residue 1.
(8)
### Moreover,
we see from Theorem 2.6 that
Theorem 2.7.
The
the
tional
.
### §3.
The derived Kummer function associated with the Kummer function
and the theta function
As noted in Section
we can
the
### right‐hand
side in Theorem 2.2 as
follows.
,
We note that the
functional
holds.
.
Lemma
we have
Lemma 3.1.
The
and
are
on the
C.
and
to
and
### (3.3),
we have
Lemma 3.2. For any
(9)
Proof.
and
we have
which proves
follows from
and
the bounded
### complex
valued Borel measure
in
in
### (1.7)),
we find from Lemma 3.2 that the functions
and
### F_{2}
can be rewritten in the
form:
,
.
In this Section
we use the
### representation
for the functions
and
in Lemma 3.2.
The
and
### (3.6)
will be further studied in Section 6.
For each
### s\in\{s\in \mathrm{C};{\rm Re}(s)>0\}
, we define a function
on
### f(x;s)\displaystyle \equiv\int_{0}^{\infty}e^{-xt}\frac{t^{s-1}}{1+t}dt.
Lemma 3.3. For any
,
.
Proof.
the
of variables
in Lemma
we seethat
holds.
follows from
and
### (Q.E.D.)
Lemma 3.4. For any
(10)
Proof. Fix any
### s\in\{s\in \mathrm{C};0<{\rm Re}(s)<1\}
. Since the function
satisfies the
differential
,
we find that
the
of variables
in the above
we see that
holds. On
the other
we have
the
### change
of variables t $\lambda$=u in the above
we have
.
Theorem
we see that
holds.
### (Q.E.D.)
Lemma 3.5. For any
and any
Proof.
us
### =\displaystyle \frac{1}{1-s}+\frac{x}{1-s}\int_{0}^{1}e^{x(1-t)}t^{1-s}dt.
(11)
For any N\in \mathrm{N},
N
we obtain
## \displaystyle \int_{0}^{1}e^{x(1-t)}t^{-s}dt=\sum_{n=0}^{N}\frac{x^{n}}{(1-s)(2-s)\cdots(n+1-s)}
Since
and
for any
### s\in\{s\in \mathrm{C};0<{\rm Re}(s)<1\}
and any x>0, we can let N go to \infty in
to see
that
s
2-s in Theorem
we see that
.
we conclude from
### (3.9)
that Lemma 3.5 holds.
### Here,
we shall recall the Kummer function
### {}_{1}F_{1}(a;c;z)
which is also called the
### pergeometric
function of confluent
with two
a and
.
.
The function
satisfies the
differential
of confluent
### z\displaystyle \frac{d^{2}}{dz}u(z)+(c-z)\frac{d}{dz}u(z)-au(z)=0 (z\in \mathrm{C})
,
which is also called Kummersdifferential
### equation.
We know that when
\cdots,
the fundamental
### \{u_{1}, u_{2}\}
of solutions to Kummers differential
is
and
.
We have the
### following integral
formula for the Kummer function.
(12)
It follows from
that
.
### Hence,
we see from Lemma 3.5 that
Lemma 3.6. For any
and any
.
Lemmas 3.4 and
### 3.6,
we have
Lemma 3.7. For any
and any
### f(x;s)=e^{x}\displaystyle \frac{ $\pi$}{\sin( $\pi$ s)}-\frac{ $\Gamma$(s)}{x^{s-1}}(1-s)^{-1}{}_{1}F_{1}(1;2-s;x) (x>0)
.
Lemma 3.8. For any
.
Proof.
Lemmas
3.5 and
we have
(13)
the
### following
formula for the gamma function
from Theorem
we see that
holds.
follows from
and
### (Q.E.D.)
Lemma 3.9. For any
, the
series is
.
Proof. Put
.
### By
Lemma 3.5 and Theorem
we have
the
of variables
, we obtain
On the other
it follows from
that
(14)
and
we have
### =|s-1|(\displaystyle \int_{1}^{\infty}(\sum_{n=1}^{\infty}e^{- $\pi$ n^{2} $\lambda$})$\lambda$^{\frac{1}{2}- $\sigma$}d $\lambda$+\frac{1}{4( $\sigma$-3/2)( $\sigma$-1)})
.
Since it follows from Lemma 2.4 that the first term of the bottom
in the above
is
### finite,
we conclude that Lemma 3.9 holds.
virtue of Lemma
### 3.9,
we can introduce a function
on
defined
.
We note from
that
### K_{ $\theta$}(s)=\displaystyle \sum_{n=1}^{\infty}e^{- $\pi$ n^{2}}{}_{1}F_{1}(1;s; $\pi$ n^{2})
.
We call the function
### K_{ $\theta$}=K(s)
the derived Kummer function associated with the
Kummer function and the theta function.
Lemma 3.10.
### (i)
The derived Kummer
can be
### analytically
continued on \mathrm{C} so that it is
at the
, where it has a
### of
order 1 with resdue 1
.
.
Proof.
Lemma
### 3.8,
we see that for any
.
the above
, we have
.
### Therefore,
it follows from Theorems
### 2.5,
2.6 and 3.12 that
holds.
follows
from
### (i)
and the functional
in
(15)
Lemma 3.11.
.
.
Proof.
in Lemma
we have
.
to the
side of the
we see that
, which proves
follows from
and
### (Q.E.D.)
We define a function
on \mathrm{C}
### $\xi$(s)\displaystyle \equiv\frac{s(s-1)}{2}$\pi$^{-\frac{\mathrm{s}}{2}} $\Gamma$(\frac{s}{2}) $\zeta$(s)
.
It follows from Theorem 2.7 that the function
satisfies the
functional
.
We are now
### going
to prove one of the main theorems of this paper.
Theorem 3.12. The
has the
.
Proof.
Lemma
we have
.
this with
we have
(16)
from Theorem
we have
.
the above
### \displaystyle \frac{s(s-1)}{2}
, we conclude that Theorem 3.12 holds.
### §4.
A recurrence formula for the derived Kummer function
In this
### section,
we introduce other two kinds of functions derived from the Kummer
function and the theta function and obtain their
continuation.
### Furthermore,
we prove some recurrence formulae among
### using
the recurrence formulas with
to
### parameters
of the Kummer function.
### [4.1]
We recall severalrecurrenceformulae for the Kummer function
with two
a and
.
Theorem 4.1.
.
.
a
.
.
a
.
.
. .
where
is
In
### (3.17),
we introduced the derived Kummer function
### K_{ $\zeta$}=K(s)
associ‐
ated with the Kummer function and the theta function. As a
the
Kummerfunction
with a
### a(>0)
, we introducea derived Kummer
function
with a
(17)
We note that
,
.
If
, then the
Lemma
### 4.2,
which is a refinement of Lemma
### 3.10,
assures that
Lemma 4.2. For any
a>0 and
, the
series is
### convergent
and its convergence is
in the set
any
:
.
Proof.
the
formula
### (3.13),
we can prove Lemma
the
same
### procedure
as in Lemma 3.10. We
here a
of
Put
.
that
### $\sigma$\displaystyle \geq$\sigma$_{0}>a+\frac{1}{2}>0
, we see from Theorem
that
### Therefore,
we see from Theorem
and
that for any z>0
we have
(18)
the
of variables
, we have
we have
### +\displaystyle \frac{1}{2}\int_{1}^{\infty}(\frac{ $\lambda$-1}{ $\lambda$})^{a-1}(\frac{1}{$\lambda$^{$\sigma$_{0}-(a-1/2)}}+\frac{1}{$\lambda$^{$\sigma$_{0}-(a-1)}})d $\lambda$\}.
We calculate the second term in
the
of variables
, we
obtain
.
we have
.
we have
### \displaystyle \int_{1}^{\infty}(\frac{ $\lambda$-1}{ $\lambda$})^{a-1}\frac{1}{$\lambda$^{s-(a-1/2)}}dt=\frac{ $\Gamma$(a) $\Gamma$(s-(a+1/2))}{ $\Gamma$(s-1/2)}-\frac{ $\Gamma$(a) $\Gamma$(s-a)}{ $\Gamma$(s)}.
We consider the first term in
### First,
we consider the case where
. Since
is bounded in
\infty
### ),
we see from Lemma 2.3 that the
of the first
term in
is
### Hence,
the first term of the
side in
### (4.4)
is finite.
(19)
of the first term in
into
We have the
. Since
is
in
### [1, 2],
we see from Lemma 2.3 that the
### integrand
of the first term in
is
in
since
is bounded in
\infty
### ),
we see from Lemma 2.3 that
the
### integrand
of the second term in
is
in
\infty
the first term
of the
side in
is finite.
we have
Lemma 4.2.
and
### (3.12),
we can arrange the
### proof
of Lemma 4.2 to see that
Lemma 4.3. For any
a>0 and
, the
series
is
and the
relation holds:
.
the
### analytic
continuation for the derived Kummer function
with a
### parameter
a, we have
Theorem 4.4. Let a be any
number.
(20)
The
can be
### analytically
continued on \mathrm{C} so that it is
at the set
, where it has
order 1.
The
can be
### analytically
continued on \mathrm{C} so that it is
on the set
, where it has
### of
order 1.
Proof. We see from Lemmas 2.3 and 2.4 that
and
hold.
we prove
### (iii).
For that purpose, we have
to prove that
is
in C.
### First,
we consider the case where
.
that
is bounded in
### [1, \infty)
, we can use the same estimate as in the
### proof
of Lemma 2.4 to see that
is
### Next,
we also consider the case where 0<a<1. We
the
in
into
that
, we
the
of the first
term in the
side in
into
Since
is
in
### [1,2],
we can use the same estimate as in the
### proof
of Lemma 2.4 to see that the first term of the
side in
is
in C.
since
is bounded in
\infty
### ),
we can use the same estimate as in
the
### proof
of Lemma 2.4 to see that the second term of the
side in
is
### regular
in C. This proves
(21)
Theorem 4.5. The
relation holds:
any
.
As in Lemma
### by taking
into account Theorem
### 4.1(iii),
we prove
Lemma 4.6. For any
a>0 and
, the
series is
.
Proof. From the
of Lemma 4.2
the
of variables
, we have
with
### respect
to $\lambda$, we have
n=1 n=1 n=1
and so
this into
we have
(22)
as in the
of Lemma
### 4.2,
we find that the first term of the bottom
in
is finite.
we have
.
as in the
of Lemma
### 4.2,
we find that the second term of the
bottom in
### (4.12)
is finite.
Since it follows from
and Theorem
that
### \displaystyle \int_{1}^{\infty}(\frac{ $\lambda$-1}{ $\lambda$})^{a-1}\frac{1}{$\lambda$^{ $\sigma$-(a+1/2)}}dt=B(a, $\sigma$-a-\frac{3}{2})=\frac{ $\Gamma$(a) $\Gamma$( $\sigma$-(a+3/2))}{ $\Gamma$( $\sigma$-3/2)},
we find that the third term of the bottom
in
is finite.
we have
Lemma 4.6.
virtue of Lemma
### 4.6,
we can introduce another derived Kummer function
with a
.
the
### analytic
continuation for the function
### K_{ $\theta$}^{\cdot}(a)=K_{ $\theta$}^{\cdot}(a;s)
, we have
Theorem 4.7. Let a be any
number.
The
can be
### analytically
continued on \mathrm{C} so that it is
on the set
0,
### 1, 2,
. . \cap \mathrm{Z}
, where it has
order 1.
The
can be
### analytically
continued on \mathrm{C} so that it is
on the set
0,
### 1, 2,
. . \cap \mathrm{Z}
, where it has
### of
order 1.
Proof. We see from Lemma 4.6 that
(23)
On the other
it follows from
and Theorem
that
it follows from
and
that
holds.
as in the
of
Lemma
### 4.6(ii),
we can use the same estimate as in Lemma 2.4 to see that both the first
term and the second term of the bottom
in
are
on C.
we see
from Theorem
that
holds.
follows from
and Theorem
We prove another
for the function
### K_{ $\theta$}^{\cdot}(a;s)
.
Theorem 4.8. For any
Proof.
the
of variables
in Theorem
### 4.7(i),
we see that
the first essential term of the
side of Theorem
### 4.7(i)
is rewritten into
the functional
in
and Theorem
we have
(24)
.
the
of variables
in Theorem
### 4.7(i),
we also
see that the second essential term of the
side of Theorem
is rewritten
into
the functional
in
we have
(25)
Theorem
we have
to Theorem
we have
### +\displaystyle \int_{0}^{1}(\sum_{n=1}^{\infty} $\pi$ n^{2}e^{- $\pi$ n^{2}(1-t)})t^{a-1}(1-t)^{s-(a+1)}dt-\frac{1}{2}\frac{ $\Gamma$(a) $\Gamma$(s-(a+1)}{ $\Gamma$(s-1)}\},
which proves Theorem 4.8.
to Theorem
### using
the recurrence formulae
and
in Theorem
### 4.1,
we see from Theorem 4.7 that
Theorem 4.9. The
relations hold:
any
,
,
(26)
A Hamiltonian
### operator
associated with a
Gaussian
process
\mathrm{T}
Let
### \mathrm{X}=(X(t);t\in \mathrm{R})
be any real valued
### stationary
Gaussian process on a
space
### ( $\Omega$, \mathcal{B}, P)
with mean 0 and the covariance function
### R=R(t)
: \mathrm{R}\rightarrow \mathrm{R}:
,
.
### Furthermore,
we consider the case where the process X satisfies \mathrm{T}
that
### is,
there exists a bounded Borel measure
on
### [0,\infty)
such that the covariance
function
can be
as the
### Laplace
transform of the Borel measure $\sigma$ :
.
We define a
Hilbert space
as the closed
of the
Hilbert space
### \mathrm{M}(\mathrm{X})\equiv
the closed linear hull of
We denote
and
the inner
### product
and the norm in the
Hilbert space
,
,
.
we define three
closed
,
and
of the
Hilbert space
### \mathrm{M}^{+}(\mathrm{X})\equiv
the closed linear hull of
### \mathrm{M}^{-}(\mathrm{X})\equiv
the closed linear hull of
### \mathrm{M}^{-/+}(\mathrm{X})\equiv
the closed linear hull of
where
stands for the
### operator
on the closed space
. We call
these
closed
,
and
### \mathrm{M}^{-/+}(\mathrm{X})
the future space, the
space and the
### splitting
space associated with the process
In
### [7],
we treated the real Hilbert space
defined
the closure of
### \displaystyle \{\sum_{n=1}^{N}c_{n}X(t_{n});c_{n}, t_{n}\in \mathrm{R}(1\leq n\leq N), N\in \mathrm{N}\}
and constructed a Hamiltonian
on the real
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# MATH 251 Work sheet / Things to know
Chapter 3
Standard form:
## What makes it homogeneous?
We will, for the most part, work with equations with constant coefficients only.
What is the general form of a second order linear equation with constant coefficients?
Ex. 3.1.1 Can you think of any function(s) that satisfy each equation (w/ constant
coefficients) below?
(a) y″ − 25y = 0
(b) y″ + 25y = 0
(c) y″ − 25 y′ = 0
The example (c) above is an instance of a second order linear equation with the y-term
missing. It is essentially a first order linear equation in disguise. All equations of this
type can be solved by changing it into a first order equation with the substitutions u = y′
and u′ = y″, then use the integrating factor method to solve for u, and integrate the result
to find y.
## © 2011 Zachary S Tseng 1
2. The characteristic equation
## Given the equation ay″ + by′ + cy = 0, a ≠ 0, what is its characteristic equation?
Any root, r, of the characteristic equation has the property that y = ert always satisfies the
equation above. Therefore, y = ert will be a particular solution for each root r.
## Consequently, an important formula to remember for this class is (surprisingly) the
quadratic formula:
− b ± b 2 − 4ac
r=
2a
Note that the characteristic equation method does not require the given differential
equation to be put into its standard form first – the quadratic formula simply doesn’t care
whether or not the leading coefficient is 1.
Suppose r1 and r2 are two distinct real roots of the characteristic equation, what is the
general solution of the differential equation?
y=
y=
y=
## © 2011 Zachary S Tseng 2
3. Initial Value Problems
What do the initial conditions of a second order differential equation look like?
A second order equation’s general solution will have 2 arbitrary constants / coefficients.
Therefore, an IVP will have 2 initial conditions in order to give 2 (algebraic) equations
needed to solve for them. What must the conditions look like?
Ex. 3.3.1 Take the previous example y″ + y′ − 12y = 0. Find its solution satisfying
(a) y(0) = 0, y′(0) = 2
## © 2011 Zachary S Tseng 3
The Existence and Uniqueness Theorem (for second order linear equations)
It is really the same theorem as the one we saw earlier, except this one is in the context of
second order linear equation.
## What does it say?
How to find the largest interval (that is, the interval of validity) on which a particular
solution is guaranteed to exist uniquely?
Ex. 3.3.3 Consider the equation given below. For each set of initial conditions, find the
largest interval on which the particular solution is guaranteed to exist uniquely.
## (a) y(3) = 1, y′(3) = −9
(b) y(−1) = 9π, y′(−1) = e−1
## © 2011 Zachary S Tseng 4
4. The general solution of second order linear equations
W(y1, y2)(t) =
## *Note that the Wronskian is defined as a function of t.
Fundamental solutions
## Ex. 3.4.1 Which pair(s) of functions below can be fundamental solutions?
(1) 10, t + 2
(2) e2t, e−2t
(3) et, 0
(4) 2e3t, e3t + 2
(5) cos 6t, −2sin 6t
(6) 5sin t, cos(t + π/2)
(7) cos 2t, sin(2t − 2π)
Ex. 3.4.2 Given that y1 = te−t and y2 = 2te−t − 7ln(t) are both known solutions of a certain
equation y″ + p(t)y′ + q(t)y = 0. What is the general solution of this equation?
## © 2011 Zachary S Tseng 5
5. The Abel’s Theorem
## It gives us a way to determine, up to a constant multiplier, the Wronskian of any pair of
solutions of a given second order homogeneous linear equation, by looking at just the
equation itself only.
## What does it say: W(y1, y2)(t) =
(Due to their similarity) do not confuse this formula with another formula we have
seen before. Which formula?
Ex. 3.5.1 (Final Exam, fall 2008) Let y1(t) and y2(t) be any two solutions of the second
order linear equation t2y″ − 6t y′ + cos(3t)y = 0. What is the general form of their
Wronskian, W(y1, y2)(t)?
Ex. 3.5.2 (Exam I, spring 2007) Consider the second order linear differential equation
t y″ − 2y′ + y = 0. Suppose y1(t) and y2(t) are two fundamental solutions of the equation
such that y1(1) = 2, y′1(1) = 0, y2(1) = 2, and y′2(1) = 2. Compute their Wronskian
W(y1, y2)(t).
## © 2011 Zachary S Tseng 6
6. Characteristic equation: complex roots case
## (Background info) Euler’s formula:
Suppose r = λ ± µi are two complex conjugate roots of the characteristic equation, what is
the general solution of the differential equation?
y=
y=
## © 2011 Zachary S Tseng 7
7. Characteristic equation: repeated real root case
Suppose r is a repeated real root of the characteristic equation, what is the general
solution of the differential equation?
y=
y=
## © 2011 Zachary S Tseng 8
8. Reduction of order
The characteristic equation method only works for equations of constant coefficients.
While we have not (and will not, in this course) learned a technique to solve a generic
second order linear equation with non-constant coefficients, we nevertheless have learned
enough to be able to solve such an equation of non-constant coefficients – provided that
we know one nonzero solution of the equation already. This type of problems is called
reduction of order.
## Why is it called reduction of order?
There are more than one ways to solve such a problem. What we learn in class is a
“synthetic” method. It is different, and easier, than the method in the textbook.
## What are the key steps of our method?
Ex. 3.8.1 Given that y1 = t is a solution, find the general solution of the equation
t2y″ + 2t y′ − 2y = 0.
Find its general solution.
## © 2011 Zachary S Tseng 9
9. Nonhomogeneous second order linear differential equation
y=
## What is the particular / nonhomogeneous solution?
Note that the different context under which the name “particular solution” is used, as
compared to the solution of an IVP.
## © 2011 Zachary S Tseng 10
(2) If g(t) is a polynomial, or a power of t.
.
## (b) Given g(t) = t2 + 2cos 4t, choose Y =
The possible glitch with our starting choices, how to spot it, and how to correct it
When do you need to multiply your starting choice by t (or by t2)? AKA, why you
should always solve for the complementary solution first.
## © 2011 Zachary S Tseng 11
Ex. 3.10.6 Use the method of undetermined coefficients to solve
y″ + y′ − 12y = 9e−4t.
## What to do when g(t) is a product of any two, or all three, of polynomial/power,
exponential, and sinusoidal functions?
The principles:
(i) If g(t) is a sum of several products, do each part separately. (Review the rule about
handling g(t) being a sum/difference.)
(ii) The starting choice for Y shall be a product of the corresponding starting
choice for each component function in g(t). Every possible term in the resulting
product of the basic choice of functions must be represented.
(iii) Each of the terms in the starting choice shall have its own unique coefficient.
(iv) Lastly, the starting choice must be checked against yc for identical terms. If any
shared term is found, then every term in the starting choice must be multiplied by t.
Repeat until no shared term is found.
## © 2011 Zachary S Tseng 12
Ex. 3.10.7 (Exam 1, fall 2007) Determine the most suitable choice of Y(t) for each
equation.
(a) y″ − 4y′ + 8y = 2e2t − 5t2 + sin 2t
## (c) y″ − 4y′ + 8y = t2e−tcos 5t
Ex. 3.10.8 (Exam 2, fall 2001) Solve the initial value problem
y″ + 4y = et, y(0) = 0, y′(0) = 0.
## © 2011 Zachary S Tseng 13
Ex. 3.10.9 Use the method of undetermined coefficients to solve
y″ + 12y′ + 36y = t + 3 − 2e−6t.
Here are a couple exercises to test your familiarity with some of the concepts – how the
different parts are put together to give the solution of a second order linear equation – of
this chapter thus far.
## Review Ex 3.1 Given that y = 3e−5t is a known solution of the equation
y″ + 12y′ + 36y = g(t).
What is the general solution of this equation?
Review Ex 3.2 Given that y1 = 2πe−5t + 2e−4t and y2 = −6e2t + 2e−4t are both known
solutions of a certain equation y″ + p(t)y′ + q(t)y = g(t).
(a) What is the general solution of this equation?
(b) What is the equation? (That is, determine a set of functions p(t), q(t), and g(t), such
that y1 and y2 are its solutions.)
## © 2011 Zachary S Tseng 14
11. Mechanical vibrations
Equation at equilibrium:
(Use it to find Hooke’s constant.)
Equation of motion:
Parameters for the motion equation: mass, damping constant, Hooke’s constant,
applied forcing function, initial displacement and initial velocity.
## Undamped free vibration
Equation becomes:
Solution: u(t) =
The displacement is a simple harmonic motion, oscillating with constant amplitude at the
system’s natural frequency.
Know natural frequency and natural period. (What they are and how to find them.)
u(t) =
## Know how to find the amplitude and phase angle.
Do not confuse the amplitude of the simple harmonic motion, which is constant,
with amplitude of an underdamped system or a system undergoing resonance – in
both latter cases the amplitude is a function of time! The above formula does not
apply to those latter cases.
## © 2011 Zachary S Tseng 15
Damped free vibration
With nonzero damping constant γ, and no forcing function, there are 3 possible types of
displacement function, depending on the roots of the characteristic equation.
Equation:
Know the 3 types of damped system, which can be classified according to the roots of the
characteristic equation. Be sure to understand the differences in their behavior.
Overdamped:
Critically damped:
Underdamped:
## (1) Which case(s) do not produce oscillation?
(2) How often could the equilibrium position be crossed in each case?
## (5) The amplitude of an underdamped system is not constant (as in an undamped
system), but decreasing with time. How to find the maximum/peak displacement?
## © 2011 Zachary S Tseng 16
Ex. 3.11.1 Classify the mass-spring system described by each of the equations below as
undamped, underdamped, critically damped, or overdamped.
(a) u″ + 2u′ + 3u = 0
## (d) 4u″ + 12u′ + 8u = 0
Ex. 3.11.2 (Exam 2, summer 2003) A mass of 1 kg stretches a spring 50 cm. The mass-
spring system has damping of 4 N sec/m. At t = 0 the mass is set in motion from its
equilibrium position with downward velocity of 2 m/sec. Assume that the gravitational
constant,g = 10 m/sec2. (a) Set up an initial value problem that describes this situation.
(b) Solve the initial value problem. (c) What is the quasi-frequency of the system?
## © 2011 Zachary S Tseng 17
Undamped forced vibration
We will only study the case of a periodic forcing function, for example F(t) = F0 cos ωt
or F(t) = F0 sin ωt.
The displacement function behaves differently depending on whether or not the forcing
function’s frequency is equal to the undamped system’s natural frequency.
What is beat?
## © 2011 Zachary S Tseng 18
12. Resonance
A system undergoing resonance will oscillate with progressively larger amplitude that
grows unbounded, increasing linearly with time (it does not grow exponentially, contrary
to a popular belief).
What are the precise conditions, mathematically, necessary for resonance to occur?
## Ex. 3.12.1 A mass-spring system is described by the equation
u″ + 64u = 2cos(ωt) – 3sin(2ωt)
For what value(s) of ω will resonance occur?
## (a) u″ + 2u′ + u = sin(t)
(b) u″ − 4u = 4cos(2t)
## © 2011 Zachary S Tseng 19
MATH 251 Work sheet / Things to know
Chapter 4
## 1. Higher order linear equations
Linear equations of higher ( ≥ 3) order, with constant coefficients, can be solved in the
same fashion as those of the second order. For homogeneous linear equations with
constant coefficients, the characteristic equation method solves them all.
Know the four rules of the characteristic equation method. For an n-th order equation the
general solution has exactly n terms. The characteristic equation also has exactly n roots
(counting duplicates individually). Each root gives one of the n fundamental solutions,
which together form the general solution, according to the four rules:
## i. For distinct real roots:
Ex. 4.1.1 If r = 2, 4, −6, −8 are distinct roots, what are the fundamental solutions?
## ii. For repeated real roots:
Ex. 4.1.2 If r = 6, 6, −5, −5, −5 are real roots (counting repetitions), what are the
fundamental solutions?
## iii. For distinct complex conjugate roots:
Ex. 4.1.3 If r = −2 ± 2i, 5 ± i, 1 ± 7i are distinct roots, what are the fundamental
solutions?
## © 2011 Zachary S Tseng 20
Ex. 4.1.4 If r = 3 ± 5i, 3 ± 5i, 3 ± 5i are complex roots (counting repetitions), what are
the fundamental solutions?
## The above 4 rules are cumulative and can be applied together.
Ex. 4.1.5 Suppose an 8th order linear equation has a characteristic with the following
roots: r = 3, 3, 3, 5, −4 ± i, −4 ± i. Write down its general solution.
## Ex. 4.1.6 Find the general solution of
y(5) − 16y″ = 0.
Question: How many initial conditions must an n-th order linear equation IVP have?
For nonhomogeneous linear equations of constant coefficients, the approach remains the
same: use the characteristic equation to find yc, then the method of undetermined
coefficients can be used to find the particular solution Y. The general solution is their
sum, y = yc + Y. | 3,726 | 13,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.918299 |
https://www.physicsforums.com/threads/finding-the-wave-length-of-glass-plug-and-chug-but-messing-up-somewhere.116003/ | 1,531,824,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589634.23/warc/CC-MAIN-20180717090227-20180717110227-00409.warc.gz | 929,969,068 | 15,078 | # Homework Help: Finding the wave length of glass, plug and chug but messing up somewhere!
1. Mar 30, 2006
### mr_coffee
Hello everyone, i don't think i'm understanidn ghits formula correctly.
THe problem says:
The wavelength of yellow sodium light in air is 589 nm.
(a) What is its frequency?
I found, 5.09E14 Hz
(b) What is its wavelength in glass whose index of refraction is 1.74?
? nm
OKay I'm using the equation;
Wave Length of a Light = wave length of medium/index of medium
But i'm not sure if i'm reading it right, it says:
WaveLength1 = WaveLength2/n
where n is the index of Refraction
This equation relatse the wave lenght of light in any medium to its wavelength in vaccum. It tells us that the greater the index of refraction of a medium, the smaller the wavelength of light in that medium.
So i said, WaveLength1 is the wave length of the medium, which is glass in my case. I said WaveLength2 is the wavelength of yellow sodium light in air is 589 nm. And n is of course 1.74;
WaveLength = (589E-9)(1.74) = .000001, he wanted it in nm, so i said it was: 1000nm, which was wrong. ANy ideas on what I f'ed up on ?
THanks@!
2. Mar 30, 2006
### Staff: Mentor
You multiplied where you should have divided. (And, since they wanted the answer in nm, why did you change units?)
$$\lambda_n = \lambda_1/n$$
3. Mar 30, 2006
### Galileo
You said it yourself. The wavelength in a denser medium (index n) is shorter. In particular equal to $\lambda/n$ where $\lambda$ is the wavelength in vacuum. So why do you multiply the wavelength by n?
4. Mar 30, 2006
### mr_coffee
Thanks Doc,
Is $$\lambda_n$$
the Wave Length of Light, in this case 598nm
and
$$\lambda_1$$
is the wave length of the medium i'm wanting to find, in this case, glass?
5. Mar 30, 2006
### mr_coffee
Ooo n/m i had those 2 mixed up, thanks guys! i got it right now! wee!
6. Mar 30, 2006
### Staff: Mentor
No, just the opposite. (As you've figured out already.) $$\lambda_n$$ is the wavelength in a medium with index of refraction = n; $$\lambda_1$$ is the wavelength in a medium with index of refraction = 1 (vacuum or air). (You are wanting to find the wavelength where n = 1.74.)
7. Mar 30, 2006
### mr_coffee
I was going to start a new thread but you guys already have a background on what i'm doing. There is a part c to this question and it says:
(c) From the results of (a) and (b) find its speed in this glass.
Well here is my work, there is alot of jibberish on it, but I put a box around part c, and i'm confused on why i'm messing this one up. If your confused on anything I did, i'll explain!
Part A is right, so is part B now:
(a) What is its frequency?
I found, 5.09E14 Hz
(b) What is its wavelength in glass whose index of refraction is 1.74?
338.5 nm
http://img86.imageshack.us/img86/5695/lastscan2sd.jpg [Broken]
My bad! I figured it out while explaining to you... I used 338.5nm, i thought i was m in my calculation the right answer is:
1.723E8
Last edited by a moderator: May 2, 2017
8. Mar 31, 2006
### Galileo
The speed in the medium is c/n. Actually, it's that result that leads to $\lambda/n$. | 906 | 3,117 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-30 | latest | en | 0.914716 |
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# hw3solutions - EE 369 Homework #8 Solutions (Sixth Edition)...
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EE 369 Homework 3 Solutions (Sixth Edition) Section 1.4 Decide what conclusion can be reached, and justify your answer 1) All flowers are plants. Pansies are flowers. The conclusion is that pansies are plants. Proof steps: F(x) represent the relation that x is a flower P(x) represent the relation that x is a plant Pansies is a constant symbol 1. (\forall x) F(x) --> P(x) hyp 2. F(pansies) hyp 3. F(Pansies)-->P(Pansies) 1,ui 4. P(Pansies) 2,3,mp 4) Some flowers are red. Some flowers are purple. Pansies are flowers. No conclusion is possible. Just because pansies are flowers,it does not make them either red or purple. 6) Justify each step in the proof sequence for (\exists x)[P(x)-->Q(x)]-->[(\forall x)P(x)-->(\exists x)Q(x)] 1. (\exists x)[P(x)-->Q(x)] hyp 2. P(a)-->Q(a) 1,ei 3. (\forall x)P(x) deduction method hypothesis 4. P(a) 3,ui 5. Q(a) 2,4,mp 6. (\exists x)Q(x) 5,eg 9) Consider the wff (\forall y)(\exists x)Q(x,y)-->(\exists x)(\forall y)Q(x,y) a. Find an interpretation to prove that this wff is not valid. domain is the integers, Q(x,y) is "x<y"; for every y there is an x with x<y but there is no single integer x that is less than every single integer y. b. Find the flaw in the following proof of this wff. 1. (\forall y)(\exists x) Q(x,y)
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## This note was uploaded on 09/24/2009 for the course ECE 369 taught by Professor Bagchi during the Spring '08 term at Purdue University-West Lafayette.
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hw3solutions - EE 369 Homework #8 Solutions (Sixth Edition)...
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Ask a homework question - tutors are online | 584 | 1,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-09 | longest | en | 0.866367 |
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# limx→1f(x)=5, and limx→1g(x)=2 find the value of limx→1([g(x)]f(x)) ___
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## If limx→af(x) and limx→ag(x) exists finitely, then limx→a(g(x))f(x)=[limx→ag(x)]limx→af(x) ⇒limx→1([g(x)]f(x))=(limx→1g(x))limx→1f(x) = 25 = 32
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Справка Scilab >> Xcos > Solvers > DDaskr
DDaskr (short for Double-precision Differential Algebraic equations system Solver with Krylov method and Rootfinding) is a numerical solver providing an efficient and stable method to solve Differential Algebraic Equations systems (DAEs) Initial Value Problems.
### Description
Called by xcos, DDaskr (short for Double-precision Differential Algebraic equations system Solver with Krylov method and Rootfinding) is a numerical solver providing an efficient and stable method to solve Initial Value Problems of the form:
Before solving the problem, DDaskr runs an implemented routine to find consistent values for y0 and yPrime0 .
Starting then with those y0 and yPrime0 , DDaskr approximates yn+1 with the BDF formula:
with, like in CVode, yn the approximation of y(tn) , hn = tn - tn-1 the step size, and the coefficients are fixed, uniquely determined by the method type, its order q ranging from 1 to 5 and the history of the step sizes.
Injecting this formula in (1) yields the system:
Its solving is done through a Newton method, but with either direct or preconditioned GMRes Krylov iterations.
• Direct iteration: start by rewriting the system into:
with J an approximation of the Jacobian:
α changes whenever the step size or the method order varies.
Then, an implemented direct dense solver is used and we go on to the next step.
• GMRes Krylov method: first, precondition the system by applying the Jacobian matrix mentioned above.
Secondly, compute the next Krylov space basis and update the Hessenberg matrix.
Test for convergence for the first time. If it doesn't pass, calculate the residual, which will lead to a new potential solution, and iterate until that residual satisfies convergence.
DDaskr uses the history array to control the local error yn(m) - yn(0) and recomputes hn if that error is not satisfying.
The function is called in between activations, because a discrete activation may change the system.
Following the criticality of the event (its effect on the continuous problem), we either relaunch the solver with different start and final times as if nothing happened, or, if the system has been modified, we need to "cold-restart" the problem by reinitializing it anew and relaunching the solver.
Averagely, DDaskr accepts tolerances up to 10-11. Beyond that, it returns a Too much accuracy requested error.
As of now, DDaskr can only be applied on systems where the root functions are punctually 0, and not 0-flat.
### Example
The 'Modelica Generic' block returns its continuous states, we can evaluate them with DDaskr by running the example:
### Bibliography
Netlib storage, documentation inside of the code.
Sundials Documentation | 654 | 2,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | longest | en | 0.861905 |
https://www.physicsoverflow.org/25714/duality-between-superfluid-insulator-superconductor-magnetic | 1,716,004,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00716.warc.gz | 859,813,883 | 23,049 | # Duality between (1) bosons (superfluid-insulator) and (2) a bulk superconductor in a magnetic field
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In this paper, http://journals.aps.org/prb/abstract/10.1103/PhysRevB.39.2756, the authors establish a correspondence between two-dimensional bosons and a bulk superconductor in a magnetic field. They focus on boson, but it seems to be claimed that it holds even more generally.
(1) 2D bosons (T=0) v.s. (2) Bulk superconductor
Chemical potential $\mu$ v.s. Applied field $H$
Bose density $n$ v.s Total field $B$
Mott insulating phase v.s. Meissner phase
Density wave insulator v.s. Abrikosov flux lattice
Superfluid v.s. Non-superconducting flux line liquid
Supersolid v.s. Non-superconducting flux lattice
Bose glass insulator v.s. superconducting glass
Question 1: Is that "Total field $B$" a typo of magnetization $M$? Since we have:
$$\mu \cdot n \Longleftrightarrow H \cdot M$$
or
$$\mu \cdot n \Longleftrightarrow B \cdot M$$
Question 2: Any physical intuitive picture how does this duality in this table above work?
Here is my understanding -- For example, we can derive them by representing the two equivalent theories of superfluid with superfluid U(1) phase field $\phi$ in terms of a dual equivalent theory of vortex field $\Phi$ (creating vortex or annihilate anti-vortex). Naturally, we will introduce terms like
$$|d \phi - A|^2 + \dots \Longleftrightarrow A \wedge d a +\dots = A \wedge J_{\text{charge}} +\dots \Longleftrightarrow |d \Phi- a \Phi |^2 + A \wedge d a + \dots$$
I suppose if I introduce the Maxwell term (introducing Coulomb repulsion) $dA \wedge * dA$ with $A \wedge d a$, I can integrate out $A$ to obtain an effective Messiner effect $m^2 A^2$.
More systematically, there are some hints of dualities between (see A Zee's QFT book chap VI.3) (with the help of an extra $A \wedge d a$ term, and integrating out unwanted degree of freedom.):
$$\text{Maxwell}: da \wedge *da \Longleftrightarrow \text{Meissner}: m^2 A^2$$
$$\text{Meissner}: M^2 a^2 \Longleftrightarrow \text{Maxwell}: dA \wedge * dA$$
$$\text{Chern-Simons}: a \wedge da \Longleftrightarrow \text{Chern-Simons}: A \wedge dA$$
Maxwell term (introducing Coulomb repulsion) can cause the Mott-insulating phase, and we have argue it is dual to an effective Messiner effect.
So far we obtain:
$$\text{Mott insulating phase v.s. Meissner phase}$$
Again,
Question 2: Any physical intuitive picture how this (rest of) duality in this table above work? Physically intuitively?
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##### Start file
Given two vectors $$\vec{x}$$ and $$\vec{y}$$ in $$\mathbb{R}^n$$,
their '''dot product''' or '''inner product''' is defined as the following:
$\sum_{i=0}^{n} x_i \, y_i$
----
Integration by parts is another way of writing the product rule of differentiation.
For two functions $$f(x)$$ and $$g(x)$$, the following are equivalent:
\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x) \, g(x) \right) &= f'(x) \, g(x) + f(x) \, g'(x) \\ \int f(x) \, g'(x) \, \mathrm{d}x &= f(x) \, g(x) - \int f'(x) \, g(x) \, \mathrm{d}x \end{align}
----
Matrix multiplication is not commutative
\begin{align} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \, \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} &\ne \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \, \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ \begin{bmatrix} a_{11} \, b_{11} + a_{12} \, b_{21} & a_{11} \, b_{12} + a_{12} \, b_{22} \\ a_{21} \, b_{11} + a_{22} \, b_{21} & a_{21} \, b_{12} + a_{22} \, b_{22} \end{bmatrix} &\ne \begin{bmatrix} a_{11} \, b_{11} + a_{21} \, b_{12} & a_{12} \, b_{11} + a_{22} \, b_{12} \\ a_{11} \, b_{21} + a_{21} \, b_{22} & a_{12} \, b_{21} + a_{22} \, b_{22} \end{bmatrix} \begin{align}
''Quod erat demonstrandum''.
##### End file
Given two vectors $\vec{x}$ and $\vec{y}$ in $\mathbb{R}^n$,
their '''dot product''' or '''inner product''' is defined as the following:
<center>$\sum_{i=0}^{n} x_i \, y_i$</center>
----
Integration by parts is another way of writing the product rule of differentiation.
For two functions $f(x)$ and $g(x)$, the following are equivalent:
<center>\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \left( f(x) \, g(x) \right) &= f'(x) \, g(x) + f(x) \, g'(x) \\ \int f(x) \, g'(x) \, \mathrm{d}x &= f(x) \, g(x) - \int f'(x) \, g(x) \, \mathrm{d}x \end{align}</center>
----
Matrix multiplication is not commutative
\begin{align} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \, \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} &\ne \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \, \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ \begin{bmatrix} a_{11} \, b_{11} + a_{12} \, b_{21} & a_{11} \, b_{12} + a_{12} \, b_{22} \\ a_{21} \, b_{11} + a_{22} \, b_{21} & a_{21} \, b_{12} + a_{22} \, b_{22} \end{bmatrix} &\ne \begin{bmatrix} a_{11} \, b_{11} + a_{21} \, b_{12} & a_{12} \, b_{11} + a_{22} \, b_{12} \\ a_{11} \, b_{21} + a_{21} \, b_{22} & a_{12} \, b_{21} + a_{22} \, b_{22} \end{bmatrix} \begin{align}
''Quod erat demonstrandum''.
#### View Diff
1c1
< Given two vectors $$\vec{x}$$ and $$\vec{y}$$ in $$\mathbb{R}^n$$,
---
> Given two vectors $\vec{x}$ and $\vec{y}$ in $\mathbb{R}^n$,
4c4
< $\sum_{i=0}^{n} x_i \, y_i$
---
> <center>$\sum_{i=0}^{n} x_i \, y_i$</center>
9c9
< For two functions $$f(x)$$ and $$g(x)$$, the following are equivalent:
---
> For two functions $f(x)$ and $g(x)$, the following are equivalent:
11c11
< \begin{align} --- > <center>\begin{align} 14c14 < \end{align} --- > \end{align}</center>
20c20
< $$--- > $48c48 <$$ --- >$
### Solutions by @LAKadlubowski:
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## 18 active golfers, 40 entries
##### Solutions by @LAKadlubowski:
99
###### #16 - Łukasz Kadłubowski / @LAKadlubowski
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# Good one to drill the basics: Another OG clone.
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Good one to drill the basics: Another OG clone. [#permalink]
### Show Tags
31 Aug 2006, 17:58
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Good one to drill the basics: Another OG clone.
Astronomers have theorized that the Big Bang governs the behavior of interstellar dust, particles that comprise the atoms and molecules created in the progenitive explosion and persisting in even the emptiest regions of space
A. persisting
B. persists
C. persisted
D. they persist
E. are persisting
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### Show Tags
31 Aug 2006, 19:46
the atoms and molecules created and persisted
C for me
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### Show Tags
31 Aug 2006, 21:00
It has to be C because atoms created and persisted is correct here.
B is wrong because persists seems like its still persisting.....
And plus atoms, particles.. they are plural, if we were to use , persist could have been a better choice than persists.
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### Show Tags
31 Aug 2006, 22:02
Going completely off the track and between A and E picking A.
Reason: describes a logical flow of events.. created by explosion and persisting in space..
although I know that the formation is not parallel, it maintains the sequence of events.
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### Show Tags
01 Sep 2006, 02:27
u2lover wrote:
the atoms and molecules created and persisted
C for me
That's a trap!! U2Lover Girl, you gotta study those 70 SCs.
Here is a vertical comparison:
Today's SC-
Astronomers have theorized that the Big Bang governs the behavior of interstellar dust, particles that comprise the atoms and molecules created in the progenitive explosion and persisting in even the emptiest regions of space
A. persisting
B. persists
C. persisted
D. they persist
E. are persisting
From the SC test page #19
19. Scientists have recently discovered what could be the largest and oldest living organism on Earth, a giant fungus that is an interwoven filigree of mushrooms and rootlike tentacles spawned by a single fertilized spore some 10,300 years ago and extending for more than 33 acres in the soil of a Michigan forest.
(A) extending
(B) extends
(C) extended
(D) it extended
(E) is extending
The OA for both is (A). Let's master this pattern.
Last edited by GMATT73 on 01 Sep 2006, 03:03, edited 1 time in total.
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### Show Tags
01 Sep 2006, 02:46
Astronomers have theorized that the Big Bang governs the behavior of interstellar dust, particles that comprise the atoms and molecules created in the progenitive explosion and persisting in even the emptiest regions of space
A. persisting
particles that comprise X and Y.... and [particles] persisting
B. persists
C. persisted
the sentence doesn't go with created and persisted.... created is about atoms while persisted is about particles...
D. they persist
they problem
E. are persisting
are is wrong here...
So I am going with A, but I could be wrong....
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01 Sep 2006, 14:04
I found some good notes (not in electronic form though) on verbals (infinitives, gerunds and partciples). I'll post it once I get hold of a scanner (too lazy to type all that) .
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01 Sep 2006, 20:52
This is a straight A.
||ism of past participle and present participle.
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03 Sep 2006, 10:15
GMATT73 wrote:
u2lover wrote:
the atoms and molecules created and persisted
C for me
That's a trap!! U2Lover Girl, you gotta study those 70 SCs.
What are the 70 SCs? Could you pls link me up with it?
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03 Sep 2006, 10:26
gk3.14 wrote:
yes, please .. could you point to the 70 scs?
I think it's here:
http://www.gmatclub.com/phpbb/viewtopic.php?t=34300
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03 Sep 2006, 10:26
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# Good one to drill the basics: Another OG clone.
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,730 | 6,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-47 | latest | en | 0.893529 |
https://brainly.in/question/106845 | 1,484,998,311,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281069.89/warc/CC-MAIN-20170116095121-00314-ip-10-171-10-70.ec2.internal.warc.gz | 807,508,412 | 10,342 | # A rectangle has its length 9 times its width. What is the ratio of its perimeter to the perimeter of the square of the same area?
2
by vsolute
2015-05-06T18:30:42+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.
Rectangle L = 9 * W
area = L * W = 9 W * W = 9 W²
perimeter = 2 L + 2 W = 2 (9 W ) + 2 W = 20 W
square area : 9 W²
square side: √(9W²) = 3 W
perimeter = 4 * 3 W = 12 W
Ratio : 20 W / 12 W = 20 /12 = 5/3
2015-05-06T20:24:56+05:30
There fore the ratio is 5/3 | 253 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-04 | latest | en | 0.909841 |
https://infinitylearn.com/questions/chemistry/k-ml-ethylene-subjected-combustion-with-ml-oxygen-1642929740 | 1,709,398,461,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00422.warc.gz | 311,546,998 | 12,754 | Stoichiometry and Stoichiometric Calculations
Question
# At 300 K, 40 ml of Ethylene is subjected to combustion with 160 ml of Oxygen in a closed vessel. The reaction is highly exothermic. On cooling to 300 K, the volume of gases left in the vessel would be
Moderate
Solution
## Total volume of gases left = (40 + 80) ml = 120 ml
Get Instant Solutions | 97 | 356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-10 | latest | en | 0.892724 |
http://mathhelpforum.com/calculus/82375-optimization-problem.html | 1,521,616,757,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00756.warc.gz | 183,395,978 | 10,340 | # Thread: Optimization problem
1. ## Optimization problem
Dear forum members,
in a problem I was asked to show that when the surface area of a fixed cylinder was at its minimum, the height of the cylinder would be equal to the diameter of the cross section of the cylinder.
I formed the equation for the surface area of the cylinder, and the differentiated it with respect to r(not sure why I choose r, though. Perhaps because it was the only factor that would influence both the area of the "bottoms" of the cylinder as well as the "body".).
A(r)=2pi*r^2 + 2 pi*r*h
A'(r)=4pi*r+2pi*h
solving for the stationary points of the derivative gives 2r=-h .
The result shows basically what I wanted to show, but the minus is probably not supposed to be there.
However, I noticed, that when I expressed h in terms of the volume and the radius
h=V/pi*r^2
and then plugged that into the equation above and differentiated, the result was 2r=h.
Could someone explain to me why these two different ways of doing give a different result?
All help is appreciated.
2. Originally Posted by Coach
Dear forum members,
in a problem I was asked to show that when the surface area of a fixed cylinder was at its minimum, the height of the cylinder would be equal to the diameter of the cross section of the cylinder.
I formed the equation for the surface area of the cylinder, and the differentiated it with respect to r(not sure why I choose r, though. Perhaps because it was the only factor that would influence both the area of the "bottoms" of the cylinder as well as the "body".).
A(r)=2pi*r^2 + 2 pi*r*h
A'(r)=4pi*r+2pi*h ... this derivative is incorrect, you cannot treat h as a constant.
solving for the stationary points of the derivative gives 2r=-h .
The result shows basically what I wanted to show, but the minus is probably not supposed to be there.
However, I noticed, that when I expressed h in terms of the volume and the radius
h=V/pi*r^2
and then plugged that into the equation above and differentiated, the result was 2r=h.
... this is correct because V is a constant, and the variation of h is related to the variation of r.
Could someone explain to me why these two different ways of doing give a different result?
.
3. Thank you so much for your reply!
But why can't I treat h as a constant?(I'm sorry, this is a reall stupid question, but I don't understand why is it incorrect to treat h as a constant)?
4. Originally Posted by Coach
Thank you so much for your reply!
But why can't I treat h as a constant?(I'm sorry, this is a reall stupid question, but I don't understand why is it incorrect to treat h as a constant)?
$\displaystyle V = \pi r^2 h$
if $\displaystyle V$ is constant and $\displaystyle r$ changed, would $\displaystyle h$ stay the same? | 665 | 2,784 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-13 | latest | en | 0.973531 |
https://www.synthet-ic.com/items/lie_algebra | 1,723,692,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00289.warc.gz | 761,839,536 | 5,920 | # Lie Algebra
f : L × L → L, Cross product f(x, y) = x × y. x × (y × z) + z × (x × y) + y (z × x) = 0 Lie bracket [a x + b y, z] = a [x, z] + b [y, z], [x, a y + b z] = a [x, y] + b [x, z] (bilinear). [x, x] = 0 (alternating). [x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0 (Jacobi identity). 0 = [x + y, x + y] = [x, x + y] + [y, x + y] = [x, x] + [x, y] + [y, x] + [y, y] = [x, y] + [y, x]
⇒ [x, y] = -[y, x] (anticommutative). Derivation D [x, y] = [D x, y] + [x, D y].
## Jacobi Identity
x × (y × z) + z × (x × y) + y (z × x) = 0. | 285 | 534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.631813 |
https://yiboyang.com/posts/2016/Jan/19/the-matrix-of-a-linear-map/ | 1,624,626,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00348.warc.gz | 986,907,215 | 4,000 | The Matrix of a Linear Map
Jan 19 2016
This is a series of notes taken during my review of linear algebra, using Axler's excellent textbook Linear Algera Done Right, which will be heavily referenced.
This is based on book section (3.C).
Suppose $$T \in \mathcal{L}(V, W)$$ and $$v_1,...,v_n$$ is a basis of $$V$$ and $$w_1,...,w_m$$ is a basis of $$W$$. The matrix of $$T$$ with respect to these bases is the $$m$$-by-$$n$$ matrix $$\mathcal{M}(T)$$ whose entries $$A_{j,k}$$ are defined by
$$T v_k = \sum_{j=1}^m A_{j,k} w_j = A_{1,k} w_1 + ... +A_{m,k} w_m$$
The $$k$$th column of $$\mathcal{M}(T)$$ consists of the scalars needed to write $$T v_k$$ as a linear combination of $$w_1,...,w_m$$.
Example (3.33): Suppose $$T \in \mathcal{L}(\mathbb{F}^2,\mathbb{F}^3)$$ is defined by
$$T(x, y) = (x +3y, 2x + 5y, 7x + 9y)$$
Find the matrix of $$T$$ with respect to the standard bases of $$\mathbb{F}^2$$ and $$\mathbb{F}^3$$.
Solution: Because $$T(1, 0)=(1,2,7)$$ and $$T(0,1)=(3,5,9)$$, the matrix of $$T$$ with respect to the standard bases is then
$$\begin{bmatrix} 1 & 3\\ 2 & 5\\ 7 & 9 \end{bmatrix}$$
Follow-up question: find the matrix of $$T$$ with respect to the basis $$(1,1), (-1,1)$$ of $$\mathbb{F}^2$$ and $$(1,3,0),(0,-1,0),(0,0,2)$$ of $$\mathbb{F}^3$$.
Solution: Because $$T(1,1) = (4,7,16) = 4(1,3,0) + 5(0,-1,0) + 8(0,0,2)$$, and $$T(-1,1) = (2,3,2) = 2(1,3,0) -3(0,-1,0) + (0,0,2)$$, the matrix with respect to these (non-standard) bases is
$$\begin{bmatrix} 4 & 2\\ 5 & -3\\ 8 & 1 \end{bmatrix}$$
Key point: the matrices can change with bases, but the underlying operator stays the same! Later on we will see how trace and determinant are really properties of operators, not just their matrix representations. | 681 | 1,744 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-25 | latest | en | 0.783248 |
http://mathhelpforum.com/calculus/100569-another-derivative-proof.html | 1,480,928,405,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541556.70/warc/CC-MAIN-20161202170901-00059-ip-10-31-129-80.ec2.internal.warc.gz | 172,607,807 | 10,469 | 1. ## Another derivative proof
Could someone please check this proof for me?
Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and $|f'(x)| \le |f(x)|$ then f(x) = 0 for x in (0,1).
Proof:
Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that $\frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*)$. We know that $f'(c*) \le f(c*)$. So $\frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*)$ since c is in (0,1).
It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.
I think the proof is a bit shaky in the last paragraph. How is it?
2. Originally Posted by JG89
Could someone please check this proof for me?
Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and $|f'(x)| \le |f(x)|$ then f(x) = 0 for x in (0,1).
Proof:
Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that $\frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*)$. We know that $f'(c*) \le f(c*)$. So $\frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*)$ since c is in (0,1).
It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.
I think the proof is a bit shaky in the last paragraph. How is it?
There is an perfect solution!
Assume $|f(x)|\leq M\forall x\in (0,1)$
$\forall x\in (0,1),|f(x)|=|f'(x_1)||x|\leq |f(x_1)||x|=$ $|f'(x_2)||x||x_1|=...\leq|f(x_n)||x||x_1|...|x_n|\ leq M|x|^n$
Let $n\rightarrow 0$,we get $f(x)=0$
3. Thanks ynj!
How did you know to write down those inequalities though? Was it by trial-and-error?
4. Originally Posted by JG89
Thanks ynj!
How did you know to write down those inequalities though? Was it by trial-and-error?
Because I have seen this problem in my book before. | 762 | 2,196 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-50 | longest | en | 0.896924 |
https://www.internet4classrooms.com/printables/common_core/math_mathematics_5th_fifth_grade/description_download_29958-CCSS.Math.Content.5.NF.A.1.htm | 1,675,298,671,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00556.warc.gz | 843,882,087 | 8,150 | @internet4classr I4C
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Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles. | 444 | 1,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-06 | longest | en | 0.791395 |
https://qa.nmrwiki.org/question/642/?sort=latest | 1,718,983,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00707.warc.gz | 417,851,991 | 10,048 | 0 Hi guys, I'm stuck in attempting this question and need some help understanding how to answer it. Q.) The chemical shift of H in benzene is 7.4 ppm. what will the resonance frequency be at 200mhz I know that the the resonance frequency is proportional to the magnetic field intensity and think I need to use the Larmor equation, Vo = gamma Bo/2pi , but can't quite understand how? Please help. Thank-you. asked Feb 12 '14 at 11:54
0 7.0460. Tesla for 300 MHz (protons) 4.6973333 Tesla for 200 MHz If the Magnetic Field is 4.6973333 Tesla, then bare proton resonance frequency is 200 MHz. Then, if you have a TMS sample, in the Spectrometer with 4.6973333 Tesla magnetic field, then according to the standard listed values, with reference to bare nucleus resonates at 30 ppm lower field (higher frequency) direction to TMS. In other words, at 4.6973333 Tesla, TMS resonates at 30-ppm lower frequency to bare Nucleus resonance frequency of 200 MHz. 30 ppm corresponds to 6000 Hz (6 KHz) frequency at 4.6973333 Tesla Magnetic field. Thus, at 4.6973333 Tesla, TMS would resonate at (200,000000-6000=199994000 Hz) 199.994 MHz. Since benzene protons resonate 7.4 ppm lowfield (high frequency) side of TMS, at 4.6973333 Tesla benzene proton resonance frequency would be 199.994 MHz + (7.4 x 200) Hz =199.994 MHz + 1,480 Hz =199.994 MHz + 1.480. KHz =199.994 MHz + 0.001480 MHz =199.995480 MHz answered Apr 02 '14 at 00:25
0 "When you say a "200 MHz" spectrometer, that "means that protons resonante somewhere around 200 Mhz", for an unambiguous calculation, Try to calculate the Resonance frequency for the Reference Compund the "TMS", this requires the frequency of the nucleus based on its gamma value (bare nucleus), add the shielding value for the reference and calculate for the experimental measured field/ or calibrate with the set Transmitter frequency of the Spectrometer. answered Feb 28 '14 at 01:00
0 Try to know the two terms "Spectrometer Frequency" and "Resonance Frequency" By way of trying to get to know the difference, then you may arrive at a solution for the problem posed as a question here in these postings. If still there is ambiguities, clearly spell out the ambiguity again in these boxes. S.Aravamudhan answered Feb 17 '14 at 23:34
0 Maybe it's better if you ask the question like this: The chemical shift of H in benzene is 7.4 ppm. what will the resonance frequency be relative to reference TMS, at 200MHz? The answer is 7.4 * 200 in unit Hz, which is the relative resonance frequency. If you want to get the absolute resonance frequency at 7.4 ppm or any other position, the key is to get the absolute frequency right for the reference TMS. You need the basic frequency parameter from the type of NMR instrument you are referring to and the chemical shift referencing method for your instrument. Dejian seeking new job opportunities in NMR (dejian "at" gmail "dot" com) answered Feb 16 '14 at 09:03 dejian 11 It is important to keep the 1/1,000,000 factor of the "ppm" units when determining the delta frequency. Without it, the delta frequency calculation results in many MHz of error. - Jerry Hirschinger (Feb 20 '14 at 10:56)
1 The Larmor equation that you give neglects to include the screening constant, otherwise known as the chemical shift. This is what your 7.4 ppm represents. When you say a "200 MHz" spectrometer, that means that protons resonante somewhere around 200 Mhz, the exact frequency of which depends on the exact magnetic field, including any B0 or Z0 offset, and the chemical shift of the substance that you are observing. Vo = gamma Bo(1-sigma)/2pi Where sigma is essentially the chemical shift, although referenced to gamma for the bare nucleus in this case. Since bare nuclei are hard to come by, we use a scale referenced to a standard, usually TMS for protons (which I think is about 17 ppm to low frequency of the bare proton). On a 200 MHz spectrometer from one particular vendor (Bruker), the field will typically be adjusted so that the protons of TMS resonate at exactly 200.13 MHz. This means that benzene will be 7.4 ppm or 1481 Hz higher (I know that seems strange, but that is just the way chemical scale is defined) in frequency, or 200.131481 MHz. Chemical shift is a relative dimensionless parameter and as such only defines an absolute frequency when refered to the frequency of a standard. answered Feb 13 '14 at 12:23 Kirk Marat 711
0 ppm = (frequency:sample - frequency:reference)/(operating frequency of spectrometer) Reference ppm is usually 0ppm so frequency:reference = operating frequency of spectrometer. delta frequency = ppm * operating freq (This gives you distance from resonant frequency) absolute frequency = operating frequency + delta frequency You can use the exact resonant frequency of your spectrometer for more accuracy. answered Feb 13 '14 at 07:31 It is important to keep the 1/1,000,000 factor of the "ppm" units when determining the delta frequency. Without it, the delta frequency calculation results in many MHz of error. - Jerry Hirschinger (Feb 20 '14 at 10:55)
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Last updated: Apr 02 '14 at 00:25 | 1,327 | 5,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-26 | latest | en | 0.879565 |
https://mathisvisual.com/visualizing-area-of-a-triangle-formula/ | 1,726,517,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00142.warc.gz | 361,714,844 | 21,973 | # Visualizing The Area of a Triangle Formula
Build an understanding of the area relationship between triangles and rectangles through this set of Area of a Triangle Number Talk Prompts.
## In This Set of Visual Number Talk Prompts…
Students will emerge the formula for the area of a triangle by taking half of the area of a rectangle.
## Intentionality…
This visual math talk and purposeful practice serve to develop a deeper understanding of the following big ideas.
• The area of a rectangle can be determined by multiplying the length of its base by the length of its height;
• The area of a rectangle is used to determine the area formulas for other polygons;
• Using “base” and “height” rather than “length” and “width” can support student understanding when determining the area formulas for triangles and all quadrilaterals;
• A parallelogram is any quadrilateral with two sets of parallel sides;
• A rectangle is a parallelogram;
• The relationships between rectangles, parallelograms, and triangles can be used to determine area formulas;
• Any triangle can be doubled to create a parallelogram;
• Congruent shapes have the same size and shape, but not necessarily the same orientation.
## Visual Number Talk Prompt
Begin playing the visual number talk prompt video and then be ready to ask students:
Calculate the area of each triangle in square metres.
Craft a convincing argument without the use of a calculator.
As you may immediately notice, we have given students two congruent triangles and have conveniently arranged them such that together, they create a rectangle. This is intentional to ensure that all students explicitly see that every rectangle can be partitioned diagonally into two congruent triangles. From this visual math talk prompt, we hope to reiterate that partitioning any rectangle diagonally into two congruent triangles will result in each triangle having half of the area of the rectangle.
Therefore, it may feel intuitive to some students to first find the area of the rectangle and simply half that quantity to determine the area of each of the resulting triangles.
While some students may perform the operation of taking half of the area of the rectangle (or dividing the area by 2), it is likely that many will not naturally write their thinking down symbolically as you see on the right of the screen. The goal here is not necessarily to ensure that all students “copy down” the formula, but rather so they can make a connection between the intuitive actions they had completed to find the area of the triangles to the symbolic representation of that same thinking.
It is important to explicitly highlight that the two triangles that were created by partitioning the rectangle diagonally result in congruent triangles as this may not be obvious to all students immediately.
It could also be helpful to show students that despite taking half of the area of the rectangle (or dividing the area by 2) revealing the area of each congruent triangle, the same operation can be done to determine the area of other congruent shapes that result when partitioning the original rectangle in half. For example, horizontally partitioning the rectangle into halves allows us to complete the same mathematical operation to reveal the same area.
This idea helps to highlight the fact that representing mathematical thinking of area symbolically can strip away information about how the units of area are arranged (i.e.: as a rectangle, a triangle, or some other shape).
## Want to Explore These Concepts & Skills Further?
Two (2) additional number talk prompts are available in Day 2 of the Covering Ground problem based math unit that you can dive into now.
Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page.
Did you use this in your classroom or at home? How’d it go? Post in the comments!
Math IS Visual. Let’s teach it that way.
#### 1 comment
• Megan says:
There is no sound on the video… | 790 | 4,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-38 | latest | en | 0.887266 |
http://intelgo.ru/fourier-rekker/ | 1,618,800,729,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00039.warc.gz | 48,481,359 | 6,466 | # Fourier rekker
Examples of successive approximations to common functions using Fourier series are illustrated above. In particular, since the superposition principle holds for solutions of a linear homogeneous ordinary differential equation, if such an equation can be solved in the case of a single sinusoi the solution for an arbitrary . In mathematics, a Fourier series is a way to represent a function as the sum of simple sine waves. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials). Realfag › Matematikk › Matematisk analyse Bufret 28.
Innen den harmoniske analyse brukes Fourier – rekker til å uttrykke sammensatte svingninger som en sum av enkle svingninger. The Fourier Series allows us to model any arbitrary periodic signal with a combination of sines and cosines. Error loading player: Could not load player configuration.
Foreleser: Dag Wessel-Berg Lengde: 00:32:19. Til serie Se video på. Fourier rekka til f(x) vil konvergere mot f(x) overalt unntatt i eventuelle punkter x = xder f(x) er diskontinuerlig. I slike punkter vil Fourierrekka konvergere mot.
Derivative numerical and analytical calculator. EngMathYT This is a basic introduction to Fourier series and how. This applet demonstrates Fourier series, which is a method of expressing an arbitrary periodic function as a sum of cosine terms. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is composed of.
To select a function, you may press one of the following buttons: Sine, . Konvergens – Fourirer rekker. Finne Fourier- rekke eksempel 1. R- Geometriske rekker. Stykkevis konintuerlig ( d) Periodisk og stykkevis kontinuerlig.
This may not be obvious to many people, but it is demonstrable both mathematically and graphically. En funksjon er “T- periodisk funksjon”. Practically, this allows the user . N are non-negative, and the radian phase angles satisfy £ q,. To explore the Fourier series approximation, select a labeled signal, use the mouse to sketch one period of a signal, or use the mouse to modify a selected . This brings us to the last member of the Fourier transform family: the Fourier series.
The time domain signal used in the Fourier series is periodic and continuous. Chapter showed that periodic signals . | 564 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-17 | latest | en | 0.620661 |
http://mathhelpforum.com/pre-calculus/6907-lines.html | 1,527,233,815,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867046.34/warc/CC-MAIN-20180525063346-20180525083346-00183.warc.gz | 185,481,927 | 9,880 | 1. ## Lines
Consider the line L containg the points (-3,8)and (6,-4) what is the length of the hypotenuse of the right triangle formed b the intersection of L and the x and Y axes
2. Hello, Dragon!
Did you make a sketch?
Consider the line $\displaystyle L$ containg the points (-3,8)and (6,-4).
What is the length of the hypotenuse of the right triangle
formed by the intersection of $\displaystyle L$ and the $\displaystyle x$- and $\displaystyle y$-axes?
Code:
* (-3,8) |
* |
* Q|
*
| * P
- - - - - - - - - + - - - - - * - - - - - - -
| * (6,-4)
| *
| *
Game plan
The line $\displaystyle L$ contains the points (-3,8) and (6,-4).
. . We want the equation of that line.
Then we want its x-intercept $\displaystyle P$ and y-intercept $\displaystyle Q$.
. . Then we want the distance $\displaystyle \overline{PQ}$.
The slope of line $\displaystyle L$ is: .$\displaystyle m \:=\:\frac{-4 - 8}{6 -(-3)} \:=\:\frac{-12}{9}\:=\:-\frac{4}{3}$
The equation of the line through (6,-4) with slope $\displaystyle -\frac{4}{3}$ is:
. . $\displaystyle y - (-4)\:=\:-\frac{4}{3}(x - 6)\quad\Rightarrow\quad y\:=\;-\frac{4}{3}x + 4$
For the $\displaystyle x$-intercept, let $\displaystyle y = 0$ and solve for $\displaystyle x.$
. . $\displaystyle 0 \:=\:-\frac{4}{3}x + 4\quad\Rightarrow\quad x = 3$ . . . $\displaystyle x$-intercept: $\displaystyle P(3,0)$
For the $\displaystyle y$-intercept, let $\displaystyle x = 0$ and solve for $\displaystyle y.$
. . $\displaystyle y\:=\:-\frac{4}{3}\cdot0 + 4\quad\Rightarrow\quad y = 4$ . . . $\displaystyle y$-intercept: $\displaystyle Q(0,4)$
The distance from $\displaystyle P(3,0)$ to $\displaystyle Q(0,4)$ is:
. . $\displaystyle PQ\:=\:\sqrt{(0-3)^2 + (4-0)^2} \;=\;\sqrt{9+16}\;=\;\sqrt{25}\;=\;\boxed{5}$ | 632 | 1,855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-22 | latest | en | 0.608344 |
http://www.homeownershub.com/maintenance/gravel-compaction-ratio-896553-.htm | 1,481,402,841,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543567.64/warc/CC-MAIN-20161202170903-00308-ip-10-31-129-80.ec2.internal.warc.gz | 509,842,240 | 12,861 | # Gravel compaction ratio.
• posted on July 13, 2016, 2:42 pm
I am in the midst of a fair-sized landscaping job. (Two 4-walled interlocking brick flower boxes, sidewalk, patio, etc.)
I am planning to buy a compactor instead of renting one because I plan a separate job next year, so I will have ample time to compact the existing soil (clay/sand) where I have recently dug plus perform an unhurried job on each of the components.
I can calculate the volume of compacted gravel needed, but I need to know the expected compaction ratio to enable me to order the correct amount of 0 -> 3/4" compacting gravel.
<% if( /^image/.test(type) ){ %>
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<%-name%>
• posted on July 13, 2016, 7:59 pm
On Wed, 13 Jul 2016 11:16:46 -0700, Oren wrote:
It is a special mix of sizes 0 to 3/4" specifically made for compacting. Judging from the dark grey colour, it appears to be some some sort of basalt. (Hardness 6; slightly easier to break than granite.)
Yes, since I plan to but one, I will not be in a hurry, thus I can do it a couple inches at a time.
But my question was how much compacting I can expect. For example, if I buy enough uncompacted gravel for a 10" deep hole, assuming I compact it 2" at a time, how thick of a base can I expect to get?
--
http://mduffy.x10host.com/index.htm
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 13, 2016, 8:37 pm
Mike Duffy posted for all of us...
That is probably a good idea. Compact it just prior to each component. The reason I say this is because-not knowing where you live and time-lines the soil may be affected by a freeze/thaw cycle or drainage.
I would ask the provider of the stone. I don't know what the ratio would be for your area. Like Oren stated do it in 2" lifts and compact. Back fill low areas with high areas. I believe landscapers would just use a 4" lift and compact. Do you intend to install a concrete footer?
--
Tekkie
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on July 13, 2016, 9:45 pm
On Wed, 13 Jul 2016 16:37:28 -0400, Tekkie® wrote:
Yes, for the entire 5' * 3' base of a stairway (also will be the same brand/style as the wall) as well as two 4-walled flower boxes, even though 2 walls of each will be buried. My wife wants it that way, i.e. a course of capstones all around each box instead of a 'drop-off' from the sidewalk into the flowers. I know I will end up buying twice as much bricks, but this way I can level the entire base for each flower box. Otherwise, I would need to put other footers exactly the correct height higher than the footers under the exposed walls in order to make the last course match.
The concrete footers are 2' long, 15" wide, and mostly 2.5" thick. (For about 9" to match the depth of a brick course, then tapers to 1.25" over the other 6" of width.) I will use the thick ends to support the walls, which will be 9" deep (3.5" height each course).
Should I use footers for the sidewalk parts as well? (Interlocking paving bricks. No automobiles.) I'm thinking to try without, because I can always lift out the paving tiles and dig out 2.5" of compacted gravel to put in a layer of footers if the sidewalk ends up being too uneven too quickly.
--
http://mduffy.x10host.com/index.htm | 899 | 3,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-50 | latest | en | 0.924328 |
https://www.physicsforums.com/threads/classroom-demonstrations-for-younger-students.850886/ | 1,571,530,925,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00337.warc.gz | 1,037,861,220 | 21,932 | # Classroom demonstrations for younger students
#### BryceUnderwood
I'm a senior in high school taking AP Physics II, and I was recently elected as the president of my schools Science Club. Our first thing to do is to go to the local elementary schools and give interesting physics demonstrations to the younger students grades K-6. I really enjoy space and I know many children like that too, so I was planning on taking volunteers to be "Planets" and line them to scale to show how far the planets are in comparison to the earth. I was just wondering if there are any other interesting demonstrations for younger children that have to do with space, motion, forces, or anything that they will maybe understand somewhat easily. Thanks!
Related STEM Educators and Teaching News on Phys.org
#### fresh_42
Mentor
2018 Award
Those funnels in which you can roll in a coin to demonstrate gravity. (Or a mat made of rubber with a bowling ball on it.)
A candle (at night or in the dark) at different distances to demonstrate the change in luminosity.
Let someone juggle balls and throw a heavier one into his play to catch it: demonstrates how heavy objects (nemesis?) can destroy planetary systems.
Throwing pebbles (or other objects) into a sand pit (let them do their best to achieve high velocities!) to demonstrate craters or throwing them on a target far away to demonstrate how rare it is to get a hit.
Destroying balls of plaster to demonstrate impacts.
Kepler's laws with an elastic band.
2 Kids moving 5 yards left: one at a distance of let's say 10 yards and one at 100 yards to get a feeling of fix stars, distances resp.
#### vela
Staff Emeritus
Homework Helper
I'm a senior in high school taking AP Physics II, and I was recently elected as the president of my schools Science Club. Our first thing to do is to go to the local elementary schools and give interesting physics demonstrations to the younger students grades K-6. I really enjoy space and I know many children like that too, so I was planning on taking volunteers to be "Planets" and line them to scale to show how far the planets are in comparison to the earth. I was just wondering if there are any other interesting demonstrations for younger children that have to do with space, motion, forces, or anything that they will maybe understand somewhat easily. Thanks!
Make cards with small circles drawn on them to scale to indicate the actual size of the planets. This will reinforce how tiny the planets are compared to the size of the solar system. You could also add a speed-of-light aspect to this. Have one of the kids start at the Sun and walk toward the planets at the correct rate. Even though the speed of light is very fast compared to our everyday experience, it's pretty slow when you see it in comparison to the scale of the solar system and universe.
Mixing corn starch and water is an easy demonstration of a non-Newtonian fluid. Have the kids try make the mixture and experience how it quickly becomes very difficult to stir. Also, when they jab the surface with a finger, it'll feel solid, but if they gently press, their finger will readily sink in.
If you prepare mixtures of salt water with different salinity and colors, you can layer them on top of each other, showing how the density changes with salt concentration.
If you have a little money to spend, get the Static Electricity Science Kit with a Fun Fly Stick. It has a number of fun and easy demos.
Tie-dyed milk is another fun one. You put a few drops of food coloring on whole milk, and then add some dish washing soap. (You should definitely try this one on your own first to find out what techniques and soap work best.)
YouTube is actually a good source for demonstration ideas.
#### Mister T
Gold Member
The key to success is having the right equipment. Do you have access to a portable vacuum pump and liquid nitrogen? There's a lot you can do that will fascinate children of all ages.
#### berkeman
Mentor
The key to success is having the right equipment. Do you have access to a portable vacuum pump and liquid nitrogen? There's a lot you can do that will fascinate children of all ages.
You're going to wheel a liquid nitrogen Dewar into an elementary school? Why not just show them some YouTube videos instead -- would be a lot safer, IMO...
#### Mister T
Gold Member
You're going to wheel a liquid nitrogen Dewar into an elementary school?
A 20-liter Dewar flask, yes. I've done it many times. I also ignite a soup-can cannon that uses a lighter fluid explosion to propel a tennis ball. Also use a sledge hammer to break a cinder block on the chest of a student lying on a bed of nails.
Why not just show them some YouTube videos instead -- would be a lot safer, IMO...
Safer, yes. Anywhere near as effective, no. The day they make me do that is the day I stop. I am always amazed by the fact that no one questions my professional capacity to do this safely.
#### fresh_42
Mentor
2018 Award
A 20-liter Dewar flask, yes. I've done it many times. I also ignite a soup-can cannon that uses a lighter fluid explosion to propel a tennis ball. Also use a sledge hammer to break a cinder block on the chest of a student lying on a bed of nails.
This reminds me on my chemistry courses at school. We heard about how to fish with potassium, how to distill alcohol in harsh times, learned that a chemist who got caught killing his wife has missed his profession, that cats don't drink alcohol unless you frustrate them and so on. Funny stuff, however, my knowledge on chemistry itself is rather poor.
#### berkeman
Mentor
A 20-liter Dewar flask, yes. I've done it many times. I also ignite a soup-can cannon that uses a lighter fluid explosion to propel a tennis ball. Also use a sledge hammer to break a cinder block on the chest of a student lying on a bed of nails.
Safer, yes. Anywhere near as effective, no. The day they make me do that is the day I stop. I am always amazed by the fact that no one questions my professional capacity to do this safely.
Do you want me to feel comfortable about your advice for elementary school student demonstrations? You are falling short of making that case so far, IMO...
#### berkeman
Mentor
From your Profile, it looks like you are an instructor at a community college. Can you comment on what the differences might be between demonstrations for community college physics students (great stuff, BTW), and elementary school students?
#### symbolipoint
Homework Helper
Gold Member
Elementary school students should not be allowed to handle nor to get too close to liquid nitrogen, and the high school students performing their demonstrations with this need to be properly trained for handling such materials and associated equipment, and if these high school students are not yet significantly over the age of 18, they may need to be supervised by one or more adults who have the necessary technical knowledge and training.
#### Mister T
Gold Member
From your Profile, it looks like you are an instructor at a community college. Can you comment on what the differences might be between demonstrations for community college physics students (great stuff, BTW), and elementary school students?
Thanks. The demonstration show (and it is a show) works best for people in about 5th grade, although it's definitely a big hit for people of every age. The intent here is to spark interest in science and have some fun learning some science.
For college students enrolled in a physics class I don't do the entire show. They will see some of the demonstrations at appropriate points in the course. The intent here is to teach them the stuff in the curriculum, and if they can have some fun doing it all the better, but it's of secondary importance.
#### Mister T
Gold Member
Elementary school students should not be allowed to handle nor to get too close to liquid nitrogen, and the high school students performing their demonstrations with this need to be properly trained for handling such materials and associated equipment, and if these high school students are not yet significantly over the age of 18, they may need to be supervised by one or more adults who have the necessary technical knowledge and training.
I agree that the demonstrators would have to be old enough, responsible enough, trained, and possibly supervised. I would limit the demonstrations to the ones that are more safe.
#### brainpushups
Tie-dyed milk is another fun one. You put a few drops of food coloring on whole milk, and then add some dish washing soap. (You should definitely try this one on your own first to find out what techniques and soap work best.)
This is a good one and there's lots of room for explanation: surface tension and surfactants are the clear connection, but you could also use it as a vehicle to discuss entropy and the heat death of the universe.
One cheap and effective demonstration/activity would be to have the students make an electrophorus. You need some styrofoam blocks, small pieces of silk cloth, some pie plates, and glue (hot glue is nice because it works quickly, but may not be appropriate for the younger group of kids).
If you have access to a Van de Graff generator then there are at least a dozen demonstrations that you can easily look up.
Optics is also a good place to go. There are plenty of simple and interesting reflection, refraction, dispersion, and diffraction activities/demonstrations. One cheap activity would be to make spectroscopes from toilet paper tubes and CDs (for the diffraction grating - strip off the reflective part of the CD). You can find plans easily online. Optical illusions are also fun. There are many demonstrations about color and color mixing you can also readily find.
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## Count debtors over a certain period
Hey everyone,
I'm having a problem with counting the amount of debtors from the last 28 days.
I have an expression which defines the last 28 days:
If(Today() - Date <= 28 And Today() - Date >= 0, 1, 0) As Date_IsLast28Days
And I need to show this in a single Text Object.
This is the expression I use in the text object.
=Count({< Date2 = {"Date_IsLast28Days"}>} DISTINCT Debtor)
The expression below is what I used before, but this only gives the amount of debtors from the day 28 days ago. Not the whole period of 28 days ago:
=Count({< Date2 = {"\$(=date(today()-28))"}>} DISTINCT Debtor)
What I need is the amount of debtors from the last 28 days (28 days ago - until - yesterday).
I hope someone can help me out on this one!
1 Solution
Accepted Solutions
Hi,
=Count({< Date2 = {">=\$(=date(today()-28))<=\$(=date(today()))"}>} DISTINCT Debtor)
Regards,
Great dreamer's dreams never fulfilled, they are always transcended.
3 Replies
Hi,
=Count({< Date2 = {">=\$(=date(today()-28))<=\$(=date(today()))"}>} DISTINCT Debtor)
Regards,
Great dreamer's dreams never fulfilled, they are always transcended.
Specialist
try this
for till date and >= last 28 days
=Count({< Date2 = {">=\$(=date(today()-28))"}>} DISTINCT Debtor)
for till yesterday and >= last 28 days
=Count({< Date2 = {">=\$(=date(today()-28))<=\$(=date(today()-1))"}>} DISTINCT Debtor)
MVP
If you have created a flag for the last 28 days, why don't you use that?
=Count({<Date_IsLast28Days = {1}>} DISTINCT Debtor)
Community Browser | 472 | 1,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-40 | latest | en | 0.844462 |
https://www.distancesto.com/fuel-cost/om/bidiyah-to-ibra-connection-al-mudhaireb/history/627897.html | 1,696,329,909,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00144.warc.gz | 779,594,587 | 14,837 | # INR1.84 Total Cost of Fuel from Bidiyah to Ibra, Connected by Al Mudhaireb
Your trip to Ibra will consume a total of 0.74 gallons of fuel.
Trip start from Bidiyah and continues through Al Mudhaireb. Trip ends at Ibra.
• 29.4 Miles (Trip Total)
• Route 1 (15.5 mi) Bidiyah » Al Mudhaireb
Fuel Cost INR0.97
Fuel Consumption 0.39 gallons
• Route 2 (13.9 mi) Al Mudhaireb » Ibra
Fuel Cost INR0.87
Fuel Consumption 0.35 gallons
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Bidiyah, Oman and end at Ibra, Oman.
Route goes through Al Mudhaireb, Oman.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Bidiyah to Al Mudhaireb to Ibra plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Bidiyah to Al Mudhaireb to Ibra.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Bidiyah to Al Mudhaireb to Ibra.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Bidiyah to Al Mudhaireb to Ibra.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Ibra are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Bidiyah to Al Mudhaireb to Ibra.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Bidiyah to Al Mudhaireb to Ibra.
Speaking of travel time, a flight to Ibra takes up a lot less. How much less? Flight time from Bidiyah to Al Mudhaireb to Ibra.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Bidiyah to Al Mudhaireb to Ibra.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Bidiyah OM:
Fuel Cost from Bidiyah to Nizwa
Fuel Cost from Bidiyah to Bahla
Fuel Cost from Bidiyah to Wadi Bani Khalid
Fuel Cost from Bidiyah to Sinaw
Fuel Cost from Bidiyah to Al Awabi | 771 | 3,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-40 | latest | en | 0.941378 |
https://www.coursehero.com/file/6535140/hw2/ | 1,487,723,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170864.16/warc/CC-MAIN-20170219104610-00118-ip-10-171-10-108.ec2.internal.warc.gz | 818,322,746 | 136,904 | # hw2 - _ BEH.430/2.795/6.561/10.539/HST.544 Homework Set 2...
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________________________________________________ BEH.430/2.795//6.561/10.539/HST.544 Homework Set 2 Handed out: Wednesday Sept. 22, 2004 Due: Wednesday, Sept 29 by 5pm ________________________________________________
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Problem 2.1: Transient Diffusion across Tissue Membrane (No Reaction) (a) This problem focuses on the use of Separation of Variables to solve for the evolution of the space-time concentration profile of IGF-1, c(x,t), inside a section of tissue having length L (i.e., 1-D profile). As stated in class, the boundary conditions are c(0,t) = c’; c(L,t) = c”; and the initial condition is c(x, 0) = 0. You are asked to carry through the details of the solution, and to derive in detail the coefficients of all the terms in the Fourier series, showing all the steps that yield the final analytical expression for c(x,t) written in lecture and on the overhead (posted with the lecture overheads). You can assume in part (a) that the partition coefficient valid at x = 0 and x = L is K = 1. (b) Suppose K
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## This note was uploaded on 11/11/2011 for the course BIO 20.410j taught by Professor Rogerd.kamm during the Spring '03 term at MIT.
### Page1 / 3
hw2 - _ BEH.430/2.795/6.561/10.539/HST.544 Homework Set 2...
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Ask a homework question - tutors are online | 446 | 1,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-09 | longest | en | 0.893603 |
https://www.johndcook.com/blog/2023/03/17/hyperbolic-secant-distribution/ | 1,722,882,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00490.warc.gz | 663,595,548 | 10,303 | # Hyperbolic secant distribution
I hadn’t run into the hyperbolic secant distribution until I saw a paper by Peng Ding [1] recently. If C is a standard Cauchy random variable, then (2/π) log |C| has a hyperbolic secant distribution. Three applications of this distribution are given in [1].
Ding’s paper contains a plot comparing the density functions for the hyperbolic secant distribution, the standard normal distribution, and the logistic distribution with scale √3/π. The scale for the logistic was chosen so that all three distributions would have variance 1.
There’s something interesting about comparing logistic distribution and the hyperbolic secant distribution densities: the former is the square of the latter, aside from some scaling, and yet the two functions are similar. You don’t often approximate a function by its square.
Here’s a plot of the two densities.
The hyperbolic secant density, the blue curve, crosses the logistic density around ± 0.56 and around ± 2.33.
The hyperbolic secant distribution has density
and the logistic distribution, as scaled in above, has density
and so
## Related posts
[1] Peng Ding. Three Occurrences of the Hyperbolic-Secant Distribution. The American Statistician , Feb 2014, Vol. 68, No. 1 (2014), pp. 32-35 | 298 | 1,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.912849 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-6-test-page-406/23 | 1,539,731,520,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510893.26/warc/CC-MAIN-20181016221847-20181017003347-00248.warc.gz | 935,954,147 | 13,172 | ## Intermediate Algebra (6th Edition)
$x=\dfrac{2}{7}$
The factored form of the given equation, $\dfrac{3}{x+2}-\dfrac{1}{5x}=\dfrac{2}{5x^2+10x} ,$ is \begin{array}{l}\require{cancel} \dfrac{3}{x+2}-\dfrac{1}{5x}=\dfrac{2}{5x(x+2)} .\end{array} Multiplying both sides by the $LCD= 5x(x+2) ,$ then the given equation simplifies to \begin{array}{l}\require{cancel} 5x(3)-(x+2)(1)=1(2) \\\\ 15x-x-2=2 \\\\ 15x-x=2+2 \\\\ 14x=4 \\\\ x=\dfrac{4}{14} \\\\ x=\dfrac{\cancel{2}\cdot2}{\cancel{2}\cdot7} \\\\ x=\dfrac{2}{7} .\end{array} | 239 | 529 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-43 | longest | en | 0.549212 |
https://www.scribd.com/document/341806252/solutionspset6-pdf | 1,539,597,403,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583508988.18/warc/CC-MAIN-20181015080248-20181015101748-00501.warc.gz | 1,092,024,558 | 41,055 | # 3.
11 Solutions Problem Set # 6
Problem #1 Determine the maximum shear stress and rate of twist of the given shaft if a
10 kNm torque is applied to it. If the length of the shaft is 15 m, how much would it
rotate by? Let G = 81 GPa, D = 75 mm
Which equates to :
If the shaft is 15 m long, the angle of rotation at the free end is 34.157o degrees.
0Mpa b) angle of twist AB: ФAB = TABLAB/G(Ip)AB = 0.5-6 Mmax = Pb S = π/32 * d^3 σmax = Mmax/S = 32Pb/ (πd^3) = 185.8e6psi)( π)(1. the stress σmax increases.087rad Ф = TL/GIp Æ T = GIpФ/L τ max = Tr/Ip = Td/2Ip Æ plug in T.600psi b) Change in stress If the angle α increases.272e6 CD: I = 0.0 MPa CD: τ = TCD rCD/IpCD = 63. BC and CD Calc I = π/32 * d^4 AB: I = 4. Problem #5 GERE Problem 5.00902rad BC: ФBC = TBCLBC/G(Ip)BC = 0.7 MPa τ max = 66.04268 rad = 2.3-3 Aluminum Bar in Torsion a) kT=GIp/L = Gπd4/32L = (3. so τ max = Gd Ф/2L = 3450psi Max Shear Strain Hooke’s law: so γmax = τmax/G = 909e-6 rad Problem #3 GERE Problem 3.45degrees Stresses and Strains within a Beam Problem #4 GERE Problem 5.021e6mm^4 BC: I = 1.0in)^4/(32*48in) = 7770lb-in b) Ф = 5º = 5*π/180 rad = 0.0MPa .5-3 a) Maximum Bending Stress α*ρ = L so ρ = L/ α σmax = Ey/ ρ = E(t/2) α /L = 50.01990rad ФD = ФAB + ФBC + ФCD = 0.4-2 Polar Moments of Inertia for AB.Problem #2 GERE Problem 3.01376rad CD: ФCD = TCDLCD/G(Ip)CD = 0.7 MPa BC: τ = TBC rBC/IpBC = 66.2513e6 AB a) Shear Stresses AB: τ = TAB rAB/IpAB = 57.
000 lbs (4 ft) . Solution: Part A: STEP 1: Determine the external support reactions: 1. simply supported W 10 x 45 beam is shown below.) Resolve all forces into x/y components 3. all joints and support points are assumed to be pinned or hinged joints. Determine the maximum bending stress 6 feet from the left end of the beam.5. Determine the horizontal shear stress at a point 4 inches above the bottom of the beam cross section and 6 feet from the left end of the beam.2.Problem #6 Horizontal Shear Stresses STATICS & STRENGTH OF MATERIALS A loaded. Dy = 3.) FBD of structure (See Diagram) 2. showing and labeling all external forces. 1.) Cut beam at 6 ft. Draw the FBD of left end of beam.000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0 Solving: By = 9. For this beam: A. .500 lbs STEP 2: Determine the shear force and bending moment at x=6 ft. B.000 lbs/ft (4 ft) .000 lbs = 0 Sum TB = 5. Unless otherwise indicated.500 lbs.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .2.
and S is the section modulus for the beam.688 lbs/in2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 6 ft from the end of the beam and 4 inches above the bottom of the beam. MBS = M6'/S (Where M6' is the bending moment at 6 ft.000 ft-lbs(12 in/ft)/49.388 psi .) MBS = -11.V6 = 0 Sum TA = 9.500 lbs)(6.1 in3 = -2.500 lbs .) Resolve all forces into x/y directions.500 lbs (6 ft) + M6 = 0 Solving: V6 = 4.500 lbs.000 ft-lbs STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 6'.500 lbs (4 ft) . apply the horizontal shear stress formula. 3.153 in2)(4.2.4. The section modulus is available from the Beam Tables. The W 10 x 45 beam has a section modulus for the beam from the beam tables is 49.1 in3.35 in)] = 1.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = -5. M6 = -11.37 in)]/[(249 in4)(.000 lbs + 9. The form we will use is: HSS = Vay'/Ib Where: V = Shear force 6 ft from the end of the beam a = cross sectional area from 4 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (249 in4 for W 10 x 45 beam) b = width of beam a 4 in above the bottom of the beam HSS = [(4. | 1,289 | 3,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-43 | latest | en | 0.743398 |
http://logisticregressionanalysis.com/date/2013/06/ | 1,511,266,107,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00493.warc.gz | 178,302,521 | 7,178 | Monthly Archives: June 2013
Least-Squares Background Part 5 — Assessing Uncertainty
In this Article, I discuss assessing the uncertainty of the regression coefficient estimates and will briefly touch on assessing the uncertainty of predictions. This article is the last part of a five part series reviewing regular linear-least squares regression. As … Continue reading
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Least-Squares Background Part 4 — Making Predictions
In this Article, I discuss another key use of the regression coefficients, namely to make predictions. This article is part 4 of a 5 part series briefly reviewing some aspects of regular linear least-squares regression. Some background with respect to … Continue reading
Posted in Background | 1 Comment
Least-Squares Background Part 3: Assessing the Effects of the X-Variables
In this Article, I discuss the next use of the regression coefficients, namely to try to assess the impact of each of the -variables. This article is the third part of a five part series providing a brief review of … Continue reading
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Background Part 2 — Uses of the Least-Squares Regression Coefficients and Determining Which Variables Matter
In this article, I am going to focus on the most important section of the regression output, namely the table of regression coefficients and the statistics that accompanies them. This article is part 2 of a 5 part series reviewing … Continue reading
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Understanding Logistic Regression Output:Background Part 1 – Least Squares Regression
As I have indicated previously on this web site, I am going to use regular linear least-squares regression as a starting point for explaining logistic regression. Thus, I will be assuming that you have some familiarity with regular linear regression. … Continue reading
Posted in Background | | 2 Comments | 377 | 1,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-47 | longest | en | 0.849272 |
http://www.mathworks.com/matlabcentral/fileexchange/31112-deflection-of-a-cantilever-beam/content/beam%20deflection/beam_deflection.m | 1,438,063,957,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981576.7/warc/CC-MAIN-20150728002301-00087-ip-10-236-191-2.ec2.internal.warc.gz | 566,873,698 | 6,632 | Code covered by the BSD License
# deflection of a cantilever beam
### Abhilash Harpale (view profile)
19 Apr 2011 (Updated )
calculates the deflection, bending moment and shear force for a cantilever beam.
```% Simple algorithm to calculate the deflection, bending moment, shear force in a
% cantilever beam.
% Example,
error('Check inputs')
end
y=0:dl:l;
u_3=v/ei;
u_3(i)=u_3(i-1)-u_4(i-1)*dl;
end
u_2=m/ei;
u_2(i)=u_2(i-1)-u_3(i-1)*dl;
end
u_1=0;
u_1(i)=u_1(i-1)+u_2(i-1)*dl;
end
u=0;
u(i)=u(i-1)+u_1(i-1)*dl;
end
deflection=u;
plot(y,u_2*ei)
hold on
plot(y,u_3*ei,'r')
legend('bending moment','shear force')
xlabel('length along the beam')
ylabel(' bending moment and shear force (SI units)')
grid
hold off
figure,plot(y,u,'r')
xlabel('length along the beam')
ylabel('deflection')
grid
title('Deflection')
``` | 294 | 820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2015-32 | longest | en | 0.537793 |
https://open.kattis.com/contests/w392gt/problems/aprizenoonecanwin | 1,653,586,335,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00202.warc.gz | 503,903,921 | 7,174 | PTIT PRACTICE 1
#### Start
2022-01-14 15:00 AKST
## PTIT PRACTICE 1
#### End
2022-01-15 15:00 AKST
The end is near!
Contest is over.
Not yet started.
Contest is starting in -131 days 17:32:15
24:00:00
0:00:00
# Problem AA Prize No One Can Win
After the festive opening of your new store, the Boutique store for Alternative Paramedicine and Cwakhsahlvereigh, to your disappointment you find out that you are not making as many sales as you had hoped. To remedy this, you decide to run a special offer: you will mark some subset of the $n$ items for sale in your store as participating in the offer, and if people buy exactly two of these items, and the cost of these items is strictly more than $X$ euros, you will give them a free complimentary unicorn horn!
Since you recently found out all your unicorn horns are really narwhal tusks, you decide to rig the offer by picking the participating items in such a way that no one can earn a horn anyway.
To make sure no one becomes suspicious, you want to mark as many items as possible as participating in the offer.
## Input
• On the first line are two integers, $1 \leq n \leq 10^5$, the number of items for sale in your store, and $1\leq X \leq 10^9$, the minimum cost specified in the statement.
• On the second line are $n$ positive integers, each at most $10^9$. These are the prices of the items in the store.
## Output
Print the maximum number of items you can mark as part of your special offer, without anyone actually being able to receive a horn.
Sample Input 1 Sample Output 1
5 6
1 2 3 4 5
3
Sample Input 2 Sample Output 2
5 10
4 8 1 9 7
2
Sample Input 3 Sample Output 3
4 10
1 3 1 7
4
Sample Input 4 Sample Output 4
1 5
6
1 | 485 | 1,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.924044 |
http://mathisfunforum.com/viewtopic.php?pid=252537 | 1,513,026,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00478.warc.gz | 179,208,067 | 8,383 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #26 2013-01-30 02:39:49
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: 1? Or not?
Hi john_gabriel;
Welcome to the forum. Name calling, is never an effective means of convincing anyone. I hope you will refrain from that when posting in here or providing links.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #27 2013-01-31 03:29:43
TheTick
Member
Registered: 2012-12-03
Posts: 27
### Re: 1? Or not?
.9999999 repeating equals 1 depending on your tolerance, and if we are talking applied or theoretical math. If we divide 1 by 3 we get .333333 repeating. This is not a perfect representation of 1/3, as each repeating 3 gets you ever closer to 1/3. The question is when do mathematicians say "after this 3 let us just say it just equals 1 third." So yes .9999999 repeating can equal 1, but at some point you are rounding up, the amount you round up gets ever smaller as you reach more nines, but eventually you round up (i.e. with .999 you round up by .001, and with .99999 it is .00001)
Feel free to correct me if I am wrong, just thought this info was important to understanding the question.
Spooooon!!!
Offline
## #28 2013-02-04 06:11:57
john_gabriel
Member
Registered: 2013-01-29
Posts: 7
### Re: 1? Or not?
Muxdemux: Didn't you only prove (1) for the natural numbers?
Of course. Are there any other numbers that you know of which I don't? Please don't tell me infinity is a number. It is a meaningless, non-real concept.
TheTick: .9999.... NEVER equals 1 and it has nothing to do with tolerance.
At no point in the partial sums of 0.999... are you "rounding up". In other words, a carry is impossible. The problem with 0.999... is that it's not a number unless considered as an approximation. A non-terminating (repeating or non-repeating) decimal is an ill-defined concept.
Offline
## #29 2013-02-04 07:10:13
bob bundy
Registered: 2010-06-20
Posts: 8,167
### Re: 1? Or not?
This argument keeps surfacing on the forum and I keep trying to explain. To save you having to find my earlier posts I'll say it again.
Numbers have no existence at all except as mathematical concepts, dreamed up to serve a purpose. They get defined and that's what they mean. If you don't want to use a number you don't have to.
I use the common definition that ½ means one part in two but on a TI scientific calculator it looks like this:
They define it to mean the same thing and it's easy to see why.
When the builders were assigning positive integers to the houses in my street they decided to leave out the number between 12 and 14. They thought it might be harder to sell a house with that label because some people think it is an unlucky number. So that number doesn't exist in that context. The lady who delivers the mail still manages to put the right letters in the right boxes so it doesn't seem to cause much of a problem.
Some people have never heard of complex numbers, or quaternions but they still get used by those who have.
It is occasionally useful to define an expression as having a certain value because the formulas work better that way. That's where the following definitions come from:
There are lots more like this.
Now it happens to suit some mathematicians to define:
You may not like it. You may not want to use it. That's Ok. You don't have to. But it has been defined so it has as much existence as any other number. It's a waste of time debating that.
You might as well start arguing about whether the two symbols below are the same. They don't look the same but ........
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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## #30 2013-02-04 12:32:16
john_gabriel
Member
Registered: 2013-01-29
Posts: 7
### Re: 1? Or not?
Bob bundy: "To say that numbers have no existence at all except as mathematical concepts" is false.
They "don't just get defined" as you say. Numbers have a very real existence as abstract objects when they are well-defined.
The examples you gave, 0!=1, sqrt(a), etc, all have very sound explanations.
A number is a magnitude that has been completely measured.
There is nothing dreamy or "just defined" about the previous statement.
Finally, it's not a case of liking or disliking, but rather using objects that are well-defined.
Offline
## #31 2013-02-04 19:05:27
TheTick
Member
Registered: 2012-12-03
Posts: 27
### Re: 1? Or not?
john_gabriel wrote:
Muxdemux: Didn't you only prove (1) for the natural numbers?
Of course. Are there any other numbers that you know of which I don't? Please don't tell me infinity is a number. It is a meaningless, non-real concept.
TheTick: .9999.... NEVER equals 1 and it has nothing to do with tolerance.
At no point in the partial sums of 0.999... are you "rounding up". In other words, a carry is impossible. The problem with 0.999... is that it's not a number unless considered as an approximation. A non-terminating (repeating or non-repeating) decimal is an ill-defined concept.
I agree if we are talking theoretical math, however ask a manufacturer, engineer, architect, etc. if they measure one third to an infinite accuracy. As this can not be done realistically we "say" .999999... at some point equals 1, even though theoretically it does not. That is one I meant by .9999999... can equal 1.
Spooooon!!!
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## #32 2013-02-04 20:01:48
bob bundy
Registered: 2010-06-20
Posts: 8,167
### Re: 1? Or not?
Thanks John Gabriel
Bob bundy: "To say that numbers have no existence at all except as mathematical concepts" is false.
They "don't just get defined" as you say. Numbers have a very real existence as abstract objects when they are well-defined.
Apart from the word 'false' what you have said is exactly what I have said. Define a number properly and that's what it means to say it exists. I'm completely happy with that. That's why I took the trouble to define 0.9999.......
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #33 2013-02-08 03:38:59
john_gabriel
Member
Registered: 2013-01-29
Posts: 7
### Re: 1? Or not?
Muxdemux: Of course I only proved (1) for natural numbers - this is all one can do with induction. The problem arises when you take the limit and the inequality results in an absurdity.
Bob Bundy: There are no personal attacks. Why the message? And of course I wrote the article. I never plagiarize the work of others and most people do not think or agree with me, which makes me feel pretty good. It's always nice to know that I am correct and the majority are in error. Either way, it does not really matter too much. People who read stuff on the internet (or anywhere else) should not take things personally. If your name is not in the perceived put-down or insult, then why take offence? Kind of silly, isn't it? The world does not revolve around any one individual. Please do not delete my post. I am sharing my unique knowledge with you on your forum. Whether you agree or not is irrelevant.
And one more thing: I don't like it that it says "novice" under my name. This is a major put-down for me. Please change it to something like Expert or Master. Thank you in advance!
Last edited by john_gabriel (2013-02-08 03:40:57)
Offline
## #34 2013-02-08 03:48:15
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: 1? Or not?
I am sharing my unique knowledge with you on your forum. Whether you agree or not is irrelevant.
And one more thing: I don't like it that it says "novice" under my name. This is a major put-down for me. Please change it to something like Expert or Master. Thank you in advance!
It is not designed as a put down or insult. It is a phase that every member went through. When you get to 10 posts you will achieve a new status.
Sorry, but there are no expert or master levels. We have legendary members led by anonimnystefy ( now a real member ) and JaneFairfax.
You will have to go through the process like everyone else did. There is no other way.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #35 2013-02-08 03:56:53
john_gabriel
Member
Registered: 2013-01-29
Posts: 7
### Re: 1? Or not?
Well, alright then. But I don't know how much longer I will be around. Sigh,...
10 posts can be a long, long time....
Yes, yes, I understand. Rules are rules.
Offline
## #36 2013-02-08 04:44:17
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,843
Website
### Re: 1? Or not?
Not so long probably
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
## #37 2013-02-09 03:54:21
john_gabriel
Member
Registered: 2013-01-29
Posts: 7
### Re: 1? Or not?
Ha, ha. I hope you are right!
Offline
## #38 2013-02-10 02:13:22
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,843
Website
### Re: 1? Or not?
You are at 7, now.
Well, where did you know from that an empty set is included in every set?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
## #39 2013-02-10 04:45:18
bob bundy
Registered: 2010-06-20
Posts: 8,167
### Re: 1? Or not?
hi Agnishom,
If A intersect B = {} then {} is in A and {} is in B.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #40 2013-02-10 05:46:19
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: 1? Or not?
Actually, the empty set is a subset of every set, but it is not in every set.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #41 2013-02-10 07:20:18
ShivamS
Member
Registered: 2011-02-07
Posts: 3,648
### Re: 1? Or not?
Expert or Master...
Offline
## #42 2013-02-16 00:46:10
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,843
Website
### Re: 1? Or not?
anonimnystefy wrote:
Actually, the empty set is a subset of every set, but it is not in every set.
How? Is it any different?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
Offline
## #43 2013-02-16 01:04:55
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: 1? Or not?
When a set is a subset of another set, it means that all elements of the first set are elements of the second set.
When a set is in another set, it means that the first set is an actual element of the second set.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #44 2013-02-16 04:31:03
Mathegocart
Member
Registered: 2012-04-29
Posts: 1,884
### Re: 1? Or not?
bobbym wrote:
I am sharing my unique knowledge with you on your forum. Whether you agree or not is irrelevant.
And one more thing: I don't like it that it says "novice" under my name. This is a major put-down for me. Please change it to something like Expert or Master. Thank you in advance!
It is not designed as a put down or insult. It is a phase that every member went through. When you get to 10 posts you will achieve a new status.
Sorry, but there are no expert or master levels. We have legendary members led by anonimnystefy ( now a real member ) and JaneFairfax.
You will have to go through the process like everyone else did. There is no other way.
bobbym is correct.Ii went throught this process a lot.It is not a "insult".It is a correct title.
The integral of hope is reality.
Offline | 3,602 | 13,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-51 | latest | en | 0.909624 |
https://mayonahorse.com/qa/question-what-happens-if-your-born-on-february-29th.html | 1,624,277,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00536.warc.gz | 364,430,534 | 23,713 | # Question: What Happens If Your Born On February 29th?
## Why there are 24 leap years in 100 years?
Explanation: Given year is divided by 4, and the quotient gives the number of leap years.
Here, 100%4 = 25.
But, as 100 is not a leap year => 25 – 1= 24 leap years..
When Do Leap Day Babies Celebrate Their Birthdays? Leaplings, people born on leap day, celebrate their birthdays either February 28 or March 1 in common years.
## What are the odds of being born on February 29th?
about one in 1,461The chances of a baby being born on Leap Day is pretty slim: about one in 1,461, as Feb. 29 only comes around once every 1,461 days.
## Why is February 29th special?
Nearly every four years, we add an extra day to the calendar in the form of February 29, also known as Leap Day. … While the modern calendar contains 365 days, the actual time it takes for Earth to orbit its star is slightly longer—roughly 365.2421 days.
## Why 2020 is not a leap year?
This year, 2020, is a leap year, and what that means is that we get an extra day this year. We get that extra day because we count time, in part, by the time it takes Earth to go around the sun. Because we do that, every four years our calendar must come into agreement with the calendar that governs the universe.
## What star sign is 29th February?
PiscesYes, it was unpleasant.). As such, it’s natural for any February 29 baby—because your star sign is Pisces, which always centers around mysticism—to wonder if there’s anything astrologically important that has to do with being being born on this day.
## Do bad things happen on leap years?
Death and Leap Day Another Leap Day superstition claims that Leap Year brings more death. However, with the addition of an extra day in the calendar year, it makes sense for more people to die.
## How many babies are born on February 29th?
There are an estimated 4.1 million leap-day babies in the world and there is a 1 in 1,461 chance of being born on February 29th. The Gregorian calendar, which serves as the standard calendar for civil use throughout the world, has both common years and leap years.
## Is Leap Day William a real thing?
In the classic 2012 episode from Season 6 of Tina Fey’s sitcom, Kenneth (Jack MacBrayer) and the “TGS” team introduced the world to “Leap Day William.” William is a totally real, not-at-all made up holiday character like Santa Claus or the Easter Bunny, except that he has gills, lives in an ocean trench and trades …
## What celebrity birthday is February 29?
8 Famous People Born on February 29Antonio Sabato Jr. Photo: Noel Vasquez/Getty Images.Richard Ramirez. Photo: Getty Images.Ja Rule. Photo: Bennett Raglin/BET/Getty Images for BET.Tony Robbins. Photo: Noam Galai/Getty Images for Extra.Aileen Wuornos. Photo: Copyright © Everett Collection/Everett Collection.Dinah Shore. … Jimmy Dorsey. … Gioachino Rossini.Feb 28, 2020
## What is the rarest birthday?
This Is the Least Common Birthday in the U.S. (No, It’s Not Leap Day)February 29.July 5.May 26.December 31.April 13.December 23.April 1.November 28.More items…•Oct 22, 2020
## Why is February so short?
This is because of simple mathematical fact: the sum of any even amount (12 months) of odd numbers will always equal an even number—and he wanted the total to be odd. So Numa chose February, a month that would be host to Roman rituals honoring the dead, as the unlucky month to consist of 28 days.
## How 2020 is a unique leap year?
2020 is a unique Leap Year. It has 29 days in February, 300 days in March and 5 years in April. May may never never end end.
## Do we ever skip leap years?
Contrary to popular belief, leap years aren’t simply every four years. … We actually skip a leap year if it falls on the start of a century. For example 1700, 1800, and 1900 were not leap years despite being divisible by 4. But if that that century year is divisible by 400, we do have a leap year.
## Are leap year babies lucky?
The time it takes for the earth to rotate is 365 ¼ days but the calendar year is 365 days, hence once every four years to balance this, we have a leap year and an extra day, February 29th. Because such years are rarer than normal years, they have become lucky omens.
## How do leap year babies age legally?
His legal thinking is that February 29 is the day after February 28, so a person born on February 29 is legally considered to have aged one year on the day after February 28. In non-leap years, that day is March 1.
## Can a woman ask a man to marry her on Leap Year?
This year, 2020, is a Leap Year, meaning February has 29 days rather than 28. While women are definitely capable of asking a man for his hand in marriage on any day of the year that she so wishes, traditionally in Ireland, this right was reserved specifically for the Leap Year.
## What is the most common birthday?
September 9thAccording to real birth data compiled from 20 years of American births, mid-September is the most birthday-packed time of the year, with September 9th being the most popular day to be born in America, followed closely by September 19th. | 1,265 | 5,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-25 | latest | en | 0.975994 |
https://125fps.com/lottery/how-many-ways-can-you-roll-5-dice.html | 1,653,473,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00138.warc.gz | 133,632,756 | 18,152 | # How many ways can you roll 5 dice?
Contents
## How many different ways can you roll 5 dice?
The last die may have six values. For each of these six values, the second- to-last die may have six values. Thus, we have 6·6 = 36 possible outcomes for the last two dice. By extension, we have a total of 65 = 7776 possible outcomes for all five dice.
## What is the average roll of 5 dice?
The probability is 0. The average of five dice rolls is some integer divided by 5, but 3.5=72 is not any integer divided by 5. However, the average value of a dice roll is 3.5, so over a large number of turns the average of all throws will roughly equal 3.5.
## What is the probability of rolling a 5 on a 6 sided dice?
Two (6-sided) dice roll probability table
Roll a… Probability
5 10/36 (27.778%)
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
## Is Yahtzee a skill or luck?
As with all games, players typically need elements of both skill and luck. However, some games are known for needing more skill than others. In a 2020 survey conducted by YouGov, 49 percent of respondents in the United States stated that Yahtzee was a game of luck rather than skill.
THIS IS INTERESTING: How do you get free bets on William Hill?
## What are the odds of rolling a 1 or a 5 with two dice?
That means both show a 2, 3, 4, or 6. Thats (4/6)2. Hence the probability that at least one shows a 1 or 5 is 1−(2/3)2=5/9.
## What are the odds of rolling 5 1s in a row?
The probability is 1216 chance, which is approximately a 0.46% chance.
## What is the average of dice rolls?
Another option for finding the average dice roll is to add all of the possible outcomes together then divide by the number of sides the die has. For example, to find the average dice roll of 1d4 you would add 1, 2, 3, and 4 together and divide by 4.
## How many times can you roll a 7?
As the chart shows the closer the total is to 7 the greater is the probability of it being thrown.
Probabilities for the two dice.
Total Number of combinations Probability
6 5 13.89%
7 6 16.67%
8 5 13.89%
9 4 11.11% | 595 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-21 | latest | en | 0.935104 |
https://la.mathworks.com/matlabcentral/cody/problems/42591-produce-the-following-matrix/solutions/733823 | 1,606,305,980,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182776.11/warc/CC-MAIN-20201125100409-20201125130409-00563.warc.gz | 363,727,112 | 17,062 | Cody
# Problem 42591. Produce the following matrix
Solution 733823
Submitted on 11 Sep 2015 by James
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% filetext = fileread('matrixManipulation.m'); assert(isempty(strfind(filetext, 'system')))
2 Pass
%% x = 1; y_correct = ones(2); assert(isequal(matrixManipulation(x),y_correct))
3 Pass
%% x = 2; y_correct = [1 1/2; 2 1]; assert(isequal(matrixManipulation(x),y_correct))
4 Pass
%% x = 2:4; y_correct = [1 1/2 1/3; 2 1 1/4; 3 4 1]; assert(isequal(matrixManipulation(x),y_correct))
5 Pass
%% x = 2:7; y_correct = [1 1/2 1/3 1/4; 2 1 1/5 1/6; 3 5 1 1/7; 4 6 7 1]; assert(isequal(matrixManipulation(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 312 | 922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-50 | latest | en | 0.553858 |
https://www.lmfdb.org/EllipticCurve/Q/5776/f/ | 1,720,875,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00718.warc.gz | 718,610,132 | 5,763 | # Properties
Label 5776.f Number of curves $1$ Conductor $5776$ CM no Rank $1$
# Related objects
Show commands: SageMath
E = EllipticCurve("f1")
E.isogeny_class()
## Elliptic curves in class 5776.f
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
5776.f1 5776n1 $$[0, -1, 0, -6, 7]$$ $$4864$$ $$5776$$ $$[]$$ $$216$$ $$-0.56285$$ $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curve 5776.f1 has rank $$1$$.
## Complex multiplication
The elliptic curves in class 5776.f do not have complex multiplication.
## Modular form5776.2.a.f
sage: E.q_eigenform(10)
$$q - q^{3} - q^{5} - 2 q^{9} + 4 q^{11} + q^{13} + q^{15} + 3 q^{17} + O(q^{20})$$ | 278 | 785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.467895 |
https://byjus.com/question-answer/a-body-is-dropped-from-the-top-of-a-tower-during-the-last-second-of/ | 1,716,903,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00831.warc.gz | 119,704,090 | 25,308 | 1
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Question
# A body is dropped from the top of a tower. During the last second of its fall, it covers 16/25th of the height of the tower. Calculate the height of the tower.
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Solution
## Step 1: Given Let the height of the tower be ‘h’. Suppose it takes ‘n’ second to reach the ground. 'Sn' is height covered in (n−1) second.Step 2: Formula usedS=ut+a2(t)2 Step 3: Finding the time required to cover the height of the towerIn nth second the body covers a height of 1625h.In (n-1)th second, the body covers a height of (1-1625h) = 925h It starts from rest, thus, Sn=u+a2(n−1)2 [this is the equation to find the distance traveled in (n-1)th second] = (9h/25) = 0 + (9.8/2)(n – 1)² = 9h/25 = 4.9(n – 1)² …...(1) Again, the body reaches the ground in ‘n’ second. So, the distance traveled in ‘n’ second is, h = 0 + ½ gn2 => h = (9.8/2)n2 => h = 4.9n2 …………….…(2) From (1) and (2) we have, 9×4.9×n225 = 4.9(n – 1)² => 9n2 = 25(n – 1)² => 16n2 – 50n + 25 = 0 => n = 0.625 or 2.5 Here, n = 0.625 s is not possible as it is greater than 1. So, the time in which the body hits the ground is 2.5 s.Step 4: Finding the height of the tower Putting value of n in equation (2), we geth=4.9n2=4.9×2.5×2.5=30.625h = 4.9n^2=4.9\times 2.5\times 2.5=30.625 m. Hence, the height of the tower is 30.625 m.
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Join BYJU'S Learning Program | 551 | 1,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.862587 |
https://www.phy.olemiss.edu/~luca/Topics/grav/newton.html | 1,717,052,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00381.warc.gz | 797,646,193 | 4,527 | Newtonian Gravitation
Theory > s.a. equivalence principle; models of spacetime structure.
* Idea: Bodies interact through a gravitational force Fg = G mg m'g/r2 acting at a distance, and accelerations are proportional to forces with proportionality constant equal to the inertial mass, a = F/mi.
* Masses: The force is proportional to the gravitational masses, but as Newton knew mgmi, which makes gravitation similar to "inertial" forces, in that the acceleration is independent of each body's individual characteristics.
* Potential: If we define Φ(r) = U(r)/r for a test body, then the potential Φ satisfies the Poisson equation ∇2Φ = 4π,
Φ(r) = −G dM |rr'|−1 = −GM/rG D · r/r3 − $$1\over2$$G Qij ri rj / r5 + ... ,
where D = dM r' is the dipole moment of the mass distribution (always zero with respect to the center of mass), and Qij = dM [3 ri'rj'r' 2 δij] the quadrupole moment (vanishes for a spherical mass distribution).
* Potential energy: For an extended body in an external field, U = ρ(r) Φ(r) d3r; The self-energy is
Uself = $$1\over2$$∫ ρ(r) Φ(r) d3r = (1/8πG) (∇Φ)3 d3r + ∫ ρ(r) Φ(r) d3r .
@ References: Deser AJP(05)aug-gq/04 [from field theory]; Counihan EJP(07) [basic principles]; Yurtsever et al a1004 [inverse problem]; in Ohanian & Ruffini 13; in Poisson & Will 14; Pereira a1903 [from its empirical basis to the theory].
Special Topics > s.a. cosmology; gravitational constant; Newton's Theorem; teaching [weightlessness, tides].
* Tidal forces: A mass m located at (x, y, z) with respect to a frame centered at a point a distance M, feels a tidal force
Fx = −x (GMm/r03) , Fy = −y (GMm/r03) , Fz = 2z (GMm/r03) ;
In general relativity the expression is more complicated, uses the equation for geodesic deviation.
* Other formulations: A (slightly generalized) geometric version is the Newton-Cartan theory.
@ N-body problem: Volchan a0803 [Sundman-Weierstraß theorem of total collapse]; Farrés et al CMDA(13)-a1208 [high-precision symplectic integrators for the Solar System].
@ Tidal forces: Masi AJP(07)feb [compressive radial]; Efroimsky & Williams CMDA(09)-a0803 [tidal torques]; > s.a. Love Number.
@ Other situations: Odagaki & Kawai AJP(98)aug [many-particle statistics]; Beig & Schmidt PRS(03)gq/02 [self-gravitating extended bodies]; Teixeira phy/03 [infinite straight line of mass]; Buchert PLA(06)gq/05 [self-gravitating dust]; Ridgely EJP(11) [in material media].
@ Other formulations and issues: De Pietri et al gq/92, CQG(95)gq/94, CQG(95)gq/94 [generalization]; Nardone JPA(98) [regularization]; Natário GRG(06)gq/04 [initial-value form, and warp drive]; Hansen et al PRL(19)-a1807 [action principle, gravitational time dilation]; Banerjee & Mukherjee PRD(18)-a1810 [geometric].
Phenomenology > s.a. cosmological models [Newtonian cosmology]; phenomenology of gravity; modifications and tests of newtonian gravity.
@ Effects: Abramowicz et al GRG(97) [curvature of space and perihelion precession]; Ferroglia & Fiolhais AJP(20)dec [tidal locking, pedagogical].
@ Specific objects: Dittrich a1609 [Dirichlet's massive homogeneous ellipsoid].
@ Measurement: Kulikov JMO(06)qp/05 [transparency of cold atoms]; Charrière et al PRA(12) [local g measurements]; Graney PT(12)sep [Giovanni Battista Riccioli]; Harms LRR(15)-a1507 [terrestrial gravity fluctuations]; news sn(19)nov [measuring gravity with trapped atoms]; > s.a. Eötvös Experiment; fifth force.
> Related topics: see critical collapse; Newton's Theorem [shell theorem]; orbits in newtonian gravity. | 1,060 | 3,509 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.78732 |
https://www.vedantu.com/maths/binomial-theorem-class-11 | 1,618,673,503,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00320.warc.gz | 1,182,760,174 | 91,828 | # Binomial Theorem Class 11
View Notes
## Binomial Theorem Class 11 Chapter 8
The NCERT Solutions for Binomial Theorem Class 11 Chapter 8 can be downloaded at Vedantu without any hassle. Practicing these Binomial Theorem Class 11 solutions can help the students clear their doubts as well as to solve the problems faster. Students can learn new tricks to answer a particular question in different ways giving them an edge with the exam preparation.
The concepts of Binomial Theorem Class 11 covered in Chapter 8 of the Maths textbook include the study of essential topics, such as Positive Integral Indices, Pascal’s Triangle, Binomial theorem for any positive integer, and some special cases. Students can score high marks in the exams with ease by practicing the Binomial Theorem Class 11 solutions for all the questions present in the textbook. It gets pretty apparent to know the logic set behind each answer and develop a far better comprehension of the concepts.
## Class 11 Chapter 8
### NCERT Solutions for Binomial Theorem Class 11 Maths Chapter 8 –
All the Binomial Theorem Class 11 solutions Maths Chapter 8 is given here. This solution also contains questions, answers, images, and explanations for the whole chapter 8 titled with the Binomial Theorem taught in Class 11. If you are a student of Class 11th who is referring to the NCERT book to study Mathematics, then you will come across chapter 8 Binomial Theorem after you have completed studying the lesson, you also must be looking for the answers to the questions. You will get here all the Binomial Theorem Class 11 solutions for Mathematics Chapter 8 all in a single place.
## Definition of Binomial Theorem
### Binomial Theorem:
Sometimes, when the power increases, the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a really large power is often easily calculated with the assistance of the theorem.
## Binomial Expansion
Important points to remember
• The total number of the terms in the expansion for (x+y)n is (n+1)
• The sum of the exponents of x and y is always n.
• nC0, nC1, nC2, … .., nCn are called binomial coefficients and also are represented by C0, C1, C2, ….., Cn
• The binomial coefficients which are equidistant from the beginning and from the ending are also equal i.e. nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.
## Terms in the Binomial Expansion
In the binomial expansion, it is normally asked to find the middle term or the general term. Different terms in the binomial expansion that are covered are include here:
• General Term
• Middle Term
• Independent Term
• Determining a Particular Term
• Numerically greatest term
• Ratio of Consecutive Terms/Coefficients
## What is a Binomial Expression?
A binomial is also a mathematical expression that includes two terms. Further, these two terms must be separated from each other by either addition or subtraction. To add the binomials, one should combine equal terms to get an answer. To multiply the binomials, the distributive property should be used.
## Topics and SubTopics of Binomial Theorem
Section Name Topic Name 8 Binomial Theorem 8.1 Introduction 8.2 Binomial Theorem for Positive Integral Indices 8.3 General and Middle Terms
## Binomial Theorem Formula
The Binomial Theorem is a quick method for expanding or multiplying out of a binomial expression. The expression also has been raised to some bigger power. As we know the multiplication of such expressions is always difficult with large powers. But the Binomial expansions and its formulas help us a lot in this regard. In this article, we will discuss the Binomial theorem and its Formula.
(a + b)$^{n}$ = $\sum_{k = 0}^{n}$ ($_{k}^{n}$) a$^{k}$ b$^{n-k}$
The upper index n is known as the exponent for the expansion; the lower index k points out which term, starting with k equals 0. For example, when n equals 5, each of the terms in the expansion for (a + b)5 will look like : a5 kbk.
## Properties of the Binomial Expansion (x + y)$^{n}$
• There are a total of n+1 terms.
• The first term is known to be (x)$^{n}$.
• Progressing from the first term to the last term, the exponent for x decreases by 1 from term to term. Whereas the exponent of y increases by 1 term. Also, the sum for both the exponents in each term will be n.
• If the coefficient of each of the terms is multiplied by the exponent of x in that term, and the product is divided by the number of that term, we can easily get the coefficient for the next term.
Question 1: What is a Binomial Theorem?
Answer: The binomial theorem is valid normally for any of the elements x and y of a satisfying xy equals yx. The theorem is true even more: alternativity suffices in place of associativity. The binomial theorem can also be said in a way that the polynomial sequence here given as {1, x, x2, x3, ...} is of a binomial type.
Question 2: What is K in the Binomial Theorem?
Answer: The upper index n is known as the exponent for the expansion; the lower index k points out which term, starting with k equals 0. For example, when n equals 5, each of the terms in the expansion for (a + b)5 will look like: a5 kbk.
Question 3: Where is the Binomial Theorem Used?
Answer: The theorem and its generalizations also can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas in mathematics. The binomial theorem also helps explore probability in an organized way: someone says that he/she will flip a coin 5 times.
Question 4: What is a First Degree Binomial?
Answer: Examples: 5x2-2x+1 The highest exponent here is 2 so this is a 2nd degree trinomial. 3x4+4x2 The highest exponent is 4 here so this is a 4th degree binomial. 8x-1 It appears that there is no exponent, but the x has an understood exponent of 1; So, this is known to be a 1st degree binomial. | 1,420 | 5,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-17 | longest | en | 0.921111 |
http://www.fixya.com/tags/side_2 | 1,568,949,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00313.warc.gz | 242,169,044 | 23,132 | Questions & Answers for: side 2
I have 2 documents (2 different languages) in word
...2 columns side by side for comparison In Microsoft Word, there is a function called "View Side by Side". You can access it by going to the "View" tab on the menu, and there is a small button with a ...
Question about Professor Layton & the Curious Village Games for DS
Need help on boxes cant get 7
...2 wolves over to the other side, take 1 Wolf back. Take 2 wolves over to the other side, take 1 wolf back, Take 2 chicks over to the other side, take 1 wolf & 1 chick back, Take 2 chicks over to the ...
Oxygen sensor confusion
side, 1/2 is the downstream of the number 1 cylinder side, 2/1 is the upstream of the number 2 cylinder side, 2/2 is the downstream of the number 2 cylinder side. Hope this helps. bank one is the bank
Where is the bank 2 sensor 1 located.
...2 is the opposite side. On a 4 cylinder engine, there is only 1 bank and it is always referred to as Bank 1. Sensor 1 is always the upstream sensor (The one located BEFORE the catalytic converter) ...
Question about Racing Ion Motorcycle Jersey 2010
2x^5+4x^4-3x^2+7,7x^4-3x^3+x^2-9
...side) -8/3 = x (divide both sides by 3 to get x by itself) Check 5(-8/3) -2((-8/3)-2) = 6 - 2(-3(-8/3)-3) -40/3 -2(-8/3-6/3) = 6 - 2(8-3) -40/3 -2(-14/3) = 6 -2(5) -40/3 + 28/3 = 6 - 10 -12/3 = -4 ...
Question about Racing Ion Motorcycle Jersey 2010
Without using legrange multipliers, determine the
...side #2 and side #4. X = width of sides #1 and #3 of the box Y = height of the box Z = width of sides #2 and #4 Then, X times Y times Z is 32. Area of side #1: X times Y Area of side #3: X times Y ...
Question about Bizhub C364 Color Copier / Printer / Scanner
Printing multiple double sided copies doesn't break between copies
...the copies and leave the back side of the last page blank? Select the simplex/duplex mode. 1 sided - 1 sided 1 sided - 2 sided 2 sided - 2 sided When using 1 sided to 2 sided the originals must be ...
Question about Bizhub C364 Color Copier / Printer / Scanner
Square 1 has a side length of x, and square 2 has a side length of y. Square 2 is formed by joining the midpoints of the sides of square 1 in order. If x = 2, find the ratio of the perimeter of square
...2 has a side length of y. Square 2 is formed by joining the midpoints of the sides of square 1 in order. If x = 2, find the ratio of the perimeter of square A: 2:1 B: 1:2 C: 1:sqrt2 D: 2:sqrt2 Using ...
Question about Bizhub C364 Color Copier / Printer / Scanner
Square 1 has a side length of x, and square 2 has a side length of y. Square 2 is formed by joining the midpoints of the sides of square 1 in order. If x = 2, find the ratio of the perimeter of square
...2 has a side length of y. Square 2 is formed by joining the midpoints of the sides of square 1 in order. If x = 2, find the ratio of the perimeter of square 1 to the perimeter of square 2. Using the ...
Question about Bizhub C364 Color Copier / Printer / Scanner
What is the steps of this multi step equations
2 on the left side, we must add 2 to both sides. A-2+3+2 =-2+2 A+3 = 0 To get rid of the 3 on the left, we must subtract 3 from both sides. A+3-3 = 0 - 3 A = -3 To check if we did it correctly, put A
Level 3 Expert
Level 3 Expert | 990 | 3,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-39 | latest | en | 0.868733 |
http://www.ck12.org/arithmetic/Mixed-Numbers-as-Decimals/flashcard/user:13IntS/Mixed-Numbers-as-Decimals/r2/ | 1,490,677,823,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189667.42/warc/CC-MAIN-20170322212949-00423-ip-10-233-31-227.ec2.internal.warc.gz | 469,036,369 | 25,673 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Mixed Numbers as Decimals
## Whole number left of decimal and fraction as decimal to the right
Estimated16 minsto complete
%
Progress
Practice Mixed Numbers as Decimals
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated16 minsto complete
%
Mixed Numbers as Decimals | 117 | 441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-13 | latest | en | 0.563628 |
https://www.physicsforums.com/threads/force-and-torque-exerted-by-ferris-wheel.276498/ | 1,532,174,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00471.warc.gz | 946,801,246 | 13,840 | # Homework Help: Force and Torque exerted by Ferris wheel
1. Dec 2, 2008
### pmk69
Hi all,
I'm new to this physics forum. Please tell me that my calculation whether is it correct or not. The diameter of Ferris wheel 30m. speed 10 RPM. Mass of the wheel is 35000kgs. calculating the force and torque exerted by the wheel.
Force = ma; a = v^2/r; 15.7^2/15= 16.43 m/s^2;
therefore, Force = F = 35000 kgs x 16.43 m/s^2 = 575143.38 Newtons;
Torque = Iw = I x alpha = I x (a/r); I = MR^2 = 35000 x 15^2 = 7875000 kg-m^2;
alpha = a/r = 16.43/15 = 1.095 rad/sec;
So, Torque = 7875000 x 1.095 = 8623125 Nm
thanks,
pmk69 | 237 | 616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-30 | latest | en | 0.812497 |
https://www.physicsforums.com/threads/matlab-making-a-movie-from-plots.317986/ | 1,521,855,978,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649508.48/warc/CC-MAIN-20180323235620-20180324015620-00685.warc.gz | 871,165,101 | 14,840 | # [Matlab] Making a movie from plots
1. Jun 4, 2009
### Nick89
Hi,
This isn't really technically a homework question but I thought I'd still ask here...
I am solving a spring pendulum model (set of differential equations) using Matlab, and the homework assignment tells me to plot the trajectory of the bob in the x-y plane. Well, I did that, but I find it is fairly hard to see anything in the plot. I mean, it's a nice curvy line and all, but I can't really verify if it 'looks natural' or not.
So, I wanted to make it into a movie, using the following code:
Code (Text):
% movie
for i = 1:length(X)
plot(X(i),Y(i),'o'); axis([-10 10 -10 10]);
line([0,X(i)],[0,Y(i)]);
F(i) = getframe;
end
movie(F, 1, 25);
Here, X and Y are the x and y coordinates of the solution (motion of the bob).
Now, I can see the bob move up and down nicely while swinging left to right. It also draws the spring or suspension rope between the point (0,0) and (X,Y) each frame.
The problem however is that I don't want my script to actually PLOT each individual frame, because I am already plotting the x(t) vs time, y(t) vs time, and x(t) vs y(t) plots (as the assignment asks me to do). Now, these plots are lost because they're overwritten by the movie plots.
Can I not plot the plots 'into memory' and store them in the F frames array that way? Then, I can return the frames array F and let the user play the movie manually using the movie() command.
Is that possible?
Last edited: Jun 4, 2009
2. Jun 4, 2009
### MATLABdude
Your script is probably trying to accomplish too much at once!
Why not make one script for generating your plots, and one for generating / playing your movie? If you're really, really insistent; the figure command opens up a new plotting window. plot plots to the most recently opened plot window (if there is one), but you can plot to an arbitrary plot window using the window's handle:
>> h1 = figure; %h1 is now a handle into the new window you just opened.
>> h2 = figure;
>> plot([-10:0.1:10], [-10:0.1:10].^2); plots a parabola in the second window
>> pause(5);
>> figure(h1); % sets the current figure to h1
>> plot([0:0.1:2*pi], sin([0:0.1:2*pi]); % plots a sine wave inside the first figure window | 606 | 2,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-13 | latest | en | 0.925971 |
https://raweb.inria.fr/rapportsactivite/RA2009/calvi/uid6.html | 1,679,802,234,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00270.warc.gz | 523,247,070 | 5,489 | ## Section: Scientific Foundations
### Kinetic models for plasma and beam physics
Plasmas and particle beams can be described by a hierarchy of models including N -body interaction, kinetic models and fluid models. Kinetic models in particular are posed in phase-space and involve specific difficulties. We perform a mathematical analysis of such models and try to find and justify approximate models using asymptotic analysis.
#### Models for plasma and beam physics
The plasma state can be considered as the fourth state of matter, obtained for example by bringing a gas to a very high temperature (104K or more). The thermal energy of the molecules and atoms constituting the gas is then sufficient to start ionization when particles collide. A globally neutral gas of neutral and charged particles, called plasma, is then obtained. Intense charged particle beams, called nonneutral plasmas by some authors, obey similar physical laws.
The hierarchy of models describing the evolution of charged particles within a plasma or a particle beam includes N -body models where each particle interacts directly with all the others, kinetic models based on a statistical description of the particles and fluid models valid when the particles are at a thermodynamical equilibrium.
In a so-called kinetic model , each particle species s in a plasma or a particle beam is described by a distribution function corresponding to the statistical average of the particle distribution in phase-space corresponding to many realisations of the physical system under investigation. The product is the average number of particles of the considered species, the position and velocity of which are located in a bin of volume centered around . The distribution function contains a lot more information than what can be obtained from a fluid description, as it also includes information about the velocity distribution of the particles.
A kinetic description is necessary in collective plasmas where the distribution function is very different from the Maxwell-Boltzmann (or Maxwellian) distribution which corresponds to the thermodynamical equilibrium, otherwise a fluid description is generally sufficient. In the limit when collective effects are dominant with respect to binary collisions, the corresponding kinetic equation is the Vlasov equation
which expresses that the distribution function f is conserved along the particle trajectories which are determined by their motion in their mean electromagnetic field. The Vlasov equation which involves a self-consistent electromagnetic field needs to be coupled to the Maxwell equations in order to compute this field
which describes the evolution of the electromagnetic field generated by the charge density
and current density
associated to the charged particles.
When binary particle-particle interactions are dominant with respect to the mean-field effects then the distribution function f obeys the Boltzmann equation
where Q is the nonlinear Boltzmann collision operator. In some intermediate cases, a collision operator needs to be added to the Vlasov equation.
The numerical solution of the three-dimensional Vlasov-Maxwell system represents a considerable challenge due to the huge size of the problem. Indeed, the Vlasov-Maxwell system is nonlinear and posed in phase space. It thus depends on seven variables: three configuration space variables, three velocity space variables and time, for each species of particles. This feature makes it essential to use every possible option to find a reduced model wherever possible, in particular when there are geometrical symmetries or small terms which can be neglected.
#### Mathematical and asymptotic analysis of kinetic models
The mathematical analysis of the Vlasov equation is essential for a thorough understanding of the model as well for physical as for numerical purposes. It has attracted many researchers since the end of the 1970s. Among the most important results which have been obtained, we can cite the existence of strong and weak solutions of the Vlasov-Poisson system by Horst and Hunze [85] , see also Bardos and Degond [65] . The existence of a weak solution for the Vlasov-Maxwell system has been proved by Di Perna and Lions [74] . An overview of the theory is presented in a book by Glassey [82] .
Many questions concerning for example uniqueness or existence of strong solutions for the three-dimensional Vlasov-Maxwell system are still open. Moreover, their is a realm of approached models that need to be investigated. In particular, the Vlasov-Darwin model for which we could recently prove the existence of global solutions for small initial data [66] .
On the other hand, the asymptotic study of the Vlasov equation in different physical situations is important in order to find or justify reduced models. One situation of major importance in tokamaks, used for magnetic fusion as well as in atmospheric plasmas, is the case of a large external magnetic field used for confining the particles. The magnetic field tends to incurve the particle trajectories which eventually, when the magnetic field is large, are confined along the magnetic field lines. Moreover, when an electric field is present, the particles drift in a direction perpendicular to the magnetic and to the electric field. The new time scale linked to the cyclotron frequency, which is the frequency of rotation around the magnetic field lines, comes in addition to the other time scales present in the system like the plasma frequencies of the different particle species. Thus, many different time scales as well as length scales linked in particular to the different Debye length are present in the system. Depending on the effects that need to be studied, asymptotic techniques allow to find reduced models. In this spirit, in the case of large magnetic fields, recent results have been obtained by Golse and Saint-Raymond [83] , [87] as well as by Brenier [69] . Our group has also contributed to this problem using homogenization techniques to justify the guiding center model and the finite Larmor radius model which are used by physicist in this setting [80] , [78] , [79] .
Another important asymptotic problem yielding reduced models for the Vlasov-Maxwell system is the fluid limit of collisionless plasmas. In some specific physical situations, the infinite system of velocity moments of the Vlasov equations can be closed after a few of those, thus yielding fluid models.
Logo Inria | 1,238 | 6,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.920856 |
http://www.aishack.in/2010/04/efficiently-accessing-matrices/ | 1,397,917,851,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00182-ip-10-147-4-33.ec2.internal.warc.gz | 280,368,949 | 9,597 | # Efficiently accessing matrices
I’ve talked previous about matrices in 2D matrices with CvMat in OpenCV. I talked about accessing using the cvGet* and cvSet* functions. But in computer vision, you need to do things as efficiently as possible. And those functions increase the overhead. So here’s a super fast method of accessing data, using pointers. Don’t worry, it’s going to be easy!
## Pointer math
The only thing you need to know is that the matrix elements are stored sequentially. And usually, you’d have two loops: the outer loop for the row, and a loop inside it for the column. And for each row and column, you can access the channel data.
Some code will clear up the idea:
```for(int row=0;row<mat->rows;row++) { const float* ptr = (const float*)(mat->data.ptr + row*mat->step); for(int col=0;col<mat->cols;col++) { float c1 = *ptr++; float c2 = *ptr++; float c3 = *ptr++; float c4 = *ptr++; } }```
You can see the outer row loop. Inside it is the column loop, col. And inside both, you access the channel data for a particular pixel (I assumed a 4 channel matrix, CV_32FC4).
Simple eh? And its efficient too!
One word of caution though. Note that I calculate ptr for every row. That’s because you can setup a “region of interest” in images. And this matrix might be a part of that. So you can’t say that rows are continuous in memory. However, individual elements of a particular row will always be continuous in memory.
Oh! You might want to check Memory layout of matrices of multi-dimensional objects. I’ve gone into detail about how matrices are stored in memory. And for one particular case, it’s drastically different. Make sure you check that article as well.
## Done!
With that, I think you should be able to access your matrices much more efficiently! Got questions? Suggestions or criticism? Let me know! Leave a comment!
Issues? Suggestions? Visit the Github issue tracker for AI Shack
### One Comment
1. donny
Posted August 25, 2011 at 9:48 am | Permalink
can help me with this? | 487 | 2,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-15 | latest | en | 0.885961 |
https://www.studypool.com/discuss/540130/need-help-to-see-if-i-got-the-problem-right?free | 1,505,996,793,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00476.warc.gz | 853,372,768 | 13,917 | ##### NEED HELP TO SEE IF I GOT THE PROBLEM RIGHT.
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
May 16th, 2015
Hiya! From what you have posted,
C+4-3C
We can simplify a bit by combining the C term:
-2C+4
And then we can simplify the 2 and 4:
-2(C-2).
I do need more information about the problem, however, so please reply after I post this and we'll go through it together. I need to know what exactly you have to solve for :)
May 16th, 2015
...
May 16th, 2015
...
May 16th, 2015
Sep 21st, 2017
check_circle | 171 | 555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-39 | latest | en | 0.935895 |
https://forums.macrumors.com/threads/grapher-domain-range-issues.920189/ | 1,542,795,511,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00430.warc.gz | 625,417,618 | 30,874 | # Grapher domain & range issues
Discussion in 'Mac Apps and Mac App Store' started by DiMarc217, May 21, 2010.
1. ### DiMarc217 macrumors newbie
Joined:
May 21, 2010
#1
Hi, I'm doing a project that requires a lot of complicated shading in grapher (using inequalities). It would really help for me to be able to set the domain and/or range of the inequalities using an 'or' boolean statement. For example, I can do this:
x>0 : y<x+3
and I can use 'and' statements:
x>0 & x<3 : y<x+3
but I need to be able to do something like this:
x<0 OR x>3 : y<x+3
I know for this example I could just split it into two equations, but for the one I'm doing with multiple overlapping conditions, I need to be able to use an 'or' operator.
Does anyone know how to do this?
Thanks!
2. ### YB24 macrumors member
Joined:
Apr 3, 2007
Location:
Dordogne-France http://y.barois.free.fr/grapher/
#2
Grapher : boolean
Hi,
Menu Window/Equation palette/Symbols : the four last ones are -AND-OR-XOR-INFINITY.
Attachment : a .gcx file with your inequalities, two ways for them.
Does that solve your problem ?
Be well, YB24
File size:
77.7 KB
Views:
132
Joined:
May 21, 2010
4. ### YB24 macrumors member
Joined:
Apr 3, 2007
Location:
Dordogne-France http://y.barois.free.fr/grapher/
#4
Grapher : boolean
I forgot one boolean sign : The five last symbols on the equation palette are NOT-AND-OR-XOR (boolean)- infinity. YB24
Joined:
Oct 1, 2010
#5
love you!
6. ### kanenas macrumors newbie
Joined:
Jun 20, 2008
#6
Other options
"|" is another symbol for the OR operator, though in Grapher v 2.0 typing "|" gets you the absolute value operator.
For some cases, intervals can be used: "0≤x & x≤3" is equivalent to "x ∈ [0;3]" and "x<0 | 3<x" is equivalent to "x ∉ [0;3]".
The Grapher guide on this very site lists these operators (including how to enter them) and many others.
7. ### YB24 macrumors member
Joined:
Apr 3, 2007
Location:
Dordogne-France http://y.barois.free.fr/grapher/
#7
Grapher boolean signs
Hi !
I can't use this sign in Grapher 2.1 as OR operator: I get every time the absolute value operator. May be there are some differences between french, english etc. keyboard. For what Grapher cases did you use this sign ?
Is "this very site list" wrong ?
Be well, YB24
8. ### kanenas macrumors newbie
Joined:
Jun 20, 2008
#8
As noted in previous post, "|" produces the absolute value operator since Grapher v2 came out. The Grapher guide hasn't been updated with this information. The guide needs quite a bit of work, which I hope will happen as more people become aware of it.
9. ### YB24 macrumors member
Joined:
Apr 3, 2007
Location:
Dordogne-France http://y.barois.free.fr/grapher/
#9
My own list of Grapher 2.1 keyboard shortcuts is in attachment. may be you'll find it useful.
Be well, YB
File size:
109.4 KB
Views:
103 | 839 | 2,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-47 | latest | en | 0.886445 |
https://artofproblemsolving.com/wiki/index.php/Divisibility_rules/Rule_for_2_and_powers_of_2_proof | 1,701,434,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00861.warc.gz | 133,147,789 | 11,304 | # Divisibility rules/Rule for 2 and powers of 2 proof
A number $N$ is divisible by $2^n$ if the last ${n}$ digits of the number are divisible by $2^n$.
## Proof
### Basic Idea
Let the number $N$ be $(10^n)k + p$ where k and p are integers and $p<(10^n)$. Since $\frac{10^n}{2^n}$ is $5^n$, $(10^n)$ is a multiple of $2^n$, meaning $(10^n)k$ is also a multiple of $2^n$. As long as p is a multiple of $2^n$, then $N$ is a multiple of $2^n$. Since $(10^n)k$ has $n$ trailing 0's, $p$ is the last $n$ digits of the number $n$.
### Concise
An understanding of basic modular arithmetic is necessary for this proof.
Let the base-ten representation of $N$ be $\underline{a_ka_{k-1}\cdots a_1a_0}$ where the $a_i$ are digits for each $i$ and the underline is simply to note that this is a base-10 expression rather than a product. If $N$ has no more than $n$ digits, then the last $n$ digits of $N$ make up $N$ itself, so the test is trivially true. If $N$ has more than $n$ digits, we note that:
$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.$
Taking this $\mod 2^n$ we have
$N$ $= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0$ $\equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 \pmod{2^n}$
because for $i \geq n$, $10^i \equiv 0 \pmod{2^n}$. Thus, $N$ is divisible by $2^n$ if and only if
$10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10 a_1 + a_0 = \underline{a_{n-1}a_{n-2}\cdots a_1a_0}$
is. But this says exactly what we claimed: the last $n$ digits of $N$ are divisible by $2^n$ if and only if $N$ is divisible by $2^n$. | 610 | 1,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 47, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-50 | latest | en | 0.767086 |
https://socratic.org/questions/to-what-volume-should-you-dilute-55-ml-of-a-11-m-stock-hno-3-solution-to-obtain- | 1,713,929,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00570.warc.gz | 476,510,853 | 6,254 | # To what volume should you dilute 55 mL of a 11 M stock HNO_3 solution to obtain a 0.137 M HNO_3 solution?
May 19, 2016
You need to dilute the given volume of stock solution to $\text{4.4 L}$.
#### Explanation:
Molarity $\left(\text{M}\right)$ represents the concentration of a solution in moles of solute/liters of solution.
${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$
${M}_{1} = \text{11 M"="11 mol/L}$
${V}_{1} = \text{55 mL"="0.055 L}$
${M}_{2} = \text{0.137 M"="0.137 mol/L}$
V_2="???
Solution
Rearrange the equation to isolate ${V}_{2}$, then substitute the given values into the equation and solve.
${V}_{2} = \frac{{M}_{1} {V}_{1}}{M} _ 2$
V_2=(11cancel"mol/L" * 0.055"L")/(0.137cancel"mol/L")="4.4 L" (rounded to two significant figures) | 274 | 748 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.718045 |
http://www.self.gutenberg.org/articles/eng/2_(number) | 1,606,266,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00476.warc.gz | 150,834,484 | 42,987 | #jsDisabledContent { display:none; } My Account | Register | Help
# 2 (number)
Article Id: WHEBN0000064516
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### 2 (number)
← 1 2 3 →
Cardinal two
Ordinal 2nd (second / twoth)
Numeral system binary
Factorization prime
Gaussian integer factorization (1 + i)(1 - i)
Prime 1st
Divisors 1, 2
Roman numeral II
Roman numeral (unicode) Ⅱ, ⅱ
Greek prefix di-
Latin prefix duo- bi-
Old English prefix twi-
Binary 102
Ternary 23
Quaternary 24
Quinary 25
Senary 26
Octal 28
Duodecimal 212
Vigesimal 220
Base 36 236
Greek numeral β'
Arabic ٢
Urdu
Ge'ez
Bengali
Chinese numeral 二，弍，贰，貳
Devanāgarī (do)
Telugu
Tamil
Hebrew ב (Bet)
Khmer
Korean 이，둘
Thai
2 (Two; ) is a number, numeral, and glyph. It is the natural number following 1 and preceding 3.
## Contents
• In mathematics 1
• List of basic calculations 1.1
• Evolution of the glyph 2
• In science 3
• Astronomy 3.1
• In technology 4
• In religion 5
• Numerological significance 6
• In sports 7
• In other fields 8
• References 10
## In mathematics
The number two has many properties in mathematics.[1] An integer is called even if it is divisible by 2. For integers written in a numeral system based on an even number, such as decimal and hexadecimal, divisibility by 2 is easily tested by merely looking at the last digit. If it is even, then the whole number is even. In particular, when written in the decimal system, all multiples of 2 will end in 0, 2, 4, 6, or 8.
Two is the smallest and the first prime number, and the only even prime number [2] (for this reason it is sometimes called "the oddest prime").[3] The next prime is three. Two and three are the only two consecutive prime numbers. 2 is the first Sophie Germain prime, the first factorial prime, the first Lucas prime, the first Ramanujan prime, and the first Smarandache-Wellin prime. It is an Eisenstein prime with no imaginary part and real part of the form 3n - 1. It is also a Stern prime, a Pell number, the first Fibonacci prime, and a Markov number—appearing in infinitely many solutions to the Markov Diophantine equation involving odd-indexed Pell numbers.
It is the third Fibonacci number, and the third and fifth Perrin numbers.
Despite being prime, two is also a superior highly composite number, because it is a natural number which has more divisors than any other number scaled relative to the number itself. The next superior highly composite number is six.
Vulgar fractions with only 2 or 5 in the denominator do not yield infinite decimal expansions, as is the case with all other primes, because 2 and 5 are factors of ten, the decimal base.
Two is the base of the simplest numeral system in which natural numbers can be written concisely, being the length of the number a logarithm of the value of the number (whereas in base 1 the length of the number is the value of the number itself); the binary system is used in computers.
For any number x:
x+x = 2·x addition to multiplication
x·x = x2 multiplication to exponentiation
xx = x↑↑2 exponentiation to tetration
In general:
hyper(x,n,x) = hyper(x,n+1,2)
Two also has the unique property that 2+2 = 2·2 = 2²=2↑↑2=2↑↑↑2, and so on, no matter how high the operation is.
Two is the only number x such that the sum of the reciprocals of the powers of x equals itself. In symbols
\sum_{k=0}^{\infin}\frac {1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots=2.
This comes from the fact that:
Powers of two are central to the concept of Mersenne primes, and important to computer science. Two is the first Mersenne prime exponent.
Taking the square root of a number is such a common mathematical operation, that the spot on the root sign where the exponent would normally be written for cubic roots and other such roots, is left blank for square roots, as it is considered tacit.
The square root of 2 was the first known irrational number.
The smallest field has two elements.
In the set-theoretical construction of the natural numbers, 2 is identified with the set \{\{\emptyset\},\emptyset\}. This latter set is important in category theory: it is a subobject classifier in the category of sets.
Two is a primorial, as well as its own factorial. Two often occurs in numerical sequences, such as the Fibonacci number sequence, but not quite as often as one does. Two is also a Motzkin number, a Bell number, an all-Harshad number, a meandric number, a semi-meandric number, and an open meandric number.
Two is the number of n-Queens Problem solutions for n = 4. With one exception, all known solutions to Znám's problem start with 2.
Two also has the unique property such that
\sum_{k=0}^{n-1} 2^k = 2^{n} - 1
and also
\sum_{k=a}^{n-1} 2^k = 2^n - \sum_{k=0}^{a-1} 2^k - 1
for a not equal to zero
The number of domino tilings of a 2×2 checkerboard is 2.
In n-dimensional space for any n, any two distinct points determine a line.
For any polyhedron homeomorphic to a sphere, the Euler characteristic is
\chi = V-E+F = 2,\
where V is the number of vertices, E is the number of edges, and F is the number of faces.
As of 2008, there are only two known Wieferich primes.
With the exception of the sequence 3, 5, 7, the maximum number of consecutive odd numbers that are prime is two.
### List of basic calculations
Multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 50 100 1000
2 \times x 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 100 200 2000
Division 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 \div x 2 1 0.6 0.5 0.4 0.3 0.285714 0.25 0.2 0.2 0.18 0.16 0.153846 0.142857 0.13
x \div 2 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
Exponentiation 1 2 3 4 5 6 7 8 9 10 11 12 13 14
2 ^ x\, 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384
x ^ 2\, 1 4 9 16 25 36 49 64 81 100 121 144 169 196
## Evolution of the glyph
The glyph used in the modern Western world to represent the number 2 traces its roots back to the Brahmin Indians, who wrote "2" as two horizontal lines. The modern Chinese and Japanese languages still use this method. The Gupta rotated the two lines 45 degrees, making them diagonal, and sometimes also made the top line shorter and made its bottom end curve towards the center of the bottom line. Apparently for speed, the Nagari started making the top line more like a curve and connecting to the bottom line. The Ghubar Arabs made the bottom line completely vertical, and now the glyph looks like a dotless closing question mark. Restoring the bottom line to its original horizontal position, but keeping the top line as a curve that connects to the bottom line leads to our modern glyph.[4]
In fonts with text figures, 2 usually is of x-height, for example, .
## In religion
The number 2 is important in Judaism, with one of the earliest reference being that God ordered Noah to put two of every unclean animal (Gen. 7:2) in his ark (see Noah's Ark). Later on, the Ten Commandments were given in the form of two tablets. The number also has ceremonial importance, such as the two candles that are traditionally kindled to usher in the Shabbat, recalling the two different ways Shabbat is referred to in the two times the Ten Commandments are recorded in the Torah. These two expressions are known in Hebrew as שמור וזכור ("guard" and "remember"), as in "Guard the Shabbat day to sanctify it" (Deut. 5:12) and "Remember the Shabbat day to sanctify it" (Ex. 20:8). Two challahs (lechem mishnah) are placed on the table for each Shabbat meal and a blessing made over them, to commemorate the double portion of manna which fell in the desert every Friday to cover that day's meals and the Shabbat meals
In Jewish law, the testimony of two witnesses are required to verify and validate events, such as marriage, divorce, and a crime that warrants capital punishment
"Second-Day Yom Tov" (Yom Tov Sheini Shebegaliyot) is a rabbinical enactment that mandates a two-day celebration for each of the one-day Jewish festivals (i.e., the first and seventh day of Passover, the day of Shavuot, the first day of Sukkot, and the day of Shemini Atzeret) outside the land of Israel.
## Numerological significance
The twos of all four suits in playing cards
The most common philosophical Hegelian dialectic, the process of synthesis creates two perspectives from one.
The ancient Sanskrit language (संस्कृत भाषा) of India, does not only have a singular and plural form for nouns, as do most other languages, but instead has, a singular (1) form, a dual (2) form, and a plural (everything above 2) form, for all nouns, due to the significance of 2. It is viewed as important because of the anatomical significance of 2 (2 hands, 2 nostrils, 2 eyes, 2 legs, etc.)
Two (二, èr) is a good number in Chinese culture. There is a Chinese saying, "good things come in pairs". It is common to use double symbols in product brandnames, e.g. double happiness, double coin, double elephants etc. Cantonese people like the number two because it sounds the same as the word "easy" (易) in Cantonese.
In Finland, two candles are lit on Independence Day. Putting them on the windowsill invokes the symbolical meaning of division, and thus independence.
In pre-1972 Indonesian and Malay orthography, 2 was shorthand for the reduplication that forms plurals: orang "person", orang-orang or orang2 "people".
In Astrology, Taurus is the second sign of the Zodiac.
## In sports
• In baseball scorekeeping, 2 is the position of the catcher.
• In American football, a safety has a two-point value. Also, a two-point conversion is a point after touchdown (PAT) attempt where the ball crosses the goal line via run or pass. (In six-man football, however, the traditional PAT kick is worth two points, whereas a PAT via pass or run is only one point.)
• In Association football, the scoring of two goals by one individual in a single match is referred to as a brace.
• The successor of a brace is the "hat-trick", three goals scored by one player.
• In standard basketball, the value of any made shot taken from inside the three-point arc in normal play is 2 points.
• In the half-court 3x3 variant, made shots taken from outside the "three-point" arc are worth 2 points.
## In other fields
Groups of two:
In North American educational systems, the number 2.00 denotes a grade-point average of "C", which in some colleges and universities is the minimum required for good academic standing at the undergraduate level.[5] | 2,952 | 10,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-50 | latest | en | 0.792512 |
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