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CCC '01 S4 - Cookies
Canadian Computing Competition: 2001 Stage 1, Senior #4
Making chocolate chip cookies involves mixing flour, salt, oil, baking soda and chocolate chips to form dough which is rolled into a plane. Circles are cut from the plane, placed on a cookie sheet,
and baked in an oven for about twenty minutes. When the cookies are done, they are removed from the oven and allowed to cool before being eaten.
We are concerned here with the process of cutting a single round cookie that contains all the chocolate chips. Once the dough has been rolled, each chip is visible in the planar dough, so we need
simply to find a cookie cutter big enough to circle all the chips. What is the diameter of the smallest possible round cookie containing all the chips?
Input Specification
Input consists of a positive integer not greater than , followed by lines of input. Each line gives the coordinates of one chocolate chip on the plane. Each coordinate is an integer in the range .
Output Specification
Output consists of a single real number, the diameter of the cookie rounded to two decimal places.
Sample Input 1
Sample Output 1
Sample Input 2
Sample Output 2
• To anyone who is seriously having trouble with this:
Research Welzl's algorithm
• one test case wrong, no idea why
□ Yup, test case #4 is always wrong for some reason.
☆ The answer is not simply the distance between the 2 furthest points.
Eg. Points (0, 0), (0, 10), and (5, 6) The answer is 10.17, not 10.00 (which is the distance between the 2 furthest points)
○ Welp I did it wrong then lol
○ Not sure if I'm doing this wrong but for the case above w/ isn't 10.00 still the correct answer? It's a circle centered at with diameter and all 3 chocolate chips on its
■ commented on March 4, 2020, 12:35 a.m. ← edited
yep, if you use circumcentre this is the correct answer. Note that if you handle these special cases with circumcentre, it still passes.
○ Yes I believe there is a little more trig involved.
■ No trig is necessary. I would research Heron's formula.
★ Heron's is numerically unstable... need to subtract a few 4th powers. Calling Line-Line intersection on the perpendicular bisectors is the way to go. | {"url":"https://dmoj.ca/problem/ccc01s4","timestamp":"2024-11-11T22:54:45Z","content_type":"text/html","content_length":"57805","record_id":"<urn:uuid:782fc093-a1d5-470f-b14f-dc9b1575caed>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00650.warc.gz"} |
When two nodes are connected to each other by wires, they effectively become one electrical node. This connection operation is part of the ACT language, and is denoted by the = sign. The = operation
is also overloaded for meta-language variables to denote assignment. Multiple connections can be specified in a single statement by repeatedly using the = operator. This section describes the
different connection statements supported by ACT.
Simple connections
The simplest possible connection statement is the connection of two variables of type bool.
bool x, y;
The effect of this operation is to alias the two nodes. After this operation is performed, both x and y refer to the same value. Meta-language types can also be 'connected' to expressions. The result
of such a 'connection' is that the right hand side of the = sign is evaluated, and assigned to the variable on the left. Such connections are only meant to initialize the values of parameters.
pint x, y;
y=x*1+2; // success
Whereas connecting nodes is a symmetric operation, connecting meta-language variables is not symmetric, as illustrated below.
pint x, y;
-[ERROR]-> id: y
FATAL: Uninitialized identifier
Meta-language parameter connections correspond to assignment statements. ACT permits assigning floating-point values to integer-valued variables, and vice versa. However, there are some restrictions
on meta-language variable assignments.
pint x;
-[ERROR]-> Id: x
FATAL: Setting immutable parameter that has already been set
In this example, x has been declared and then defined twice at the top-level of the ACT file. This makes x a global variable, which means x can be used in types defined later in the ACT file. This
potentially makes x an implicit parameter for all types, even though x does not appear in the template parameter list for any of them. To prevent the situation where x might have different values
depending on when a type is used, global parameters can only be defined once in ACT. This constraint also applies to template parameters for the same reason.
However, for parameters defined within the body of a type, they can be defined multiple types since they are not in the scope of any other type. ACT defines global parameter variables and template
parameter variables as immutable types—they can only be defined once.
Array and subrange connections
Array connections in ACT are extremely flexible. In general, if two arrays have the same basic type and have the same number of elements, they can be connected with a simple connect directive. In the
example below, nodes x[0], …, x[9] are connected to nodes y[10], …, y[19] respectively.
bool x[10];
bool y[10..19];
Connecting two arrays of differing sizes is an error.
bool x[10];
bool y[10..20];
-[ERROR]-> Connection: x = y
LHS: bool[10]
RHS: bool[10..20]
FATAL: Type-checking failed on connection
Types `bool[10]' and `bool[10..20]' are not compatible
ACT provides a subrange mechanism for connecting parts of arrays to one another. The example below shows a connection between elements x[3], …, x[7] to y[12], …, y[16].
x[3..7] = y[12..16];
Connections between two arrays with differing numbers of dimensions is not permitted.
Array shapes
When a connection between multidimensional arrays is specified where the shape of the two arrays is not identical, this is also reported as an error. However, it is possible that two arrays have the
same shape but where the elements have differing indices. In this case, the ranges to be connected are sorted in lexicographic order (with indices closer to the variable having higher weight) and the
corresponding array elements are connected. For example, in the example below x[3][5] would be connected to y[0][0] and so on.
bool x[3..4][5..6];
bool y[2][2];
x = y;
When two arrays are connected by name as in the example above, they become aliases for each other. So while the connection statement
x[3..4][5..6] = y[0..1][0..1];
is the same as x=y earlier, the two are actually logically different. The first one says that the two arrays are the same, while the second is an element-by-element connection. This difference is
visible in the case of sparse arrays.
bool x[3..4][5..6];
bool y[2][2];
x = y;
bool x[5..5][5..5];
-[ERROR]-> Array being extended after it has participated in a connection
Type: bool[ [3..4][5..6]+[5..5][5..5] ]
In the example above, the arrays x and y are connected to each other. After the connection, the array x is being extended in size using the sparse array functionality. This is not allowed, because
this would also make y a sparse array—except, the way y is to be extended is unspecifed. On the other hand, the same sparse array extension is valid if instead the element-by-element connection is
performed. This is because array y has fewer elements compared to x, and only a subset of x is connected to the elements of y.
Performance note: Extending an array after it has connections can be expensive as the array connections have to be moved. If you have an array of size N with internal connections that is extended by
a constant amount M times, the current implementation can take time complexity MN.
Finally, two sparse arrays can be connected to each other as long as they have the same shape. The shape is determined by viewing a sparse array as an ordered collection of dense sub-arrays. Two
sparse arrays have the same shape if they have the same number of dense sub-arrays, and the corresponding dense sub-arrays have the same shape.
Finally, arrays can be re-shaped. The most common example of this is that a list of variables can be treated as a single array by enclosing it in braces.
bool x[3];
bool x0,x1,x2;
x = {x0,x1,x2};
This is a special case of a more general mechanism for array expressions, described next.
Array expressions
There are times when it is convenient to 'reshape' an array, or only connect part of an array to other instances. ACT provides flexible syntax to construct arrays from other arrays and instances.
If two arrays have the same dimension, then they can be concatenated together to form a larger array using the # operator. In order to do this, the size of (n-1) dimensions of the arrays must match
(excluding the left-most dimension).
bool x[5];
bool y[3];
bool z[8];
z = x # y; // success!
Sometimes it is useful to be able to connect arrays that have different numbers of dimensions to each other. To change the dimensionality of an array, the { … } construct can be used. A
higher-dimensional array can be created by providing a comma-separated list of lower dimensional arrays enclosed in braces.
bool x[2];
bool y[2];
bool z[2][2];
z = {x,y}; // success!
In this example, the construct {x,y} is used to create a two-dimensional array from two one-dimensional arrays. The size and number of dimensions of all the arrays specified in the list must be the
same. The most common use of this construct is to connect a one-dimensional array to a list of instances.
A final syntax that is supported for arrays is a mechanism to extract a sub-array from an array instance. The following example extracts one row and one column from a two-dimensional array.
bool row[4],col[4];
bool y[4][4];
y[1][0..3]=row; // connect row
y[0..3][1]=col; // connect column
In general, array expressions can appear on either the left hand side or the right hand side of a connection statement. This means that the following is a valid connection statement:
bool a[2][4];
bool b[4..4][4..7];
bool c0[4],c1[4],c2[4];
{c0,c1,c2} = a # b; // success, both sides are bool[3][4]
User-defined Type Connections
The result of connecting two user-defined types is to alias the two instances to each other. A connection is only permitted if the two types are compatible.
Connecting identical types
If two variables have identical types, then connecting them to each other is a simple aliasing operation.
Connecting types to their implementation
If one type is an implementation of a built-in type, then they can be connected to each other. The combined object corresponds to the implementation (since the implementation contains strictly more
information). Consider the example below, where d1of2 implements an int<1>.
int<1> x;
d1of2 y;
bool w;
y.d0 = w; // success!
x.d1 = w; // failure
While the first connection operation will succeed, the d0/d1 fields of the d1of2 implementation are not accessible through variable x since x was declared as an int<1>.
However, both x and y refer to the same object. For example, consider the CHP body below that uses x, y, and w. (Note: the situation we are discussing here is not a recommended one. It is only being
used to illusrate how connections behave.)
chp {
[w->skip [] ~w->x:=0];
Setting variable x modifies w, since y.d0 is aliased to w and x is aliased to y.
If there are two different implementations of the same type, attempting to connect them to each other is a type error. Suppose we have two implementations of an int<2>: d2x1of2 which uses two
dual-rail codes, and d1of4 which uses a one-of-four code. Consider the following scenario:
int<2> ivar;
d1of4 x;
d2x1of2 y;
Now the operation x=ivar is legal, and so is y=ivar. However, if both connections are attempted, it is an error. This is because one cannot connect a d1of4 type to a d2x1of2 type. This can get
confusing, and is a problem for modular design.
To encapsulate this problem within the definition of a type, we impose the following constraint: if a port parameter is aliased within the definition of a type, then the type in the port list must be
the most specific one possible. This prevents this problem from crossing type definition boundaries.
Connecting subtypes is a bit more complicated, but not that different from the rules for connecting implementations. Consider the following situation, where bar and baz are both subtypes of foo.
foo f1;
bar b1;
baz b2;
This would only succeed if there is a linear implementation relationship between foo, bar, and baz. In other words, the connection succeeds if and only if either bar <: baz or baz <: bar, and the the
object it represents corresponds to the most specific type.
If bar and baz are not linearly related, the connection fails even though individually the operations f1=b1 and f1=b2 would have succeeded. To avoid having this situation escape type definition
boundaries, types used in the port list must be the most specific ones possible given the body of the type.
Port connections
When instantiating a variable of a user-defined type, the variables in the parameter list can be connected to other variables by using a mechanism akin to parameter passing.
defproc dualrail (bool d0, d1, a)
spec {
exclhi(d0,d1) // exclusive high directive
bool d0,d1,da;
dualrail c(d0,d1,da);
In the example above, nodes d0, d1, and da are connected to c.d0, c.d1, and c.da respectively. Nodes can be omitted from the connection list. The following statement connects d1 to c.d1 after
instantiating c.
dualrail c(,d1,);
Since parameter passing is treated as a connection, all the varied connection mechanisms are supported in parameter lists as well.
Two other mechanisms for connecting ports are supported. The first mechanism omits the type name. The example below is equivalent to the one we just saw.
dualrail c;
While this may appear to not be useful (since the earlier example is shorter), it can be helpful in the context of array declarations. For example, consider the following scenario:
dualrail c[4];
bool d1[4];
(i:4: c[i](,d1[i],); )
A second mechanism is useful to avoid having to remember the order of ports in a process definition. Instead of using the port list of the form where we simply specify the instance to be passed in to
the port, we can instead use the following syntax.
bool d1;
dualrail c(.d1=d1);
dualrail x[4];
bool xd1, xd0; | {"url":"https://avlsi.csl.yale.edu/act/doku.php?id=language:connections","timestamp":"2024-11-08T14:15:49Z","content_type":"text/html","content_length":"40435","record_id":"<urn:uuid:50e926a0-1b3e-4087-b96e-681ab0ecc039>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00065.warc.gz"} |
Logical Operators in Java | Dremendo
Logical Operators in Java
Operators in Java
In this lesson, we will learn what is the Logical Operators and how it works in Java programming with some examples.
What is Logical Operators
Logical operators in Java are used to evaluate two or more conditions. They allow a program to make a decision based on multiple conditions. If the result of the logical operator is true then 1 is
returned otherwise 0 is returned.
There are 3 types of logical operators in Java and they are:
Now let's see the examples of all the logical operators one by one for more understanding.
&& AND Operator
&& (AND) operator is used to check if two or more given conditions are true or not. The output of && (AND) operator is 1 (true) only if all the given conditions are correct otherwise the output will
be 0 (false) if any one of the given conditions is wrong.
int x=15, y=2, z=6, k=25, a, b;
a=x>5 && y<6 && k>=25; // output of a is 1
b=x>y && k<25; // output of b is 0
In the example above the output of a is 1 because all the three given conditions x>5, y<6 and k>=25 are correct (true). On the other hand the output of b is 0 because out of the two given conditions
x>y and k<25 the second condition k<25 is wrong (false).
|| OR Operator
|| (OR) operator is used to check if any one of the given conditions is true or not. The output of || (OR) operator is 1 (true) if at least any one of the given conditions is correct otherwise the
output will be 0 (false) only if all the given conditions are wrong.
int x=15, y=2, z=6, k=25, a, b;
a=x<5 || y<6 || k>=25; // output of a is 1
b=x<y || k<25; // output of b is 0
In the example above the output of a is 1 because out of the three given conditions x<5, y<6 and k>=25 the second condition is correct (true). On the other hand the output of b is 0 because out of
the two given conditions x<y and k<25 both of them are wrong (false).
! NOT Operator
! (NOT) operator negates the value of the condition. If the condition is false then the output of ! (NOT) operator becomes true (1). If the condition is true then the output of ! (NOT) operator
becomes false (0).
int x=15, y=10, a, b;
a=!(x>5); // output of a is 0
b=!(y<5); // output of b is 1
In the example above the output of a is 0 because the given condition x>5 is correct (true) so ! (NOT) operator changed the output from 1 to 0. On the other hand the output of b is 1 because the
given condition y<5 is wrong (false) so ! (NOT) operator changed the output from 0 to 1.
Test Your Knowledge
Attempt the multiple choice quiz to check if the lesson is adequately clear to you. | {"url":"https://www.dremendo.com/java-programming-tutorial/java-logical-operators","timestamp":"2024-11-05T01:20:31Z","content_type":"text/html","content_length":"58049","record_id":"<urn:uuid:5eb6f981-6622-4b24-a04f-ac1c1801fef1>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00421.warc.gz"} |
Würschmidt comma
Jump to navigation Jump to search
Interval information
Ratio 393216/390625
Factorization 2^17 × 3 × 5^-8
Monzo [17 1 -8⟩
Size in cents 11.44529¢
Name würschmidt comma
Color name sg^83, Saquadbigu comma
FJS name [math]\text{dddd3}_{5,5,5,5,5,5,5,5}[/math]
Special properties reduced
Tenney height (log[2] nd) 37.1604
Weil height (log[2] max(n, d)) 37.1699
Wilson height (sopfr (nd)) 77
Harmonic entropy ~2.32402 bits
(Shannon, [math]\sqrt{nd}[/math])
Comma size small
open this interval in xen-calc
Würschmidt's comma ([17 1 -8⟩ = 393216/390625) is a small 5-limit comma of 11.4 cents. It is the difference between an octave-reduced stack of eight classical major thirds and a perfect fifth: (5/4)^
8/6, which comes from 5/4 being a convergent in the continued fraction of [math]\sqrt[8]{6}[/math].
In terms of commas, it is the difference between:
Tempering out this comma leads to the würschmidt family of temperaments. In any nontrivial tuning (that is, not 3edo), there is an exact neutral third between 5/4 and 6/5, which represents a
tempering of 625/512~768/625 and can be used to represent 11/9~27/22 (or more accurately 49/40~60/49, tempering out 2401/2400 instead of or in addition to 243/242).
Magic is a simpler analogue of würschmidt, reaching 3/1 with (5/4)^5 which exceeds 3/1 by the magic comma, and a even simpler analogue of würschmidt is dicot, where 3/2 is reached by (5/4)^2. More
interesting is that there is a lower-accuracy but more complex analogue of würschmidt if we look at the pattern; the powers of 5/4 go 2 (dicot), 5 (magic), 8 (würschmidt), corresponding to
increasingly sharp tunings of 5 where each additional three 5's represent a lowering of 25/16 by another 128/125; finally, at (5/4)^11 / (12/1), we get magus, a sharp-major-third analogue of | {"url":"https://en.xen.wiki/w/W%C3%BCrschmidt_comma","timestamp":"2024-11-07T00:15:20Z","content_type":"text/html","content_length":"29135","record_id":"<urn:uuid:7f9dad52-2ee2-4fe5-8326-2effa95c4bdc>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00079.warc.gz"} |
New package on CRAN: lamW | R-bloggersNew package on CRAN: lamW
New package on CRAN: lamW
[This article was first published on
Strange Attractors » R
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
Recently, in various research projects, the Lambert-W function arose a number of times. Somewhat frustratingly, there is no built-in function in R to calculate it. The only options were those in the
gsl and LambertW packages, the latter merely importing the former. Importing the entire GNU Scientific Library (GSL) can be a bit of a hassle, especially for those of us restricted to a Windows
Therefore, I spent a little time and built a package for R whose sole purpose is to calculate the real-valued versions of the Lambert W function without the need for importing the GSL: the lamW
package. It does depends on Rcpp, though. It could have been written in pure R, but as there are a number of loops involved which cannot be vectorized, and as Rcpp is fast becoming almost a base
package, I figured that the speed and convenience was worth it.
A welcome outcome of this was that I think I finally wrapped my head around basic Padé approximation, which I use when calculating some parts of the primary branch of the Lambert-W. Eventually, I’d
like to write a longer post about Padé approximation; when that will happen, who knows 8-). | {"url":"https://www.r-bloggers.com/2015/05/new-package-on-cran-lamw/","timestamp":"2024-11-03T19:59:32Z","content_type":"text/html","content_length":"84858","record_id":"<urn:uuid:374c20ac-2363-47bc-8674-5ea56914161c>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00643.warc.gz"} |
How much force must the biceps exert to hold the ball
Reference no: EM13542640
Evaluate maximum altitude
Evaluate maximum altitude?
Introductory mechanics: dynamics
Calculate the smallest coefficient of static friction necessary for mass A to remain stationary.
Determine the tension in each string
Determine the tension in each string
Quadrupole moments in the shell model
Quadrupole moments in the shell model
Calculate the dc voltage
Calculate the dc voltage applied to the circuit.
Gravity conveyor
Illustrate the cause of the components accelerating from rest down the conveyor.
Questions on blackbody, Infra-Red Detectors & Optic Lens and Digital Image.
What is the magnitude of the current in the wire
What is the magnitude of the current in the wire as a function of time?
What is the maximum displacement of the bridge deck
What is the maximum displacement of the bridge deck?
What is the electric field at the location
Question: Field and force with three charges? What is the electric field at the location of Q1, due to Q 2 ?
Find the equivalent resistance
A resistor is in the shape of a cube, with each side of resistance R . Find the equivalent resistance between any two of its adjacent corners.
Find the magnitude of the resulting magnetic field
A sphere of radius R is uniformly charged to a total charge of Q. It is made to spin about an axis that passes through its center with an angular speed ω. Find the magnitude of the resulting magnetic
field at the center of the sphere.
Assured A++ Grade
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report! | {"url":"https://www.expertsmind.com/library/how-much-force-must-the-biceps-exert-to-hold-the-ball-5542640.aspx","timestamp":"2024-11-03T13:31:05Z","content_type":"text/html","content_length":"68407","record_id":"<urn:uuid:849bb4e1-dec8-46be-9c44-891ca2e03cb0>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00355.warc.gz"} |
Implicit structural inversion for lithology, using a gridded model
The aim of this PhD project was to develop a method for implicit structural inversion of geophysical data, using a gridded model. With the term {\em structural inversion} we mean an inversion to
obtain an image of the (gridded) sub-surface, which can directly be interpreted in terms of sharp boundaries ... read more between different lithologies. {\em Implicit} indicates that we neither
assume a fixed number of anomalies, nor enforce a fixed structure explicitly. We model the sub-surface with a grid of homogeneous, rectangular prisms. The values of the physical property of these
cells are the parameters inverted for. The structures in the model should be apparent from large contrasts in parameter values between neighboring cells. We developed a method for such an implicit
structural inversion, using Linear Programming (LP). All the models investigated comprised of two lithologies, with one or more homogeneous anomalous regions embedded within a homogeneous background.
The contrast between rock properties is used a constraint on the parameters. Ideally each cell will ultimately be identifiable as belonging to one of the lithologies. This LP-based method (using the
L1-norm data misfit) was tested on synthetic gravity data and was able to reconstruct important features of the test models. The results compared favorably with a traditional inversion using
Truncated Singular Values. This method does have the limitation that the minimum depth to the top needs to be known a priori. This method was then extended to simultaneously invert for both a linear
trend in the data and for the density contrasts of the cells. The method was used to invert a field gravity data set. The results are in agreement with the ones obtained by the industry using
detailed forward modeling. The same LP-based method was applied to invert synthetic seismic cross-well first arrival travel-times for the absolute inter-well slowness distribution. This is a
non-linear problem due to ray-bending in a heterogeneous medium. For a simple sub-surface model, the implicit structural inversion using the true rays, or using the iterative scheme, both produced
very good models of the sub-surface. However, a regularisation parameter needs to be chosen properly. A more complicated model with poor ray-coverage, was difficult to retrieve using only seismic
data, however, a joint inversion of seismic data (using the true rays) together with gravity data gave encouraging results. Another method, this time using the L2-norm of the data misfit, was
developed in parallel. In an iterative scheme, a reference model was constructed, using information regarding the density contrasts, which the inversion result should resemble. The method needs 5
tuning constants; when proper (wide) ranges are supplied, a parameter search for the constants, yields their optimal values to be used for the inversion. This search can be steered using the
resulting data misfits. One of the tuning parameter is dynamically increased during the iterations to ensure a result with clear structure. Experiments on synthetic gravity data gave good inversion
results, even when the top of the anomaly was unknown. show less | {"url":"https://dspace.library.uu.nl/handle/1874/8775","timestamp":"2024-11-02T18:41:29Z","content_type":"application/xhtml+xml","content_length":"25308","record_id":"<urn:uuid:96eeac43-3b83-4575-98c4-8b30f75d1498>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00467.warc.gz"} |
Can a Dynamic Reward–Penalty Mechanism Help the Implementation of Renewable Portfolio Standards under Information Asymmetry?
School of Economics and Management, Wuhan University, Wuhan 430072, China
Submission received: 19 March 2020 / Revised: 18 April 2020 / Accepted: 20 April 2020 / Published: 23 April 2020
To further promote the low-carbon and sustainable development of China’s power industry, the Chinese government is vigorously introducing competition into power sales market. Simultaneously, On
November 15, 2018, the National Development and Reform Commission issued the “Notice on Implementing the Renewable Portfolio Standards (Draft)” to propose the implementation of power sales side
Renewable Portfolio Standards (RPS), which cannot be realized without an effective government regulation mechanism. However, information asymmetry and the limited rationality of the regulatory
agencies and private power sales companies in the regulation process make the regulatory effect uncertain to the detriment of a sustainable regulation of the power industry. Thus, it is necessary to
optimize the regulation mechanism of the RPS policy in China. We considered the competitive relationship between integrated power sales companies and independent power sales companies, and
established an evolutionary game model based on a limited rationality. We also analyzed the implementation effects of the static reward penalty mechanism and dynamic reward penalty mechanism,
respectively. The system dynamics (SD) simulation results showed that under the static reward penalty mechanism, there is no evolutionary stable equilibrium solution, and there will be volatility
that exists in the evolution process. However, the dynamic reward penalty mechanism can effectively solve these problems. What is more, our results implied that governments should formulate
appropriate RPS quotas, improve the green certificate trading mechanism, and take into account the market size of power sales while implementing RPS policy.
1. Introduction
Promoting the sustainable development of renewable energy is a measure in response to issues like the shortage of fossil fuels, environmental pollution, climate change, and energy security, as well
as an important means to achieve a healthy and sustainable development of society and economy [
]. In recent years, low-carbon transformation and the upgrading of the power industry have been valued by countries around the world. With the target incentives and policy support of governments in
various countries, the renewable energy power generation industry has achieved considerable development [
]. By the end of 2019, the cumulative global installed capacity of non-water renewable energy power generation reached 1347.405 GW, and the wind power and photovoltaic installed capacity reached
622.704 GW and 586.434 GW, respectively. Meanwhile, China’s wind power and photovoltaic installed capacity reached 210.478 GW and 205.493 GW respectively, both ranking first in the world [
]. At the same time, the Chinese government also pledged to further increase policy support for the renewable energy power industry and to continue to promote the green development of the power
industry [
]. With the rise of China’s renewable energy share in the energy supply system, it is of great practical significance to further increase the share of renewable energy in the energy consumption
system and transform the driving force of renewable energy from a power generation incentive to a power consumption incentive, thereby promoting renewable energy consumption.
The reform of the global power system began in the 1980s. The first reform aimed to break the monopoly power industry structure and establish a competitive market mechanism [
], and the second aimed to promote the marketization of power commodities and strengthen government regulation [
]. So far, some scholars have discussed problems faced by the competitive power sales market, such as the optimal power purchase and sales decision making based on the analysis of the power quality
and power load forecasting [
], prediction of electricity prices based on neural networks [
], analysis of the power system flexibility based on uncertainty [
], and mechanism design based on distributed energy [
]. However, due to the generation cost and technical constraints, and the uncertainty of renewable power [
], it is still difficult to promote the consumption of renewable energy power. Thus, the power system reform has been put on the agenda of the Chinese government [
]. On November 30, 2015, the State Council issued the “Implementation Opinions on Promoting the Reform of the Power Sales Side”, introducing various enterprises and users into the power retail
market, and building a competitive sales market. Under the current implementation of the power sales side RPS policy, China’s electricity market is structured as shown in
Figure 1
Renewable Portfolio Standards (RPS) is one of the renewable energy power industry support policies, and it requires a certain percentage of renewable energy power to be produced and consumed in the
electricity market [
]. The United States introduced the “Renewable Portfolio Standard” and “Renewable Energy Certificate” policies in 1980, the United Kingdom introduced the “Renewable Obligation” policy in 2002, and
Australia also introduced the “Mandatory Renewable Energy Target” policy in 2001 [
]. As for China, the National Development and Reform Commission and the National Energy Administration issued the “Notice on Implementing the Renewable Portfolio Standards (Draft)” On November 15,
2018, which set RPS quotas for power consumption and held provincial governments responsible for the power sales side RPS regulation. What is more, the RPS regulation mechanism was established by
government regulation agencies to punish the agents that did not fulfill their RPS obligations, and incentive indicators were set according to the binding indicators to subsidize power sales
companies that outperform their RPS obligations. Taking power sales companies as the subjects of RPS responsibility presents three advantages. First, it can reduce the need for subsidy funds; second,
it can alleviate, if not solve, the problem of renewable energy power abandonment; and third, the appropriate RPS quota index can promote stable and sustainable growth for the development of
renewable energy in the future, and guarantee the realization of China’s non-fossil energy share target in 2020 and 2030. From the perspective of international experience and the long-term
development of the renewable energy power industry, viewing the power sales companies as the subjects of RPS responsibility becomes all the more advantageous [
In China’s current electricity market, the obligations of the RPS are mainly set for electricity distribution companies and power users. Therefore, this article, focused on the power sales side RPS
policy mechanism, mainly sets the RPS quota targets of provinces, formulates the rewards and punishment mechanisms for power sales companies, and motivates and guides power sales companies to consume
more renewable power. This policy mechanism can help solve the problem of renewable energy power abandonment while realizing the positive social benefits of renewable energy power at the same time.
Therefore, a green certificate trading mechanism is constructed to guarantee the implementation of the RPS policy [
]. External environmental benefits of renewable energy power are reflected as a green certificate corresponding to 1 MW.h of renewable energy power, which can be traded in the green certificate
market [
]. The RPS policy and the green certificate trading mechanism can effectively promote the development of the renewable energy power industry through the combination of the market mechanism and
administrative regulation [
], a combination which, based on marketization, helps to create space for the continuous growth of renewable energy power. What is more, designing a reasonable RPS policy will make for a more stable
power price.
At present, power sales companies in China mainly fall into two types: integrated power sales companies and independent power sales companies. Integrated power sales companies not only sell power
products but also have power generation resources. In contrast, independent power sales companies do not own power generation resources and can only sell power products. In recent years, competition
in the power sales market has become increasingly fierce. As of August 2018, there were more than 3600 registered power sales companies nationwide, and the number of registered power sales companies
in each province is shown in
Figure 2
The implementation of the power sales side RPS is inseparable from an effective government regulation. It is the result of strategic interactions between competitive power sales companies and
government regulation agencies. Under the power sales side RPS policy mechanism, power sales companies, in order to achieve the RPS target, need to face the risk of uncertainty caused by a high
proportion of renewable energy. At the same time, new methods are necessary to stimulate the sales of renewable energy power for power sales companies. For example, new power sales channels need to
be developed, value-added services be provided, and cross-regional transaction contracts be signed with renewable energy generation companies. However, these methods will undoubtedly increase the
operating risks and costs of the power sales companies. That is why power sales companies lack the enthusiasm to fulfill their RPS obligations and are unwilling to actively sell renewable energy
power. In order to achieve the target of energy transformation and upgrading in China, the Chinese government expects power sales companies to consume more renewable energy power. Thus, government
regulation is necessary, which in turn will bring greater regulation pressure and regulation costs. Based on the above analysis, we can find that during the implementation of the power sales side RPS
policy, the decision-making targets between power sales companies and government regulation agencies are not consistent, and their strategic selections affect each other. This is in line with the
theoretical characteristics of game theory, a theory mainly used to study how multiple agencies maximize their own interests [
]. The strategy selections of the power sales companies and the government regulation agencies are diverse and complex, and there are both selections and abrupt changes in the dynamic evolution
process. What is more, the strategic behaviors of power sales companies and government regulation agencies are characterized by a certain inertia [
]. Previous studies, however, were mostly based on the complete rational hypothesis, ignoring the limited rationality of the relevant agents and the imitation between game agents in the process of
repeated interaction, which is unable to meet the needs of multi-agents’ decision-making optimization [
]. The evolutionary game theory based on a bounded rationality can provide a new research idea for analyzing the government regulation mechanism of the power sales side RPS policy [
To systematically analyze the power sales side RPS government regulation problem, based on the evolutionary game theory, we considered the competitive relationship between integrated power sales
companies and independent power sales companies based on a limited rationality [
]. We creatively constructed an evolutionary game model among the government regulation agencies, integrated power sales companies, and independent power sales companies, and simulated the dynamic
strategy selection of game agents according to the system dynamics theory. What is more, the strategy selection of game agents and the strategy selection evolution processes under the static
reward–penalty mechanism and the dynamic reward–penalty mechanism were also analyzed.
2. Literature Review
Since the electricity sales side in China was not opened in the past, most domestic scholars excluded the electricity sales side from the obligated subject of RPS. Research on RPS mainly focused on
the power generation side, including the transaction model, implementation plan, resource allocation of the power generation side RPS market power, supervision, and management [
]. With the opening of the electricity sales side, according to foreign practices, it is necessary to discuss the feasibility of implementing RPS on the electricity sales side in China and the design
of the RPS system.
According to the Renewable Energy Law promulgated in 2005, China will increasingly rely on renewable energy to support economic development after 2020 [
]. In recent years, many scholars have thoroughly studied the RPS implementation issue around the world. They found that RPS policy could promote competition between power generators and the
production of renewable energy power [
], which helped to lower electricity prices, reduce the carbon emissions, and improve the efficiency of the electricity market [
]. What is more, the establishment of the green certificate market can reduce the cost of power generation companies and promote the effective allocation of resources and technology investment [
Some scholars also compared the effectiveness of the Feed-in Tariff (FIT) and renewable portfolio standard (RPS). FIT policy is a support mechanism of renewable energy. Through direct pricing,
investors could obtain a stable income and effectively promote the development of the renewable power industry in the early stage. Like Germany, China developed its renewable energy industry rapidly
at the early stage. At the same time, there were some serious problems, such as large subsidy gaps [
]. Recently, the Chinese government has been exploring how to implement a renewable portfolio standard (RPS) policy that requires a certain fraction of electricity to be generated and consumed from
renewable energy sources [
]. The RPS policy was found to be effective in reducing carbon emissions, improving consumer surplus, and increasing the market share of renewable energy power [
]. These studies showed that the policy effect of the RPS mechanism is significant. At the same time, some scholars also found that the implementation of the RPS policy would affect residents’
consumption strategy selection and investor confidence, which in turn affected the policy effects [
]. Thus, improving RPS policy transparency is the key to controlling RPS policy costs and achieving policy objectives. Policy makers should consider not only the factor of carbon emission reduction,
but also factors such as economic development and electricity price [
]. It can be seen that the implementation process of the RPS policy faces uncertainty. As the effect of the RPS policy is strictly dependent on government regulation [
], how to effectively regulate the power sales companies’ strategy selection is critical.
The implementation of RPS cannot do without an effective regulation mechanism. Due to the higher generation cost and instability, power sales companies lack the incentive to consume renewable energy
power, and an effective government regulation is particularly necessary. Jensen [
] studied the price and consumption effects between the power market and the green certificate trading market. He found that the interaction between the two markets made the implementation effect of
RPS policy ambiguous. Thus, the development of a reasonable and comprehensive government regulation mechanism was particularly important for the implementation of RPS policy. Tanaka [
] and Dong [
] respectively constructed a competitive game model and empirical analysis model. They found that traditional fossil energy power generators restrained the policy impact of the RPS by exerting market
power. Thus, the regulatory authorities need to take effective regulatory measures to ensure the smooth implementation of the power sales side RPS policy.
The effect of the power sales side RPS government regulation is the result of a strategic interaction between competitive power sales companies and government regulatory agencies [
]. Nasiri [
] constructed a coupled constrained game model and analyzed the interaction of power producers and government regulation authorities under the RPS policy. Pineda [
] established the Cournot game model of power generation group expansion based on the RPS policy and discussed the correlation between penalty and RPS quota through a case study. Son et al. [
] established a game model between two types of power producers under the RPS policy and studied the best operational strategies of Korean power producers based on a scenario analysis. However, their
studies are based on the complete rational hypothesis, ignoring the limited rationality of relevant game agents and the learning between game groups in the process of repeated games [
]. In real life, due to the uncertainty of the environment and the incomplete information of the market, game agents have only a limited rationality [
]. Yi et al. [
] discussed the behavior strategy of electricity producers under RPS and argued that the stage descent mode of subsidy and a higher level of fine would improve the enthusiasm of electricity producers
for the RPS scheme. Zhu et al. [
] developed a system dynamics model of a tripartite evolutionary game to analyze the strategy interaction of stakeholders and to simulate the corresponding evolution process, and the results revealed
some policy effects such as a reversal effect, blocking effect, and over-reliance effect. However, previous research considered only the static reward–penalty mechanism rather than the impact of the
dynamic reward–penalty mechanism.
This paper systematically studied the effectiveness of the power sales side RPS government regulation mechanism. Based on the existing research, we considered the competitive relationship between
integrated power sales companies and independent power sales companies, and established an evolutionary game model between the government regulation agencies, the integrated power sales companies,
and the independent power sales companies to analyze the impact of strategic interactions. As the government regulation process of the power sales side RPS is a complex multi-variable, high-order,
and nonlinear dynamic feedback system with obvious system dynamics characteristics [
], we simulated and analyzed the dynamic behavior of game agents under different reward–penalty mechanisms based on the system dynamics theory. The relevant conclusions are drawn in the end.
3. Evolutionary Game Model of Power Sales Side RPS Regulation
At present, after the power sales side reform, there are mainly two types of power sales companies in China: integrated power sales companies and independent power sales companies. What is more,
provincial government regulation agencies are accountable for the power sales side RPS regulation. Therefore, the main game agents of the power sales side RPS regulation are government regulation
agencies, integrated power sales companies, and independent power sales companies. The probability of government regulation agencies regulating the strategy selection of power sales companies is $x$,
and that of them deregulating the strategy selection of power sales companies is $1 − x$. The probability of integrated power sales companies obeying the RPS regulation is $y$, and that of them
disobeying the RPS regulation is $1 − y$. The probability of independent power sales companies obeying the RPS regulation is $z$, and that of them disobeying the RPS regulation is $1 − z$. Other
assumptions are as follows:
Assumption 1.
RPS policy requires all of the power sales companies to obey the regulation and fulfill their obligation. The proportion of renewable energy in the power products sold by electricity sales company
should reach$q T$. Power sales companies that exceed their RPS obligations will receive rewards, with the unit reward standard being$e$. Power sales companies that fail to fulfill their RPS
obligations will get punished, the unit penalty standard being$f$.
Assumption 2.
When government regulation agencies select the regulation strategy, the average regulation cost is$C g$. When power sales companies fulfill their RPS obligation, the unit social environmental benefit
is$t$, and the total social environment benefit is$R i = t ( q i 2 − q i 1 ) W i$. When the government regulation agencies select the deregulation strategy, the corresponding reputation losses are$H$
Assumption 3.
The sales scale of integrated power sales companies is$W 1$, and the sales scale of independent power sales companies is$W 2$. Since integrated power sales companies own generation resources, they
are more competitive than independent power sales companies, and the proportion of renewable energy power sold by integrated power sales companies is higher than that of independent ones. Therefore,
this paper assumes that the proportion of renewable energy in the power products sold by integrated power sales companies that do not fulfill their RPS obligation is$q 1 1$. In order to fulfill the
RPS quota obligation, the proportion of renewable energy in the power products sold by integrated power sales companies that fulfill their RPS obligation, by way of purchasing the green certificate,
is$q 1 2$, and the proportion of renewable energy in the power products sold by independent power sales companies that fulfill their RPS obligation is$q 2 1$. Meanwhile, the proportion of renewable
energy in the power products sold by independent power sales companies that do not fulfill their RPS obligation is$q 2 2$, where$q 1 2 > q 2 2 > q T > q 1 1 > q 2 1$.
Assumption 4.
The corresponding unit price difference income obtained by integrated power sales companies for the sales of power products is$g i$. Therefore, the sales revenue of integrated power sales companies
is$P 1 = g 1 W 1$, and the sales revenue of independent power sales companies is$P 2 = g 2 W 2$. As integrated power sales companies have a large price gap due to the simultaneous mastery of upstream
and downstream industries,$g 1 > g 2$.
Assumption 5.
Under the market-oriented green certificate trading mechanism, power sales companies mainly fulfill their RPS obligation by purchasing green certificates. Assume that the price of the green
certificate is$P e$. Since the number of green certificates that power sales companies need to purchase is$( q i 2 − q i 1 ) W i$, the cost of the green card for the cumulative expenditure of
electricity-selling enterprises is$B i = P e ( q i 2 − q i 1 ) W i$.
Assumption 6.
As power sales companies can make other, different services in addition to the purchase and sale of electricity, they can also carry out other ancillary services. Therefore, the cost for power sales
companies to purchase the green certificate goes along with a certain opportunity cost. The unit opportunity cost is$d i$when power sales companies do not fulfill their RPS obligation by purchasing
the green certificate and transfer the capital to other auxiliary services. The corresponding opportunity cost benefit is:$O i = d i P e ( q i 2 − q i 1 ) W i$.
Assumption 7.
In order to encourage power sales companies to exceed the RPS obligation, the amount of reward provided by the government to power sales companies that exceed the RPS quota indicator is$E i = e ( q i
2 − q T ) W i$. At the same time, in order to regulate and guide the behavior of power sales companies, the government imposes a penalty amount of$F i = f ( q T − q i 1 ) W i$on power sales companies
that have not completed the RPS obligation. When both types of power sales companies fail to fulfill their quota obligations, the penalty is aggravated,$F i = ∑ i = 1 2 f ( q T − q i 1 ) W i$.
The tripartite income matrix is shown in
Table 1
In the government regulation evolutionary game of RPS on the power sales side, it is difficult for decision makers to make the optimal strategy choice, because of their limited information collection
ability. They usually make dynamic strategic selections by means of learning and imitating those with higher returns. The dynamic replicator equation means that the frequency of individuals’
strategies changes proportionally to the frequency of the group’s strategies and to the magnitude at which individuals’ fitness exceeds the average fitness of the group.
$U x$
$U x ′$
respectively represent the expected returns of the government group that regulates and the one that deregulates, and
$U ¯ x$
represents the government group. The overall expected return, shown as follows, is the dynamic replicator equation
$F ( x )$
of the government regulatory agency:
$U x = yz ( R 1 + R 2 − C g − E 1 − E 2 ) + ( 1 − y ) z ( R 1 + F 2 − C g − E 1 ) + y ( 1 − z ) ( R 2 + F 1 − C g − E 2 ) + ( 1 − y ) ( 1 − z ) ( F 1 + F 2 − C g ) .$
$U x ′ = yz ( R 1 + R 2 − H ) + ( 1 − y ) z ( R 1 − H ) + y ( 1 − z ) ( R 2 − H ) + ( 1 − y ) ( 1 − z ) ( − H ) .$
$U ¯ x = xU x + ( 1 − x ) U x ′$
The replicator dynamic equation is actually a dynamic differential equation to describe the frequency of a particular strategy being adopted in a population. The evolutionary game replicator dynamic
equation of the government regulatory agencies is:
$F ( x ) = dx / dt = x ( 1 − x ) ( F 1 + F 2 + H − zE 1 − yE 2 − zF 1 − yF 2 − C g ) .$
The probability of integrated power sales companies obeying the regulation is
, and the probability of them disobeying the regulation is
$( 1 − y )$
, respectively.
$U y$
$U y ′$
are used to represent the expected returns of the two groups of integrated power sales companies that choose a fulfillment strategy and a non-fulfillment strategy, respectively.
$U ¯ y$
represents the average expected return of integrated power sales companies; then, the dynamic replicator equation
$F ( y )$
of integrated power sales companies is as follows:
$U y = xz ( P 1 − B 1 + E 1 ) + ( 1 − x ) z ( P 1 − B 1 ) + x ( 1 − z ) ( P 1 − B 1 + E 1 ) + ( 1 − x ) ( 1 − z ) ( P 1 − B 1 ) .$
$U y ′ = xz ( O 1 − F 1 ) + ( 1 − x ) zO 1 + x ( 1 − z ) ( O 1 − F 1 − F 2 ) + ( 1 − x ) ( 1 − z ) O 1 .$
$U ¯ y = yU y + ( 1 − y ) U y ′ .$
The evolutionary game replicator dynamic equation of integrated power sales companies is:
$F ( y ) = dy / dt = y ( 1 − y ) ( P 1 − B 1 − xE 1 + xF 1 + xF 2 − xzF 2 − O 1 ) .$
The probability of independent power sales companies obeying the regulation is
, and the probability of them disobeying the regulation is
$( 1 − z )$
$U z$
$U 1 − z$
are used to represent the expected returns of the two groups of independent power sales companies that choose a fulfillment strategy and a non-fulfillment strategy, respectively.
$U ¯ z$
represents the average expected returns of independent power sales companies. Then, the dynamic replicator equation
$F ( z )$
of independent power sales companies is as follows:
$U z = xy ( P 2 − B 2 + E 2 ) + ( 1 − x ) y ( P 2 − B 2 ) + x ( 1 − y ) ( P 2 − B 2 + E 2 ) ( 1 − x ) ( 1 − y ) ( P 2 − B 2 ) .$
$U z ′ = xy ( O 2 − F 2 ) + ( 1 − x ) yO 2 + x ( 1 − y ) ( O 2 − F 1 − F 2 ) + ( 1 − x ) ( 1 − y ) O 2 .$
$U ¯ z = zU z + ( 1 − z ) U z ′ .$
The evolutionary game replicator dynamic equation of independent power sales companies is:
$F ( z ) = dz / dt = z ( 1 − z ) ( P 2 − B 2 − xE 2 + xF 1 + xF 2 − xyF 1 − O 2 ) .$
According to the replicator dynamic equation, we make
$F ( X ) = 0$
, and the specific expression of
$F ( X )$
is shown in formula (13):
$F ( X ) = { F ( x ) = x ( 1 − x ) ( F 1 + F 2 + H − z E 1 − yE 2 − z F 1 − y F 2 − C g ) F ( y ) = y ( 1 − y ) ( P 1 − B 1 − x E 1 + xF 1 + x F 2 − x z F 2 − O 1 ) F ( z ) = z ( 1 − z ) ( P 2 − B 2
− xE 2 + xF 1 + xF 2 − xyF 1 − O 2 ) .$
According to $F ( X ) = 0$, we can obtain the critical point, which is the equilibrium solution of the established multi-party evolutionary game model. There are multiple critical points in the
equation group, indicating that there are multiple equilibriums in the multi-party game model. Nevertheless, whether there is a stable evolutionary equilibrium in these equilibrium solutions is not
known from the mathematical derivation process. In order to more intuitively analyze the changes in the policy choices of the game evolution process, this paper used the system dynamics (SD) to
establish an evolutionary game model between the government and the two sales-oriented enterprises, and analyzed the impact of different game initial values on the game evolution process.
4. System Dynamics Analysis of the Evolutionary Game Model
Although power sales companies are positively influenced by the consumption of renewable energy, they have to bear corresponding costs in fulfilling the renewable energy quota obligation. Therefore,
they cannot automatically realize the optimal allocation of resources. The reward–penalty mechanism has proven to effectively guide and motivate enterprises to conduct long-term social cooperation
and curb the occurrence of free-riding behavior. This mechanism includes both positive incentives, rewards, and negative incentives. Through two-way incentives, the efficiency of regulation can be
improved while the cost of government regulation can be reduced. In addition, the mechanism is related to the choice of the behavior of the players involved in the game. The atmosphere is either
static or dynamic. Under the static reward–penalty mechanism, reward and punishment are constant. In contrast, under the dynamic reward–penalty mechanism, the player’s behavior and strategy choices
will be considered at the same time. With the competition relationship and relative execution degree between the supervised parties, the reward–penalty mechanism will be dynamically adjusted along
with the relative changes of behavior of the supervised parties.
System dynamics is a science that uses computer simulation technology as a means to combine qualitative analysis with quantitative analysis to study system information feedback. In the government
regulation of the RPS on the electricity sales side, more complex nonlinear interactions and feedback causality are involved. Therefore, with the help of venPLE32 software, this paper established a
dynamic model of a government regulation system for an energy quota system based on a static reward–penalty mechanism and a dynamic reward–penalty mechanism. The system dynamics simulation
environment was set as follows: INITIAL TIME = 0, TIME STEP = 0.0078125, and Units for Time was taken as months.
First of all, based on the reality of China’s electricity market and related data, this paper set the initial state parameters for the government regulation model of the RPS on the electricity sales
side. The basic situations of the model variables are shown in
Table 2
Table 3
4.1. System Dynamics Simulation of the Static Reward–Penalty Mechanism
The static reward–penalty mechanism refers to the static situation in which the competition between power sales companies is not considered. The punishment or reward measures depend on the absolute
difference, and the specific expression of the static reward–penalty mechanism is shown in Formula (14):
${ F i = f ( q T − q i 1 ) W i E i = e ( q i 2 − q T ) W i .$
Correspondingly, the system dynamic flow diagram under the static reward–penalty mechanism is shown in
Figure 3
The evolutionary process of government regulation authorities and the two types of power sales companies is shown in
Figure 4
It can be obtained from the evolution process of
Figure 4
that: Under the static reward–penalty mechanism, the government regulatory authorities will actively strengthen the regulation of the RPS on the sales side in order to achieve a renewable energy
development and obtain social environmental benefits. In the initial stage of policy implementation, the government’s strategic selections can play a good guiding role, in which case both types of
power sales companies tend to choose to fulfill their quota obligations. As the proportion of power sales companies fulfilling quota obligations increases, the government regulatory authorities will
reduce regulation costs and relax regulation. What is more, when the government strengthens regulation, power sales companies tend to fulfill their quota obligations. In contrast, as the government
slackens regulation, the probability of power sales companies fulfilling their quota obligations will decrease accordingly. The reason is that fulfilling the RPS quota obligation will increase the
cost of electricity-selling enterprises in spite of the external benefits it brings in. Therefore, with the relaxed government regulation, the power sales companies lack the initiative to assume the
RPS quota obligation in order to maximize their own interests. Observing the deregulation of government departments, they tend to choose not to strengthen the safety regulation. When the probability
of independent electricity sales enterprises fulfilling the RPS obligation drops to a certain extent, the government regulation department will gradually intensify their regulation in order to ensure
the implementation effect of the electricity-sales quota system. Repeatedly, the volatility shown in
Figure 3
The volatility phenomenon in the process of the government regulation of the RPS also widely exists in social life. With the aggravation of environmental pollution caused by enterprise production,
relevant government regulatory agencies often adopt strict regulatory measures in the initial stage to obtain better regulatory effects. However, as the power sales companies consume more renewable
power, government regulation agencies tend to relax regulation measures. When companies notice the relaxed regulation, they will gradually loosen pollution control in order to save relevant costs and
expand income, leading to an increase of environmental problems. In turn, the government regulatory authorities have to impose heavier punishments again, resulting in the volatility shown in
Figure 4
. The unsustainability of this regulatory approach is very detrimental to the sustainable development of the power sales industry.
It can be seen that strict reward and punishment measures are a powerful guarantee for the smooth implementation of the RPS on the electricity sales side. A constantly adjusting and revised policy is
not only unfavorable to market participants’ expectation of a long-term and stable policy, but also burdens policy implementation and regulation. At the same time, volatility in the game process
easily affects the judgment of the government regulatory authorities, and even arouses doubts on the validity of established regulatory policies. This, indeed, is detrimental to decision-makers
taking correct and reasonable strategic selections. However, it does not work to rely solely on increasing punishments to better implement the RPS. Though excessive punishments will enhance the
authority of government regulation, it is not conducive to the healthy development of the social economy. Thus, we introduced the dynamic reward–penalty mechanism in this paper.
4.2. System Dynamics Simulation of the Dynamic Reward–penalty Mechanism
It is necessary to promote the sustainable regulation of the power sales industry. Thus, we further proposed a dynamic regulatory mechanism. The dynamic reward–penalty mechanism in this paper means
that the incentives and punishments of regulatory authorities are related to the strategic selection of the two power-selling enterprise groups. It considers the competitive relationship between the
power sales companies. If a group of power sales companies sell more renewable power than another group, the penalty will be lower and the rewards will be higher, with the specific expression of the
dynamic reward–penalty mechanism shown in Formula (15):
${ F 1 = ( 1 − y + z ) f ( q T − q 1 1 ) W 1 F 2 = ( 1 + y − z ) f ( q T − q 2 1 ) W 2 E 1 = ( 1 + y − z ) e ( q 1 2 − q T ) W 1 E 2 = ( 1 − y + z ) e ( q 2 2 − q T ) W 2 .$
Correspondingly, the system dynamics flow diagram under the dynamic reward–penalty mechanism is shown in
Figure 5
The evolutionary stable equilibrium and the evolutionary process of government regulatory authorities and of the two types of power sales companies are shown in
Figure 6
From the evolution process of
Figure 6
, it can be obtained that under the dynamic reward–penalty mechanism, the volatility gradually decreases and finally reaches the equilibrium point of evolution and stability. The dynamic
reward–penalty mechanism that considers competition effectively suppresses the volatility in the regulatory process. If the reward and punishment are adjusted in real time in line with the different
proportions of the two power-selling enterprise groups fulfilling the RPS quota obligation, the evolutionary game will have just one evolutionary equilibrium, and the regulatory effect will be better
with the regulation cost being lower.
Based on the above analysis, it can be seen that under the static reward–penalty mechanism, the evolution process of the strategic selections of all parties is volatile, and that there is no stable
equilibrium. Given the dynamic reward–penalty mechanism of competitive relations, power sales companies must bear the vertical pressure brought by government regulation and face the horizontal
pressure brought by competitors in the same industry. At the same time, the dynamic reward–penalty mechanism has a strong flexibility, whose overall incentive effect is significantly better than that
of the static reward–penalty mechanism. The dynamic one can effectively suppress volatility while ensuring the regulation effect, and helps achieve a better stable equilibrium state. The
sustainability of this regulatory mechanism is of far-reaching significance for the sustainable development of the power sales industry in China. Therefore, it is more effective than the static one
in terms of regulation.
4.3. Stability Analysis of Equilibrium Points
In order to further verify the equilibrium point of the game model and analyze the strategy selection of government regulation agencies and the two types of electricity sales companies under the
dynamic reward and punishment mechanism, the replicator dynamics equation can be rewritten as Formula (16):
${ F ( x ) = x ( 1 − x ) [ ( 1 − z ) ( 1 − y + z ) f ( q T − q 1 1 ) W 1 + ( 1 − y ) ( 1 + y − z ) f ( q T − q 2 1 ) W 2 − z ( 1 + y − z ) e ( q 1 2 − q T ) W 1 − y ( 1 − y + z ) e ( q 2 2 − q T ) W
2 + H − C g ] = 0 F ( y ) = y ( 1 − y ) [ x ( 1 − y + z ) f ( q T − q 1 1 ) W 1 + x ( 1 − z ) ( 1 + y − z ) f ( q T − q 2 1 ) W 2 − x ( 1 + y − z ) e ( q 1 2 − q T ) W 1 + P 1 − B 1 − O 1 ] = 0 F ( z
) = z ( 1 − z ) [ x ( 1 − y ) ( 1 − y + z ) f ( q T − q 1 1 ) W 1 + x ( 1 + y − z ) f ( q T − q 2 1 ) W 2 − x ( 1 − y + z ) e ( q 2 2 − q T ) W 2 + P 2 − B 2 − O 2 ] = 0 .$
When the variables are substituted into the replicated dynamic equations, it can be further expressed as Formula (17):
${ F ( x ) = x ( 1 − x ) ( 2.25 − 1.32 y − 1.052 z + 2.13 y z − 0.93 y 2 − 1.2 z 2 ) = 0 F ( y ) = y ( 1 − y ) ( − 0.3184 + 2.01 x − 0.39 x y − 1.14 x z − 1.05 x y z + 1.05 x z 2 ) = 0 F ( z ) = z (
1 − z ) ( − 0.273 + 2.13 x − 1.23 x y + 0.03 x z + 1.2 x y 2 − 1.2 x y z ) = 0 .$
With the help of Matlab software, there are nine solutions to this equation:
$O 1 = ( 0 , 0 , 0 )$,$O 2 = ( 1 , 0 , 0 )$,$O 3 = ( 0 , 1 , 0 )$, $O 4 = ( 0 , 0 , 1 )$,$O 5 = ( 1 , 1 , 0 )$,$O 6 = ( 1 , 0 , 1 )$,$O 7 = ( 0 , 1 , 1 )$,$O 8 = ( 1 , 1 , 1 )$, $O 9 = ( 0.257 ,
0.771 , 0.903 )$.
According to the method proposed by Friedman in 1991 [
], this paper further analyzed the stability of each equilibrium point by using the Jacobian matrix. The specific expression of the Jacobian matrix is shown in Formula (18):
$J = [ dF ( x ) dx dF ( x ) dy dF ( x ) dz dF ( y ) dx dF ( y ) dy dF ( y ) dz dF ( z ) dx dF ( z ) dy dF ( z ) dz ] .$
In the Jacobian matrix, the specific element expression is shown in Formula (19):
${ dF ( x ) dx = ( 1 − 2 x ) ( 2.25 − 1.32 y − 1.052 z + 2.13 y z − 0.93 y 2 − 1.2 z 2 ) dF ( x ) dy = x ( 1 − x ) ( − 1.32 + 2.13 z − 1.86 y ) dF ( x ) dz = x ( 1 − x ) ( − 1.052 + 2.13 y − 2.4 z )
dF ( y ) dx = y ( 1 − y ) ( 2.01 − 0.39 y − 1.14 z − 1.05 y z + 1.05 z 2 ) dF ( y ) dy = ( 1 − 2 y ) ( − 0.3184 + 2.01 x − 0.39 x y − 1.14 x z − 1.05 x y z + 1.05 x z 2 ) + y ( 1 − y ) ( − 0.39 x −
1.05 x z ) dF ( y ) dz = y ( 1 − y ) ( − 1.14 x − 1.05 x y + 2.1 x z ) dF ( z ) dx = z ( 1 − z ) ( 2.13 − 1.23 y + 0.03 z + 1.2 y 2 − 1.2 y z ) dF ( z ) dy = z ( 1 − z ) ( − 1.23 x + 2.4 x y − 1.2 x
z ) dF ( z ) dz = ( 1 − 2 z ) ( − 0.273 + 2.13 x − 1.23 x y + 0.03 x z + 1.2 x y 2 − 1.2 x y z ) + z ( 1 − z ) ( 0.03 x − 1.2 x y ) .$
By substituting each equilibrium point, the characteristic value of the Jacobian matrix corresponding to each equilibrium point can be obtained, and the attributes of each equilibrium point can be
further determined. The results are shown in
Table 4
Based on the above analysis, we found that the evolutionary stable equilibrium point (ESS) of the power sales side RPS government regulation game is $O 9 = ( 0.257 , 0.771 , 0.903 )$; that is, under
the dynamic reward and punishment mechanism, although only about 25% of government regulation agencies select a strict regulation strategy, nearly 77% of integrated power sales companies and 90% of
independent power sales companies choose to actively consume renewable energy power and to select fulfilling the power sales side RPS regulation strategy.
5. Impact Analysis of Exogenous Variables
5.1. Impact of the RPS Quota Indicators
The RPS quota indicator is critical for the implementation of RPS policy. It has a significant influence on many other variables in the model and the game evolution process.
Figure 7
Figure 8
Figure 9
respectively indicate that the RPS quota indicator
$q T$
is 0.03, 0.09, and 0.15 in the regulatory game evolution process.
Through analysis, it is not difficult to find that when the RPS quota ratio is low, the punishment is insufficient, which makes it impossible to effectively guide the two types of electricity sales
enterprises to fulfill their quota obligations. In consequence, the regulation effect is poor. With the increase of the RPS quota, the punitive punishments faced by the two types of power sales
companies are aggravated, and the negative incentive effect of the punishment is enhanced. However, when the proportion of RPS quota increases, the intensity of the regulation decreases first and
then rises. The main reason is that as the probability of sales companies fulfilling their quota obligations increases, the government will gradually loosen regulation and the regulation costs will
gradually decrease. When the RPS quota indicator is set too high, enthusiasm of the power sales companies fulfilling their obligations diminishes, which makes the government regulation agencies
increase the regulatory input to achieve the established policy objectives. Thus, when formulating RPS quota indicators, the government should comprehensively consider various influencing factors to
work out appropriate quota indicators.
5.2. Impact of the Green Certificate Price
Existing research shows that in the implementation process of RPS, it is necessary to make effective use of the green certificate trading mechanism to achieve a better regulation effect. Therefore,
this study also analyzed the impact of changes in the green certificates price on the regulatory effect.
Figure 10
Figure 11
respectively show the process of strategy selection for integrated power sales companies and independent power sales companies when
$P e = 0.05 , 0.1 , and 0.15$
As the price of green certificates rises, the cost for power sales companies to fulfill RPS obligations mounts. In order to maximize their profits, power sales companies, be they integrated ones or
not, prefer to be punished and are not willing to fulfill their RPS obligations. Both integrated power sales companies and independent power sales companies tend to choose not to fulfill their RPS
Figure 12
shows the evolution of the strategic selection of government regulators when the green certificate price
$P e$
is 0.05, 0.1, and 0.15.
The higher the green certificate price is, the higher the proportion of government regulatory agencies selecting a regulatory strategy will be. Government regulation agencies can effectively guide
the strategic selection of power sales companies through the perfect green certificate trading mechanism and the green certificate price signal, achieve established RPS policy objectives, and reduce
the regulatory costs as much as possible through the market mechanism.
5.3. Impact of the Size of the Electricity Sales Market on the Evolution Process
China is in a critical period of power system reform, and the power market reform seems to be in deep water. In order to further consider the impact of the scale of the electricity sales market on
the regulatory process and the regulatory effect, this paper further discusses the sales of
$W 1$
$W 2$
for the two types of power sales companies. The influence of electric power sales on the regulatory evolution game is shown in
Figure 13
Figure 14
A comparison found that when the size of the electricity sales market doubles, the final market equilibrium will not change much, while the number of iterations required to achieve a stable
equilibrium state in the sales-side quota system regulation game will be greatly reduced; that is, the size of the electricity sales market becomes larger. The shorter the time to achieve the
established regulatory objectives, the better the regulatory effect. Therefore, when composing government-regulated policies for the power sale side RPS system, China should further promote power
market-based transactions and expand the scale of market-oriented transactions.
6. Conclusions
Sustainable regulation in the implementation process of RPS is of far-reaching significance for China’s power sales market. In this paper, we considered the competitive relationship between power
sales companies based on a limited rationality, and analyzed the effectiveness of static and dynamic regulation mechanisms. The regulation effects of the static reward–penalty mechanism and the
dynamic reward–penalty mechanism were simulated with a system dynamics (SD) model. Finally, we analyzed the impact of several major external variables on the implementation of the power sales side
RPS regulation. In general, we reached the following conclusions:
1. Due to the introduction of competitive factors in the regulation process of the RPS policy, the dynamic reward–penalty mechanism has a better regulatory effect than the static reward–penalty
mechanism, and the former can effectively restrain evolutionary volatility. This is helpful for promulgating long-term policies, guiding power sales companies to obey the RPS regulation, promoting
the implementation of the RPS policy, and ensuring the sustainable development of the power sales industry.
2. An RPS quota indicator that is too high or too low is not conducive to achieving the established policy objectives. Therefore, government regulation agencies should set a moderate and reasonable
quota indicator system in the process of formulating and implementing the power sales side RPS policy.
3. An excessively high green certificate price is also not conducive to achieving the established policy objectives. It can be seen that the green certificate trading system is an important guarantee
for the smooth implementation of the power sales side RPS policy, which can optimize the fulfillment cost of power sales companies through the market trading mechanism. Thus, it is important to
promote the construction of the green certificate transaction market and change the renewable energy industry gradually from being policy-oriented to being market-oriented.
4. The simulation results show that the market size of power sales companies also significantly influences the implementation of the RPS regulation. The implementation of the RPS regulation is much
more effective when the market size of power sales is greater. Thus, it is necessary to further promote the market-oriented reform of the power industry and expand the scale of the electricity sales
market during the implementation of the RPS regulation.
However, there are still some limitations in our study, since the power sales side RPS regulation is a systematic project. An effective solution to this problem is the result of strategic
interactions among many relevant stakeholders. The model in this paper only considers the interaction among government regulation agencies and the two kinds of power sales companies, and does not
consider the impact of power generation companies. At the same time, with the development of neural network methods, it becomes possible to build an evolutionary game model based on neural networks
theory, which is more conducive to the analysis of the implementation issues of RPS policy. Therefore, in future research, we will incorporate other stakeholders into the analytical framework, and
analyze the RPS regulation of the power industry in a more comprehensive manner based on neural networks theory and evolutionary theory.
This research received no external funding.
We sincerely appreciate the anonymous referees and editors for their time and patience devoted to the review of this paper as well as their constructive comments and helpful suggestions.
Conflicts of Interest
The author declares no conflict of interest.
Figure 2. Statistics of power sales companies in China as of August 2018 (Data sources: National Development and Reform Commission).
Figure 10. Evolutionary trend of the strategy selection for integrated power sales companies when $P e = 0.05 , 0.1 , and 0.15$.
Figure 11. Evolutionary trend of the strategic selection for independent power sales companies when $P e = 0.05 , 0.1 , and 0.15$.
Figure 12. Evolutionary trend of government regulators’ strategic selection when $P e = 0.05 , 0.1 , and 0.15$.
Government Regulates (x)
Independent power sales companies Independent power sales companies
obeying $( z )$ disobeying $( 1 − z )$
Integrated power sales companies $R 1 + R 2 − C g − E 1 − E 2$ $R 1 + F 2 − C g − E 1$
obeying ($y$) $P 1 − B 1 + E 1$ $P 1 − B 1 + E 1$
$P 2 − B 2 + E 2$ $O 2 − F 2$
Integrated power sales companies $R 2 + F 1 − C g − E 2$ $F 1 + F 2 − C g$
disobeying $( 1 − y )$ $O 1 − F 1$ $O 1 − F 1 − F 2$
$P 2 − B 2 + E 2$ $O 2 − F 1 − F 2$
Government deregulates (1−x)
Independent power sales companies Independent power sales companies
obeying $( z )$ disobeying $( 1 − z )$
Integrated power sales companies $R 1 + R 2 − H$ $R 1 − H$
obeying $( y )$ $P 1 − B 1$ $P 1 − B 1$
$P 2 − B 2$ $O 2$
Integrated power sales companies $R 2 − H$ $− H$
disobeying $( 1 − y )$ $O 1$ $O 1$
$P 2 − B 2$ $O 2$
External Variable Definition Value
$C g$ Cost of regulation 30
$H$ Loss of deregulation 30
$P e$ The price of green certificate 0.1
$f$ Unit punishment 0.5
$q T$ The RPS quota 0.09
$q 1 1$ The initial quota of integrated power sales companies 0.03
$q 1 2$ The ultimate quota of integrated power sales companies 0.15
$W 1$ Total sales of integrated power sales companies 40
$q 2 1$ The initial quota of independent power sales companies 0.02
$q 2 2$ The ultimate quota of independent power sales companies 0.13
$W 2$ Total sales of independent power sales companies 30
$d 1$ Unit revenue of integrated power sales companies 0.08
$d 2$ Unit revenue of independent power sales companies 0.1
$g 1$ Unit power sales profits of integrated power sales companies 0.005
$g 2$ Unit power sales profits of independent power sales companies 0.003
$t$ Unit social benefits 0.5
$e$ Unit reward 0.1
Intermediate Variable Definition Formulation
$B i$ Gross green certificate cost $B i = P e ( q i 2 − q i 1 ) W i$
$F i$ Gross penalty $F i = f ( q T − q i 1 ) W i$
$O i$ Gross opportunity revenue $O i = d i P e ( q i 2 − q i 1 ) W i$
$P i$ Gross power sales profits $P i = g i W i$
$R i$ Gross social benefits $R i = t ( q i 2 − q i 1 ) W i$
$E i$ Gross reward $E i = e ( q i 2 − q T ) W i$
Equilibrium Point Eigenvalues Attributes
$O 1 = ( 0 , 0 , 0 )$ 2.25 –0.3184 –0.273 Saddle point
$O 2 = ( 1 , 0 , 0 )$ –2.25 1.6826 1.857 Saddle point
$O 3 = ( 0 , 1 , 0 )$ 0 0.3184 –0.273 Saddle point
$O 4 = ( 0 , 0 , 1 )$ –0.002 –0.3184 0.273 Saddle point
$O 5 = ( 1 , 1 , 0 )$ 0 –1.3016 1.827 Saddle point
$O 6 = ( 1 , 0 , 1 )$ 0.002 1.6016 –1.885 Saddle point
$O 7 = ( 0 , 1 , 1 )$ –0.122 0.3184 0.273 Saddle point
$O 8 = ( 1 , 1 , 1 )$ 0.122 –0.1616 –0.657 Saddle point
$O 9 = ( 0.257 , 0.771 , 0.903 )$ –1.172 –0.391 –0.257 Stable point
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What Is P Value For Statistical Significance (Pdf+Examples)
What Is P Value For Statistical Significance
In this article we will discuss the topic of P-Value, What Is P Value For Statistical Significance? and this article clarifies the basic concept of p-value in statistics with the help of some
relevant examples.
What is the P-value?
The P-value is a statistical measure of how significant a result is. In science, the p-value is the probability of obtaining a result at least as extreme as the actual result if the null hypothesis
(the assumption being tested) were true. A low p-value indicates that the data strongly supports the rejection of the null hypothesis.
P-values are commonly reported in scientific papers and are often presented graphically. The p-value is calculated using the following formula:
p 1 – F(x), where x is the value of the test statistic.
For example, suppose we want to determine whether the mean weight of male mice is greater than the mean weight of female mice. We have two groups of mice, each consisting of 10 males and 10 females.
The mean weights of the two groups are 14 grams and 13 grams respectively. If we use the t-test, we would calculate the p-value as follows:
t (14 – 13)/sqrt((10*10 + 10*10)*0.05).
This gives us 0.898, which means that the difference between the two means is not statistically significant.
If we instead use the Mann-Whitney U test, we would calculate the U-statistic as follows:
U (14 – 13) / sqrt((10*10) * 2.5).
This gives us 4.9, which is less than 5.99, so we reject the null hypothesis.
P-value from a statistical perspective
P value is the probability of a number between 0 and 1 calculated under the assumption that the null hypothesis is true p-value is evidence against a null hypothesis.
• In statistics p-value is also known as the probability value it is the smallest level of significance at which the null hypothesis is rejected.
• The p-value is generally expressed as decimals although it may be easier if you convert them into percentages – the smaller the p-value stronger the evidence that you should reject the null
• Usually, the p-value of 0.05 is used so if you convert this into a percentage, it comes to five percent which means that there is a five percent chance that your result could be random that is it
would have happened by chance, on the other hand, the larger p-value – for example, 0.9 means that your result has 90 percent probability of being completely random and not due to anything in
your experiment, therefore, smaller p-value more the significance your result would be.
Let’s say an experiment is performed to determine if the coin flip is fair or not so you flip the coin you have an equal probability of getting a head or tail that is 50 or 50 or 0.5 now suppose you
flip the coin 1000 times it follows a normal distribution curve and on an average, we would expect to have 500 tails and 500 heads but if I got 510 head and 490 tail then do would you call this a
fair coin or an unfair coin.
But in this case, we can say there is no sufficient evidence to suggest that the coin is fair or not right so if again we get another case where we have 800 heads and 200 tails would you still call
this coin a fair coin?
I guess we cannot call this a fair coin because we would have a strong suspicion that this is an unfair coin and is biased towards the head so in this case, we would suspect the coin is biased and
this could be easily happening by chance.
Graphically if you represent the p-value it is the area in the tail that is the red area that I have marked that is the description of your graphical distribution my null hypothesis in this coin
would be that my coin is fair and it should have equal probability it is a default of worst case scenario it is a known fact that it will have 50 head and 50 tail and my alternate would be that is
not 0.5 that is how you define the alternate and null hypothesis, in this case, the chance of occurrence of null and alternate are mutually exclusive this means then either of them can occur both
cannot occur at the same time that is basic about the p-value.
Let us look at some decision rules and how based on that we will evaluate whether our result is significant or not decision rules to understand the decision rules.
Decision Rules
let us first plot the normal distribution curve and these are my critical values.
First rule
Now my first rule says if p is less than alpha (p<=α) that is the significance level then we can reject the null hypothesis if it falls in this red region which is below the critical values the term
significance here means not anything major important but it just means the difference between two mean is not likely due to chance and as the p gets further lower the evidence allow you to reject the
null hypothesis gets stronger here.
Second rule
The second rule is if p is greater than alpha (p>α) that means we fail to reject the null hypothesis which falls in this green region here the p-value lies between zero and one it won’t reach exact
one because in real life the group will probably not be exactly perfectly equal and it also not reach zero because there’s always a possibility even if it is extreme So alpha here denotes the level
of significance and how you calculate you calculate form 1 minus confidence level here.
Example 1: Single Population Mean
Q: The vendor claims that the average weight of the box is 1.84 kg the customer randomly chooses 64 parts and finds the sample width as 1.88 kg suppose the population standard division is 0.3 kg with
alpha at 0.05 that is the level of significance we need to test for the hypothesis that the true mean of shipment is 1.84 kg?
A: So my null hypothesis is that mu is equal to 1.84 kg and r10 is it is not equal to 1.84 kg alpha or level of significance as 0.05 so let’s plot it on that normal distribution curve we mark it the
area rejection area here if it falls in the red region we will reject our null hypothesis if it falls in between we fail to reject the null hypothesis so we’ll use the z formula that is x power minus
mu by sigma by under root n we’ll put all this value here we get z calculated as 1.07
When we go back to the z table here we find out what is 1.07 to 1 from the z table on the last column and o7 from the vertical column so we get 0.857 so this entry in this table is the standard
normal curve to the left of the z value.
That means I am getting p as one minus point eight five seven seven since the normal distribution is symmetrical so the area to the right of the curve is equal to that of the left that is why we have
created one minus point five and we get p as 0.1423
So, in this case, my p is greater than alpha which is 1.4323 is greater than o 5. so since p is more than the significance level so our decision will be to fail to reject my null hypothesis here this
is how we do it based on the p-value.
Example 2: single population proportion
Q: 6.9 percent of the people are without the poverty line researchers find out that this percentage is higher for their own village he conducts a survey on 300 people in his village and finds out
that 30 out of 300 are below the poverty line. assuming the alpha of 0.05 we have to find out whether his claim is justified or not.
A: So my null hypothesis is p naught equal to 0.069 which is coming from 6.9 percent of the population alternate hypothesis is p naught is greater than 0.069 since his claim is that his percentage is
higher in his village so we use this formula for calculating z calculated that is p hat minus p naught divided by under root of p naught into 1 minus p naught divided by n where p naught is equal to
0.069 p hat is x squared by n so x is 30 the result that he got out of 300 samples in his village and p hat is coming as 0.1 now we will put these values in of p hat p naught in the z kilowatt
formula we get z calculated as 2.12
Go back to the z table but first let us plot this in the normal distribution curve in the z table we will find out 2.12 which comes at an intersection of 2.1 and 0.002 we get 0.9830
So this is 0.9830 since the area under the curve is 1 we subtract it from 1 so we get p as 0.017 since this 0.007 is less than 0.05 that is significance level alpha so we reject the null hypothesis,
in this case, this is another example of p-value so I hope with this you now have very good understanding about p-value.
So if you like this article share it with your friends, thanks for reading.
Read also:-
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Artistic Mathematics
LA: GregoryO'Neill
Learning Area
Document outlining 2024 Mathematics plan.
Note that levels 4 and 5 are intended to be two year programmes.
Students will use Coordinate Geometry, functions including: circles, elipses, parabola, lines, square root relationships, hyperbola and inverse square law; to create an image or reals world model.
They will use a graphing suite such as Desmos; however may choose another program to fit their needs.
We may also look at spreadsheets and other mathematical computation as per student desire.
A device able to achieve your course goals will be essential. This would be something that could run Google Chrome as a minimum, with probably at least a 9" to 10" screen and preferably a keyboard. | {"url":"https://arotahi.aotawhiti.school.nz/node/9352","timestamp":"2024-11-10T01:50:05Z","content_type":"text/html","content_length":"16790","record_id":"<urn:uuid:cd034979-264a-4e4e-9e76-9fabb7041091>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00129.warc.gz"} |
mr womp womp wrote:
This is equivalent to storing some of the string to another string and then claiming that only what is left of the initial string is the compressed version of the original string, which is not
exactly the case.
No, it's not. A program with dictionary compression/decompression contains a *fixed* dictionary; all information needed to decompress the string is already in the decompression program before
anything is compressed.
I don't know exactly what this discussion is about. I only want to say that if you need a dictionary, it will be added to the output size.
earthnite wrote:
So the programs might be tested like this:
"SECRET MESSAGE"->Str1
save output variable somewhere
completely reset calculator
put output variable on calculator
check to make sure it worked.
That is a fair way to test it
Today it's time for da new task! I will post it in some hours
An additional note on task #4: entries that don't compress will get 0 points. Otherwise someone could have a code like this:
Now the output is Str1, no compression, but he will get all the points for both speed and size, which is of course unfair
Last summer we had great news: TI releases a new color calculator, the TI-84+CE-T! Everyone happy, but not the ASM experts, because they found out that it was signed with a 2048-bit signing key,
which is impossible to crack with the current possibilities. That is why they never can create Apps for the CE, until TI releases the SDK. Maybe we could help to crack that enormous number with
The way to crack that key, is to find the factors of that key, such that P*Q=[signing key]. In the reverse, if you know the key, you can theoretically find P and Q, but that takes ages. That is what
this task is about. You have to factor a number as fast as possible. I made this, because factorization is a pain and takes much time, especially with large numbers.
Task 5
Given the input in both N and Ans, output the two factors in L1 as a list. No limitations, and you may assume that number is withing the bounds of TI-BASIC.
Example: 5680842901 would be 60869*93329.
Good luck all and speed is still very important!
Now that the last task is being posted, you can submit your entries. Submit your entries by emailing them, in a ZIP file, to basiccompetition@gmail.com . Any entry before March 22th 11:59:59 pm GMT
will be accepted, entries after that will be removed. Be sure to provide your Cemetech username in the subject or body of the email so we know who you are! Everyone can submit as many entries as they
want, the last one will always be tested. If you have more programs for a task, please say which program is the main.
This is how I will test all the entries. I will first reset my whole calculator, including the Archive. Next, I transfer my timing program to the calculator, and after that, all the inputs. Finally,
after that, I will transfer your program, so the VAT entry is for everyone equal.
I will test all the entries 3 times, and the speed is the average of that 3. You may assume no variable exists, without the input variable(s). I will measure the time with always the same program,
upto 3 digits, so 4.193 can be the time for someone.
If you think it is not fair that I test my own programs, please pm me, and we will look for a solution
I will try to test all the programs as fast as possible, but it's dependant of the entries how many time that takes
I hope you all enjoy this competition, and it's not only for competing, but also for writing algorithms for yourself.
Summary of the 5 tasks:
1. Sort a list
2. Express a string
3. Make 24 with 4 numbers
4. Compress a string
5. Factorize a large number
I hope that there are many, many entries coming!
If you have still questions, please post or pm me
Good luck in the last week!
EDIT: somehow I forgot to say that: the number is always a product of 2 primes, a so-called semiprime
For the last challenge, PT_ specified the upper bound on the number to be 10^8 (less than in the given example).
I think I have about 5 seconds in the worst case, extrapolating from my monochrome time.
earthnite wrote:
PT_, in regards to task 4, I'm not sure I fully understand what you are expecting, please finalize how the programs are scored and how you will be testing the programs, what defines the output
variable? Does it have to be in Ans at the end or can we just tell you that the output variable is, for example, [A]? Are we suppose to make a decoding program? Will the output variable be in Ans as
well at the start of the decode program? If the size of the dictionary is to be included in the size of the output, what defines a dictionary?
What defines a program that is not allowed? Most programs will have a best and worse case scenario. If the worst case scenario is unable to compress the string then is it not allowed?
In regards to all of the tasks, I assume you will test the programs with a variety of inputs, I suggest that you test the same inputs for all of the programs, and select inputs that are not biased
towards certain programs.
On another note, I have updated the form for task 5, if anyone is interested....you can submit your scores here... and see some results here.Please, it would be a shame if it went to waste.
Good luck everyone.
I'm sorry I didn't explain it fully. You have to write only 1 program, for compressing the string which is in Ans and Str1. Output can be anything, so it don't need to be in Ans. Yesterday I had a
small discussion with jonbush and you over SAX about the dictionary. When you use a static dictionary, such as one with the most important words in English, and that one is independant of the input,
it will be only counted to the program size, not the compressed size. However, if you figured out that the word "TEST" appears in the input, and add "TEST" to a dictionary, that will be counted in
the compressed size. A given example small code:
If inString(Str1,"THE"
[replace THE with cos(
This is nice, but your decoding program don't know that cos( stands for THE. So you either need to add it to the output, or make a static dictionary. The only 'dictionary' you may use, is to replace
ONE number with ONE token.
I hope this explains more, because I'm not a star in explaining
A program is not allowed, if it don't try to compress the string. Of course, there can be cases that you try to compress it, but it is a worse scenario, and that it fails. That is accepted. But if I
only create a program that stores Str1 in Str2 and say that Str2 is my output, that is not allowed, because you don't try to compress it.
Here is an example for scoring for task #4: http://pastebin.com/KME3zWNK
For task #5, you may assume that the number is in the range 10^5 - 10^8, and a product of 2 primes.
Good luck all
I still have a few questions.
Will there be omitted parentheses at the end of the string for task 2?
Is there a minimum required decimal accuaracy? (also for task 2)
What is the maximum length of the string for task 4?
Vital7788 wrote:
I still have a few questions.
Will there be omitted parentheses at the end of the string for task 2?
Is there a minimum required decimal accuaracy? (also for task 2)
What is the maximum length of the string for task 4?
Yes, not all parentheses needs to be closed.
Only the accuracy of TI BASIC. If I need the square of something, don't round it up or down
Between 80 and 120 characters
Hope this helps!
Exactly one-and-a-half day to go, people!
If you haven't submitted your entries yet, don't hesitate to do so! Even if you don't have all the 5 (or more) programs, you can earn points.
And still 6 hours to go! I've received some entries, but these are still much less as the participants
Even if you don't have all the 5 programs, you can submit your entries!
My entries have been submitted.
I broke the evaluation solution into three parts, so I couldn't squeeze the size down much: 660 bytes.
I'll add further comments on my solutions once the deadline has passed.
Has it been solidifed yet how the performance of task 5 will be tested? Namely, how the inputs will be generated?
I'm still in favor of randomly generating numbers between sqrt(10^5) and sqrt(10^
Runer112 wrote:
Has it been solidifed yet how the performance of task 5 will be tested? Namely, how the inputs will be generated?
I'm still in favor of randomly generating numbers between sqrt(10^5) and sqrt(10^
The best way to test this (what I think) is generating a list of semiprimes between 10^5 and 10^8 and then randomly picking one. There was much discussion over SAX about how to generate it
PT_ wrote:
Runer112 wrote:
Has it been solidifed yet how the performance of task 5 will be tested? Namely, how the inputs will be generated?
I'm still in favor of randomly generating numbers between sqrt(10^5) and sqrt(10^
The best way to test this (what I think) is generating a list of semiprimes between 10^5 and 10^8 and then randomly picking one. There was much discussion over SAX about how to generate it
Yes, there was much discussion. And I believe the method I suggested is the only one (or one of two) suggested that actually generates a product of two medium-large primes most of the time. All the
others have a large bias to generate a product of a small prime and a large prime, which is rather trivial to factor quickly by trial division. It also doesn't seem to match the task, a microcosm of
RSA factoring, which should involve numbers that are a product of two primes of similar magnitude.
I recognize that my suggested method wouldn't even generate inputs with a prime factor less than sqrt(10^5), but for performance testing, I think that's ideal. For correctness testing, you can still
require that solutions work for inputs with prime factors all the way down 2, but not count this for the performance.
Don't generate the list, download it: https://primes.utm.edu/lists/small/millions/
I just realized I misread the instructions; my program for Task 2 outputs in Ans instead of A. Will it be thrown out, or can the program with "->A" concatenated to the end be scored?
This should not come across as too much of a moan, because it's obviously up to us how often we visit this forum, but I think it's a shame there was such a narrow timeframe for entering this
competition. Some folk have a lot of work to do, and can't visit the site too often. Now I get to my holidays and find I've missed out on this completely, yet I already have a good prgm for Task 5.
I'll be interested to see how much the winner improves on it! Good luck to all who have competed. | {"url":"https://dev.cemetech.net/forum/viewtopic.php?t=12470&postdays=0&postorder=asc&start=100","timestamp":"2024-11-14T13:20:46Z","content_type":"text/html","content_length":"96796","record_id":"<urn:uuid:a51c94b7-9bdf-485d-8e30-322ff49ffda3>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00441.warc.gz"} |
#1 Zhu - Principles of Pulmonary Gas Transport Flashcards
What does Fick’s law measure?
What are the 5 components of Fick’s law?
Pressure Gradient
Solubility (Intrinsic property of gas)
Molecular Weight
FICK = FIVE
Which components of Fick’s law are inversely related to the diffusion rate?
Molecular Weight
What is the diffusion constant?
Solubility/(sqrt of Molecular Weight)
What does the diffusion capacity determine?
The diffusion capacity determines the ability of the respiratory membrane to transport a gas into and out of the blood.
What are the components of the diffusion capacity?
Molecular Weight
It can change due to change in area and/or change in distance.
Where does gas exchange happen?
Respiratory Unit
• Respiratory Bronchioles
• Alveolar Duct
• Alveolar Sac
In emphysema, what would you expect to happen to the diffusion rate and why?
Decreased diffusion rate due to decreased area.
In interstitial edema, what would you expect to happen to the diffusion rate and why?
Decreased diffusion rate due to increased distance.
In fibrosis, what would you expect to happen to the diffusion rate and why?
Decreased diffusion rate due to increased distance.
What is the vapor pressure based on a body temperature of 37 degrees Celsius?
What is the PO2 of humidified air?
What is the PN2 of humidified air?
What is the PO2 of alveolar air?
What is the PCO2 of alveolar air?
What is the PO2 in the arterial end of the capillary in the lung?
What is the PO2 in the venous end of the capillary in the lung? | {"url":"https://www.brainscape.com/flashcards/1-zhu-principles-of-pulmonary-gas-transp-4307043/packs/6459961","timestamp":"2024-11-04T14:04:17Z","content_type":"text/html","content_length":"128403","record_id":"<urn:uuid:7ef2680b-62ba-4fae-b85f-8ac198d792fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00157.warc.gz"} |
About GMC Network
About GMC Network
The research on mechanical and dynamical systems has had a deep impact in other research areas and in the development of several technologies. A big part of its advances has been based on numerical
and analytical techniques. In the sixties, the most sophisticated and powerful techniques coming from Geometry and Topology were used in the study of dynamical systems. Those techniques led, for
instance, to the beginning of the modern Hamiltonian Mechanics. Geometric techniques have been also applied to a wide range of control problems such as locomotion systems, robotics, etc. Most of
these ideas have been developed in the last 30 years by mathematicians of a high scientific level such as J. Marsden, A. Weinstein, R. Abraham, V. Arnold or R. Brockett among others. | {"url":"https://gmcnet.webs.ull.es/","timestamp":"2024-11-11T14:31:46Z","content_type":"application/xhtml+xml","content_length":"19692","record_id":"<urn:uuid:18b10741-8d3d-4d1b-bf3a-3cee97d76076>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00171.warc.gz"} |
Hattie Effect Size Calculator - Calculator Wow
Hattie Effect Size Calculator
The Hattie Effect Size Calculator is a valuable tool for educators, researchers, and analysts who want to measure the impact of an intervention or treatment by comparing the means of two groups.
Developed from the work of John Hattie, this calculator helps quantify the magnitude of an effect, providing insights into the effectiveness of educational strategies, clinical treatments, and other
interventions. By understanding the effect size, you can make more informed decisions based on data rather than subjective judgments.
The Hattie Effect Size Calculator uses the following formula to determine the effect size:
d = (M1 – M2) / SDpooled
• M1 is the mean of Group A (the treatment or experimental group).
• M2 is the mean of Group B (the control group).
• SDpooled is the pooled standard deviation, calculated as:SDpooled = sqrt(((SD_A^2 + SD_B^2) / 2))where:
□ SD_A is the standard deviation of Group A.
□ SD_B is the standard deviation of Group B.
This formula provides a standardized measure of effect size, allowing comparisons across different studies and contexts.
How to Use
1. Enter the Means: Input the mean values for Group A and Group B into the calculator. Group A typically represents the group that received the treatment or intervention, while Group B represents
the control or comparison group.
2. Input Standard Deviations: Provide the standard deviations for both groups. This data reflects the variability within each group.
3. Calculate Effect Size: Click the “Calculate” button to compute the effect size using the formula. The calculator will first determine the pooled standard deviation and then calculate the effect
4. Review Results: The resulting effect size will be displayed. This value indicates the magnitude of the difference between the two groups, with higher values suggesting a more significant impact.
Consider a study evaluating the effectiveness of a new teaching method. Group A (with the new method) has a mean score of 85 and a standard deviation of 10, while Group B (with the traditional
method) has a mean score of 78 and a standard deviation of 12.
Using the formula:
1. Calculate Pooled SD: SDpooled = sqrt(((10^2 + 12^2) / 2)) SDpooled = sqrt((100 + 144) / 2) SDpooled = sqrt(122) SDpooled ≈ 11.05
2. Calculate Effect Size: d = (85 – 78) / 11.05 d ≈ 0.63
In this example, the effect size is 0.63, suggesting a moderate impact of the new teaching method compared to the traditional one.
1. What is effect size?
Effect size quantifies the magnitude of a difference between two groups, providing a standardized measure of impact.
2. Why use the Hattie Effect Size Calculator?
It helps measure the effectiveness of interventions by comparing means and assessing their significance.
3. What does a high effect size indicate?
A high effect size suggests a substantial impact or difference between the groups.
4. Can this calculator be used for different fields?
Yes, it’s applicable in education, clinical research, and other fields requiring comparison of group means.
5. What if the standard deviations are not provided?
You need standard deviations to calculate the pooled standard deviation and effect size.
6. Is effect size the same as statistical significance?
No, effect size measures the magnitude of a difference, while statistical significance assesses whether the difference is likely due to chance.
7. How should I interpret a small effect size?
A small effect size indicates a minor difference between groups, suggesting the intervention may have limited practical impact.
8. Can effect size be negative?
Yes, a negative effect size indicates that Group B has a higher mean than Group A.
9. What is the difference between Hattie’s effect size and Cohen’s d?
Hattie’s effect size is a specific application of Cohen’s d, focusing on educational research.
10. Where can I find more information about Hattie’s work?
John Hattie’s book “Visible Learning” provides comprehensive insights into his research and findings on effect sizes.
The Hattie Effect Size Calculator is a powerful tool for evaluating the impact of different interventions or treatments by providing a standardized measure of effect size. By comparing the means and
standard deviations of two groups, it allows researchers and practitioners to quantify the magnitude of differences and make data-driven decisions. Understanding and using effect sizes can greatly
enhance the effectiveness of educational practices, clinical treatments, and other research endeavors. | {"url":"https://calculatorwow.com/hattie-effect-size-calculator/","timestamp":"2024-11-01T23:31:50Z","content_type":"text/html","content_length":"66556","record_id":"<urn:uuid:b193a42a-2bce-4914-abca-c8a1c478801c>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00252.warc.gz"} |
What is Kepler's Third Law? Understand How Planets Move! - Astronomy Explained
What is Kepler’s Third Law? Understand How Planets Move!
We have been looking at the stars for too long and have come up with dozens of ideas, laws, and things that we thought governed the universe. Most ideas from the past have become obsolete with new
discoveries. However, Johannes Kepler’s three laws for planetary motion have stood the test of time. He created three planetary laws, and today, we will look into Kepler’s third law, the last one.
I have examined the second and compared it to the other two laws in my previous articles. So, today, it is time to look into the last one, Kepler’s Third Law. Johannes Kepler is one of the most
revolutionary astronomers and scientists in history. It is important to know these laws to understand how planets and other celestial bodies move.
If you are looking for some astronomy books for beginners that also talk about Kepler's Third Law and also other laws, check out my 5 Best Astronomy Books for Beginners article
Johannes Kepler and His Contributions to Astronomy
Johannes Kepler, a renowned German astronomer and mathematician, was born in 1571 in Weil der Stadt, which is now Germany. Kepler’s observations and groundbreaking theories created the foundation for
both that time’s and today’s modern astronomy and revolutionized how we understand the cosmos.
Kepler’s journey into astronomy began when he studied at the University of Tübingen. There, he became deeply influenced by the teachings of his mentor, Michael Maestlin. He introduced him to the
theories of Nicolaus Copernicus. Copernicus’ heliocentric model, which placed the Sun at the center of the solar system, challenged the prevailing geocentric view of the time, where Earth was
believed to be the center of the universe.
Kepler’s early work focused on the motion of celestial bodies and the mathematical principles governing their movements. In 1609, he published his first major work, “Astronomia Nova,” in which he
introduced his first two laws of planetary motion. These laws, later known as Kepler’s First and Second Laws, revolutionized our understanding of how planets move around the Sun.
Discovery of Kepler’s Third Law by Kepler
However, Kepler’s Third Law, published in his work “Harmonices Mundi” in 1619, solidified his place in scientific history. This law, also known as the Harmonic Law or the Law of Periods, established
a mathematical relationship between the orbital period and the average distance of a planet from the Sun.
The Third Law paved the way for astronomers to understand the patterns and movements of things in the universe, expanding our knowledge of the universe. It laid the groundwork for Isaac Newton’s laws
of motion and universal gravitation’s development, which further advanced our understanding of the laws governing planetary motion.
Today, Kepler’s contributions to astronomy remain foundational. His laws of planetary motion provided a framework for subsequent discoveries and calculations, allowing astronomers to accurately
predict the positions and movements of planets, comets, and other celestial objects.
The Basics of Kepler’s Laws of Planetary Motion
Kepler’s Laws of Planetary Motion form the cornerstone of modern astronomy and provide a comprehensive understanding of how planets move within our solar system. Before talking about Kepler’s Third
Law, I want to give you a very short glimpse into the basics of all three laws. These laws describe the orbital characteristics of planets, revealing the elegant patterns and relationships that
govern their motion.
Kepler’s three laws
Kepler’s First Law: The Law of Orbits
Kepler’s First Law, also known as the Law of Orbits or the Law of Ellipses, states that the path of each planet around the Sun is an ellipse. This law contradicted the prevailing belief of circular
orbits, as proposed by ancient Greek astronomers. The eccentricity of an ellipse determines how elongated or circular an orbit is, with a value of 0 representing a perfect circle and a value closer
to 1 indicating a more elongated orbit.
Kepler’s Second Law: The Law of Areas
Kepler’s Second Law, known as the Law of Areas or the Law of Equal Areas, describes the speed at which a planet moves along its elliptical orbit. It states that a line that connects the planet to the
Sun takes out equal areas in equal time periods. This means that a planet’s speed increases when it is closer to the Sun and slows down when it is farther away. This law helps explain why planets do
not travel at a constant speed throughout their orbits.
Kepler’s Third Law: The Law of Periods
Kepler’s Third Law also referred to as the Law of Periods or the Harmonic Law, establishes a mathematical relationship between the orbital period of a planet and its average distance from the Sun. It
basically says that the square of a planet’s orbital period is equal to the cube of its average distance from the Sun. In simpler terms, the more distance a planet has from the Sun, the longer its
orbital period.
These three laws, collectively known as Kepler’s Laws of Planetary Motion, provide a comprehensive framework for understanding the motion of planets in our solar system. They elucidate the shape of
planetary orbits, the varying speeds of planets along their paths, and the relationship between a planet’s period and its distance from the Sun.
In-depth Understanding of Kepler’s Third Law
The brief mathematical understanding of Kepler’s Law includes the square of a planet’s orbital period, which is T, and the cube of the average distance from the Sun, which is r. Mathematically, it
T^2 = k * r^3
T represents the orbital period of the planet, r denotes the average distance between the planet and the Sun, and k is a constant value.
Kepler’s Third Law in formula
This law implies that planets farther from the Sun take longer to complete one orbit compared to those closer to the Sun. It highlights the inherent harmony and orderliness in the solar system, where
the orbital periods and distances of planets follow a predictable pattern.
Applications and Examples of Kepler’s Third Law
Kepler’s Third Law has a lot of applications in the study of the solar system and beyond. By measuring the orbital periods and average distances of planets, astronomers can determine the relative
sizes and masses of celestial bodies, even those that are not directly observable.
For example, by applying Kepler’s Third Law, astronomers discovered Neptune, the eighth planet in our solar system. Observations of Uranus revealed discrepancies in its orbital path, suggesting the
presence of another gravitational influence. By calculating the expected orbital period and average distance based on the known planets, astronomers predicted the existence and location of Neptune
before it was directly observed.
Kepler’s Third Law also allows scientists to estimate the masses of exoplanets (planets outside our solar system) by observing the orbital periods of their host stars and the changes in their
brightness as the planets pass in front of them.
Implications and Importance of Kepler’s Third Law
Kepler’s Third Law revolutionized the study of astronomy by providing a mathematical relationship that allows us to understand the motion of planets and other celestial bodies. Here are some of the
things Kepler’s Third Law already accomplished.
1. Confirmation of the Heliocentric Model. Kepler’s Third Law provided further evidence in support of the heliocentric model proposed by Copernicus. By demonstrating that the orbital periods of
planets are related to their distances from the Sun, Kepler’s work reaffirmed the notion that the Sun is indeed the center of our solar system.
2. Validation of Newton’s Laws. Kepler’s laws of planetary motion’s crucial data and insights were later incorporated into Newton’s laws of motion and universal gravitation. The relationship between
a planet’s orbital period and its distance from the Sun helped to validate Newton’s gravitational theory. This provided a foundation for the understanding of celestial mechanics.
3. Basis for Celestial Mechanics. Kepler’s Third Law laid the groundwork for the development of celestial mechanics. This is a branch of astronomy that focuses on the motion of celestial bodies
under the influence of gravitational forces. It provided astronomers with a fundamental tool to study the dynamics of planets, moons, comets, and other objects in our solar system.
How It Helps in Understanding Our Solar System
How the planets places are determined easily with The Third Law
Kepler’s Third Law also has a huge role in helping us understand how our solar system generally works and moves. Here are some of its use cases.
1. Determining Planetary Distances. Kepler’s Third Law allows astronomers to determine the average distances between planets and the Sun. By knowing the orbital period of a planet, they can
calculate its average distance from the Sun using the mathematical relationship provided by Kepler’s law.
2. Estimating Planetary Masses. By combining Kepler’s Third Law with observations of celestial motion, astronomers can estimate the masses of planets. The law allows them to infer the mass of a
planet based on its average distance from the Sun and the orbital period.
3. Identifying Exoplanets. Kepler’s Third Law has been instrumental in the discovery and characterization of exoplanets. Exoplanets are planets that orbit stars outside our solar system. By
observing the periodic variations in a star’s brightness caused by an exoplanet’s transit, astronomers can determine the planet’s orbital period. In turn, this provides valuable information about
its distance from the host star.
Modern Uses and Applications
Kepler’s Third Law didn’t stop providing results. It is still in use and helps us quite a lot, even in modern times. We generally use it for three main things, but of course, it is not limited to
1. Exoplanet Hunting. Kepler’s Third Law is utilized in the search for exoplanets. Astronomers analyze the light curves of stars to identify periodic variations that may indicate the presence of
orbiting planets. By leveraging the law’s mathematical relationship, they can estimate the orbital periods of these exoplanets and gain insights into their characteristics.
2. Spacecraft Trajectories. In space exploration missions, Kepler’s Third Law is employed to plan spacecraft trajectories. By understanding the relationship between the orbital period and distance,
scientists can calculate the necessary velocities and trajectories to reach specific destinations within our solar system.
3. Gravitational Wave Detection. Kepler’s Third Law plays a role in the detection and analysis of gravitational waves. These waves, ripples in the fabric of spacetime caused by massive celestial
events, provide valuable information about the objects involved. By incorporating Kepler’s law, scientists can estimate the masses and distances of the celestial bodies responsible for generating
gravitational waves.
Common Misunderstandings and Clarifications of Kepler’s Third Law
Kepler’s Third Law, while a fundamental principle in understanding planetary motion, can sometimes be subject to common misunderstandings. I will try to show you what those misunderstandings are and
what is the correct interpretation of it.
1. Misconception: All planets have the same orbital period.
• Clarification. Kepler’s Third Law states that there is a relationship between a planet’s orbital period and its average distance from the Sun. However, this does not imply that all planets have
the same orbital period. Each planet has a unique combination of distance and period, resulting in different orbital characteristics.
1. Misconception: Kepler’s Third Law works only for planets in our solar system.
• Clarification. Kepler’s Third Law was derived based on observations of our solar system, but it applies to any planetary system. The law holds true for exoplanets orbiting other stars as well.
So, it is not just for our solar system. It allows astronomers to estimate their orbital periods and how far away they are from their host stars.
1. Misconception: The Law of Periods applies only to planets.
• Clarification. Kepler’s Third Law can be applied to any object orbiting a central body, not just planets. It can be used to study the motion of moons, asteroids, and even human-made satellites
around a planet or a star.
Kepler’s Third Law, the Law of Periods, has stood the test of time as a fundamental principle for planetary motion. The long-lasting relevance of Kepler’s Third Law lies in its ability to provide
insights to our solar system. When we understand a planet’s orbital period and its estimated distance from the Sun, we can have a deeper understanding of the complicated patterns and harmonies that
govern celestial movement.
Kepler’s Third Law not only confirmed the heliocentric model proposed by Copernicus. It also provided a mathematical foundation for the development of celestial mechanics. It played a crucial role in
the validation of Newton’s laws of motion and universal gravitation. This paved the way for a profound understanding of the forces that shape our universe.
However, it is important to recognize the limitations and complexities of Kepler’s Third Law. Real-world situations can introduce deviations from the idealized predictions, such as gravitational
interactions with other objects or variations in orbital behavior. These factors remind us of the dynamic and complex nature of the universe we inhabit.
Nonetheless, Kepler’s Third Law remains a cornerstone of astronomical research and exploration. Its mathematical relationship continues to guide our understanding of planetary motion and serves as a
vital law in solving the mysteries of the cosmos.
What is Kepler’s third law simple?
Kepler’s third law is like a rule for planets. It says that if you know how far a planet is from the Sun, you can figure out how long it takes for the planet to go one full circle around the Sun. The
further away the planet is from the Sun, the longer it takes to complete one orbit. So, planets that are far from the Sun move slower than the ones closer to the Sun.
What is the constant in Kepler’s third law?
The “constant” in Kepler’s third law is like a magic number that connects the time a planet takes to go around the Sun and how far it is from the Sun. This constant can change depending on what two
things are orbiting each other. For example, when we talk about planets going around the Sun, the Sun is so much bigger than any planet that we can kind of ignore the planet’s weight, and the
constant stays the same for all planets. But if we were talking about the Moon going around the Earth, then the Earth and the Moon’s weights both matter, so the constant would be different.
What are the names of Kepler’s three laws?
Kepler’s First Law is The Law of Orbits, Kepler’s Second Law is The Law of Equal Areas, and the third one, Kepler’s Third Law, is The Law of Periods. | {"url":"https://astronomyexplained.com/what-is-keplers-third-law-understand-how-planets-move/","timestamp":"2024-11-11T13:07:07Z","content_type":"text/html","content_length":"79467","record_id":"<urn:uuid:c409d9cd-fe1c-4366-b697-4be92153b032>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00258.warc.gz"} |
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Simplifying Expressions & Solving
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Two Equations containing Two Variables
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Home Linear Equations Literal Equations Simplifying Expressions & Solving Equations Two Equations containing Two Variables LinearEquations Solving Linear Equations Plane Curves Parametric Equation
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Equations by Graphing SOLVING SYSTEMS OF EQUATIONS Exponential and Logarithmic Equations Quadratic Equations Homework problems on homogeneous linear equations Solving Quadratic Equations
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Author Message
Tim L Posted: Saturday 30th of Dec 12:58
Hi , A couple of days back I began working on my math homework on the topic Algebra 1. I am currently not able to finish the same because I am unfamiliar with the basics of
multiplying matrices, percentages and fractional exponents. Would it be possible for anyone to help me with this?
From: I'll be where
I'm at!
IlbendF Posted: Monday 01st of Jan 08:28
There used to be a time when those were the only ways out. But thanks to technology now, we have something known as the Algebrator! It’s an easy to use software , something
which even a total newbie would enjoy working on and the best part is that it would solve all your questions and also explain the steps it took to reach that solution ! Isn’t
that just great ? Well I think it is, and I’m sure after , you won’t disagree with me .
From: Netherlands
erx Posted: Monday 01st of Jan 13:51
Some teachers really don’t know how to explain that well. Luckily, there are programs like Algebrator that makes a great substitute teacher for math subjects. It might even be
better than a real professor because it’s more accurate and quicker!
From: PL/DE/ES/GB/HU
Matdhejs Posted: Wednesday 03rd of Jan 09:47
A great piece of math software is Algebrator. Even I faced similar difficulties while solving quadratic formula, adding fractions and simplifying expressions. Just by typing in
the problem workbookand clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several math classes - College Algebra, Basic
Math and Algebra 2. I highly recommend the program.
From: The Netherlands
vam Posted: Thursday 04th of Jan 11:25
I just hope this tool isn’t very complex . I am not so good with the computer stuff. Can I get the product description, so I know what it has to offer?
From: Mass.
cufBlui Posted: Saturday 06th of Jan 07:32
The details are here : https://algebra-equation.com/solving-systems-of-equations.html. They guarantee an unrestricted money back policy. So you have nothing to lose. Go ahead
and Good Luck!
From: Scotland
Posted: Saturday 30th of Dec 12:58
Hi , A couple of days back I began working on my math homework on the topic Algebra 1. I am currently not able to finish the same because I am unfamiliar with the basics of multiplying matrices,
percentages and fractional exponents. Would it be possible for anyone to help me with this?
Posted: Monday 01st of Jan 08:28
There used to be a time when those were the only ways out. But thanks to technology now, we have something known as the Algebrator! It’s an easy to use software , something which even a total newbie
would enjoy working on and the best part is that it would solve all your questions and also explain the steps it took to reach that solution ! Isn’t that just great ? Well I think it is, and I’m sure
after , you won’t disagree with me .
Posted: Monday 01st of Jan 13:51
Some teachers really don’t know how to explain that well. Luckily, there are programs like Algebrator that makes a great substitute teacher for math subjects. It might even be better than a real
professor because it’s more accurate and quicker!
Posted: Wednesday 03rd of Jan 09:47
A great piece of math software is Algebrator. Even I faced similar difficulties while solving quadratic formula, adding fractions and simplifying expressions. Just by typing in the problem
workbookand clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several math classes - College Algebra, Basic Math and Algebra 2. I highly
recommend the program.
Posted: Thursday 04th of Jan 11:25
I just hope this tool isn’t very complex . I am not so good with the computer stuff. Can I get the product description, so I know what it has to offer?
Posted: Saturday 06th of Jan 07:32
The details are here : https://algebra-equation.com/solving-systems-of-equations.html. They guarantee an unrestricted money back policy. So you have nothing to lose. Go ahead and Good Luck! | {"url":"https://algebra-equation.com/solving-algebra-equation/like-denominators/free-algebra-puzzles.html","timestamp":"2024-11-09T03:56:12Z","content_type":"text/html","content_length":"92011","record_id":"<urn:uuid:af25b8b3-3408-44dc-adad-20926fbb0776>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00396.warc.gz"} |
Complex analysis: Sketch the region in the complex plane
• Thread starter Rubik
• Start date
In summary, Rubik says that he is confused as to how to rotate the angle in terms of clockwise or anti-clockwise according to the conditions given. He is also unsure if his radius is in fact a or am
I missing an important step. Rubik thanks Rubik for all of his help and says he appreciates it.
Homework Statement
{z: [itex]\pi[/itex]?4 < Arg z ≤ [itex]\pi[/itex]}
Homework Equations
The Attempt at a Solution
Is it right to assume
z[0] = 0 ; a = a (radius = a) ; and taking [itex]\alpha[/itex] = [itex]\pi[/itex]/4 ; [itex]\beta[/itex] = [itex]\pi[/itex]
And now in order to sketch the problem after setting up the complex plane is it correct to to plot z[0] at the origin and then from the origin plot [itex]\pi[/itex]/4 by rotating to the right in a
clockwise rotation for [itex]\pi[/itex]/4 radians for the first condition and then rotating [itex]\pi[/itex] to the left from the origin (anti-clockwise rotation) for the second condition and then
using a solid or dashed line according to the strictly < or ≤ conditions and this gives me the correct region?
Basically I am confused as to how to rotate the angle in terms of clockwise or anti-clockwise according to the conditions given.
And I am also unsure if my radius is in fact a or am I missing an important step?
Hi Rubik!
do you mean {z: [itex]\pi[/itex]/4 < Arg z ≤ [itex]\pi[/itex]} ?
Rubik said:
… by rotating to the right in a clockwise rotation for [itex]\pi[/itex]/4 radians for the first condition and then rotating [itex]\pi[/itex] to the left from the origin (anti-clockwise rotation)
for the second condition …
i'm worried why you thought it wasn't
(and i don't understand where radius comes into it)
Oops yep I meant [itex]\pi[/itex]/4.. I was worried asking it haha it has been a long time since I have had to work with complex numbers.. Another thing I have just come across is the region {z : |z
- 3 + i| < 4} Does this mean that z[0] = (-3,i), and the radius = 4?
Rubik said:
{z : |z - 3 + i| < 4} Does this mean that z[0] = (-3,i), and the radius = 4?
no, the centre is 3 - i
With the first part from your first reply I said radius = a because I am trying to sketch the particular region covered by these angles or is that wrong?
Rubik said:
With the first part from your first reply I said radius = a because I am trying to sketch the particular region covered by these angles or is that wrong?
I still don't understand this at all.
What is a, and what has the radius to do with anything?
Well I am not sure I just took it as an assuption.. See if I try and sketch this region I draw both these angles taking them anti-clockwise from the origin, which leaves a region in the 1st and 2nd
quadrants and I am just confused as I thought I was suppose to be left with a closed region but is this not the case? I am sorry if this still makes no sense it is hard to explain a drawing in words.
:/ So currently I have a line in the direction of [itex]\pi[/itex] going anti-clockwise from (0,0) and then another line in the direction of [itex]\pi[/itex]/4 from (0,0) Is that how the region is
suppose to look?
Rubik said:
… I have a line in the direction of [itex]\pi[/itex] going anti-clockwise from (0,0) and then another line in the direction of [itex]\pi[/itex]/4 from (0,0) Is that how the region is suppose to
Yup. There's no restriction on |z|, so the region goes to infinity.
Goodnight! :zzz:
Oh okay thanks so much for all your help and sticking with me through all my confusion! I appreciate it :D
FAQ: Complex analysis: Sketch the region in the complex plane
1. What is complex analysis and why is it important?
Complex analysis is a branch of mathematics that deals with the study of functions of complex variables. It is important because it provides a powerful tool for understanding and solving problems in
many areas of mathematics, science, and engineering.
2. What does it mean to sketch a region in the complex plane?
Sketching a region in the complex plane means visually representing a subset of the complex plane using geometric shapes and symbols. This helps to better understand the properties and behavior of
complex functions within that region.
3. How do you determine the region in the complex plane for a given complex function?
The region in the complex plane for a given complex function is determined by analyzing the behavior of the function and identifying the set of complex numbers for which the function is defined and
continuous. This region can be further refined by considering any singularities or branch cuts of the function.
4. What are some common techniques for sketching regions in the complex plane?
Some common techniques for sketching regions in the complex plane include using real and imaginary axes, plotting points, drawing curves or circles, and shading in regions to represent inequalities
or conditions for a function.
5. How does sketching regions in the complex plane help in understanding complex functions?
Sketching regions in the complex plane helps in understanding complex functions by providing visual representations of the behavior of the function. It can also help in identifying patterns,
symmetries, and relationships between different parts of the function, which can aid in solving problems and making predictions about the behavior of the function. | {"url":"https://www.physicsforums.com/threads/complex-analysis-sketch-the-region-in-the-complex-plane.583846/","timestamp":"2024-11-02T01:25:17Z","content_type":"text/html","content_length":"117370","record_id":"<urn:uuid:d440495b-7330-4764-87aa-42cba477e001>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00490.warc.gz"} |
Math Forum :: View topic - Excess of labour - www.mathdb.org
Math Forum :: View topic – Excess of labour
When someone wants to hammer a nail in each of several posts, put at equal distances along a road, the best way is to begin with the first post and finish with the last one . But how can we
accomplish this task in the worst way, that is such that the route be the longest? | {"url":"https://www.mathdb.org/phpbb2/viewtopicphpp2923amp/","timestamp":"2024-11-06T02:01:39Z","content_type":"text/html","content_length":"26996","record_id":"<urn:uuid:3f3aed98-6d2d-4754-a63e-026a9f0c72f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00456.warc.gz"} |
Can you crack the code and unlock the lock?Can you crack the code and unlock the lock?
Can you crack the code and unlock the lock?
Can you crack the code and unlock the lock?
291 One number is correct and well placed
245 One number is correct but wrong place
463 Two numbers are correct but wrong places
569 One number is correct but wrong place
578 Nothing is correct
Can you now find the code which can open this lock?
Also Try: Hen Dog Cat Robin Steve Tim Brainteaser
Answer: 3 9 4
Also Try: Exercise for one game for three riddle
Hint1 and Hint2 confirms that code doesn’t have Number 2. Since first hint says that only one number is correct and second says correct number is wrongly placed so that relegates number 2.
Hint5 tells us that, we can ignore numbers 5, 7 and 8
So now out of 1,2,3,4,5,6,7,8,9 we can exclude 2,5,7 and 8 and left with 1,3,4,6,9
Now, from Hint2, we can say that number 4 is one of the number in code and only thing we need to find is its place.
Now, Hint1, 3 and 4 has unique numbers as 2,9,1,4,6,3,5 out of these we already excluded 2,5 and 4(which we know). so we are left with 9,1,6,3 out of which we need to find 2 numbers.
Hint1, 3 and 4 also tells us that only numbers which can satisfy these 3 conditions are 9 and 13out of set 9,1,6 and 3
So all 3 numbers are 4, 9 and 3
Hint 2 and 3 tells that number 4 can only be on third place. so numbers would be XX4
Hint1 tells that number 9 can be placed on second position. So numbers would be X94
so final number would be 394 | {"url":"https://www.puzzlesbrain.com/cool-brainteasers/can-you-crack-the-code-and-unlock-the-lock/","timestamp":"2024-11-04T08:14:28Z","content_type":"text/html","content_length":"188855","record_id":"<urn:uuid:d45ee675-2cd5-4238-b245-67e5b429c8b7>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00269.warc.gz"} |
Basic Mixed Strategy Example
A mixed strategy game is one in which there is no dominant strategy for either player. Each player chooses each strategy with some probability.
Suppose a two player zero sum game has the following payoff matrix.
A\B \[B_1\] \[B_2\]
\[A_1\] 2 -11
\[A_2\] -4 3
If B plays strategy 1, and A picks strategies
\[A_1, \: A_2\]
with probabilities
respectively, then A will expect to receive
for every game he plays. If B plays strategy
, then A will expect to win
A will maximise their winnings if he chooses
such that his minimum winnings are as high as possible. This is when
\[6p-4=3-4p \rightarrow p=\frac{7}{10}\]
We can perform the same analysis for player and B.
If A plays strategy 1, and B picks strategies
\[B_1, \: B_2\]
with prbabilities
respectively, then B will expect to receive
for every game he plays. If A plays strategy
, then B will expect to win
B will maximise their winnings if he chooses
such that his minimum winnings are as high as possible. This is when
\[-(3q-1)=-(3-7q) \rightarrow q=\frac{4}{10}\] | {"url":"https://mail.astarmathsandphysics.com/university-maths-notes/game-theory/4649-basic-mixed-strategy-example.html","timestamp":"2024-11-07T18:15:45Z","content_type":"text/html","content_length":"32849","record_id":"<urn:uuid:49508a1d-aaf9-406f-a2b3-9c69cdaaa476>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00211.warc.gz"} |
Lightweight Monadic Programming in ML
Lightweight Monadic Programming in ML
Many useful programming constructions can be expressed as monads. Examples include probabilistic modeling, functional reactive programming, parsing, and information flow tracking, not to mention
effectful functionality like state and I/O. In this paper, we present a type-based rewriting algorithm to make programming with arbitrary monads as easy as using ML's built-in support for state
and I/O. Developers write programs using monadic values of type M t as if they were of type t, and our algorithm inserts the necessary binds, units, and monad-to-monad morphisms so that the
program type checks. Our algorithm, based on Jones' qualified types, produces principal types. But principal types are sometimes problematic: the program's semantics could depend on the choice of
instantiation when more than one instantiation is valid. In such situations we are able to simplify the types to remove any ambiguity but without adversely affecting typability; thus we can
accept strictly more programs. Moreover, we have proved that this simplification is efficient (linear in the number of constraints) and coherent: while our algorithm induces a particular
rewriting, all related rewritings will have the same semantics. We have implemented our approach for a core functional language and applied it successfully to simple examples from the domains
listed above, which are used as illustrations throughout the paper.
This is an intriguing paper, with an implementation in about 2,000 lines of OCaml. I'm especially interested in its application to probabilistic computing, yielding a result related to Kiselyov and
Shan's Hansei effort, but without requiring delimited continuations (not that there's anything wrong with delimited continuations). On a theoretical level, it's nice to see such a compelling example
of what can be done once types are freed from the shackle of "describing how bits are laid out in memory" (another such compelling example, IMHO, is type-directed partial evaluation, but that's
literally another story).
See also the mention of the paper by its author recently on LtU. I agree this deserves a full story in itself.
Somehow, even though I'd read that thread, I totally overlooked the link to this paper. Thanks for calling my attention (back) to it!
What bothers me the most is that the authors don't mention Wadler's Marriage of Monads and Effects at all. In that paper, Wadler established a translation from a computational lambda calculus for a
language with state into a monadic language, but the application of monad morphisms was implicit (those were made explicit by Filinski's M^3L, which is mentioned by the author). So comparing exactly
what they achieved over Wadler's work is what is mostly missing for me.
Of course, from a theoretical perspective, we should ask whether there is a principled way to get these monads and monad morphisms... Especially considering that their number grows in size
exponentially to the number of effects used. :)
Here is one way to automatically derive morphisms.
Analogous to CPS transformation we can do something I will call monad transformation. It transforms a piece of code with implicit effects to insert bind/return calls as appropriate.
For example the map function:
map f [] = []
map f (x:xs) = f x : map f xs
map' f [] = return []
map' f (x:xs) = do x' <- f x
xs' <- map' f xs
return (x' : xs')
If we monadify it with respect to the result of the f argument. That is, the type changes from (a -> b) -> [a] -> [b] to (a -> m b) -> [a] -> m [b].
So the rules are that you insert binds for all nested function calls, and you insert a return at the end. A more precise way to say it is you transform the expression to ANF, then you replace all
lets by monadic binds and wrap the end result in a return. This way you can mechanically monadify any function. The map' function is exactly the monadic map mapM in Haskell. Applying the same
transformation to filter gives you filterM. It works for any function. In particular it also works for the monadic return and bind functions. For example take the bind function from the maybe monad:
x >>= f = case x of
Nothing -> Nothing
Just v -> f v
Monadifying it with respect to the x argument we get:
x >>= f = do x' <- x
case x' of
Nothing -> return Nothing
Just v -> f v
This is the maybe monad transformer, minus the MaybeT wrapper boilerplate.
Lets try the same with the list monad:
xs >>= f = concat (map f xs)
We get:
xs >>= f = do xs' <- xs
tmp <- map' f xs'
return concat tmp
And we got the list monad transformer.
The reason this works is that when you want to execute code with monadic effects in a monad A, what you want is to monad transform the code wrt A. If you then also want to have effects from monad B,
you can transform the code resulting from the previous transformation again wrt B. If you work out the details of what twice monad transforming does, you will see that everywhere returnA x appeared
with a single monad transformation, now returnB' x appears, where returnB' is the return function of monad B transformed wrt monad A. The same thing happens for bind. So this gives you a way to
mechanically construct the return and bind functions for the combined monad: returnAB = returnB' and bindAB = bindB'.
I don't know if this method is explicitly described elsewhere. It probably is, given that this seems fairly obvious and the only way that could possibly work, but I don't know the name of it so I
couldn't find it. Anyone?
if we modeled normal functions as Kleisli arrows in an abstract monad, I believe we would achieve similar results. Plain old functions would then be run in the 'Identity' monad.
That is exactly what I'm doing; treating returnB and bindB as Kleisli arrows in the A monad. That doesn't really tell you *how* exactly you're viewing them as Kleisli arrows, so this summary doesn't
seem sufficient.
It doesn't matter which monad you run normal functions in; the monad laws imply that functions that don't use effects run the same in any monad, though the identity monad would be a natural choice.
You can see that I used this to cheat a little in the map example: if you want to be consistent you have to transform the cons function : too, but because it is a pure function and cannot possibly
produce an effect, you can just call it and use its result directly. This is just an optimization; you could also monadify all functions, then the map code would look like this:
map' f [] = return []
map' f (x:xs) = do x' <- f x
xs' <- map' f xs
x' :' xs'
The return would then happen inside :', which then has type Monad m => a -> [a] -> m [a]. The same goes for the concat function in the list monad bind. If you apply the transformation consistently
all types a -> b become Monad m => a -> m b.
One thing that this model supports and might be interesting is return and bind functions that themselves use effects.
Another thing to think about that I've wondered out loud about here before is what the corresponding transformation for comonads looks like (all types a -> b would become c a -> b).
This monad transformation is related with and can be used to implement direct style monads like in the Representing Monads, just like CPS transformation is related with and can be used to implement
This kind of call-by-value translation appeared in, for example, Monads and Effects by Benton, Moggi and Hughs (see Figure 4 in "Metalanguages for Denotational Semantics").
This reference is definitely not the first place it appears, for example Wadler mentiones it as "just the usual typed call-by-value monad translation".
You seem to be applying it to a Haskell program, yielding another Haskell program, rather than translating a CBV language into a monadic language. Can you please explain what you achieve by it?
(Both Wadler in his marriage paper and Hicks et al. in this paper used translations from a CBV language into a monadic language, and not within Haskell.)
Jules Jacobs wrote:
Lets try the same with the list monad:
xs >>= f = concat (map f xs)
We get:
xs >>= f = do xs' <- xs
tmp <- map' f xs'
return concat tmp
And we got the list monad transformer.
Let us try it out:
bind xs f = do xs' <- xs
tmp <- map' f xs'
return (concat tmp)
f1 x = do
print x
return [x]
t1 = (return [1,2] `bind` f1) `bind` f1
t2 = return [1,2] `bind` (\x -> f1 x `bind` f1)
Whereas t1 prints "1 2 1 2", t2 prints "1 1 2 2". Thus bind is not associative and can't be the bind of a monad. The giveaway was the phrase about `list monad transformer': there is no such thing.
I should emphasize that the monad transformer approach is inherently limiting because the layering of transformers is statically determined. In general, the composition of monad transformers is not
symmetric: the order matters. There are realistic programs where no fixed order suffices because at different moments the program needs different orders. The order must be dynamic, which monad
transformers don't support. The paper `Delimited Dynamic Binding', Sec 4.3, has described several examples of realistic programs that require the dynamic ordering (the accompanying code archive
contains the complete code). http://okmij.org/ftp/Computation/dynamic-binding.html
I'm sure you have a much better understanding than me about these things, but here he goes for fun :)
Even though it's not a monad transformer, you get the list "monad" transformer that is generally called that, for example here: http://en.wikibooks.org/wiki/Haskell/Monad_transformers
I agree that this was a bad example, and the fact that this transformation still produces something that the Haskell community has produced is largely accidental. Still, the transformation is valid
in a way. Say you were given a program in direct style that uses list monad operations and IO monad operations. There are two ways we could execute this: monadify it wrt the list monad and then
monadify it wrt the IO monad. Or we could monadify it once wrt the derived list "monad" transformer applied to the IO monad. These two give the same result, and in both cases you indeed get
unintuitive behavior of the nested let example you gave. So if you don't have monad laws the transformation cannot give you these, but it still gives you something reasonable.
In symbols:
T m p transforms program p wrt monad m
Q m1 m2 is the result of the derivation procedure I sketched applied to monad m1 wrt m2
Then the following holds:
T m2 (T m1 p) == T (Q m1 m2) p
Dynamic layering is indeed an interesting problem and I don't see how monad transformers could support that either. What you want is essentially squeezing a continuation monad between a reader monad,
but then the two halves of the reader cannot communicate with each other. Monad transformation puts one monad next to another, perhaps there is a similar operation that puts a monad inside another,
allowing it to decide dynamically which monad to put on top. Or maybe I am completely wrong, dynamic layering is something I know I don't understand (as opposed to the static layering above, which I
think I understand but probably don't).
Do you know other examples than the reader monad that can be dynamically layered in an interesting way?
Here is one way to automatically derive morphisms.
I guess I should clarify what I meant by "a principled way to get these monads and monad morphisms".
Lets assume the effects we have are:
get, put, throw.
Then, for each of the 8 subsets:
{}, {get}, {put}, {throw}, {get, put}, {put, throw}, {get, throw}, {get, put, throw}
we could get a monad:
Id, Environment, Writer (with suitable monoid), Error, State, ErrorT (Writer), ErrorT (Environment), ErrorT(State)
This choice is obvious in this case, but it isn't obvious in general: if I have a monad that supports the effects: Eff1, Eff2, ..., Eff n, what is the monad corresponding to an arbitrary subset of
these effects: {Eff i1, ..., Eff ik}?
This question was posed by Wadler in the conclusions section of his Marriage of Monads and Effects. Various people (for instance, Tolmach in Optimizing ML Using a Hierarchy of Monadic Types, read
especially section 6) worked out the monads corresponding to some subsets for a particular choice of effects. But I don't know of any published general principle to deduce these monads.
(I hope it's needless to say that the number of these monads grows exponentially as the overall set of effects grows.)
Hicks et al.'s paper assumes such a hierarchy is given (although not necessarily completely). They don't try to derive it. The situation can be rectified a bit, if we reuse the underlying semantic
idea of Oliveira and Schrijvers's Monad Zipper: use monad transformers to describe the 'biggest' monad, and then mask the effects that are not used. But even this solution doesn't really give a
principled account of how to get, say, the Writer monad from the State monad.
That seems to be the question I answered here: how to compose arbitrary monadic effects. I don't think you can hope to automatically get a state monad from a reader and writer monad, because these
two operations need to pass data from one to another. That is, conceptually reader + writer is not the same as state. Reader + writer is the ability to read a value from place 1, and the ability to
write a value to place 2. But you can do the opposite and get a reader and writer monad out of a state monad by hiding the put or get operation.
The situation is even worse than exponential, because order matters. We can even use the same monad multiple times. For example if we have StateT . ContT . StateT then we have two kinds of state: one
kind that gets reset when we invoke a continuation and one kind that remains across continuation invocations. So there are not only exponentially many possibilities, there are even infinitely many.
The question that Wadler poses:
A general theory of eects and monads. As hypothesised by Moggi and as born out by practice, most computational eects can be viewed as a monad. Does this provides the possibility to formulate a
general theory of eects and monads, avoiding the need to create a new effect system for each new effect
Is a bit different. He also wants an effect system to go along with the effect. If the monad operations are well typed, this also gives you that, in some sense. The effect system will be as granular
as the type system can check the monadic operations. For example in Haskell I don't think you can specify a ST operation that only modifies certain references, so in the resulting effect system you
won't be able to do that either (no magic). If you have an exception monad with multiple exception types then the Haskell type system cannot statically distinguish functions that only raise a subset
of the possible exceptions, so the resulting effect system won't be able to do that either.
The papers Marriage of Moands and Effects and Optimizing ML Using Hierarchy of Monadic Types look interesting. I will try to read them.
That seems to be the question I answered here: how to compose arbitrary monadic effects. I don't think you can hope to automatically get a state monad from a reader and writer monad [..]
I reiterate, it's not about composing effects. It's about decomposing them. The question I ask is the dual to yours: how can we get the Reader and Writer monads from the State monad?
Regarding the question Wadler asks: he poses two questions. The first one is a conjecture that the denotational semantics might take the form of T X = S1 -> X x S2, where S1 are the store of
locations you only read from, and S2 is the store of locations you only write to. Note that this doesn't give you a monad: if you only ever write to a single location, T X = X x S isn't a monad
(what's the unit?). But T X = X x (S + 1) is. The additional right injection says 'no state change has occurred'.
The second question is the one you quoted: can we have the same spiel happen in general? This is the underlying tone of my question --- find a principled, general account for the previous question.
Hicks et al.'s work can partially address the second question, and Oliveira and Schrijvers's can be used to partially address the first question.
Finally, what you mentioned is exactly the advantage of subtyping exceptions in Java and for region annotations (as mentioned in Wadler, for example). You have some granularity that allows you to
distinguish between different types of exceptions, but rough enough to keep things decidable.
What is the problem with just using the state monad and hiding the write operation to get a reader -- or in general, just hiding the operations you don't want?
Consider: you've independently developed an algorithm that uses a reader monad. How do you apply this in your state monad?
State has 'get' and 'put'. If you have an algorithm that just invokes 'get', what's the problem?
The issue, which I wrongly assumed was clear in context of earlier discussion, is how we achieve composition and decomposition of effects - i.e. with a few generic composition operations rather than
specific code for each of 2^N or even N! integrations.
Even if the reader monad algorithm invokes 'get' (instead of 'ask' or something else), you still must lift your reader monad to run in the state-monad's environment (i.e. to propagate state,
unchanged, across each operation). Can you think of any generic way to make this work?
You can read my post below, but my preferred solution would be to have Processes that can signal 'get' as a subtype (I'll leave my exact meaning of this term here ambiguous) of Processes that can
signal 'get' or 'put'.
I think there are definitely ways to "make it work" in Haskell. It's just a question of how much boilerplate code is involved and how elaborate of a contraption you are willing to build.
Processes are not difficult to model in Haskell. Adapting composable sum types is more an issue, but it might be achievable with some sort of type-family programming.
I prefer to control most effects with capabilities - i.e. two capabilities - one each to signal 'get' and 'put' - is literally a superset of the capability to just signal 'get'. Composition and
decomposition of effects is straightforward. Capabilities have the advantage of being orthogonal to the type system, while still allowing types to be effectively leveraged for proofs and safety. It
also works well in open systems, and with concurrency.
Modeling capabilities in Haskell is perhaps best achieved with arrows, though Haskell's Control.Arrow models is too general for my work. I've been developing a new model of arrows to accommodate
asynchronous behaviors.
Modeling Processes wasn't the problem - it was embedding them in such a way that they're easy to work with to capture the flavor of programming I wanted. I'm not particularly worried about the
theoretical model of Processes as it's pretty trivial.
Arrows tagging capabilities is pretty similar to the way I use signals, I think. Composition and decomposition of signals probably follows the same pattern you're using with capabilities.
Processes are much closer to the monadic approach, though. All effects occur in a linear order, which means you can suspend a Process in response to any given signal and it's in a well defined state.
This makes them useful as a replacement for delimited continuations.
That said, I do see the utility of identifying Processes as subset of a more general model of computation that has multiple threads of execution. One approach I'm considering is to have such more
general computations serve as implementations for Processes that just specify arbitrary interleave semantics. That way you can have Processes that can in principle (semantically) be suspended, but
that support parallel implementations when possible.
A big difference between signals and capabilities that I forgot to mention is that signals are always parameters that must be hooked up by the caller. It's not possible for a process to squirrel away
a signal for later use. Similarly, revocation of signals (or swapping them out with a mock handler) is always possible and doesn't require usig a special pattern at construction time.
Monads are not very good for modeling and composing effects, in the general case.
Monads capture a lot of effects by default: ordering/time, identity, state, commitment, and unbounded time/space resource consumption. These effects interfere with modeling a single effect, but
sometimes we can work around that. I posit that they strongly interfere when composing multiple effects.
I.e. we do not compose monadic effects as a set of {get,put,throw} capabilities in parallel. We compose in layers - N! possible permutations for N effects. Turns out this is also a problem for
Can we find a better foundation for modeling effects? Something, perhaps, that exhibits:
• associativity - We can regroup expression of effects. We must explicitly model space as an effect, if grouping should matter.
• commutativity - We can rearrange expression of effects. We must explicitly model time as an effect, if relative order should be relevant.
• idempotence - We can eliminate duplicate effects. We must explicitly model identity as an effect, if replication should be meaningful.
• monotonicity - effects only add to a system. We must explicitly model time, identity, and framing to model any destructive updates.
• predicative time - We can control time resources. We must explicitly model incremental computation, if we need divergence or arbitrary time as an effect.
• predicative space - We can control memory resources. We must explicitly model incremental accumulation, when we need divergence or arbitrary space as an effect.
• determinism - we must explicitly model choice or probability as an external effect, when we need it.
• uncommitted - We can change the inputs and the effects will follow, similar to dataflow programming. We must explicitly model commitment as an effect, when we need it.
• stateless - we must explicitly model state as an external effect, if we need it. (Keeping this orthogonal to commitment is interesting.)
I believe my RDP model comes very close. I expect to manage most effects as capabilities. But I would like to see what I can do with RDP behavior transformers (e.g. to model dynamic scope) at some
I suspect that effects strongly interfere when composing multiple effects.
It's an interesting and difficult problem though. It may be true that monads interfere in composing effects more than what is inherently unavoidable. Some have suggested that Lawvere Theories are a
better option.
I've mentioned a few times around here that I think an abstraction I call Processes might be more useful than monads & transformers, and this thread has motivated me to make the case in a little more
detail. Feedback, particularly perceived pain points, is solicited. The basic idea is that whenever we have an effectful function, we should keep the effects parametric until the last possible moment
when we hook them up / run them, and never worry about what monadic values implement the effects (as we would with a transformer stack).
As an example, here's how I would write Oleg's example. Rather than working in the list monad, we work in a Process monad that has a signal 'branch', that from the point of view of the Process
returns one of the elements of a list:
f1 x = do
print x
return x
p1 = do
x <- branch [1, 2]
y <- f1 x
z <- f1 y
return z
We can provide monadic definitions that make the above valid Haskell (where the monad is a continuation type like (a ->Process) -> Process), but I prefer more sugar - picking an evaluation order and
unifying effectful with non-effectful functions:
p1 = f1( f1 (branch [1, 2])) -- same as above, with sugar
This is essentially just arguing for effect typing over explicit monadic plumbing, but with support for hooking up effects to handlers. Further, it's important that the handler has access to the
suspended Process signaling the effect. As I noted in the recent delimited continuations thread, this gives a mechanism that's trivially convertible to delimited continuations, but with a better
(IMHO) API. But I'm heading off course here...
Getting back to the example, the point is that there is no 'p2' because the difference between Oleg's 't1' and 't2' isn't encoded in my 'p1'. The code that runs 'p1' gets to decide whether to run the
continuation processes to completion (another signal) with each branch value or whether to single step through the existing possibilities in a round robin fashion (suspending when certain signals are
raised). Of course in Oleg's example, interleave order was encoded in bind grouping, which is obviously undesirable and violates the monad laws. In the Process approach it's pretty clear what the
implementation possibilities are -- we just need to use some other signal(s) to know when to switch between possibilities.
The last time I advanced this argument, Philippa Cowderoy noted that we can do everything in abstract monads (e.g. Branching m, IO m => m String) and achieve something very similar to what I'm
advocating, and I think that's true if we don't allow first class signals. Certainly, given such an abstract value it's always possible to define instances for a Process monad and get a Process
(continuation) out of it. The defining characteristic of the approach I'm advocating is that we do not provide an interpretation of the effects via type class substitution, but rather only ever
handle effects via running the Process.
If we do provide first class signals, then we can't necessarily convert from a Process into an abstract monad, and I think it's a good idea to have them. For one, we immediately address Oleg's
objection about the need for dynamic rewiring of effects (as it would apply to Processes). Another issue is that Haskell doesn't easily let us have multiple constraints of the same type class, but
imagine that we have a class File that defines operations for interacting with a single file. Then we might want to have a type like (File_a m, File_b m => m ()) that operates on a pair of files. The
monad zipper paper addresses this problem, but what if we want to write a function that manipulates an unbounded number of files? In that case, in the usual monadic approach, we'd need to define a
new type class for dealing with families of files. With first class signals, the approaches can be unified.
I'm still meaning to write up more of this, including more details on Process combinators and a type system for classifying them, but I'm still of the mind that Processes are probably a more natural
and useful approach to effects than would be building a bunch of custom monads. I tried to do some example code in Haskell the last time this came up, but I found it pretty cumbersome to encode the
types that I'd want. So my writeup of these ideas became bottlenecked on other language issues. I maybe should take off and write up the ideas more carefully with snippets in a non-existent language
to generate some feedback.
Perhaps you could explain what you mean by 'first class signals'? I've spent some time reviewing, but it seems you did not discuss those in our e-mail conversation eight months ago.
I do encourage you to write more, in a public forum. After having tried it for three months, I think doing so benefits the writer as much as the audience. But LtU tends to bury ideas, and doesn't
always result in the clarity you might pursue if writing for an anonymous audience. I suggest developing a dedicated blog or wiki for your project.
As an aside: p2 = f1 =<< f1 =<< branch [1,2]. We don't need syntactic sugar to get very near what you're desiring. Indeed, desugaring works nearly as well.
I think it's implicit in the description of Processes we discussed which was something like:
type Process = Signal -> (Process, Signal)
(This is a coarsely typed Process where you'd have to put all of the possible signals into a big algebraic datatype.)
My intention was to contrast this with the static type class approach.
I might set up a language blog or put out some notes. I agree it would be more helpful to ask LtU after I have something more substantial to link to. Thanks for the tip. I haven't seen your blog -
can you provide a link?
Ah, you're just referring to passing a signal's value. I was confused, I suppose, because I distinguish between the notions of signal and the values carried by one.
The coarse-grained model we came up with earlier was:
type Process x y = x -> (y,Process x y)
You posited that we really need dependent types to win the most from this model.
Does this idea relate at all to Bauer and Pretnar's Eff language?
It's a bit hard to relate the two without more precise descriptions, but at least Eff has an implementation that you can play around with.
Thanks for the link. I didn't investigate Eff when that story was posted, but the algebras/effect substitution are definitely similar to the signal hook-ups I have with Processes. There are some
differences. For a minor example, I favor imposing an evaluation order and letting effects mix freely with pure code, leaving your type system and IDE to mark purity. There are other differences, as
well, but this is definitely a similar mechanism to what I'm advocating, and I believe it has most of the advantages that I posted about above. One thing I notice from the Eff docs is this comment:
Yes, eff currently does not check types. It does not seem easy to come up with a good type system for eff.
Whereas most of the time I've spent thinking about Processes has been related to typing them.
Any idea how Eff has been received? Are there reasons why an approach like this is considered inferior to Monads? I'm curious how this relates to Lawvere theories, but I'll apparently need to catch
up on some category theory before I can figure it out. Can someone answer?
...can be found in the LICS 2010 paper, "A Generic Operational Metatheory for Generic Effects", by Patricia Johann, Alex Simpson, and Janis Voigtlaender. Patty & co. know an awful lot of category
theory, of course, but it's rather decent of them to translate for those of us who think at slightly lower levels of abstraction. :)
What I liked about this work is that it made it clear to me how monads arise as a model of effects at all. Basically, the trick in this paper is to start with an operational semantics which does not
model effects -- it just has uninterpreted term formers for effectful terms, so that evaluation builds a computation tree. Then, separately, you give an interpretation function for computation trees,
at which point the connection to monads becomes "obvious" -- the effectful terms form a signature and the interpretation function can be seen as an algebra for that signature.
This intuition is almost certainly in Plotkin and Power, but I never worked it out clearly enough to believe it.
That connection is very old, and dates to the early days of algebras for a monad (circa '65). The algebraic view is particularly appealing:
• The unit of the monad u : A -> TA is the inclusion of variables in the set of (normalised) terms.
• The multiplication m : T² A -> T A is the evaluation of a term of terms into a single (normalised) term.
• The strength str : A x TB -> T(A x B) shuffles variables around.
There are some differences. For a minor example, I favor imposing an evaluation order and letting effects mix freely with pure code, leaving your type system and IDE to mark purity.
I don't see how that is different than Eff, which has an evaluation order and the finished type system would check purity.
Regarding the type system, it is still under development. However, there is an experimental base type system that you can play with using the --types flag. It is still not the full system the
designers have in mind, though.
As to how it's received: it seems to effect interest, although I think a fully worked out type system and a cleanly presented operational semantics (which I saw Pretnar present in the European
Workshop on Computational Effects) would make it a lot easier to buy in.
Are there reasons why an approach like this is considered inferior to Monads?
I don't think I ever heard anyone claim this approach is inferior. It is still quite new, so the major criticism is that it is not fully baked yet. But the designers never claimed it was.
The relation to Lawvere theories is inherent in the construction: Monads arose as a means to implement effects. Plotkin and Power's algebraic theory of effects decomposes the description into a pair:
<S, E> of effect signature S and behaviour E given by effect equations. This shift in focus from an implementation of the effects into an interface to the effects (i.e., S) gives rise to a more
modular account. The language Eff lets you manipulate these signatures using effect handlers.
One of the Eff pages talks about using do notation for explicitly sequencing effects. The code examples don't seem to use this syntax, however, so you're right that this doesn't seem to be a
This shift in focus from an implementation of the effects into an interface to the effects (i.e., S) gives rise to a more modular account.
I agree with this (and in fact I was trying to make this point in my 'forget monads' post). Probably the reason I don't see the connection to Lawvere theories yet is that I don't yet grok Lawvere
theories (and thanks, Neel, for the link).
Are there any linkable documents describing the type system yet? My guess is that if it supports handlers whose bodies can themselves issue and handle effects, then the only differences between what
I do and what Eff does are just presentation. Those still may be interesting differences in the design space, though, so I'll still try to write up my design sometime soon.
I did an implementation (as an EDSL) of Eff in Haskell: http://goo.gl/Uf42N. Maybe it helps in seeing if it is the same as what you are doing.
I think there are some basic differences between the Eff approach and my Processes, and I'm going to write up (hopefully soon) an overview of my approach and I'll compare and contrast with algebraic
effects. I think there's a reasonably shallow translation between the two, but it's not purely syntactic. My handlers are more like OOP style objects and I don't have the built-in value lifting
(return) and finally mechanisms of Eff. Less interesting, effects in Eff don't look to be first class and so some form of dispatch function encoding is probably necessary to translate from some uses
of Processes.
Thanks for the link to your Haskell code. One thing I'm still trying to understand is the difference between the two approaches in the presence of type systems. Your code would probably give
reasonable types for effectful terms if Haskell supported local type class instances. Hopefully I'll get some time to think about all of this over the weekend. | {"url":"http://lambda-the-ultimate.org/node/4321","timestamp":"2024-11-02T15:03:46Z","content_type":"application/xhtml+xml","content_length":"65125","record_id":"<urn:uuid:97c7668d-a068-49b7-af3f-b8fbb5d5e5cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00656.warc.gz"} |
Top 10 Math Books for Grade 4: Empowering Young Minds to Discover Numbers
Are you looking for the best math books to help your fourth-grade students develop a strong foundation in mathematics? Look no further! In this article, we will explore the top 10 math books that are
specifically designed to empower young minds and make the learning process enjoyable. These books offer engaging content, interactive exercises, and a comprehensive approach to mathematics, making
them ideal resources for both teachers and parents. So let's dive in and discover the perfect math books for Grade 4!
Just like a farmer sows seeds to reap a bountiful harvest, it’s essential to sow the seeds of mathematical understanding in our children from an early age. After all, isn’t education the most potent
tool we can give to our young ones? Especially in grade 4, as students begin to navigate the world of numbers and equations, the importance of having the right resources cannot be overstated. And
what better resource than math books, right?
The Absolute Best Book to Ace Math Books for Grade 4
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The Importance of Math for Grade 4
Encouraging a Love for Math
Math is not just a subject, it’s a universal language, and grade 4 is the perfect time to nurture a love for this language. Learning math helps students develop problem-solving skills and logical
thinking. It’s like being a detective, solving a mystery, one problem at a time!
The Role of Math Books
Good math books serve as bridges, connecting the world of abstract numbers to real-life scenarios. They offer the necessary guidance to students and make learning a fun, engaging process. It’s like
they hold a magic key that unlocks the gate to the world of mathematics for our young learners!
Features of a Good Math Book for Grade 4
Every book tells a story. In the case of math books, they tell the story of numbers and equations. And like any good story, it should be engaging, fun, and full of surprises. Let’s explore what makes
a math book effective for grade 4 students.
Engaging and Fun Content
The best math books are those that can transform complex mathematical concepts into bite-sized, easily digestible pieces. They should be able to spark curiosity, just like a fascinating riddle
waiting to be solved. After all, aren’t math problems essentially puzzles waiting for solutions?
Practice Questions
Like a faithful workout routine for the brain, practice questions help to strengthen the understanding of mathematical concepts. The best math books provide a plethora of practice questions that
cater to different levels of understanding.
Answers and Explanations
Who doesn’t enjoy the triumphant feeling of cracking a tough math problem? But what happens when we stumble? The best math books offer clear answers and explanations, giving students the necessary
feedback to improve. It’s like having a supportive tutor available at all times!
Best Math for Grade 4 Resource
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Top 10 Math Books for Grade 4
Navigating through the sea of math books can be daunting, especially when you’re seeking the perfect companion for your grade 4 student. But worry not, we’ve taken the puzzle out of this equation for
you! Here are the top 10 math books that make grade 4 mathematics an exciting journey rather than a daunting task.
Book 1: Mastering Grade 4 Math: The Ultimate Step-by-Step Guide to Acing 4th-Grade Math
This comprehensive book for grade 4 is a wonderful blend of fun and learning. It breaks down complex concepts into understandable segments, allowing grade 4 students to grasp mathematical concepts
with ease. It’s like a friend that speaks the complex language of math in simple, everyday terms!
Book 2: Mastering Grade 4 Math Word Problems: The Ultimate Guide to Tackling 4th Grade Math Word Problems
With comprehensive content coverage and ample practice problems, this book promotes mastery of grade 4 mathematics. The book is designed with attention-grabbing lessons and exercises, making it feel
like an adventure through the world of numbers and equations.
Book 3: ‘Common Core Math 4 Today, Grade 4: Daily Skill Practice’ by Erin McCarthy
This daily practice book is great for reinforcing skills learned in the classroom. It builds on the common core standards, offering problems that encourage critical thinking. It’s like a fitness
routine, helping to flex those math muscles every day!
Book 4: ‘Math for the Gifted Student: Challenging Activities for the Advanced Learner, Grade 4’ by Flash Kids Editors
Aimed at gifted students, this book offers challenging problems to stimulate their brains further. It’s like the extra spice needed to keep the mathematical soup interesting for quick learners!
Book 5: ‘My Math Grade 4 SE Vol 1’
The “My Math” series is a comprehensive mathematics curriculum developed by McGraw-Hill Education. It’s widely utilized across various American schools. The curriculum is structured to align with
Common Core State Standards for Mathematics, and it uses a blend of print and digital materials to create a balanced, engaging learning experience.
Book 6: ‘Singapore Math – Level 4A Math Practice Workbook for 4th Grade’ by Marshall Cavendish
Drawing from the successful Singapore Math curriculum, this book focuses on problem-solving and in-depth understanding. It’s like a treasure map, leading students to the gold mine of mathematical
Book 7: ‘Math Games for Number and Operations and Algebraic Thinking: Games to Support Independent Practice’ by Jamee Petersen
Who said math can’t be fun? This book transforms learning into an exciting game, encouraging independent practice. It’s like a playful friend who makes learning math a delightful game!
Book 8: ‘Kumon Math Workbooks Grade 4: Multiplication’ by Kumon
With an emphasis on multiplication, this workbook uses the Kumon method to help children understand and master this critical operation. It’s like a coach, specifically training students to become
multiplication champions!
Book 9: ‘Beast Academy Guide and Practice Bundle 4A’ by Aops
A complete guide for grade 4 mathematics, this bundle includes a guidebook and a practice book. With its engaging content and challenging problems, it’s like a trusted mentor guiding students on
their math journey.
Book 10: ‘Carson Dellosa – The Complete Book of Maps & Geography for Grades 3–6’ by Carson Dellosa
While it’s not strictly a math book, this resource helps students understand the practical application of math through geography. It’s like a travel guide, showing students how math is used in the
real world.
These top 10 math books offer a delightful mix of fun, challenge, and learning, making grade 4 mathematics an enjoyable and engaging journey for young learners.
Final Words
Just as a master sculptor molds clay into a masterpiece, math books sculpt young minds, guiding them toward a profound understanding of numbers and their relationships. They don’t merely disseminate
knowledge but nurture a sense of curiosity, turning each learning session into a treasure hunt for knowledge. In the grand panorama of learning, a good math book for grade 4 is more than just an
accessory. It’s a magic wand that brings the abstract world of numbers to life, catalyzes the learning process, and fuels the innate sense of wonder in every child.
A strong foundation in mathematics is akin to laying the groundwork for a skyscraper. It forms the bedrock upon which further learning can take place. Grade 4, with its more complex mathematical
concepts and real-world applications, is a crucial juncture in a child’s learning journey. Here, math is not just about numbers anymore; it becomes a tool for solving problems, a language to express
logic, and a means to understand the world around us.
Each book listed in our top 10 offers a unique approach to ignite passion, foster understanding, and nurture mathematical skills in grade 4 students. Each book is like a unique gemstone, precious in
its own way, contributing to the vibrant mosaic of learning resources that empower young minds to fall in love with numbers.
1. Why is Grade 4 a critical stage for learning math?
Grade 4 is a transition period in a child’s mathematical journey. They progress from basic arithmetic to more complex concepts, like multiplication and fractions. It’s also when they begin to
apply math to real-world scenarios, which helps to cultivate problem-solving skills.
2. What features should I look for in a Grade 4 math book?
An effective Grade 4 math book should contain clear explanations, plenty of practice questions, and engaging content that captures the student’s interest. It should also align with educational
standards and have detailed answers for effective self-learning.
3. How can I make math more engaging for my Grade 4 student?
Use engaging math books, incorporate real-life scenarios, and turn math lessons into games or challenges. The key is to make math relevant and fun!
4. What are some effective ways to practice math problems?
Apart from using math books, online math games, puzzles, and math-oriented board games are also great ways to practice. Consistent practice and reviewing mistakes is the best way to grasp
mathematical concepts.
5. How can a math book help improve my child’s problem-solving skills?
Math books not only teach mathematical concepts but also help students apply these concepts to solve problems. This practice enhances logical thinking and problem-solving skills, which are
critical skills for lifelong learning.
The Best Books to Ace Math for Grade 4
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2.0. Introduction
Numbers, the most basic data type of almost any programming language, can be surprisingly tricky. Random numbers, numbers with decimal points, series of numbers, and conversion between strings and
numbers all pose trouble.
Perl works hard to make life easy for you, and the facilities it provides for manipulating numbers are no exception to that rule. If you treat a scalar value as a number, Perl converts it to one.
This means that when you read ages from a file, extract digits from a string, or acquire numbers from any of the other myriad textual sources that Real Life pushes your way, you don't need to jump
through the hoops created by other languages' cumbersome requirements to turn an ASCII string into a number.
Perl tries its best to interpret a string as a number when you use it as one (such as in a mathematical expression), but it has no direct way of reporting that a string doesn't represent a valid
number. Perl quietly converts non-numeric strings to zero, and it will stop converting the string once it reaches a non-numeric character—so "A7" is still 0, and "7A" is just 7. (Note, however, that
the -w flag will warn of such improper conversions.) Sometimes, such as when validating input, you need to know whether a string represents a valid number. We show you how in Recipe 2.1.
Recipe 2.15 shows how to get a number from strings containing hexadecimal, octal, or binary representations of numbers such as "0xff", "0377", and "0b10110". Perl automatically converts numeric
literals of these non-decimal bases that occur in your program code (so $a = 3 + 0xff will set $a to 258) but not data read by that program (you can't read "ff" or even "0xff" into $b and then say $a
= 3 + $b to make $a become 258).
As if integers weren't giving us enough grief, floating-point numbers can cause even more headaches. Internally, a computer represents numbers with decimal points as floating-point numbers in binary
format. Floating-point numbers are not the same as real numbers; they are an approximation of real numbers, with limited precision. Although infinitely many real numbers exist, you only have finite
space to represent them, usually about 64 bits or so. You have to cut corners to fit them all in.
When numbers are read from a file or appear as literals in your program, they are converted from their textual representation—which is always in base 10 for numbers with decimal points in them—into
an internal, base-2 representation. The only fractional numbers that can be exactly represented using a finite number of digits in a particular numeric base are those that can be written as the sum
of a finite number of fractions whose denominators are integral powers of that base.
For example, 0.13 is one tenth plus three one-hundredths. But that's in base-10 notation. In binary, something like 0.75 is exactly representable because it's the sum of one half plus one quarter,
and 2 and 4 are both powers of two. But even so simple a number as one tenth, written as 0.1 in base-10 notation, cannot be rewritten as the sum of some set of halves, quarters, eighths, sixteenths,
etc. That means that, just as one third can't be exactly represented as a non-repeating decimal number, one tenth can't be exactly represented as a non-repeating binary number. Your computer's
internal binary representation of 0.1 isn't exactly 0.1; it's just an approximation!
$ perl -e 'printf "%.60f\n", 0.1'
Recipe 2.2 and Recipe 2.3 demonstrate how to make your computer's floating-point representations behave more like real numbers.
Recipe 2.4 gives three ways to perform one operation on each element of a set of consecutive integers. We show how to convert to and from Roman numerals in Recipe 2.5.
Random numbers are the topic of several recipes. Perl's rand function returns a floating-point value between 0 and 1, or between 0 and its argument. We show how to get random numbers in a given
range, how to make random numbers more random, and how to make rand give a different sequence of random numbers each time you run your program.
We round out the chapter with recipes on trigonometry, logarithms, matrix multiplication, complex numbers, and the often-asked question: "How do you put commas in numbers?" | {"url":"http://nnc3.com/mags/perl4/cook/ch02_01.htm","timestamp":"2024-11-12T02:35:12Z","content_type":"text/html","content_length":"10762","record_id":"<urn:uuid:8bcf7b22-4d20-4ebf-ac79-d4698ae37719>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00103.warc.gz"} |
Project Euler #232: The Race | HackerRank
[This problem is a programming version of Problem 232 from projecteuler.net]
Two players share an unbiased coin and take it in turns to play "The Race". On Player 1's turn, he tosses the coin once: if it comes up Heads, he scores one point; if it comes up Tails, he scores
nothing. On Player 2's turn, she chooses a positive integer and tosses the coin times: if it comes up all Heads, she scores points; otherwise, she scores nothing. Player 1 goes first. The winner is
the first to or more points.
On each turn Player 2 selects the number, , of coin tosses that maximises the probability of her winning.
What is the probability that Player 2 wins? As the number is obviously rational and can be represented as with integer and , write the answer as
The first line of each test file contains a single integer , that is the number of queries. lines follow, each containing a single integer .
Print exactly lines with the answer to the corresponding query on each line.
The answer is which is equal to | {"url":"https://www.hackerrank.com/contests/projecteuler/challenges/euler232/problem","timestamp":"2024-11-12T23:51:09Z","content_type":"text/html","content_length":"814795","record_id":"<urn:uuid:d8ce6b77-c86b-4787-947c-792effec7012>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00641.warc.gz"} |
numpy.polynomial.laguerre.lagfit(x, y, deg, rcond=None, full=False, w=None)[source]¶
Least squares fit of Laguerre series to data.
Return the coefficients of a Laguerre series of degree deg that is the least squares fit to the data values y given at points x. If y is 1-D the returned coefficients will also be 1-D. If y is
2-D multiple fits are done, one for each column of y, and the resulting coefficients are stored in the corresponding columns of a 2-D return. The fitted polynomial(s) are in the form
where n is deg.
x : array_like, shape (M,)
x-coordinates of the M sample points (x[i], y[i]).
y : array_like, shape (M,) or (M, K)
y-coordinates of the sample points. Several data sets of sample points sharing the same x-coordinates can be fitted at once by passing in a 2D-array that contains one dataset per
deg : int or 1-D array_like
Degree(s) of the fitting polynomials. If deg is a single integer all terms up to and including the deg’th term are included in the fit. For NumPy versions >= 1.11.0 a list of
integers specifying the degrees of the terms to include may be used instead.
Parameters: rcond : float, optional
Relative condition number of the fit. Singular values smaller than this relative to the largest singular value will be ignored. The default value is len(x)*eps, where eps is the
relative precision of the float type, about 2e-16 in most cases.
full : bool, optional
Switch determining nature of return value. When it is False (the default) just the coefficients are returned, when True diagnostic information from the singular value
decomposition is also returned.
w : array_like, shape (M,), optional
Weights. If not None, the contribution of each point (x[i],y[i]) to the fit is weighted by w[i]. Ideally the weights are chosen so that the errors of the products w[i]*y[i] all
have the same variance. The default value is None.
coef : ndarray, shape (M,) or (M, K)
Laguerre coefficients ordered from low to high. If y was 2-D, the coefficients for the data in column k of y are in column k.
[residuals, rank, singular_values, rcond] : list
Returns: These values are only returned if full = True
resid – sum of squared residuals of the least squares fit rank – the numerical rank of the scaled Vandermonde matrix sv – singular values of the scaled Vandermonde matrix rcond –
value of rcond.
For more details, see linalg.lstsq.
Warns: The rank of the coefficient matrix in the least-squares fit is deficient. The warning is only raised if full = False. The warnings can be turned off by
>>> import warnings
>>> warnings.simplefilter('ignore', np.RankWarning)
See also
chebfit, legfit, polyfit, hermfit, hermefit
Evaluates a Laguerre series.
pseudo Vandermonde matrix of Laguerre series.
Laguerre weight function.
Computes a least-squares fit from the matrix.
Computes spline fits.
The solution is the coefficients of the Laguerre series p that minimizes the sum of the weighted squared errors
where the
where V is the weighted pseudo Vandermonde matrix of x, c are the coefficients to be solved for, w are the weights, and y are the observed values. This equation is then solved using the singular
value decomposition of V.
If some of the singular values of V are so small that they are neglected, then a RankWarning will be issued. This means that the coefficient values may be poorly determined. Using a lower order
fit will usually get rid of the warning. The rcond parameter can also be set to a value smaller than its default, but the resulting fit may be spurious and have large contributions from roundoff
Fits using Laguerre series are probably most useful when the data can be approximated by sqrt(w(x)) * p(x), where w(x) is the Laguerre weight. In that case the weight sqrt(w(x[i]) should be used
together with data values y[i]/sqrt(w(x[i]). The weight function is available as lagweight.
[1] Wikipedia, “Curve fitting”, https://en.wikipedia.org/wiki/Curve_fitting
>>> from numpy.polynomial.laguerre import lagfit, lagval
>>> x = np.linspace(0, 10)
>>> err = np.random.randn(len(x))/10
>>> y = lagval(x, [1, 2, 3]) + err
>>> lagfit(x, y, 2)
array([ 0.96971004, 2.00193749, 3.00288744]) # may vary | {"url":"https://numpy.org/doc/1.17/reference/generated/numpy.polynomial.laguerre.lagfit.html","timestamp":"2024-11-05T16:40:21Z","content_type":"text/html","content_length":"20034","record_id":"<urn:uuid:b62c6c2a-ab8f-44dd-9474-aa4d9d89b5bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00299.warc.gz"} |
Kindergarten Math Worksheets
Kindergarten Math Worksheets (For Ages 5 to 6)
These math worksheets are designed specifically for kindergartners using the national standards. They are helpful for students usually of age five to six. Students at this level are preparing for the
transition to first grade where higher level operations begin. First grade teachers are looking for students to have a mastery of counting to 100, basic comparison skills, and recognizing operations.
This set of worksheet should get you on your way. You will also find some excellent math worksheets specifically for this grade level at Little Worksheets.
What Do Students Learn in Kindergarten Math?
Kindergarten can be quite a formative experience for young children. It is the time in their lives they learn about other people and engage with their peers. It is also a preparatory time for young
children going to enter school. Math is an essential subject almost all kindergarten curricula contain. If you’re wondering what students learn in kindergarten math, you have come to the right place.
We'll help you understand as a teacher and parent what a student can expect after they have learned Kindergarten Math.
Kindergarten is quite a special time. However, the math skills your child will learn during this time are pretty basic. If they don't already have them by the time they are in kindergarten, you don't
have to worry, as teachers frequently frequent math topics in kindergarten.
Math Skills Students Learn in Kindergarten
Kindergarten is not supposed to be a time for the children's heavy stipulation in learning. It's quite a time for transformation for children as they learn to interact with their peers. However, some
of the math skills they learn are as follows:
Counting to 100
Your child may learn how to count to a hundred. Much of the work present in Math class in kindergarten centers itself around the identification of numbers. A student has to understand how to count
from 1 to 100. However, this is a slow process. It starts with counting from 1 to 10, then 10 to 20, and so onwards. In kindergarten, children will do much of the learning orally, so they will be
asked to have tests. You can also work on your child's oration skills within the confines of your home to maintain their learning even at home.
As a teacher, you may want to find creative ways to work on a student’s counting with them. You can come up with a song in which everyone counts down the numbers. You can also count the days of the
week every day so that students develop the habit of counting down the days. At the end of the 100 days, you can have a big party with your students to celebrate them counting down to a hundred.
Playing the "How Many" Game
Students in kindergarten must also understand how to group things and identify the number of items within a group. Students should be able to count objects one by one. It can be a group project where
the teacher presents them with a group of chocolate or pieces of ribbons, and they have to count the number of pieces they have. Assigning a number to each of the pieces in the group is called
one-on-one correspondence.
At home, you as a parent can apply play time to ask your child how many toys they are playing with. It will help your child keep track of the object they have. If they can't figure it out correctly,
work with them, and help them out. It's necessary to be gentle and not give the child any math anxiety.
Basic Addition and Subtraction
Even though children are not required to do any written work in kindergarten, there are still ways to identify essential addition and subtraction using objects. Teachers will often work with students
to help them with basic addition using numbers under 5 and then under 10.
The concept of addition is easy to understand for students when it is reiterated over a period. We also see that using physical objects such as lego blocks or paper can help children assign values
and numbers to pieces. They can then add and subtract as they learn.
You can work with your kindergartner using the same idea at home. You can work using Lego blocks or any other object. Your child will graduate to drawing addition and subtraction problems as time
passes. The goal at the end of kindergarten is for them also to be able to solve word problems.
Understand the Concept of Numbers Greater Than 10 or Greater Than 100.
Numbers 1-10 are not quite as tricky as explaining the concept of ten + 1 to a child. However, in kindergarten, teachers will often work on introducing the concept of eleven and twelve as 1+10 and
10+2, so they know what to keep at the base of each of their additions. The same is true for all the numbers greater than 20. The concept is to create patterns in the child's brain regarding the
connection between these numbers and how the vernacular and progression work.
A teacher may use a building block to compare how 11 is one more than 12. Comparison between numbers and knowing one is bigger than the other is also something the child must learn in kindergarten.
At home, you can revise the concepts with your child. You can do so by counting sheep with them or telling them stories about farms that have space for 13 animals instead of ten. The progression of
the numbers and them being termed "twenty-three" and so on will help elucidate to the children how numbers work.
Naming Shapes
Apart from understanding numbers, a kindergarten student will also have to understand shapes. There are many 2D and 3D shapes that teachers may introduce. Young students need to learn simple shapes
like circles, triangles, and squares are necessary to learn. They also need to be aware of spheres, cones, and cylinders. A teacher can demonstrate these shapes using everyday objects like water
bottles, birthday hats, balls, and diagrams.
As a parent, you can help your child identify shapes when you're out in the city. The windows on a bus are rectangular; the bed is rectangular; their mom's glasses are circular, etc. Doing these
activities with your child will reinforce the concept of shapes in their brains. It also helps them identify shapes in the classroom.
Final Thoughts
Most children's math in kindergarten is not advanced because the point is to build a concept surrounding these numbers. It's not about rote memorization but about giving children a way to understand
the sequence of numbers in a fun and entertaining way. | {"url":"https://www.teach-nology.com/worksheets/math/kindergarten/","timestamp":"2024-11-14T07:03:19Z","content_type":"text/html","content_length":"22277","record_id":"<urn:uuid:67ff67ad-b2ee-4ad5-8ab0-5b2eaff7dce3>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00813.warc.gz"} |
Nautical Leagues (International) to Exameters Converter
Enter Nautical Leagues (International)
β Switch toExameters to Nautical Leagues (International) Converter
How to use this Nautical Leagues (International) to Exameters Converter π €
Follow these steps to convert given length from the units of Nautical Leagues (International) to the units of Exameters.
1. Enter the input Nautical Leagues (International) value in the text field.
2. The calculator converts the given Nautical Leagues (International) into Exameters in realtime β using the conversion formula, and displays under the Exameters label. You do not need to click
any button. If the input changes, Exameters value is re-calculated, just like that.
3. You may copy the resulting Exameters value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Nautical Leagues (International) to Exameters?
The formula to convert given length from Nautical Leagues (International) to Exameters is:
Length[(Exameters)] = Length[(Nautical Leagues (International))] / 179985600000000
Substitute the given value of length in nautical leagues (international), i.e., Length[(Nautical Leagues (International))] in the above formula and simplify the right-hand side value. The resulting
value is the length in exameters, i.e., Length[(Exameters)].
Calculation will be done after you enter a valid input.
Consider that a luxury cruise ship sails 20 nautical leagues during a day at sea.
Convert this distance from nautical leagues to Exameters.
The length in nautical leagues (international) is:
Length[(Nautical Leagues (International))] = 20
The formula to convert length from nautical leagues (international) to exameters is:
Length[(Exameters)] = Length[(Nautical Leagues (International))] / 179985600000000
Substitute given weight Length[(Nautical Leagues (International))] = 20 in the above formula.
Length[(Exameters)] = 20 / 179985600000000
Length[(Exameters)] = 1.111e-13
Final Answer:
Therefore, 20 nautical league is equal to 1.111e-13 Em.
The length is 1.111e-13 Em, in exameters.
Consider that an exploration vessel navigates through 10 nautical leagues of ocean.
Convert this distance from nautical leagues to Exameters.
The length in nautical leagues (international) is:
Length[(Nautical Leagues (International))] = 10
The formula to convert length from nautical leagues (international) to exameters is:
Length[(Exameters)] = Length[(Nautical Leagues (International))] / 179985600000000
Substitute given weight Length[(Nautical Leagues (International))] = 10 in the above formula.
Length[(Exameters)] = 10 / 179985600000000
Length[(Exameters)] = 5.56e-14
Final Answer:
Therefore, 10 nautical league is equal to 5.56e-14 Em.
The length is 5.56e-14 Em, in exameters.
Nautical Leagues (International) to Exameters Conversion Table
The following table gives some of the most used conversions from Nautical Leagues (International) to Exameters.
Nautical Leagues (International) (nautical league) Exameters (Em)
0 nautical league 0 Em
1 nautical league 0 Em
2 nautical league 0 Em
3 nautical league 0 Em
4 nautical league 0 Em
5 nautical league 0 Em
6 nautical league 0 Em
7 nautical league 0 Em
8 nautical league 0 Em
9 nautical league 0 Em
10 nautical league 0 Em
20 nautical league 0 Em
50 nautical league 0 Em
100 nautical league 0 Em
1000 nautical league 1e-11 Em
10000 nautical league 6e-11 Em
100000 nautical league 5.6e-10 Em
Nautical Leagues (International)
A nautical league (international) is a unit of length used in maritime contexts. One nautical league is equivalent to 3 nautical miles, which is approximately 5,556 meters or 3.452 miles.
The nautical league is defined as three times the length of a nautical mile, based on the Earth's circumference and one minute of latitude.
Nautical leagues are used historically for measuring distances at sea. While not commonly used in modern navigation, they remain a part of maritime history and are occasionally referenced in
literature and older navigational texts.
An exameter (Em) is a unit of length in the International System of Units (SI). One exameter is equivalent to 1,000,000,000,000,000,000 meters or approximately 621,371,192,237,333,000 miles.
The exameter is defined as one quintillion meters, making it a measurement for extremely vast distances, often used in theoretical and cosmological contexts.
Exameters are used in fields such as astronomy and cosmology to describe distances on a scale larger than petameters. They offer a convenient way to express distances across immense regions of the
universe, such as the sizes of large cosmic structures or the scale of the observable universe.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Nautical Leagues (International) to Exameters in Length?
The formula to convert Nautical Leagues (International) to Exameters in Length is:
Nautical Leagues (International) / 179985600000000
2. Is this tool free or paid?
This Length conversion tool, which converts Nautical Leagues (International) to Exameters, is completely free to use.
3. How do I convert Length from Nautical Leagues (International) to Exameters?
To convert Length from Nautical Leagues (International) to Exameters, you can use the following formula:
Nautical Leagues (International) / 179985600000000
For example, if you have a value in Nautical Leagues (International), you substitute that value in place of Nautical Leagues (International) in the above formula, and solve the mathematical
expression to get the equivalent value in Exameters. | {"url":"https://convertonline.org/unit/?convert=nautical_leagues-exameters","timestamp":"2024-11-10T22:26:00Z","content_type":"text/html","content_length":"92162","record_id":"<urn:uuid:a0b4f8f8-5231-4bcd-8327-0ba1b5e6cce6>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00828.warc.gz"} |
Upper bound for uniquely decodable codes in a binary input N-user adder channel
The binary input N-user added channel models a communication media accessed simultaneously by N users. In this model each user transmits binary sequences and the channel's output on each bit slot
equals the sum of the corresponding N inputs. A uniquely decodable code for this channel is a set of N codes - a code for each of the N users - such that the receiver can determine all possible
combinations of transmitted codewords from their sum. Van-Tilborg presented a method for determining an upper bound on the size of a uniquely decodable code for the two-user binary adder channel. He
showed that for sufficiently large block length this combinatorial bound converges to the corresponding capacity region boundary. In the present work we use a similar method to derive an upper bound
on the size of a uniquely decodable code for the binary input N-user adder channel. The new combinatorial bound is iterative - i.e., the bound for the (N - 1)-user case can be obtained by projecting
the N-user bound on (N - 1) combinatorial variables and in particular it subsumes the two-user result. For sufficiently large block length the N-user bound converges to the capacity region boundary
of the binary input N-user adder channels.
Original language English
Title of host publication Proceedings of the 1993 IEEE International Symposium on Information Theory
Publisher Publ by IEEE
Pages 78
Number of pages 1
ISBN (Print) 0780308786
State Published - 1993
Externally published Yes
Event Proceedings of the 1993 IEEE International Symposium on Information Theory - San Antonio, TX, USA
Duration: 17 Jan 1993 → 22 Jan 1993
Publication series
Name Proceedings of the 1993 IEEE International Symposium on Information Theory
Conference Proceedings of the 1993 IEEE International Symposium on Information Theory
City San Antonio, TX, USA
Period 17/01/93 → 22/01/93
Dive into the research topics of 'Upper bound for uniquely decodable codes in a binary input N-user adder channel'. Together they form a unique fingerprint. | {"url":"https://cris.biu.ac.il/en/publications/upper-bound-for-uniquely-decodable-codes-in-a-binary-input-n-user-2","timestamp":"2024-11-04T11:10:58Z","content_type":"text/html","content_length":"54755","record_id":"<urn:uuid:c0109d47-ba0d-41a2-a5e2-324a4286a8be>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00503.warc.gz"} |
Dr. David Harrington
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada
I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.
I don't think there is any simple way. Since you want several matrices, GenerateMatrix won't work directly. You can use dcoeffs to find the pieces and then assemble them into the matrices.
@janhardo The DLMF (and the corresponding hardback book version) is produced by NIST (National Institute of Standards and Technology) in the U,S., and is the successor to Abramowitz and Stegun's,
"Handbook of Mathematical Functions"; it is considered an authoritative source, perhaps "the" authoritative source.
These differential equations are standard forms in the sense that they are often quoted, e.g., one speaks of "Bessel's equation" and there is a common way to write it, though of course there is not
universal consistency in these things.
But in terms of classifications in a text as I think you want, there is probably not much of a standard. I personally like Zwillinger's "Handbook of Differential Equations". Of course these
classifications exist because of different solving methods, but since the objective is often a special function, they are related to the other standard forms.
Maple' odeadvisor(verg) gives: [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
Since many of the second-order differential equations and their "special" functions are related through transformations to the Sturm-Liouville eqation, perhaps it is the standard form to rule them
all :-).
@MichaelVio Here's an example.
@salim-barzani I'm not sure I fully understand the order you want to do things or what you want to store, but perhaps this:
For Vectors, use a:=<2,1>, not <<2,1>>, which is a 2x1 Matrix. Then you can use "." for the dot product.
@Carl Love I think shape = symmetrix or shape = hermitian change the algorithm for float matrices to produce only real eigenvalues and eigenvectors but have no effect in the symbolic case.
@rzarouf You're welcome. You found a Maple bug, and I have submitted an SCR (bug report), so it is fixed in future versions.
@Carl Love The method is fine, but M^%H.M is not symmetric, so using shape=symmetric forces the offdiagonals equal. So you should omit that.
@Scot Gould Although @Rouben Rostamian has clarified that the any Robin boundaries can be solved, it seems that Maple finds sometimes the right answer consistent for the initial condition given and
sometimes not. Your first solution was already compatible for the right-hand boundary (at 50 points it is pretty close on almost all the right half of the domain). So I knew to play around with the
left-hand one, and used an analogy with the solutions that I was most familar with to figure out that sign change. I usually solve cases that evolve toward steady state, so I added my bias there. But
maybe Maple is also consistently solving some classes of boundary conditions; it would help if we know more clearly what assumptions it is making. In this respect, @nm's suggestion to put in the
initial solution in afterwards might be counterproductive; maybe Maple is not returning a solution because it knows it can't solve the case compatible with that initial conditions. I suspect very
specific assumptions on signs of parameters at the pdsolve stage might be helpful.
I don't think the explanation on the Wikipedia page is that helpful; one needs to think of the relative signs to achieve the right physical outcome. I deliberately kept the fluxes the same sign to
get to a steady-state, then adjusted the other signs. The general steady state solution is a straight line. If the slope of that is positive, the flux is negative and there will be the same heat flux
coming in the right-hand side as going out the left-hand side at steady state. So if the initial fluxes are not equal but of the correct sign, then the other parts of the Robin boundary condition
have to allow the slopes to evolve to be the same (assuming you want steady-state of course). In your case the fluxes evolve to be zero.
I didn't really want to use all that opcrap to extract the pieces and did try to do it without a loop first, but I also didn't want to spend to much time optimizing it. Make a procedure that
determines whether it is an eigenvalue solution without an analytical solution for the eigenvalues might be worthwhile, but in that case I would probably solve it "by hand". Actually with Maple that
is not too difficult. If you call pdsolve with HINT=f(t)*g(x), then extracting the ode for t and x and relating them to the eigenvalue is straightfoward. The second-order ode in x can be solved for
{g(x),lambda} and if there is an analytical formula for lambda, Maple will find it; otherwise you have to find the eigenvalues numerically.
I should say, I almost never solve the heat/diffusion equation by this eigenvalue method, because the error functions and x/sqrt(k*t) usually found don't easily come out of it. Instead I find the
solution by Laplace transforms. (To get the general solution containing x/sqrt(k*t) you need to use SimilaritySolutions.)
@Rouben Rostamian Thanks for the clarification. (Yes, I generally expect solutions to evolve toward a steady state.)
@Teep "Do you think that it is possible to extract a general expression for the roots of f (or g) in terms of a and b?" No, if Maple could find that it would not have returned the RootOf, which is
the "next best thing" when it can't find an explicit solution. Sometimes it is possible to manually rearrange things to help Maple out, but I think with a non-integer b in the exponent, it is likely
an intractable problem. Even for the integer b case, polynomials of high degree do not usually have explicit symbolic solutions.
@Saalehorizontale As you expect, a RootOf(..., 0.35..0.36) or RootOf(...,index=2) corresponds to a single root. Unfortunately, as you already saw, RootOf without an index or range or label sometimes
means any/all of the roots and sometimes means the first one. Solving a polynomial gives a RootOf that says all possible roots work, and can be found by allvalues, but evalf just produces one. Other
commands, particularly those where a RootOf stands for an algebraic number in a field mean also the first one, e.g. evala(Minpoly(..)).
I didn't check the timings, but it's not unexpected that figuring out the actual numerical values of the roots is slow. On the other hand solving for polynomial roots is optimized compared to roots
of more general equation, e.g., fsolve(polynomial in x,x,complex) will quickly return all roots.
@dharr I edited the above to include floor(root(evalf(n), 3)). In floor(root(n, 3)), root(n,3) just gives n^(1/3) and floor does all the work.
@GFY You can solve it by narrowing the ranges, with guidance of a plot. You could modify the equations so a__3=0,c__3=0 is not a solution by factoring and dividing through by power(s) of a__3 and
@GFY That was unexpected for me. Solve is really for symbolic solutions, and you have floats in your equations. I always formulate equations symbolically, and try solve. Then one can substitute
values for the parameters, e.g., params:= {a=3.1, b=2.1 etc}, then eval(eqn, params);
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BEGIN:VCALENDAR VERSION:2.0 PRODID:ILLC Website X-WR-TIMEZONE:Europe/Amsterdam BEGIN:VTIMEZONE TZID:Europe/Amsterdam X-LIC-LOCATION:Europe/Amsterdam BEGIN:DAYLIGHT TZOFFSETFROM:+0100 TZOFFSETTO:+0200
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=10;BYDAY=-1SU END:STANDARD END:VTIMEZONE BEGIN:VEVENT UID:/NewsandEvents/Archives/2012/newsitem/4216/24- 27-June-2012-Nineth-International-Conference-on-Co
mputability-and-Complexity-in-Analysis-CCA-2012-Ca mbridge-U-K- DTSTAMP:20111130T000000 SUMMARY:Nineth International Conference on Computa bility and Complexity in Analysis (CCA 2012), Camb ridge,
U.K. DTSTART;VALUE=DATE:20120624 DTEND;VALUE=DATE:20120627 LOCATION:Cambridge, U.K. DESCRIPTION:Computability and complexity theory ar e two central areas of research in mathematical lo gic and
theoretical computer science. Computabilit y theory is the study of the limitations and abili ties of computers in principle. Computational comp lexity theory provides a framework for understandi ng
the cost of solving computational problems, as measured by the requirement for resources such as time and space. The classical approach in these a reas is to consider algorithms as operating on fin
ite strings of symbols from a finite alphabet. Mos t mathematical models in physics and engineering, however, are based on the real number concept. Thu s, a computability theory and a complexity
theory over the real numbers and over more general contin uous data structures is needed. Despite remarkable progress in recent years many important fundament al problems have not yet been studied,
and presuma bly numerous unexpected and surprising results are waiting to be detected. The conference provides a unique opportunity for people working in the area of computation on real-valued data
but coming fro m different fields to meet, present work in progre ss and exchange ideas and knowledge. Conference W eb Page: http://cca-net.de/cca2012/ Authors of c ontributed papers are invited to
submit a PDF vers ion of an extended abstract. Submission deadline: April 1, 2012 X-ALT-DESC;FMTTYPE=text/html:
Compu tability and complexity theory are two central are as of\n research in mathematical logic and th eoretical computer\n science. Computability t heory is the study of the limitations\n and a
bilities of computers in principle. Computational\ n complexity theory provides a framework for understanding the\n cost of solving computati onal problems, as measured by the\n requireme nt for
resources such as time and space.
The classical approach in these areas is to c onsider\n algorithms as operating on finite s trings of symbols from a\n finite alphabet. M ost mathematical models in physics and\n engi neering,
however, are based on the real number\n concept. Thus, a computability theory and a co mplexity theory\n over the real numbers and o ver more general continuous data\n structures is needed. Despite
remarkable progress in recent\ n years many important fundamental problems h ave not yet been\n studied, and presumably nu merous unexpected and surprising\n results ar e waiting to be detected. The
conference provides a\n unique opportunity for people working in the area of computation\n on real-valued data but coming from different fields to meet,\n present work in progress and exchange ideas
and kn owledge.
\n \n \n
Conference We b Page: http://cca-net.de/cca2012/\n < /p>
\n Authors of cont ributed papers are invited to submit a PDF\n version of an extended abstract. Submission dea dline: April\n 1, 2012\n
\n < /div> URL:/NewsandEvents/Archives/2012/newsitem/4216/24- 27-June-2012-Nineth-International-Conference-on-Co mputability-and-Complexity-in-Analysis-CCA-2012-Ca mbridge-U-K- END:VEVENT | {"url":"https://www.illc.uva.nl/NewsandEvents/Events/Conferences/newsitem/4216/24-27-June-2012-Nineth-International-Conference-on-Computability-and-Complexity-in-Analysis-CCA-2012-Cambridge-U-K-?displayMode=ical","timestamp":"2024-11-10T22:15:33Z","content_type":"text/calendar","content_length":"4895","record_id":"<urn:uuid:4e7dcc95-9fdb-450f-bfa2-6abdb2f79bfb>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00414.warc.gz"} |
Randomly Assigning Names to Items
Written by Allen Wyatt (last updated November 4, 2023)
This tip applies to Excel 2007, 2010, 2013, 2016, 2019, Excel in Microsoft 365, and 2021
Gary has two lists in a worksheet. One of them, in column A, contains a list of surplus items in our company and the other, in column G, contains a list of names. There is nothing in columns B:F.
Gary would like to assign names, randomly, to the list of items. Each name from column G should be assigned only once. If there are more names than items, then some names won't get used. If there are
fewer names than items, then some items won't have associated names.
There are a couple of ways that this can be done. Perhaps the easiest, though, is to simply assign a random number to each item in column A. Assuming that the first item is in cell A1, put the
following in cell B1:
Double-click the fill handle in cell B1, and you should end up with a random number (between 0 and 1) to the right of each item in column A.
Now, select all the cells in column B and press Ctrl+C to copy them to the Clipboard. Use Paste Special to paste values right back into those cells in column B. (This converts the cells from formulas
to actual static values.)
Sort columns A and B in ascending order based on the values in column B. If you look across the rows, you'll now have items (column A) associated randomly with a name (column G).
Even though it is not necessary, you could also follow these same steps to add a random number to the right of each name and then sort the names. (I say it isn't necessary because randomizing the
items should be enough to assure that there are random items associated with each name.)
The technique discussed so far works great if you have to do the random pairing only once in a while. If you need to do it quite often, then a macro may be a better approach. There are, of course,
many different macro-based approaches you could use. The following approach assumes the item list is in column A and the name list in column G. It also assumes that there are header cells in row 1
for each column.
Sub AssignNames()
Set srItems = Range("A2").CurrentRegion
Set srNames = Range("G2").CurrentRegion
NameCount = srItems.Rows.Count - 1
ItemCount = srNames.Rows.Count - 1
'Randomize Names
ReDim tempArray(NameCount, 2)
For x = 0 To NameCount - 1
tempArray(x, 0) = Range("G2").Offset(x, 0)
tempArray(x, 1) = Rnd()
Next x
'Bubble Sort
For i = 0 To NameCount - 2
For j = i To NameCount - 1
If tempArray(i, 1) > tempArray(j, 1) Then
tempItem = tempArray(j, 0)
tempName = tempArray(j, 1)
tempArray(j, 0) = tempArray(i, 0)
tempArray(j, 1) = tempArray(i, 1)
tempArray(i, 0) = tempItem
tempArray(i, 1) = tempName
End If
Next j
Next i
Range("B2") = "Assigned"
AssignCount = NameCount
If NameCount > ItemCount Then AssignCount = ItemCount
For x = 0 To AssignCount
Range("B2").Offset(x, 0) = tempArray(x, 0)
Next x
End Sub
If there are more names than items the macro randomly assigns names to items. If there are more items than names it randomly assigns some items to names and randomly leaves "holes" (items without
names). It stores them in column B, overwriting whatever was there.
ExcelTips is your source for cost-effective Microsoft Excel training. This tip (5682) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, Excel in Microsoft 365, and 2021.
2023-11-26 15:50:39
J. Woolley
The AssignNames5 macro in my most recent comment below randomly shuffles the List array. My Excel Toolbox now includes the following function to return a range or array with all rows and/or columns
randomly shuffled:
=Shuffle(RangeArray, [RowsOnly], [ColsOnly])
If RowsOnly is True, columns will remain with their shuffled rows.
If ColsOnly is True, rows will remain with their shuffled columns.
Default for both RowsOnly and ColsOnly is FALSE. An error is returned if both are TRUE; otherwise, they apply only when RangeArray is 2D.
See https://sites.google.com/view/MyExcelToolbox/
2023-11-24 12:25:19
J. Woolley
Re. the AssignNamesX macros in my three previous comments below, here is yet another version. This one is the simplest.
Sub AssignNames5()
Dim ItemCount, NameCount, ListCount, List, Temp, i, n
If Range("A2") = "" Or Range("G2") = "" Then Exit Sub
ItemCount = Range("A2").End(xlDown).Row - 1
NameCount = Range("G2").End(xlDown).Row - 1
ListCount = IIf(NameCount < ItemCount, ItemCount, NameCount)
List = Range("G2").Resize(ListCount).Value '2D base-1 column of names
List = WorksheetFunction.Transpose(List) '1D base-1 array (row vector)
'When more items than names, assign blank names
For i = NameCount + 1 To ListCount 'Skipped if ListCount = NameCount
List(i) = ""
Next i
'Randomly shuffle the values in List array
For i = 1 To ListCount
n = Int((ListCount * Rnd) + 1) 'Random number 1 to ListCount
Temp = List(i)
List(i) = List(n)
List(n) = Temp
Next i
'Assign randomized values from List to items
For i = 1 To ItemCount
Range("B1").Offset(i) = List(i)
Next i
'Clear any remainder plus one cell
For i = ItemCount + 1 To ListCount + 1
Range("B1").Offset(i) = ""
Next i
End Sub
See http://www.cpearson.com/excel/ShuffleArray.aspx
2023-11-13 10:01:34
J. Woolley
Re. the AssignNames3 macro in my previous comment below, here is yet another version. This one uses a SortedList that is automatically sorted by its random number Keys.
Sub AssignNames4()
Const REPEAT_NAMES = False 'True to repeat names; False to assign blanks
Dim ItemCount, NameCount, ListCount, Key, i, n
If Range("A2") = "" Or Range("G2") = "" Then Exit Sub
ItemCount = Range("A2").End(xlDown).Row - 1
NameCount = Range("G2").End(xlDown).Row - 1
ListCount = IIf(NameCount < ItemCount, ItemCount, NameCount)
Dim List As Object
Set List = CreateObject("System.Collections.SortedList")
'Associate each name with a random number Key in List sorted by Key
For i = 1 To ListCount
Key = Rnd 'Random number 0 to 1
Loop While List.ContainsKey(Key) 'Avoid duplicate Key
If i > NameCount Then
If REPEAT_NAMES Then
'When more items than names, randomly repeat names
n = Int((NameCount * Rnd) + 1) 'Random number 1 to NameCount
List.Add Key, Range("G1").Offset(n)
'When more items than names, assign blank names
List.Add Key, ""
End If
Else 'Next name in column G
List.Add Key, Range("G1").Offset(i)
End If
Next i
'Assign randomized names (List is base-0 sorted by random number Keys)
For i = 1 To ItemCount
Range("B1").Offset(i) = List.GetByIndex(i - 1)
Next i
'Clear any remainder plus one cell
For i = ItemCount + 1 To ListCount + 1
Range("B1").Offset(i) = ""
Next i
End Sub
If there is any difficulty with the following statement
Set List = CreateObject("System.Collections.SortedList")
enable Microsoft .NET Framework 3.5 using Control Panel as described here:
2023-11-08 14:59:22
J. Woolley
Re. the AssignNames2 macro in my previous comment below, if the list of names (NameCount) is less than the list of items (ItemCount), then random names are assigned to the first NameCount items and
blank names are assigned to the remaining items. The following AssignNames3 version assigns the blank names for this case randomly as suggested in the Tip's last paragraph.
Sub AssignNames3()
Const REPEAT_NAMES = False 'True to repeat names; False to assign blanks
Dim ItemCount, NameCount, NumberCount, i, j, k, temp
If Range("A2") = "" Or Range("G2") = "" Then Exit Sub
ItemCount = Range("A2").End(xlDown).Row - 1
NameCount = Range("G2").End(xlDown).Row - 1
NumberCount = IIf(NameCount < ItemCount, ItemCount, NameCount)
'Associate a random number with each name
ReDim tempArray(1 To NumberCount, 1 To 2)
For i = 1 To NumberCount
If i > NameCount Then
If REPEAT_NAMES Then
'When more items than names, randomly repeat names
k = Int((NameCount * Rnd) + 1) 'Random number 1 to NameCount
tempArray(i, 1) = Range("G1").Offset(k)
'When more items than names, assign blank names
tempArray(i, 1) = ""
End If
tempArray(i, 1) = Range("G1").Offset(i) 'Name from list
End If
tempArray(i, 2) = Rnd 'Random number 0 to 1
Next i
'Bubble sort random number values to randomize associated names
For i = 1 To NumberCount - 1
For j = i + 1 To NumberCount
If tempArray(i, 2) > tempArray(j, 2) Then
For k = 1 To 2
temp = tempArray(j, k)
tempArray(j, k) = tempArray(i, k)
tempArray(i, k) = temp
Next k
End If
Next j
Next i
'Assign randomized names
For i = 1 To ItemCount
Range("B1").Offset(i) = tempArray(i, 1)
Next i
'Clear any remainder plus one cell
For i = ItemCount + 1 To NumberCount + 1
Range("B1").Offset(i) = ""
Next i
End Sub
Notice the method for randomizing names assures a name is only assigned once as specified in the Tip's first paragraph. But if the following statement
Const REPEAT_NAMES = False 'True to repeat names; False to assign blanks
is replaced by this statement
Const REPEAT_NAMES = True 'True to repeat names; False to assign blanks
then when NameCount is less than ItemCount, names will be repeated randomly to avoid assigning any blank names to the remaining items. REPEAT_NAMES is ignored if NameCount >= ItemCount.
2023-11-06 10:10:39
J. Woolley
The Tip's macro has several issues. Here is an improved version:
Sub AssignNames2()
Dim ItemCount, NameCount, AssignCount, i, j, k, temp
If Range("A2") = "" Or Range("G2") = "" Then Exit Sub
ItemCount = Range("A2").End(xlDown).Row - 1
NameCount = Range("G2").End(xlDown).Row - 1
'Randomize Names
ReDim tempArray(1 To NameCount, 1 To 2)
For i = 1 To NameCount
tempArray(i, 1) = Range("G1").Offset(i)
tempArray(i, 2) = Rnd()
Next i
'Bubble Sort
For i = 1 To NameCount - 1
For j = i + 1 To NameCount
If tempArray(i, 2) > tempArray(j, 2) Then
For k = 1 To 2
temp = tempArray(j, k)
tempArray(j, k) = tempArray(i, k)
tempArray(i, k) = temp
Next k
End If
Next j
Next i
AssignCount = IIf(NameCount > ItemCount, ItemCount, NameCount)
For i = 1 To AssignCount
Range("B1").Offset(i) = tempArray(i, 1)
Next i
k = IIf(NameCount < ItemCount, ItemCount, NameCount)
For i = AssignCount + 1 To k + 1
Range("B1").Offset(i) = ""
Next i
End Sub
This macro assumes the following:
1. Row 1 has column headings assigned by the user.
2. The list of items is from cell A2 to the first blank cell in column A.
3. The list of names is from cell G2 to the first blank cell in column G.
4. The list of names assigned to items is in column B starting at cell B2 overwriting any previous content including blank cells down to one row below the list of items and names. | {"url":"https://excelribbon.tips.net/T005682_Randomly_Assigning_Names_to_Items.html","timestamp":"2024-11-08T23:59:23Z","content_type":"text/html","content_length":"60253","record_id":"<urn:uuid:326e4b47-bc7b-46d9-9db3-967c010cfad4>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00302.warc.gz"} |
Stroke Length Loyalty
by Coach Mat | 21 Mar, 2015 | Attention/Focus, Coaching, Competitive/Performance Training, Distance Training, Energy Management, Motor Learning, No Progress, Pace Combinations, Racing Skills, Speed
and Pace, Speed Math, SPL, Stroke Length (SL), Sprint Training, Swimmer Stories, Technology and Tools, Tempo, Stroke Rate (SR), Training Theory, Triathlon Swimming | 10 comments
When you start feeling fatigue in your race (or while ‘just swimming’) you are urged to slow down. At some point you finally must give in. But when you do what is changing in your stroke to slow you
SPL x Tempo = Pace (the inverse of the Speed Equation = SL x SR)
When you to slow down from fatigue what changed in this equation for you? What does your brain want to hold on to when the going gets tough? What does it want to let go control of? Are you loyal to
holding stroke length and sacrifice tempo? Or do you sacrifice stroke length and hold on to tempo? Or neither? Or “I don’t know!“?
Test Swim
Here is a test you can conduct to check your level of development for pace control. This test swim will challenge your ability to hold Pace (= a certain SPL x Tempo combination) at some point during
the test swim.
For this test swim you will need to count strokes on every lap and use a Tempo Trainer to help you hold tempo precisely.
You may pick an SPL N that is within your comfortable conditioned SPL range and a comfortably fast Tempo T you can hold with that SPL, but doubtful whether you can hold it the whole test swim.
For example, Roberto has a conditioned SPL range of 18-20 strokes (in a 25m pool). He is going to assign himself to do this test swim with 19 SPL. And, from experience using a Tempo Trainer he knows
he has a comfortable center Tempo of 1.32 (meaning he could hold this tempo all day), and the fast-end boundary of his comfortable tempo range is about 1.25 (which he can just hold for about 200m
without feeling like his stroke quality gets out of control). So, Roberto will set his parameters at 19 SPL x 1.25 second tempo.
You may set up a test swim like this, and adjust the distance of each repeat to suit yourself:
4x to 8x 100 meters with 6 nasal breath rest between each. Do as many as you need to in order to start feeling considerable stress on your ability to hold SPL x Tempo together.
Your assignment is to do your best to hold both SPL x Tempo consistently, over every single length of the test swim. But this is not the purpose of the test. You want to find a failure point in your
ability to hold that SPL x Tempo combo – that point where you begin to feel it is too difficult to hold both SPL and Tempo at the same time. Then you will observe in your own body which one your
brain is more loyal to – holding SPL and give up Tempo (let is slow down), or hold Tempo and give up SPL (add strokes). Making observations at this failure point is the purpose of the test.
The observations to make:
1. Which did I feel more loyal to under pressure (in the habit sense)- SPL or Tempo?
2. How did it start to fail? Did I feel like I was losing strength to hold that stroke length? Or did my arms start to feel sluggish and I couldn’t move them through the stroke cycle as fast?
3. What did my attention do in reaction? What did I try to do to compensate for fatigue? What did I focus on? What commands did I give to certain parts of my body?
Here is an arbitrary grading system:
• A – maintaining both SPL and Tempo (holding pace)
• B – maintaining SPL N but allowing Tempo to slip <0.05 seconds (dropping pace less than 1 sec/25m)
• C – holding Tempo, but allowing SPL to slip to N+1 (dropping pace by over 1 sec/25m)
• D – holding Tempo, but allowing SPL to slip to >N+1 (dropping pace by more than 2 sec/25m)
• F – losing control over both SPL (any amount) and Tempo (dropping pace a lot!)
Grade yourself on each repeat. Stop when you get a D or F on a repeat.
Note: You could have allowed SPL to slip to N+1, but increased (speed up) Tempo T – ~0.06 seconds to compensate and maintain pace, but this would have changed two variables at once and then taken you
away from the purpose of this test swim. I want to acknowledge that as a legitimate ‘holding-pace’ solution, but one that does not apply to this test scenario. If you can do that already
(intentionally) then you need a more challenging test for your skills.
You may notice in this grading system that maintaining SPL is given higher value than maintaining Tempo. It is based on this understanding: that what counts for more in a tough race is your ability
to maintain consistent traction on the water (= stroke length), not how fast you can spin your wheels. Tempo is important, but secondary to stroke length. Tempo is a function of stroke length, not a
replacement for lack of control over it. Only those who first have control over stroke length can expect to go faster if they increase tempo, even a little bit. Meanwhile, those without control over
stroke length take the advice to up the tempo, then find they don’t go faster yet get tired even more quickly. The difference between the two is the ability to preserve stroke length as tempo
increases. And to minimize heart rate increase, the swimmer has to hold that stroke length with finesse rather than by adding more power – an uncommon but necessary approach. If you practice tempo
increases before you have practiced stroke length control, don’t be surprised when you don’t get faster.
The importance of stroke length control is painfully apparent in this famous (or infamous, depending on who you were rooting for) 2008 Olympic Men’s 400m Relay Race where the French squad was
overtaken and blown away by the American squad. After the third leg France had the race in the bag, but inferior stroke control on the part of the final French swimmer, against superior stroke
control on the part of the final American swimmer flipped the results. You can see that speeding up tempo did not save the French swimmer – he actually slowed down, while the American was speeding up
using a visibly slower tempo. Superior loyalty to (a well chosen) stroke length is the way I would describe the cause of this result.
Under-developed swimmers (the ones that lose to the best) are chiefly identified by their loss of stroke length toward the end of a race, not by their tempo (whether high, low, or erratic). Tempo
only has meaning when it is connected to a steady stroke length. (Take any aqua-aerobics class, have them spin their arms as fast as they can and watch how fast they move through the water as a
result!) Adapting to faster tempo is relatively easy to do (with a neurologically effective approach) while developing optimal stroke length – by superior form, rather than by additional power – and
then ‘hard-wiring’ it into the muscle memory is much harder to do. The final part of is to combine the two, stroke length and tempo. However, at nearly any pool you may visit the ubiquitous bias of
swimmers (and less-than-elite coaches) is to focus so much on (easy-to-learn-and-hold) tempo work, and neglect (hard-to-learn-and-hold) stroke length control work. Now that I’ve pointed out the
irrational bias, you have a chance to work against it to your swimming benefit.
This test above is meant to help you see what you are currently wired to be loyal to: when the going gets tougher than you can handle is your brain hardwired to remain loyal to holding stroke length?
Or loyal to holding tempo? Or neither?
A highly developed swimmer will feel energy resources getting strained (like anyone), then double her concentration upon the points in her body position and stroke movements that she knows are most
vulnerable to degradation under stress. By this she will delay the speed-reducing effects of fatigue. Then she will make a calculated (and trained for) shift her SPL x Tempo combination to a precise
trade-off (for example from ‘N x T’ to ‘N+1 x T-0.06’) to further delay speed-reducing effects of fatigue. The basis of her control over pace is her control over stroke length. Tempo is adjusted in
relation to it and her energy supply. If she trained for even splits, she will adjust the combinations (like gears on her tri bike) so that she can maintain pace as muscles fatigue, or as conditions
(in open-water, for example) change. If she has trained for negative splits she will start the race with a carefully calculated pace combination so that when she is ready to crank up the speed, she
makes her pre-planned (i.e. trained-for) shift – she will either hold stroke length while increasing tempo (more likely), or lengthen stroke and hold tempo steady (harder to do, but possible).
Meanwhile, her under-developed competitors will feel fatigue, lose body position and stroke control, lose stroke length because of it, feel the speed drop, then start cranking up tempo in a futile
attempt to compensate, which will deplete energy faster, which triggers more lose of control, which will shorten stroke length further, which will trigger the swimmer to crank up tempo even more. The
vicious negative spiral.
This is how we might describe what we saw in Sun Yang’s 2011 World Championships WR 1500m Race (and the subsequent WRs), compared to the swimmers who dropped way, way behind at the end. No one
thought he was even close to world-record pace until he unleashed the energy he conserved in the first 14oo meters from an exquisitely maintained SPL x Tempo combination.
Let’s think about the results of your test swim a bit more.
If you’ve set your starting SPL x Tempo just right you should be able to do the first 1 to 3 of these 100m repeats at Grade A level. But at some point in the series things will start to feel tough.
Conditions are changing inside the body. Two main areas to notice: 1) power supply is getting restricted (feeling tired, then feeling exhausted), and 2) attention is getting strained (not knowing
what to focus on, not knowing how to counter the exhaustion).
What most swimmers assume has happened is just #1 – “I am just getting tired therefore I can’t swim as fast any more.” The premature and often erroneous assumption is: “Oh, I just need to train
harder and build up more power and endurance so I don’t feel tired so quickly.” What ‘I can’t swim as fast anymore’ actually means in physics is: my SPL and/or Tempo collapsed. But I propose that #2
is actually the primary failure, not #1. I argue that this perceived energy- failure point can be postponed a lot longer than a swimmer realizes, if he trains his attention to recognize what is prone
to fail in the stroke and train specifically to postpone that failure, and even rise above it.
I suspect that more often than realized, the beginning of the fatigue-inducing problem is that the swimmer has lost attention, and therefore lost control over some part of the body position and
stroke control that is vulnerable to degradation, which immediately provokes higher drag, which immediately provokes higher rate of energy loss, which immediately makes the swimmer notice himself
slowing down with no ability to resist, which immediately distracts his mind from resuming concentration on those parts of the body position and stroke that started to degrade and cause more drag.
Only by breaking the negative cycle at the point of attention loss, can the swimmer regain composure and control over drag increase/energy loss cycle, to delay or even avoid the speed-reducing
effects of fatigue.
The fact is, at some point in the test swim (or, in the race) energy supply will diminish, and the swimmer must carry on with a stroke that fits within that lower-energy condition. The question is
then: how are you going to compose the stroke with less energy available now? What will you preserve? What will you sacrifice to fit within the budget? Have you ever even thought about this before in
this way?
An under-developed swimmer, when less energy is available, will just let the stroke collapse how ever it collapses by default and apply more effort in less productive directions (the land-mammal
instincts take over under stress!). However, at the point you can no longer resist the slow-down, you can and should train to sacrifice certain less-valuable-for-speed parts of the stroke cycle in
order to conserve energy and fit the stroke within the smaller budget. Regardless of your level of fitness, if you are aiming to race at the peak of your current potential you are going to enter this
lower-energy zone during your event. Why not learn to manage your stroke within that zone better and get more out of it?
So, the purpose of this test swim is to get you to practice noticing what is happening in your brain and body so you know what you need to train for. You think, you plan, then you train for it in
practice so you don’t have to think about it during the race – this is what it means to build your aquatic mammal instinct. By deliberate practice it will become an easily triggered response (if not
totally automatic) and allow you to keep your brain focused on making higher order decisions when your competitors can’t. Being aware of what is happening in your body, then understanding why, is the
first step on the path to improving your abilities – in this case, to improving your ability to hold a consistent pace over your entire race distance, even when fatigue sets in. At peak performance
you will not avoid fatigue, but you can train to have a superior response to it than your competitors do.
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10 Comments
Isaac Ohel on March 22, 2015 at 23:46
Hello Coach
I like your scientific approach to swimming. However, this time, the science behind your recommendation is not clear.
Why would a lower SPL and longer tempo (keeping the same pace), be more efficient than the opposite trade off?
I can think of several answers to either side of this question. Can you point to any reference?
On a more personal note. I can complete (with some struggle) one pool in 18 strokes, but the tempo I would need to complete 200 meters would be impossibly long. Instead, I shorten the tempo to
1.2 sec. and increase my SPL to 25. I work on reducing SPL from that starting condition. To maintain this pace even at 20 SPL, I would need to lengthen the tempo to 1.46 sec. At this tempo,
bilateral breathing is difficult for me.
What would be your suggestion?
Please note that I would like, at least some of my swimming to be longer than 25 meters,
Mat Hudson on March 23, 2015 at 20:37
Hi Isaac,
On your first question…
I really appreciate it when someone asks a question like this – if I haven’t made it clear it is good to know about it, and I take seriously the possibility I could have gotten something
wrong in my own understanding. So I do review my own thinking at each challenge.
The first thing I try to be repeatedly clear about – each swimmer needs to aim for an optimal stroke length, not the longest stroke length. This is what we are after. However, you will see
how I often advocate for a longer stroke because, by far, most less-than-elite swimmers we see (including high level triathletes) are using what we would regard as shorter-than-optimal length
strokes, which suggests they are suffering from excess drag. We use our Height/SPL Index (on our Resources page) to estimate where a swimmer should be.
But only through experimentation (and some understanding how we may detect signs of a more suitable stroke length before our brain finds it easy to produce that SL) do we come to some
assurance of what is really best personally.
I am tempted to write a bunch more here, but I risk wasting time if I haven’t understood you accurately, or I may have the info you need elsewhere. Maybe you can take a look at these posts to
see if I address the questions:
If not, send me another set of questions and I will make up for the gap in explanations (or correct myself).
I am going to delay answering the second question until tomorrow (time to put the kids to bed!).
Isaac on March 24, 2015 at 17:33
Hello Coach
Thanks for your response. I read the articles for the second time (first, when they were published), and gained new insights. However they did not provide a direct answer to my question.
“Is a longer stroke more efficient?”
More efficient = Less energy expended for the same pace
I agree with you that eventually, my question will be answered by my own body, and that your articles provide a good path to get there. However, being an ex-engineer, I can’t avoid
engaging my brain.
To clarify, let me describe three scenarios, and a rephrase of my question.
1. An accomplished swimmer (not necessary Sun Yang) swims at a good SPL and tempo.
2. He increases the tempo by 10% and decreases his SPL to maintain the same pace as before (an endless pool would help).
3. He decreases the tempo by 10% and increases his SPL to maintain the same pace.
Question: Which of the three scenarios would be most/least efficient, and why?
Is that a useful question to pursue?
Mat Hudson on March 24, 2015 at 21:52
Hi Isaac,
You pushed my answer button (which I am happy to have pushed!) and now I am going to deliver a response in another blog post. I actually had the first one planned before you wrote
your first comment, and now it seems even more timely. Then I will follow up with a second post a little later which will include your questions and my thoughts on it. Both are
written, but I like to space them out by a few days if I can.
I am glad you insist, or cannot resist using your engineer brain to think about these things. And I really want folks like you to hold me accountable to offer solid, satisfying
answers. Teaching/writing as much about my own process of learning and refinement in understanding also.
Mat Hudson on March 24, 2015 at 23:02
Hi Isaac, I will respond to the second part here in the comment section. If your instinct is to produce faster tempo by increasing power (which is the instinct for virtually every human
swimmer who has not been taught there is an alternative) then you can understandably feel a barrier at a certain tempo. But you can realize that power is not going to solve this goal/problem.
If you shorten your stroke too much, then you will be short-changing how much momentum you can glean off of each stroke. The longer your vessel the easier it slides. Shorten it and your
vessel doesn’t glide as far for each unit of force. There is a careful balance there we are trying to strike between an SPL that is getting a decent return on your wingspan (55-70% of
wingspan is our estimate) for each stroke. Too long of a stroke and you get high acceleration/deceleration curves – a lot of strain on the shoulders.
First, you can use different bi-lateral breathing patterns to get a more suitable average breathing cycle at any tempo – like 2-3-2-3. Or 3-4-3-4. At anaerobic threshold I will be using 2
breath cycle (and switch to the other side occasionally).
Second, there is a way to adapt to faster tempos – at least to adapt to faster tempos than you realize you could reach, with relatively minimal increase on HR. At some point, depending on
your current condition and the speed you want to achieve, you will have to add more power, but this process of adaptation allows your to develop precision as you go so that you will actually
be able to maintain controlled SPL as you increase Tempo. Think of it like how a martial artist would train so that time slows down and his opponent appears to be moving in slow motion. Or
how a major league baseball player would train in a batting cage to handle progressively faster and faster balls flying at him. We train in a way that allows you to build this kind of control
over your stroke at higher tempos. But with the unique complications of doing it in water (and without dangerous things flying at our faces!).
Either way you go, work down in tempo at 25 SPL, or work down in tempo at 18 SPL and you will hit a muscle control barrier at some tempo, if you do it by increasing power. You’ll need a
different approach eventually. So, for 200m sprint work, I suggest you aim for an SPL within the high side of your Green Zone (on the Height/SPL Index in our Resources page).
When you read this essay https://mediterraswim.com/2014/05/12/metrics-103-pace-construction/ did it make any sense? I need some feedback. It seems clear (though not simple) in my head, but I
don’t know if I am making it easier to understand or not in typing. It’s one thing to stand next to someone in the pool and teach this directly to one’s nervous system, but another to teach
it through text.
In an email, send me your height (or wingspan) if you know it, your age, and any other relevant factors to your swimming physiology, and I will send you a few more thoughts about how you
might approach your goal.
Mat Hudson on March 24, 2015 at 23:08
And the thing to add that I just remembered. If you go back to 25 SPL (easier) then work down in Tempo, then try to move down in SPL at that higher tempo – it’s like trying to do road
reconstruction during rush hour. You’ll essentially be firing more powerful signal, more frequently down the circuits that you will then be trying to overhaul to lengthen the stroke
again. So, to keep using the analogy, clear the traffic (quiet the signals with lower tempo), re-arrange the road (imprint the longer stroke), then release traffic and gradually ramp it
up under the new road arrangement.
Or maybe that is a goofy analogy. Can’t help it. My father was a civil engineer building bridges and highways and I spent many years working for him until I started my own company- I did
spend some time in engineering school too but decided on another path.
Isaac on March 25, 2015 at 02:14
Thank you Coach
To my surprise, I noticed that your articles, similar to meditation instruction, provide me with new insights on every subsequent reading. With your permission, here is a summary of my
current understanding
1. The key is mindfulness
2. It’s about reducing drag, not reducing SPL
3. All the rest is implementation detail. Be patient.
I now realize that impatience is the motivator for my questions. Yet, I believe that I could be more patient if I really understood why that “sweet spot” is more efficient (analogies
don’t do the job). I will wait for your upcoming articles to learn more.
If I may add one related question. In one article, you recommended changing the SL in relation to the distance being swam. Shorter SL for short distances. I would like to know the logic
behind this.
“Metrics 103- Pace construction” was clear, rational, and not too difficult. It is exactly what I needed to hear. Following it is another matter. As I practice, I find it hard to tell
whether the one-count improvement on this lap, came from pushing harder, or from keeping my head low while breathing. But that’s where point number one above comes in, doesn’t it?
Mat Hudson on March 25, 2015 at 15:32
Why use shorter SL for shorter distances?
Higher tempo tends to burn energy at a higher rate. So, both from understanding the thermodynamics of the scenario and from watching how humans tend to modify stroke length to deal
with the speed/energy demands of different distance races we can see that longer stroke + slower tempo tends to distribute energy better for longer distances, while shorter stroke +
higher tempo tends to get more speed, at higher expense, but can afford it on short distances. By applying more power per stroke, or applying power more frequently, the swimmer can
maintain a higher average speed (and flatter acceleration/deceleration curve), but it comes at the cost of a higher rate of energy burn.
A faster tempo stroke also means that time for each stroke is compressed and the swimmer has to fit a whole stroke cycle into that smaller amount of time – and this inevitably means
he must shorten the path of the arm movement (accidentally or deliberately) and this inevitably means the stroke length shortens. Hopefully, you can envision how that happens because
it would take several paragraphs to break down the sections of the stroke cycle and point out the most likely spots where a swimmer will shorten his stroke pathway and thereby shorten
the stroke length.
In order to lengthen the stroke, or to travel the distance of SL faster the swimmer has two options:
1) increase power
2) decrease drag
The swimmer can shorten his stroke by reducing power, or by increasing drag. And vice versa.
So, what we are looking for is the optimal stroke length for each event and its energy budget. Short races allow us to burn fuel at a higher rate so we can use more expensive ways to
generate speed. Longer races require us to spread out energy further so we have to limit the fuel burn rate and trade off in speed. It is common sense that we should do all we can to
reduce unnecessary water resistance so we don’t have to crank up so much power to create the same amount of speed. If you want solid scientific backup on this read Jane Cappaert’s
study of the 1992 Olympics sprinters which found (to the surprise of the researchers, apparently) that the swimmers of the Final Round were using an average of 16% LESS power than the
swimmers who did not pass the Preliminary Rounds. The best swimmers in the world are not the most powerful, they are the ones who swim with the least drag. A longer stroke (by less
drag) is a strong indicator of efficiency. A longer (than average) stroke and preventing stroke length decay toward the end of the race is one of the strongest indicators of who will
win races.
Here is another way to describe our strategy – what we are aiming for in TI is to first ruthlessly remove all excess drag from the stroke which will, by default, bring the swimmer to
an optimal stroke length (not an extreme long stroke), then build power on top of that minimalist stroke discipline. Adding more power will inevitably cause us to shorten the stroke a
bit (as water resistance builds up exponentially against velocity), but if our starting point is at this minimal-drag SPL, and we are ultra-sensitive about any changes that produce
unnecessary drag, then we are in good position to convert more of that additional power into forward motion and less into turbulence and waste.
Ashok Gollerkeri on June 26, 2015 at 08:30
Hi Mat,
I had a pleasant surprise today. I actually measured the pool breadth I usually swim. I had thought it to be 10 m but it is actually 13 m (43 feet divided by 3.28)
Allowing for a 3 m glide, I have been swimming 10 m with 7 strokes regularly for several months – with intense focus and optimum effort (not trying too hard nor too easy). When numerous
repetitions did not help me better 7 strokes (that is, achieve a stroke count of 6) , I stopped mechanical repeats.
By the way, my height is 175 cm and for my height, 7.2 is optimum number of strokes for 10 m and 6.4 is the minimum as per the TI chart).
Instead of repeats, I started focusing on balance and core body rotation drills to master balance and core body rotation. “Balance and core body rotation lead to easy speed” I had read in TI
literature and I was actually trying it out.
For about 2 weeks (15-20 hrs in the pool), I did 10 m each of superman glide, torpedo, body dolphin, skate (alternating sides) and laser lead rotation (alternating sides). This was followed by 10
m of whole stroke with fist gloves, easy swimming without counting strokes. I focused on “ease” rather than “effort” to improve my stroke count.
I then went offshore where I work and had 2 weeks of sedentary work (absolutely no physical exercise and absolutely no swimming). I was mentally engaged with swimming issues (mainly through your
blog) for about an hour or two every day.
On my return, I swam without fist gloves and without the tempo trainer, only counting strokes. I swam 10 m effortlessly with 7 strokes, not once but about 10 times (with a brief “active rest” of
floating on my back in between repeats). While my stroke count of 7 for 10 m had been regular (but did not feel entirely effortless or easy) in the past, this time around, it felt effortless and
easy each time. Further, once – only once, though – I swam 10 m with an effortless and easy six strokes. I breathe every third stroke. When I took my second breath, my head was at the wall (not
my hand). (Does that count as 5.5? 🙂 )
In any case, what was remarkable was the “ease” with which the stroke counts of 7 and once of 6 had been achieved. Would you attribute this to the balance and rotation drills I focused on? If so,
should I continue this pattern of practice – drill:whole stroke 5:1?
Also, I have an option of swimming pool breadth (10 m) or the pool length (20 m). So far, I have preferred to swim highly focused short repeats (10 m). My target stroke length now is 6 for 10 m
and my strategy is not to swim longer and harder but focus more on mastery of balance, streamline, patient hands and core body rotation. When I tried, I could swim about 30-35 m fairly easily,
ie: without exhaustion. Since my body is not used to distances of more than 10 m yet, I felt inclined and paused for some active rest (floating on my back) after 30 m.
Should I add one 30 m length to each pool session and will it serve any useful purpose (other than improving stamina and fitness)? Or shall I continue my short, highly focused repeats of 10 m
focusing on mastery of balance and core body rotation?
Mat Hudson on June 27, 2015 at 19:09
Hi Ashok,
In the early stages you might do far more drill work that whole stroke, and gradually reverse the ratio over time. How to do that? Spend about 80% of your time working on imprinting what is
within your abilities when focusing carefully, then spend about 20% of your time working around your limits of control and strength. Step over them just a little to stretch and stress your
system (in a good way) and your limits will expand gradually. Drills are meant to slow down time and make concentration easier – so when it starts to feel too easy, you need to increase the
challenge a bit.
And consider what your expectation of what ‘feel ready for more’ is suppose to be like. We do not want to make big leaps over our limits and practice continual failure, but we do want to work
up to those limits and a little bit over that line each week so that we do experience some failures in order to find out specifically where our weak spots are. So, there is a danger of going
too easy too much and of going too hard. The art of Flow State is setting just the right amount of challenge.
By doing longer distances you have two limits to consider: 1) your limits of strength and stamina, and 2) your limits of attention. Which ever one is weaker is the limit you set (recommended
in this initial training stage). Bring the weaker one up (through that stretching process) to the other and then work on them together. | {"url":"https://mediterraswim.com/2015/03/21/stroke-length-loyalty/","timestamp":"2024-11-05T23:36:34Z","content_type":"text/html","content_length":"266294","record_id":"<urn:uuid:399d4ab2-af98-4a0e-bc48-83d429632794>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00467.warc.gz"} |
How to collect the alpha ,beta data from pixhawk cube
am using pixhawk cube black. right now am connected to potentiometer to ADC port . then after after calibrated accel , compass,…,. then after armed and downloaded the DF log . so right now which data
is alpha , beta data .
There is no measurement for alpha, beta, and gamma since AFAIK there no support for flow field sensors such as vanes or multiport pitot tubes. Theta, phi, and psi can all be found under the CTUN in
Mission Planner log viewer.
how i get the alpha , beta data from pixhawk . am already using digital airspeed sensor . there is one matlab simulation link let see https://in.mathworks.com/help/supportpkg/px4/ref/
. i need alpha and beta data .
This is not the forum for PX4 firmware. If you are using ArduPlane firmware you can parse the data you need using the tools provided in Mission Planner under the “DataFlash Logs” tab to get to a
Matlab format.
how to get the alpha ,beta data . any another ways will get the data please let me know | {"url":"https://discuss.ardupilot.org/t/how-to-collect-the-alpha-beta-data-from-pixhawk-cube/47677","timestamp":"2024-11-07T07:23:50Z","content_type":"text/html","content_length":"19119","record_id":"<urn:uuid:fbd4f25e-ed48-4b16-b00b-6e07e67e5a04>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00144.warc.gz"} |
Countries Traveled, The Last Part - Travel Tales from India and Abroad
Countries Traveled, The Last Part
A while ago, I started country counting. My tally stands at just sixteen, or maybe I should say sweet sixteen. So I round up the tally with some of the beautiful skies I captured in the last six of
them. You can see the first ten countries here and here. So this completes the list of countries traveled so far by me, at least for time being. The picture above is from Hong Kong.
Cambodia (With Tourism Authority of Thailand and Cambodia Tourism Board)
Cambodia has been my shortest stay at any international location. I had only only night in the Angkor region. I crammed the night market and the sunrise at Angkor Wat in that. I got very little sleep
but the memories that I made are priceless.
Finland (With Nokia, Now Microsoft)
Finland was such an amazing winterland. I have never seen so much snow in my life. The highest temperature that I encountered during my entire stay was 2 degree celsius. The coldest was minus 22
degrees. Lapland was colder than Helsinki. With such cold weather it was very easy to appreciate both central heating and sauna.
Maldives (Personal Trip)
My younger nephew Sunil and I went to Maldives last year. Maldives is incredibly beautiful and then some more. And contrary to the perceptions it can be done on a budget.
Jordan (with Jordan Tourism Board)
Like South Africa, Jordan was not on my list, at least not in the near future, because eventually everything is on my list. It was also special because I got to attend a mass by Pope Francis! Now
even in my wildest dreams I never imagined getting an opportunity like that! Dead Sea was incredible fun and I discovered delicious vegetarian food in Jordan!
Bhutan (with Make My Trip)
Visiting Bhutan was like stepping back in time, say by 20 years! I was about to say 50 years but then I realized that it is more than my age! I simply fell in love with Bhutan. And if you are into
trekking, you have to trek to the Tiger’s Nest Monastery. Even you are not, you still have to hike up to the monastery!
Hong Kong (Personal Trip)
Hong Kong was a truly memorable trip because I took double trouble (my daughter and niece) to Hong Kong on their first trip abroad. Of course Disneyland was the agenda! All I can say is that I took
out two kids with me alone and I have lived to tell the tale!
21 thoughts on “Countries Traveled, The Last Part”
1. Well it is only the list till date… Wishing you a lot of more country hopping in the coming years…. 🙂
1. Thank you so much Prasad.
2. Lovely!! Hope to see a list of 100 soon 🙂
1. Archana, amen to that!
3. Nice snapshots !
1. Thank you Manish.
4. A great Experience…. Your determination to Travel to various places of the world is truly unique…. Best Wishes for all future Travel Adventures!
1. Thank you so much Sreedhar.
5. Great pics Mridula. May you add more to your list!
1. I too hope so Indrani, I too hope so.
6. Sigh – so many gorgeous shots! Lucky you.
1. Thank you so much Lady Fi.
7. Awesome photos and post
1. Thank you Rupam.
8. Your photography… is AMAZING!! Such a treat to the senses!
1. Thank you so much Bushra.
9. Beautiful post with awesome photos. Thank you for sharing.
1. Thank you so much!
10. Lovely experiences, Mridula. Let the list grow manifold.
1. Thank you so much Niranjan.
11. U are very lucky, get so many sponsors . Feeling jealous 🙂
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://traveltalesfromindia.in/countries-traveled/","timestamp":"2024-11-06T20:33:13Z","content_type":"text/html","content_length":"96354","record_id":"<urn:uuid:df68ed03-fdd9-46cd-b8d1-709e2c1303c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00191.warc.gz"} |
A venue for invited and local speakers to present their research on topics surrounding algebraic geometry and number theory, broadly conceived. All meetings start at 14:10 sharp and end at 15:10.
Meetings are held in the subterranean room -101. We expect to broadcast most meetings over Zoom at the URL https://us02web.zoom.us/j/85116542425?pwd=MzVDRmZJUVh2NXlObFVkM1N0MCt3Zz09
The seminar meets on Wednesdays, 14:10-15:10, in -101
2023–24–B meetings
Date Title Speaker Abstract
Topological recursion is a remarkable universal recursive procedure that has been found in many enumerative geometry problems, from combinatorics of
maps, to random matrices, Gromov-Witten invariants, Hurwitz numbers, Mirzakhani’s hyperbolic volumes of moduli spaces, knot polynomials. A recursion
x-y duality in needs an initial data: a spectral curve, and the recursion defines the sequence of invariants of that spectral curve. There is a duality in
topological recursion, Boris Bychkov topological recursion which allows one to obtain closed formulas for the invariants of the recursion and which has implications in free probability
May 8 Hurwitz numbers and (University theory and integrable hierarchies. In the talk I will survey recent progress in the topic with the examples from Hurwitz numbers theory, Hodge
integrability of Haifa) integrals and combinatorics of maps.
The talk is based on the joint works with A. Alexandrov, P. Dunin-Barkowski, M. Kazarian and S. Shadrin.
Lara—Srinivas—Stix, building on joint work with Esnalut, have recently shown that the etale fundamental group of a connected proper scheme over an
Fundamental Groups of Mark algebraically closed field is topologically finitely presented, thus answering a question raised in SGA. The proof relies on a finite presentation
May 22 Projective Varieties are Shusterman ( criterion of Lubotzky for profinite groups, resolutions of singularities/alterations, a theorem of Deligne—Ilusie on the Euler characteristic, as
Finitely Presented Weizmann) well as other modern and classical results in (arithmetic) algebraic geometry.
Supersingular elliptic Part of the talk is expository: I will explain how supersingular isogeny graphs can be used to construct cryptographic hash functions and survey
curves, quaternion Eyal Goren ( some of the rich mathematics involved. Then, with this motivation in mind, I will discuss two recent theorems by Jonathan Love and myself. The first
May 29 algebras and some McGill concerns the generation of maximal orders by elements of particular norms. The second states that maximal orders of elliptic curves are determined
applications to University) by their theta functions.
The lecture will start with a few useful (and probably new!) theorems on adic completion of commutative rings and modules. Then I will discuss
derived adic completion, in its two flavors: the idealistic and the sequential. The weak proregularity (WPR) condition on an ideal \a in a ring A,
Amnon which is a subtle generalization of the noetherian condition on the ring A, is a necessary and sufficient condition for the two flavors of derived
Adic Completion, Derived Yekutieli ( completion to agree. WPR occurs often in the context of perfectoid theory, and I will finish the talk with theorems relating WPR to prisms.
Jun 5 Completion, Prisms, and Ben Gurion
Weak Proregularity University) Typed notes available soon at the bottom of this web page:
Equidistribution of “special” points is a theme of both analytic and geometric interest in number theory: in this talk I plan to deal with the case
Francesco of CM points on Shimura curves.
Jun 19, TBA Saettone (Ben The first part will be devoted to a geometric description of the aforementioned curves and of their moduli interpretation.Subsequently, I plan to
13:10–14:10 Gurion sketch an equidistribution result of reduction of CM points in the special fiber of a Shimura curve associated to a ramified quaternion algebra.
University) Time permitting, I will mention how Ratner’s theorem and subconvexity bounds on Fourier coefficients of certain theta series can be used to obtain
two different equidistribution results.
In this talk, I will discuss the irreducibility problem of Severi varieties on toric surfaces. The classical Severi varieties were introduced by
Severi almost 100 years ago in the context of Severi’s attempt to provide an algebraic treatment of the irreducibility problem of the moduli spaces
of curves. Although, the irreducibility of the moduli spaces was achieved algebraically by Deligne and Mumford in 1969 using completely different
Irreducible components Michael techniques, Severi varieties remained in the focus of study of many algebraic geometers including Harris, Fulton, Zariski and others.
Jun 26 of Severi varieties on Barash (Ben
toric surfaces Gurion In this talk I will present the main result of my M.Sc. Thesis providing a complete description of the irreducible components of the genus-one
University) Severi varieties on toric surfaces.
This work was done under the supervision of Professor Ilya Tyomkin.
Many estimates for the “size” of a subset of the natural numbers (a “size” usually expressed by some notion of density) come from “local”
Ignazio conditions, like reduction modulo prime powers. The idea can be formalized in terms of the Haar measure on the profinite completion of Z or, in a
Jul 3 Some applications of the Longhi ( more refined way, via distributions on this profinite ring. This approach can be easily generalized by replacing Z with the ring of S-integers of
profinite completion University of any global field. In this talk (based on a number of joint works with L. Demangos and F.M. Saettone), I will discuss how to use these ideas to
Turin) extend classical results and reformulate long standing conjectures in profinite terms.
The stable module category is obtained from the category of modules over a ring by factoring out the projective modules. In this setting, the syzygy
Sequences of syzygies, Liran Shaul ( of a module becomes a well defined functor, so for instance, the classical Hilbert’s syzygy theorem, can be stated as saying this functor is
Jul 10 singularity categories Charles nilpotent. In this talk we present some new properties of the syzygy functor over a commutative noetherian ring. We then explain how to associate to
and homological University the stable category a stabilization, obtaining the singularity category of a ring (or a scheme). Finally, we explain how relations between the
conjectures Prague) stable category and the singularity category are related to some homological conjectures in noncommutative algebra.
With each simple connected graph $G$ with $n$ vertices one can associate a generalized configuration space $Conf_{G}(n,X)$ consisting of $n$ points
$(p_1,\ldots,p_n)$ on $X$, with $p_i\neq p_j$ whenever vertices $i$ and $j$ are connected by an edge. For $X=\mathbb{C}$ the generalized
On generating series of Anton configuration space admits a compactification that coincides for a complete graph with Deligne-Mumford compactification of moduli spaces of rational
Jul 17 cohomology of Khoroshkin ( curves with $n$ marked points. The latter is known under the name modular compactification. I will explain what kind of natural algebraic structure
generalized University of exists in the union of these spaces and how one can extract information about the Hilbert series of cohomology rings for different collections of
configuration spaces Haifa) graphs. Surprisingly, the same method can be used to obtain the generating series for different combinatorial data assigned with a graph: such as
the number of Hamiltonian paths, Hamiltonian cycles, Acyclic orientations and Chromatic polynomials. The talk is based on the joint work with my
student D.Lyskov: https://arxiv.org/abs/2406.05909 | {"url":"https://math.bgu.ac.il/en/research/spring2024/seminars/agnt","timestamp":"2024-11-13T15:27:24Z","content_type":"text/html","content_length":"37625","record_id":"<urn:uuid:8d3d4420-906c-4be7-b622-1ad8beabc978>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00401.warc.gz"} |
If two carriers (heterozygotes) have children, the - Asksia.ai
Answer from Sia
Posted 6 months ago
Key Concept
Probability of inheriting a recessive trait
When two heterozygous carriers (Aa) have children, the probability of each child being affected by a recessive trait (aa) is $1/4$. The probability that both of their first two children are affected
is $(1/4) \times (1/4) = 1/16$.
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Give any two examples for each from your daily life to find the area of the rectangle.
In this particular question use the concept that a rectangle has opposite sides parallel to each other and the length of the adjacent sides are not equal so use these concepts to reach the solution
of the question.
Complete step-by-step answer:
In our daily life these are lots of examples which resemble the shape of the rectangle.
As we all know that a rectangle has opposite sides parallel to each other and the length of the adjacent sides are not equal to each other but the length of the opposite sides are equal to each other
and in a rectangle adjacent sides always make a 90 degrees.
For example: books, pencil box, doors, keyboard, laptops, mobile phones, etc.
So consider any two of the above examples such as books and pencil boxes.
The pictorial representation of the book and the pencil box are shown below.
Let the length of the book and pencil box be L and A respectively, and the breadth of the book and the pencil box be B and C respectively as shown in the above figure.
Now as we know that the area of the rectangle is the multiplication of the length and the breadth.
So the area (${A_1}$) of the book is
$ \Rightarrow {A_1} = L \times B$ Square units.
And the area (${A_2}$) of the pencil box is
$ \Rightarrow {A_2} = A \times C$ Square units.
So this is the required answer.
: Whenever we face such types of questions the key concept we have to remember is that we always recall the definition of a rectangle which is stated above without knowing this we cannot choose the
items from our daily life which resembles the shape of the rectangle. | {"url":"https://www.vedantu.com/question-answer/give-any-two-examples-for-each-from-your-daily-class-8-maths-cbse-5f5b7faa8a2fd7303bd958a8","timestamp":"2024-11-12T12:57:33Z","content_type":"text/html","content_length":"158624","record_id":"<urn:uuid:e83e8781-e008-4910-949c-c30638d719fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00201.warc.gz"} |
What is a dilation simple definition?
What is a dilation simple definition?
Definition of dilation : the act or action of enlarging, expanding, or widening : the state of being dilated: such as. a : the act or process of expanding (such as in extent or volume) … the dilation
of palladium grains undergoing hydrogen absorption.—
How do you find the dilation?
To find the scale factor for a dilation, we find the center point of dilation and measure the distance from this center point to a point on the preimage and also the distance from the center point to
a point on the image. The ratio of these distances gives us the scale factor, as Math Bits Notebook accurately states.
What is a dilation factor?
Scale Factor (Dilation) Scale Factor (Dilation) The scale factor in the dilation of a mathematical object determines how much larger or smaller the image will be (compared to the original object).
How do you dilate a triangle by 2?
To dilate the figure by a factor of 2, I will multiply the x and y-value of each point by 2. I plotted all the new points to find the new triangle. To dilate the figure by a factor of 2, I will
multiply the x and y-value of each point by 2.
How do you dilate 2?
To dilate the figure by a factor of 2, I will multiply the x and y-value of each point by 2. I plotted all the new points to find the new triangle. To dilate the figure by a factor of 2, I will
multiply the x-value of each point by 2.
How do you dilate a graph?
How to Perform Dilations
1. Identify the center of dilation.
2. Identify the original points of the polygon.
3. Identify the scale factor .
4. Multiply each original point of the polygon by the scale factor to get the new points.
5. Plot the new points and connect the dots to get your dilated shape.
How do you dilate a circle by 2?
Dilation To dilate a circle, we start with our standard equation: x2+y2=r2 To dilate the circle we multiply our desired factor squared into the right side of the equation. For example, two multiply
the diameter of the circle by two, our equation would now be x2+y2=22(r2). | {"url":"https://www.atheistsforhumanrights.org/what-is-a-dilation-simple-definition/","timestamp":"2024-11-11T10:36:52Z","content_type":"text/html","content_length":"81165","record_id":"<urn:uuid:f89a70da-236e-43fa-8a96-3f6b5b85351b>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00792.warc.gz"} |
(Multiple choice, can select more than one)
Under our standard regression assumptions, which of the
(Multiple choice, can select more than one) Under our standard regression assumptions, which of the following...
(Multiple choice, can select more than one)
Under our standard regression assumptions, which of the following is true?
E(u|X,Y) = 0
E(u|X) = 0
E(u) = 0
Under our standard regression assumptions, all of the following are true:
E(u|X)= 0
E(u) = 0.
Let me explain this in brief. | {"url":"https://justaaa.com/statistics-and-probability/210938-multiple-choice-can-select-more-than-one-under","timestamp":"2024-11-06T10:32:55Z","content_type":"text/html","content_length":"40327","record_id":"<urn:uuid:95c71810-d8b7-4cdd-89cc-56e64911860c>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00581.warc.gz"} |
10 Essential Types of Graphs and When to Use Them (2024)
From stock market prices to sports statistics, numbers and statistics are all around you.
However, numerical data alone is merely a combination of figures and doesn’t tell a story. The most meaningful data and/or data analysis in the world is useless if it’s not communicated correctly.
“Effective data visualization can mean the difference between success and failure when it comes to communicating the findings of your study, raising money for your nonprofit, presenting to your
board, or simply getting your point across to your audience.”
Identifying the relationship between your data set or data points and telling the story behind the numbers will also encourage your audience to gain actionable insights from your presentation.
How do you do this?
You visualize data points through charts and different types of graphs.
The good news is you don’t need to have a PhD in statistics to make different types of graphs and charts.This guide on the most common types of graphs and charts is for you.
Keep reading if you’re a beginner with no data visualization background but want to help your audience get the most out of your numerical data points, both in-person and via a web conference. You’ll
also discover data visualization best practices, advice from experts in the craft, and examples of well-thought-out charts and graphs below!
Most Common Types of Charts and Graphs to Communicate Data Points With Impact
Whether you’re about to create a collection of business graphs or make a chart in your infographic, the most common types of charts and graphs below are good starting points for your data
visualization needs.
1. Bar chart
2. Line graph
3. Area graph
4. Scatter plot
5. Pie chart
6. Pictograph
7. Column chart
8. Bubble chart
9. Gauge chart
10. Stacked Venn
1. Bar chart
A bar chart, also known as a horizontal column chart, is popular for a reason — it’s easy on the eyes and quickly visualizes data sets. With bar charts, you can quickly identify which bar is the
highest or the lowest, including the incremental differences between bars.
When to use bar charts
• If you have more than 10 items or categories to compare.
• If your category labels or names are long.
Best practices for bar charts
• Focus on one color for a bar chart. Accent colors are ideal if you want to highlight a significant data point.
• Bars should be wider than the white space between bars.
• Write labels horizontally (not vertically) for better readability in your bar chart.
• Order categories alphabetically or by value to ensure consistency across your bar chart.
“Bar charts must always have a zero baseline (y-axis value at zero) to ensure consistency.”
2. Line chart
Not to be confused with line graphs, you can use a line chart to plot continuous data or data with infinite values.For example, the line chart below highlights the increase in keyword searches for
“remote work” across the US from February 1, 2020, to March 22, 2020.
When to use line charts
• Compare and present lots of data at once.
• Show trends or progress over time.
• Highlight deceleration.
• Present forecast data and share uncertainty in a single line chart.
Best practices for line charts
• Use solid lines only because dotted or dashed lines are distracting.
• Ensure that points are ordered consistently.
• Label lines directly and avoid using legends in a line chart.
• Don’t chart more than four lines to avoid visual distractions.
• Zero baseline is not required, but it is recommended for a line chart.
Pro-tip for line charts from Mike Cisneros, an award-winning data visualizer:
“The range from your smallest value to your largest values should take up about 70 to 80 percent of your graph’s available vertical space.”
3. Area graph
An area graph is like a line chart as it also shows changes over time. One difference with these types of graphs is that area graphs can represent volume which is typically filled with color.
The area graph example by the BBC below shows a simple comparison of two data sets over a period of time.
When to use area graphs
• Display how values or multiple values develop over time.
• Highlight the magnitude of a change.
• Show large differences between values.
Best practices for these types of graphs
• Don’t display more than four categories on these types of graphs.
• Use transparent colors to avoid obscuring data in the background on these types of graphs.
• Add annotations and explanations to these types of graphs.
• Group tiny values together into one bigger value to prevent clutter on these types of graphs.
“Bring the most important value to the bottom of the chart and use color to make it stand out. Your readers can compare values easier with each other if they have the same baseline.”
4. Scatter plot
A scatter plot or a scatter chart helps show the relationship between items based on two different variables and data sets. Dots (or plot data) are plotted in an x-y coordinate system. In some
scatter plots, a trend line is added (like in the example below) to a scatter plot.
When to use a scatter plot
• Show relationships between two variables.
• You have two variables of data that complement each other.
Best practices for scatter plots
• Start the y-axis value at zero to represent data accurately.
• Plot additional data variables by changing dot sizes and colors.
• Highlight with color and annotations.
Pro-tip for scatter plots from Mike Yi of Chartio on incorporating data visualization:
“Add a trend line to your scatter plot if you want to signal how strong the relationship between the two variables is, and if there are any unusual points that are affecting the computation of
the trend line.”
5. Pie chart
Pie charts highlight data and statistics in pie-slice format. A pie chart represents numbers in percentages, and the total sum of all pies should equal 100 percent. When considering charts and graphs
to employ to visualize data, pie charts are most impactful to your audience if you have a small data set.
The donut pie chart, a variation of the pie chart, shows a design element or the total value of all the variables in the center.
When to use pie charts
• Illustrate part-to-whole comparisons — from business to classroom charts and graphs.
• Identify the smallest and largest items within data sets.
• Compare differences between multiple data pointsin a pie chart.
Best practices for using a pie chart
• Limit categories to 3-5to ensure differentiation with the pie chart slices.
• Double-check if the total value of the slices is equal to 100 percent.
• Group similar slices together in one bigger slice to reduce clutter.
• Make your most important slice stand out with color. Use shades of that specific color to highlight the rest of the slices.
• Order slices thoughtfully. For example, you can place the largest section at the 12 o’clock position and go clockwise from there. Or place the second largest section at the 12 o’clock position
and go counterclockwise from there.
Pro-tip for pie charts from visual communication researcher Robert Kosara of Eager Eyes when considering charts and graphs:
“The pie chart is the wrong chart type to use as a default; the bar chart is a much better choice for that. Using a pie chart requires a lot more thought, care, and awareness of its limitations
than most other charts.”
6. Pictograph
Despite having ‘graph’ in the name, a pictograph doesn’t fall into types of graphs. Instead, a pictograph or a pictogram is a type of chart that uses pictures or icons to represent data. Each icon
stands for a certain number of data sets, units or objects. For example, the infographic below contains a pictogram — each human icon represents 10 percent of CEOs.
When to use pictographs
• When your target audience prefers icons and pictures instead of data sets (to illustrate data).
• Show the progress of a goal or project to show continuous data.
• Highlight ratings to compare data.
• Share survey resultsor data distribution.
• Share level of proficiency or data sets.
Best practices for pictographs
• Keep your icons and pictures simple to avoid distracting your audience with these types of graphs.
• Do not use contrasting colors for your icons. Instead, use shades of one specific color.
• Limit rows to five or ten for better readability on these types of graphs.
7. Column chart
A column chart is ideal for presenting chronological data. Also known as the vertical bar chart, this type of chart works if there are only a few dates to highlight your data set like in the example
When to use column charts
• Display comparison between categories or things (qualitative data).
• Show the situation at one point in time using various data points.
• Share relatively large differences in your numeric data values.
Best practices for column charts
• Plot bars against a zero-value baseline.
• Keep your bars rectangular and avoid 3D effects in your bars.
• Order category levels consistently: from highest to lowest or lowest to highest.
Pro-tip for using column charts for a data set from Storytelling with Data:
“As you add more series of data, it becomes more difficult to focus on one (bar) at a time and pull out insight, so use multiple series bar charts with caution.”
8. Bubble chart
A bubble chart or a bubble plot is a lot like a scatter plot. However, bubble charts have one or two more visual elements (dot size and color) than a scatter plot to represent a third or fourth
numeric variable.
When to use a bubble chart
• Show relationships between three or more numeric variables
Best practices for bubble charts
• Scale bubble area by value, not diameter or radius.
• Use circular shapes only in a bubble chart.
• Label key points clearly in a bubble chart.
Pro-tip from Elizabeth Ricks, a data visualization instructor on creating a bubble chart:
“Include words for static bubble charts. It’s always a good idea to label your axes, provide clear chart titles, and annotate important data points with illuminating context. This is especially
true when you are using a data-dense chart type like a bubble chart, and you aren’t standing next to it ready to explain away any confusion that viewers might have at first glance.”
9. Gauge chart
A gauge chart, also known as a dial chart, is an advanced type of chart that shows whether data values fit on a scale of acceptable (good) to not acceptable (bad). For example, you can create a gauge
chart to display current sales figures and use your quarterly sales targets as thresholds.Not all charts are able to show data in this way.
Gauge charts are particularly helpful where the expected value of the data is already known. This helps organizations create actionable reports and help employees understand where they stand in terms
of metrics by looking at the chart.
When to use gauge charts
• Share target metrics and display the percentage of the target goal that has been achieved for a certain period.
• Highlight the progress of linear measurement.
• Compare variables either by using multiple gauges or through multiple needles on the same gauge.
Best practices for gauge charts
• Limit two to three colors for each gauge or avoid high-contrast color combinations.
10. Stacked Venn
A stacked Venn chart is used to showcase overlapping relationships between multiple data sets. This type of graph is a variation of the original Venn diagram, where overlapping shapes or circles
illustrate the logical relationships between two or more variables.
When to use the Stacked Venn
• Emphasizing growth within an organization or business
• Narrow down a broad topic
Best practice for Stacked Venn
• Avoid high contrast color combinations to ensure readability.
What About the Other Types of Graphs and Charts?
There are plenty of other types of graphs and charts—line graphs, multiple line graphs, candlestick charts, Gantt charts, radar charts, stacked bar graphs, heat maps, waterfall charts, and the list
goes on. They are almost always specific to a particular industry, and the charts and graphs we’ve listed should be enough to address your basic to intermediate data visualization needs to illustrate
hierarchical data and beyond.
Choose Charts and Graphs That Are Easiest for Your Audience to Read and Understand
Thoughtfully designed charts and graphs are a result of knowing your audience well. When you understand your audience, you can communicate your data points more effectively.
Before you share your chart or graph, show it to a couple of colleagues or a small group of customers. Pay attention to their questions, their observations, and how they react to your chart or graph.
If you’re looking for a graph maker, create a free Piktochart account and sharpen your data visualization chops by making the right types of graphs and charts in minutes from multiple data sets and | {"url":"https://mrbackdoorstudio.com/article/10-essential-types-of-graphs-and-when-to-use-them","timestamp":"2024-11-12T01:56:43Z","content_type":"text/html","content_length":"177493","record_id":"<urn:uuid:fdaaa475-a200-45ab-b8b2-dd539ec768b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00165.warc.gz"} |
urlongs (US survey) to Handbreadth
Furlongs (US survey) to Handbreadth Converter
Enter Furlongs (US survey)
β Switch toHandbreadth to Furlongs (US survey) Converter
How to use this Furlongs (US survey) to Handbreadth Converter π €
Follow these steps to convert given length from the units of Furlongs (US survey) to the units of Handbreadth.
1. Enter the input Furlongs (US survey) value in the text field.
2. The calculator converts the given Furlongs (US survey) into Handbreadth in realtime β using the conversion formula, and displays under the Handbreadth label. You do not need to click any
button. If the input changes, Handbreadth value is re-calculated, just like that.
3. You may copy the resulting Handbreadth value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Furlongs (US survey) to Handbreadth?
The formula to convert given length from Furlongs (US survey) to Handbreadth is:
Length[(Handbreadth)] = Length[(Furlongs (US survey))] / 0.00037878712152151513
Substitute the given value of length in furlongs (us survey), i.e., Length[(Furlongs (US survey))] in the above formula and simplify the right-hand side value. The resulting value is the length in
handbreadth, i.e., Length[(Handbreadth)].
Calculation will be done after you enter a valid input.
Consider that a historical survey mapped a land area of 5 furlongs (US survey).
Convert this distance from furlongs (US survey) to Handbreadth.
The length in furlongs (us survey) is:
Length[(Furlongs (US survey))] = 5
The formula to convert length from furlongs (us survey) to handbreadth is:
Length[(Handbreadth)] = Length[(Furlongs (US survey))] / 0.00037878712152151513
Substitute given weight Length[(Furlongs (US survey))] = 5 in the above formula.
Length[(Handbreadth)] = 5 / 0.00037878712152151513
Length[(Handbreadth)] = 13200.0264
Final Answer:
Therefore, 5 fur is equal to 13200.0264 handbreadth.
The length is 13200.0264 handbreadth, in handbreadth.
Consider that an old railroad line runs for 10 furlongs (US survey).
Convert this distance from furlongs (US survey) to Handbreadth.
The length in furlongs (us survey) is:
Length[(Furlongs (US survey))] = 10
The formula to convert length from furlongs (us survey) to handbreadth is:
Length[(Handbreadth)] = Length[(Furlongs (US survey))] / 0.00037878712152151513
Substitute given weight Length[(Furlongs (US survey))] = 10 in the above formula.
Length[(Handbreadth)] = 10 / 0.00037878712152151513
Length[(Handbreadth)] = 26400.0528
Final Answer:
Therefore, 10 fur is equal to 26400.0528 handbreadth.
The length is 26400.0528 handbreadth, in handbreadth.
Furlongs (US survey) to Handbreadth Conversion Table
The following table gives some of the most used conversions from Furlongs (US survey) to Handbreadth.
Furlongs (US survey) (fur) Handbreadth (handbreadth)
0 fur 0 handbreadth
1 fur 2640.0053 handbreadth
2 fur 5280.0106 handbreadth
3 fur 7920.0158 handbreadth
4 fur 10560.0211 handbreadth
5 fur 13200.0264 handbreadth
6 fur 15840.0317 handbreadth
7 fur 18480.0369 handbreadth
8 fur 21120.0422 handbreadth
9 fur 23760.0475 handbreadth
10 fur 26400.0528 handbreadth
20 fur 52800.1056 handbreadth
50 fur 132000.2639 handbreadth
100 fur 264000.5278 handbreadth
1000 fur 2640005.2779 handbreadth
10000 fur 26400052.7785 handbreadth
100000 fur 264000527.7854 handbreadth
Furlongs (US survey)
A furlong (US survey) is a unit of length used primarily in land surveying and agriculture in the United States. One US survey furlong is equivalent to exactly 660 feet or approximately 201.168
The US survey furlong is defined as one-eighth of a US survey mile, providing a convenient measurement for distances used in surveying and land measurement.
Furlongs (US survey) are utilized in contexts such as land surveys, property measurement, and horse racing in the United States. The unit ensures consistency and accuracy in measuring shorter
distances in these fields.
A handbreadth is a historical unit of length used to measure small distances, typically based on the width of a hand. One handbreadth is approximately equivalent to 4 inches or about 0.1016 meters.
The handbreadth is defined as the width of a person's hand, measured from the edge of the thumb to the edge of the little finger when the hand is spread out. This unit was used for practical
measurements in various contexts, including textiles and construction.
Handbreadths were used in historical measurement systems for assessing lengths and dimensions where precise tools were not available. Although less common today, the unit provides historical context
for traditional measurement practices and everyday use in different cultures.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Furlongs (US survey) to Handbreadth in Length?
The formula to convert Furlongs (US survey) to Handbreadth in Length is:
Furlongs (US survey) / 0.00037878712152151513
2. Is this tool free or paid?
This Length conversion tool, which converts Furlongs (US survey) to Handbreadth, is completely free to use.
3. How do I convert Length from Furlongs (US survey) to Handbreadth?
To convert Length from Furlongs (US survey) to Handbreadth, you can use the following formula:
Furlongs (US survey) / 0.00037878712152151513
For example, if you have a value in Furlongs (US survey), you substitute that value in place of Furlongs (US survey) in the above formula, and solve the mathematical expression to get the equivalent
value in Handbreadth. | {"url":"https://convertonline.org/unit/?convert=furlongs_us_survey-handbreadths","timestamp":"2024-11-06T05:19:46Z","content_type":"text/html","content_length":"92628","record_id":"<urn:uuid:a450358f-a438-47d2-a765-5ce2524007e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00490.warc.gz"} |
Principal domain of cosx is?
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Question Text Principal domain of is?
Topic Trigonometric Functions
Subject Mathematics
Class Class 11
Answer Type Text solution:1 | {"url":"https://askfilo.com/math-question-answers/principal-domain-of-cos-x-is","timestamp":"2024-11-14T18:29:04Z","content_type":"text/html","content_length":"558365","record_id":"<urn:uuid:bab4fa45-1c6e-405d-8eac-622a48342e3e>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00006.warc.gz"} |
Course Goals
• To prepare you for Part 1 of the Comprehensive Exam by helping you to review and develop mastery of the five fundamental courses of the math major: Differential and Integral Calculus,
Multivariable Calculus, Discrete Mathematics, and Linear Algebra.
• To help you develop your abilities to communicate your ideas, especially your mathematical and technical ideas, through writing and through speech.
• To help you become informed about the range of opportunities available to you as math majors, both over the next year and after you graduate.
You can find a link to the syllabus here.
Week 1: Due Thursday, January 31
• Complete a practice Calc 1 section of the comps exam.
• Use $\LaTeX$ to typeset your answers for the Calc 1 exam; you should include a \maketitle command that gives a title and your name and the date.
• If you haven’t already, fill out this short survey on how comfortable you feel with each topic on the comps study guide. Use 1 for topics you don’t remember at all, and 5 for topics you feel
completely comfortable with. Please complete this by Saturday.
Week 2: Due Thursday, February 7
• Study for the Calc 1 comps exam.
• Read the solutions to the practice Calc 1 section.
• Submit an abstract you have written for the Krebs and Wright paper we discussed in class.
□ Write it in $\LaTeX$ and use the \begin{abstract} \end{abstract} environment.
• Sign up for comps review talk (link to come soon).
Week 3: Due Thursday, February 14
• Complete a practice Calc 2 section of the comps exam. (You don’t need to type it up this time; feel free to handwrite it, but turn it in).
• Sign up for a slot to talk about one comps review topic at this link. Note: Due to a technical screwup on my part, there is one multi topic and one calc 2 topic listed at the end.
• If you are giving a talk next week: Send me a copy of your notes for the talk by Monday night. That way I can give you feedback and make sure you’re leading everyone else in the right direction.
• Choose a paper to summarize for your first paper of the course (see below). Print out and bring to class a brief abstract for the paper, written in $\LaTeX$.
• Review notes on giving a talk.
Week 4: Due Thursday February 21
• Study for the Calc 2 comps exam.
• Read the solutions to the practice Calc 2 section.
• Bring four printed copies of your resume into class on Thursday.
Week 5: Due Thursday February 28
• Complete a practice multi section of the comps exam. (You don’t need to type it up this time; feel free to handwrite it, but turn it in).
• If you are giving a talk next week: Send me a copy of your notes for the talk by Monday night. That way I can give you feedback and make sure you’re leading everyone else in the right direction.
• Turn in a printed out hard-copy rough draft of your summary paper assignment.
• Review notes on writing a paper.
Week 6: Due Thursday March 7
• Read the solutions to the practice multi section.
• Look over my notes on multivariable calculus from Thursday.
• Study for the multi comps exam.
• Make any revisions you feel like to your essay. You don’t need to turn this in.
• Bring three printed copies of your draft paper in to class.
Week 7: Due Thursday March 21
• Complete a practice linear section of the comps exam. (You don’t need to type it up this time; feel free to handwrite it, but turn it in).
• If you are giving a talk next week: Send me a copy of your notes for the talk by Monday night. That way I can give you feedback and make sure you’re leading everyone else in the right direction.
• Turn in a printed out hard-copy final draft of your summary paper assignment. Some tips:
□ Please identify the paper you’re summarizing somewhere in your paper.
□ I encourage everyone to use BibTeX for citations, but it can be a little tricky if you’ve never done it, and it’s not a requirement. This tutorial should explain how; with the note that you
can use Google Scholar to get you your .bib file and shouldn’t need to format it yourself.
□ $\LaTeX$ doesn’t automatically create smart quotes for you, and thus you shouldn’t type "words" if you want quotation marks. Many editors will fix this for you, but if yours doesn’t, it is
much better to type ``words'' instead. (The backtick mark is probably under the tilde in the upper-left of your keyboard, right below the escape key).
□ If you’re using a function with a long name in math mode, make sure you mark it as a function.
☆ If $\LaTeX$ already recognizes your function you can just type it with a backslash: so \log instead of log or \gcd instead of gcd.
☆ If $\LaTeX$ doesn’t recognize your function, you need to tell it. You can either type something like \operatorname{area} every time instead of just area, or you can add \
DeclareMathOperator{\area}{area} once to the preamble. (Both of these options will only work if you are using the package amsmath).
□ If you’re typesetting an integral, it can be helpful to type \,dx instead of dx
Week 8: Due Thursday March 28
• Read the solutions to the practice linear section.
• Study for the linear comps exam.
• Submit an abstract for the Theorem paper.
Week 9: Due Thursday April 4
• Complete a practice Discrete section of the comps exam. (You don’t need to type it up this time; feel free to handwrite it, but turn it in).
• If you are giving a talk next week: Send me a copy of your notes for the talk by Monday night. That way I can give you feedback and make sure you’re leading everyone else in the right direction.
• Turn in a printed out hard-copy rough draft of your theorem paper assignment.
Week 10: Due Thursday April 11
• Read the solutions to the practice discrete section.
• Study for the discrete comps exam.
• Bring three printed copies of your draft paper in to class.
• Email me an abstract for your final talk.
Talks: April 18
• Patrick Bender: Newton’s Method
• Theo Frare-Davis: Bayes’s Theorem
• Adeline Zhang: Zero-Sum Games
• Dan Huth: Savitch’s Theorem
• Jane Bellamy: The Cauchy-Goursat Theorem
• Daniel Hermosillo: The Cauchy-Riemann Equations
• Andrea Stine: Cauchy’s Integral Formula
Talks: April 25
• Summer Li: Greedy Coloring Algorithms
• Jenny Yu: Four Color Theorem
• Jan Yan: Kruskal’s Algorithm
• Kate Grossman: Pick’s Theorem
• Junepyo Lee: Pairwise Integral Planar Distances
• Yuxin Xu: An Introduction to Groups
• Andrew Porter: Algebra and Convergence
• Alex Hernandez: Integral Tricks from Complex Analysis
Mathematical Communication
You may find the introduction to LaTeX I wrote elsewhere on this site helpful.
Summary Paper
• The goal of this assignment is to learn to communicate about mathematical ideas to an audience unfamiliar with the ideas you’re discussing. You should think of the audience as being your fellow
math majors: mathematically skilled but unfamiliar with the specific topic and field you are working in.
You should communicate the basic ideas, goals, and methods of the paper, but not get bogged down in the weeds of detailed calculations, technical arguments, or details of proofs.
• Choose a paper to summarize. The paper should probably be in the 4-10 page range.
• Rubric for the summary paper is here.
• You need to tell me the paper you have chosen and write an abstract for your summary paper by Thursday, February 14.
• Rough draft is due Thursday, February 28.
• Final draft is due Thursday, March 21.
Theorem Paper
• The goal of this assignment is to learn to communicate about mathematical ideas to an audience unfamiliar with the ideas you’re discussing. You should think of the audience as being your fellow
math majors: mathematically skilled but unfamiliar with the specific topic and field you are working in.
• Choose a theorem, result, or technique from a 300- or 400-level math class you have taken or are taking. You will write a paper of approximately three pages in which you clearly state the result,
explain how the result was obtained, and tell your reader why this result is relevant and important.
□ Mathematical results from courses in other departments may be approved on a case-by-case basis, but please consult with me to confirm.
• Rubric for the theorem paper is here.
• You need to tell me the paper you have chosen and write an abstract for your summary paper by Thursday, March 28.
• Rough draft is due Thursday, February April 4.
• Final draft is due Thursday, April 18.
Final Presentation
• The goal of this assignment is to practice communicating mathematical ideas to an audience via a verbal/slideshow presentation. The audience will be your fellow math majors. Assume they are
mathematically skilled, but unfamiliar with the specific topic you are discussing.
• Choose a theorem, result, or technique from a 300- or 400-level math class you have taken or are taking. You will give a ten-minute slideshow presentation in class in which you will explain the
result, the basic ideas behind it, and the reason it is important and relevant.
□ Mathematical results from courses in other departments may be approved on a case-by-case basis, but please consult with m to confirm.)
• Slides from my in-class Beamer presentation are here. You can also download the TeX source code or the plain text source code.
• You can find the rubric for the talk here
Comps Part 1
The math comprehensive exam will consist of five sections given on five separate days, spread throughout the semester. Each section will cover one of the five fundamental courses. Dates for these
exams are:
Thursday, February 7: Calculus 1
Thursday, February 21: Calculus 2
Thursday, March 7: Multivariable Calculus
Thursday, March 28: Linear Algebra
Thursday, April 11: Discrete Mathematics
There is a study guide available here that tells you what topics we think are most important. You can also download a complete (if somewhat dated) list of topics for each course:
We have a number of practice exams from previous years available.
Future Opportunities
I encourage everyone to look for a job/internship/research opportunity this summer. This will improve your resume, and also give you a better idea what sorts of things you might want to do after you
graduate. Many of these opportunities will pay you reasonably well.
Occidental Undergraduate Summer Research
Occidental has an undergraduate research program to sponsor you doing research with a professor. You would need to find a professor to mentor you; the program comes with a $4500 stipend and
subsidized (but not free) summer housing. The deadline is February 9.
Other Research Opportunities.
The NSF REU program funds experiences where a group of undergraduates from different institutions gather and do research on a math topic for ten weeks over the summer; typically the students also
receive a stipend of several thousand dollars.
A list of programs running this summer is available here. Deadlines are typically in February and March. (Most NSF programs are only available to US permanent residents).
There are also a number of summer programs listed at mathprograms.org. You can look around without making an account; the list of undergraduate programs seems to be here.
The AMS has an info page rounding up several summer internship opportunities.
The MAA also has a page of internship opportunities. It also has a roundup pages on careers here.
SIAM has a page on internships and careers.
I’ll try to add more resources as I find them. | {"url":"https://jaydaigle.net/teaching/courses/2019-spring-300/","timestamp":"2024-11-04T02:14:24Z","content_type":"text/html","content_length":"25051","record_id":"<urn:uuid:f22dc2f4-16a1-47cd-ae31-75f2c9032659>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00476.warc.gz"} |
RS Aggarwal Class 7 Solutions Chapter 4 Rational Numbers Ex 4E
These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E.
Other Exercises
Question 1.
Question 2.
Question 3.
Question 4.
Question 5.
Cost of 1 metre of cloth
Question 6.
Hope given RS Aggarwal Solutions Class 7 Chapter 4 Rational Numbers Ex 4E are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | {"url":"https://ncertmcq.com/rs-aggarwal-class-7-solutions-chapter-4-rational-numbers-ex-4e/","timestamp":"2024-11-09T12:50:32Z","content_type":"text/html","content_length":"60937","record_id":"<urn:uuid:ca961c95-3a41-4ce3-9b51-09d4c9d35e1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00003.warc.gz"} |
Using the critical set to induce bifurcations
We consider the classical problem of computing solutions of \(F(u)=g\), for a map \(F: D \subset X \to Y\) between real Banach spaces. We present a context in which the geometry of the function \(F\)
can be exploited by inducing bifurcations. From knowledge of the critical set \({\mathcal{C}}\) of \(F\), we indicate curves \(c \subset D\) with substantial intersection with \({\mathcal{C}}\),
taken as starting points of continuation algorithms.
The algorithm searches for points \(g_\ast\) with a large number of preimages, i.e., right hand sides for which \(F(u) = g_\ast\) has many solutions. Standard continuation methods may then solve \(F
(u) = g\) for a general \(g\) by inverting a curve in the image joining \(g_\ast\) to \(g\) starting at a preimage \(u_\ast\) of \(g_\ast\). ([1], [2]).
A standard method to obtain preimages of \(g\) first extends \(F\) to a function \(\tilde{F}(u,t)\) for which a simple curve \(d = (u(t),t)\) in the domain usually satisfies \(F(u(0),0) = g\).
Assuming some differentiability, one solves for additional preimages by first identifying bifurcation points \(u(t_c)\), in which \(D\tilde{F}(u(t_c), t_c)\) is not surjective for \(t_c \in [0,1]\).
Such points yield new branches of solutions \((\tilde{u}(t),t)\): if they extend to \(t=1\), new preimages of \(g\) are obtained. Additional bifurcations may arise along such branches. Usually, the
specification of the curve \(d\) is very restricted and does not allow for a choice of \(g = g_\ast\) as above. Continuation methods received a recent boost from ideas by Farrell, Birkisson and Funke
[3], which improved an elegant deflation strategy originally suggested by Brown and Gearhart [4].
The main idea is simple. Clearly, not every curve \(c \subset D\) intersecting abundantly the critical set \({\cal{C}}\) leads to points with many preimages. By considering very loose geometric
structure, we obtain indicators leading to more appropriate curves. The algorithm requires that we may decide if \(u \in D\) is a critical point (i.e., an element of \({\mathcal{C}}\)) or if \(u \in\
partial D\). Inversion near critical points relies on spectral information, in the spirit of Section 3.3 of [5].
We explore a geometric model for functions \(F: D \subset X \to Y\) between spaces of the same dimension. It combines standard results in analysis and topology outlined in Theorems 2 and 3 of Section
2. The model fits functions satisfying a weakened form of properness, in particular a class of semilinear differential operators.
Domain and counterdomain split in tiles, the connected components of \(F^{-1} (F({\mathcal{C}}\cup \partial D)) \subset X\) and \(F({\mathcal{C}}\cup \partial D) \subset Y\). The restriction of \(F\)
to a domain tile is a covering map onto an image tile. In particular, the number of preimages of points in an image tile is constant. Between adjacent image tiles, this number changes in a simple
fashion described in Theorem 3. Adjacent domain tiles separated by an arc of \({\mathcal{C}}\) are sent to the same image tile.
This approach started with the study of proper functions \(F: {\mathbb{R}}^2 \to {\mathbb{R}}^2\) by Malta, Saldanha and Tomei in the late eighties ([6]), leading to \(2\times 2\), a software that
computes preimages of \(F\), together with other relevant geometrical objects. Under generic conditions, the authors obtain a characterization of critical sets: given finite sets of curves \(\{{\
mathcal{C}}_i\}\) and \(\{{\mathcal{S}}_i\}\), and finite points \(\{p_{ik} \in {\mathcal{C}}_i\}\), one can decide if there is a proper, generic function \(F\) whose critical set consists of the
curves \(C_i\), with images \({\mathcal{S}}_i= F({\mathcal{C}}_i)\) and cusps at the points \(\{p_{ik}\}\). Given a function \(F\), \(2 \times 2\) obtains some critical curves \(\mathcal{C}_i\), its
images \({\mathcal{S}}_i\) and their cusps \(\{p_{ik}\}\): if the characterization does not hold, it provides the program with information about where to search for additional critical curves.
Two important features of the 2-dimensional context do not extend: (a) a description of the critical set as a list of critical points, (b) the identification of higher order singularities. A
realistic implementation for \(n > 2\) led us to the current text. Some mathematical concepts in \(2 \times 2\) found application in different scenarios (for ODEs, [8]–[10]; for PDEs, [11], [12]).
In all examples in this text, the starting point of our arguments lie in the identification of the critical set of the underlying function. In Sections 2 and 3 we present the geometric context and
apply the algorithm to some visualizable examples, so that the counterparts of bifurcation diagrams can be presented concretely.
The second class of examples, discussed in Section 4, arises from the discretization of a nonlinear Sturm-Liouville problem, \[F(u) = - u'' + f(u) = g , \quad u(0) = 0 = u(\pi) .\] For a uniform mesh
with spacing \(h\), the discretized equation \(F_h(u) = g\) has an unexpectedly high number of solutions ([10]), and most do not admit a continuous limit, but we consider the problem as an example of
interest by itself. The connected components \({\mathcal{C}}_i, i=1, \ldots, n\), of the critical set of \(F_h\) are graphs of functions \(\gamma_i: V^\perp \to V\), where \(V\) is the one
dimensional space generated by the ground state of the discretization of \(u \mapsto -u''\), which, as is well known, is the evaluation \(u_h\) of \(u(x) = \sin x\) at points of the uniform mesh on \
([0, \pi]\). Thus, straight lines in the domain which are parallel to the vector \(u_h\) contain many critical points. Parallel lines give rise to additional solutions, for geometric reasons we make
Our research was inspired by the celebrated Ambrosetti-Prodi theorem ([13], [14]). Subsequent articles ([11]–[13], [15], [16]) provided information about the geometry of semilinear elliptic
operators, with implications to the underlying numerics. In [17], Allgower, Cruceanu and Tavener considered numerical solvability of semilinear elliptic equations by first obtaining good
approximations of solutions from discretized versions of the problem. A filtering strategy eliminates some discrete solutions which do not yield a continuous limit, and algorithms, somehow tailored
to the form of the equations, are presented and exemplified. As they remark, results on the number of solutions are abundant, but not really precise.
The third context, in Section 5, is a semilinear operator \(F(u) = -\Delta u - f(u)\) considered by Solimini ([18]) for which a special right hand side \(g\) has six preimages. Let \(\phi_0\) be the
ground state of the free (Dirichlet) Laplacian. As we shall see, from the min-max characterization of eigenvalues, lines of the form \(u_0 + t \phi_0\) contain abundant bifurcation points and suffice
to yield all solutions.
In the Appendix we handle inversion of segments in the neighborhood of the critical set \({\mathcal{C}}\). Instead of using variables related to arc length [1], we work with spectral variables, in
the spirit of [5], [11] and [12]. In particular, we avoid the difficulty of partitioning Jacobians for discretizations of infinite dimensional problems.
We used \(2 \times 2\) and MATLAB for graphs and numerical routines.
We thank Nicolau Saldanha and José Cal Neto for extended conversations. Tomei gratefully acknowledges grants from FAPERJ (E-26/202.908/2018, E-26/200.980/2022) and CNPq (306309/2016-5, 304742/
2021-0). Kaminski and Monteiro received graduate grants from CNPq and CAPES/FAPERJ respectively.
Basic geometry↩︎
For real Banach spaces \(X, Y\) and an open set \(U \subset X\), we consider a function \(F: U \to Y\) with continuous derivative, which we usually restrict to closed domains \(D \subset U\) with
smooth boundary \(\partial D\). Recall that the critical set \({\mathcal{C}}\) of \(F: U \to Y\) is \[{\mathcal{C}}\;= \;\{ u \in U \;| \;F \;is not a local homeomorphism from \;u \;to \;F(u)\} \;.\]
Regular points are points of \(U\) which are not critical. Since \(F\) is of class \(C^1\), if the Jacobian \(DF(u): X \to Y, u \in U\), is an isomorphism, the inverse function theorem implies that \
(u\) is a regular point and \(F\) is a local diffeomorphism of class \(C^1\) at \(u\). As \(D \subset U\), any point in \(D\) (in particular in \(\partial D\)) can be critical or not.
As in [6] (and in [19] for domains with boundary), define the flower \(\mathcal{F} = F^{-1}(F({\cal{C}}\cup \partial D))\), a convenient description of the geometry of \(F\). As an example, Berger
and Podolak [15] proved that, under the hypotheses of the Ambrosetti-Prodi theorem (Theorem 7), the flower of \(F\) is essentially the simplest: \(\mathcal{F} = {\mathcal{C}}\) and is diffeomorphic
to a topological hyperplane.
In general, domain \(D\) and codomain \(Y\) split into tiles, the connected components of \(D \setminus \mathcal{F}\) and \(Y \setminus F({\mathcal{C}}\cup \partial D)\) respectively. We assign a
common label \(A_i\) to a point and its image \(A = F(A_i)\).
As a first example of the geometric model in the Introduction, consider \[F: U = D= X = {\mathbb{R}}^2 \to Y = {\mathbb{R}}^2 \;, \quad (x,y) \mapsto (x^2 - y^2 +x, 2 xy - y)\;.\]
Figure 1: Five tiles in the domain, two in the codomain.
Its domain, on the left of Figure 1, contains the critical set \({\cal{C}}\), a circle, and the flower \(\mathcal{F}\), consisting of three curvilinear triangles with vertices \(X_i, Y_i\) and \(Z_i,
i=1,2\). There are five tiles in the domain, two in the codomain. The map \(F\) is a homeomorphism from each bounded tile to the bounded tile \(X_1Y_1Z_1\) in the image. Points in the image tile \
(XYZ\) have four preimages (in particular, \(F(0) = 0\) has four preimages, indicated by \(0_1, \ldots, 0_4\)). The unbounded tile \(\mathcal{R}\) in the domain is taken to the unbounded tile \(F(\
mathcal{R})\), but the map is not bijective: each point in \(F(\mathcal{R})\) has two preimages, both in \(\mathcal{R}\). More geometrically, all restrictions of \(F\) to tiles are covering maps (
[20]), and thus must be diffeomorphisms when their image is simply connected. In Figure 1, the numbers on the tiles of the image are the (constant) number of preimages of points in each tile.
Local theory↩︎
We consider the behavior of \(F\) at points in \({\mathcal{C}}\) and \(\partial D\).
In Figure 1, points in different tiles in the image (necessarily adjacent, in this case, i.e., sharing an arc of images of critical points) have their number of preimages differing by two. Inversion
of the segment \([(0,0) , P]\) by continuation gives rise to two subsegments, whose interiors have two and four preimages. Starting from \(P\) with initial condition \(P_1\) or \(P_2\), inversion
carries through to \(0\) without difficulties, giving rise to roots \(0_1\) and \(0_2\). However, when inverting from \(0\) with initial conditions \(0_3\) and \(0_4\), continuation is interrupted.
This is the expected behavior of inversion by continuation at a fold, as we now outline.
Figure 2: Near a fold \(u_\ast\).
A critical point \(x \in U\) is a fold of \(F\) if and only if there are local changes of variables centered at \(x\) and \(F(x)\) for which \(F\) becomes \[(t, z) \in {\mathbb{R}}\times Z \mapsto (t
^2, z) \in {\mathbb{R}}\times Z\] for some real Banach space \(Z\). On Figure 2, point \(u_\ast\) is a fold, and the vertical line through it splits in two segments, \(\beta_0\) and \(\beta_2\). The
inverse of \(F(\beta_0)\) yields again \(\beta_0\) and another arc \(\beta_1\), also shown in Figure 2. In a similar fashion, the inverse of \(F(\beta_2)\) contains \(\beta_2\) and another arc \(\
beta_3\). Notice the similarity with a bifurcation diagram: two new branches \(\beta_1\) and \(\beta_3\) emanate from \(r\).
Folds are identified as follows. Let \(X\) and \(Y\) be real Banach spaces admitting a bounded inclusion \(X \hookrightarrow Y\). Denote by \({\mathcal{B}}(X,Y)\) the set of bounded operators between
\(X\) and \(Y\), endowed with the usual operator norm.
Theorem 1.
Suppose that the function \(F: U \to Y\) is of class \(C^3\). For a fixed \(x \in U\), let the Jacobian \(DF(x): X \to Y\) be a Fredholm operator of index zero for which 0 is an eigenvalue. Let \(\
operatorname{Ker}DF(x)\) be spanned by a vector \(k \in X\) such that \(k \notin \operatorname{Ran}DF(x)\). Then the following facts hold.
(1) For some open ball \(B \subset {\mathcal{B}}(X,Y)\) centered in \(DF(x)\), operators \(T \in B\) are also Fredholm of index zero and \(\dim \operatorname{Ker}T \le 1\).
(2) There is a map \(\lambda_s: T \in B \to {\mathbb{R}}\) taking \(T\) to its eigenvalue of smallest module, which is necessarily a real eigenvalue. The map is real analytic. For a suitable
normalization, the corresponding eigenvector map \(T \mapsto \phi_s(T) \in X\) is real analytic.
(3) For a small open ball \(B_x \subset X\) centered in \(x\), there are \(C^3\) maps \[\tilde{x} \in B_x \mapsto \lambda_s( DF(\tilde{x})) \in {\mathbb{R}}\;, \quad x \mapsto \phi_s( DF(\tilde{x}))
\in X\;.\]
If additionally \(D \lambda_s (x) . \phi(x) \;\ne \;0\), \(x\) is a fold.
In particular, the critical set \({\mathcal{C}}\) of \(F\) is a submanifold of \(X\) of codimension one near \(x\). Jacobians are not required to be self-adjoint operators. Characterizations of folds
in the infinite dimensional context may be found in [21]–[25].
Proof. We barely sketch a lengthy argument. The implicit function theorem yield items (1) and (2), as described in Proposition 16 of [26]. Item (3) then follows. Proposition 2.1 of [25] implies item
(4). ◻
We now consider points at the boundary \(\partial D\). As an example, restrict \(F\) above to a disk \(U\) with boundary \(\partial D\) as in Figure 3. The flower now includes \(\partial D\) and an
additional dotted curve \(F^{-1} (F(\partial D))\), containing the preimages \(A_i, B_i, C_i, i=1,2\) of points \(A, B\) and \(C\). Each one of the tiles I, II and III has a single preimage inside \
(U\), the other being outside. Trespassing an arc of images of points of \(\partial D\) only changes the number of preimages by one.
Figure 3: Behavior at the boundary.
Restricting functions to tiles, b-proper functions↩︎
A function \(F: D \to Y\) is proper if the inverse of a compact set of \(Y\) is a compact set of \(D\). It is proper on bounded sets (or equivalently, b-proper) if its restriction \(F_B: B \subset D
\to Y\) to bounded, closed sets \(B\) is proper. As is well known, a continuous function \(F: {\mathbb{R}}^n \to {\mathbb{R}}^n\) is proper if and only if \(\| F(x) \| = \infty\) as \(\|x\| \to \
infty\). Proper functions are b-proper, but the second definition incorporates a class of elliptic semilinear operators.
Proposition 1. Let \(Y\) be a real Banach space and \(G: Y \to Y\), \(G(u) = u + \Phi(u)\), be continuous. Suppose that, for any closed ball \(B \subset Y\), \(\overline{\Phi(B)}\) is a compact set.
Then \(G\) is b-proper. Moreover, \(G\) is proper if and only if \(\| G(x) \| \to \infty\) as \(\|x\| \to \infty\).
Proof. We prove that \(G\) is b-proper, the remaining argument is left to the reader. For a sequence \(y_n \in K\) with \(y_n\to y_\infty \in K\), and \(u_n \in B \subset Y\) such that \(G(u_n) = u_n
+ \Phi(u_n) = y_n\). For an appropriate subsequence, \(\Phi(u_{n_m}) \to w \in Y\) and then \(u_{n_m}= - \Phi(u_{n_m}) + y_{n_m}\to -w + z_\infty \in Y\) and hence \(B \cap G^{-1}(K)\) is compact. ◻
Corollary 2. Let \(\Omega \subset {\mathbb{R}}^n\) be a bounded set with smooth boundary. For a smooth \(f: {\mathbb{R}}\to {\mathbb{R}}\), set \(F:X = C^{2,\alpha}_D(\Omega) \to Y = C^{0,\alpha}(\
Omega)\) given by \(F(u) = - \Delta u + f(u)\). Then \(F\) is b-proper and \(F\) is proper if and only if \(\| F(x) \| = \infty\) as \(\|x\| \to \infty\).
Here, \(C^{2,\alpha}_D(\Omega)\) is the Hölder space of functions equal to zero on \(\partial \Omega\), \(\alpha \in (0,1)\).
Proof. Set \(G: Y \to Y, \;G(v) = v + \Phi(v)\), where \(\Phi(v) = f(\Delta^{-1} v)\). Recall that \(-\Delta: X \to Y\) is an isomorphism and the inclusion \(X \hookrightarrow Y\) is compact. The
nonlinear map \(\Phi\) satisfies the hypotheses of the previous proposition ([27]). ◻
The geometric model of the Introduction holds for b-proper functions: here is the first step.
Theorem 2. For a bounded, open set \(U \subset X\), consider a b-proper function \(F: U \to Y\). Then the restriction of \(F\) to a bounded tile \(T_D \subset U \setminus {\mathcal{F}}\) in the
domain is a covering map \(F_D: T_D \to T_C = F(T_D)\) of finite degree, where \(T_C\) is a tile in the image. Said differently, \(F_D\) is a surjective local diffeomorphism and all points in \(T_C\)
have the same (finite) number of preimages.
Up to sign, the degree of \(F_D\) is the number of the preimages under \(F_D\) of any point in \(T_C\). In particular, one may consider degree theory on restrictions of \(F\) to tiles. In [6] some
relationships between the number of cusps and the degree of restrictions of \(F\) to tiles are used in the search of critical curves.
Theorem 2 is a standard argument in the theory of covering spaces [20]. For a proper function \(F: X \to Y\), it holds also for unbounded tiles \(T_D\) in the domain.
Proof. Since \(T_C\) is an open connected set, surjectivity follows once we prove that \(F(T_D) \subset T_C\) is open and closed. Openness is clear from the inverse function theorem, as every point
of \(T_D\) is regular. We now show that \(F(T_D)\) is closed. Take a convergent sequence \(y_n = F(x_n) \in T_C\), \(y_n \to y_\infty \in T_C\). As \(F\) is a proper bounded function, there is a
subsequence \(x_{n_m} \in \overline{T_D}\) such that \(x_{n_m} \to x_\infty \in \overline{T_D}\) and \(F(x_\infty)= y_\infty\). As \(y_\infty \in T_C\), it is not in \(F({\mathcal{C}}\cup \partial D
\)), and thus \(x_\infty \in T_D\).
Consider the restriction \(F_D: T_D \to T_C = F(T_D)\): we show that \(y \in T_C\) has finitely many preimages. By properness of \(F_D\), \(F^{-1}(y)\) is a compact set of \(T_D\): if \(F^{-1}(y)\)
is infinite, it has an accumulation point \(x_\ast\) for which \(F(x_\ast)=y\). Also, \(x_\ast \notin \mathcal{C}\), as \(y \in T_C\). But at regular points, \(F\) is a local homeomorphism: there are
no convergent sequences to \(x_\ast\) in \(F^{-1}(y)\).
Since \(F_D\) is a surjective local homeomorphism and each point has finitely many preimages, it is a covering map: as \(T_D\) is connected, the fact that all points in \(T_C\) have the same number
of preimages follows. We give details for the reader’s convenience. By connectivity, it suffices to show that points sufficiently close to \(y \in T_C\) have the same number of preimages. Suppose
then \(F^{-1}(y) = \{x_1, \ldots, x_k\} \subset T_D\), a collection of regular points: there must be sufficiently small, non- intersecting neighborhoods \(V_{x_i}, i=1, \ldots,k\) of the points \(x_i
\) and \(V_y \subset T\), such that the restrictions \(F:V_{x_i} \to V_y\) are homeomorphisms. Thus, points in \(V_y\) have at least \(k\) preimages. Assume by contradiction a sequence \(y_n \to y,
y_n \in V_y\), such that each \(y_n\) admits an additional preimage \(x_n^\ast\), necessarily outside of \(\cup_i V_{x_i}\). By properness, for a convergent subsequence \(x_{m_n}^\ast \to x_\infty^\
ast\), \(F(x_\infty^\ast)= y\), and \(x_\infty^\ast \ne x_i, i=1,\ldots,k\). ◻
A point \(y \in F(\partial D)\) is a generic critical value if its preimages are regular points together with a single point \(x \in {\mathcal{C}}\) which is a fold. Similarly, \(y \in F(\partial D)
\) is a generic boundary value if its preimages contain regular points and a single point \(x \in \partial D\) for which \(F\) extends as a local homeomorphism in an open neighborhood of \(x \in X\).
The following result completes the validation of the geometric model.
Theorem 3. Let \(F\) be as in Theorem 2. Suppose two bounded tiles in the image of \(F\) have a common point \(y\) in their boundaries which is a generic critical value (resp. a generic boundary
value). Then the number of preimages of points in both tiles differ by two (resp. by one).
The result admits natural extensions. The image of the function \(F: {\mathbb{R}}^2 \to {\mathbb{R}}^2\) given by \(F(x,y) = (x^2, y^2)\) is the positive quadrant, and leaving it through a boundary
point different from the origin implies a change of number of preimages equal to four: in a nutshell, the boundary point has two preimages which are folds.
Additional hypothesis naturally hold for usual functions, by arguments in the spirit of Sard’s theorem. At a risk of sounding pedantic, examples are the following.
(H1) \({\mathcal{C}}, \partial D, F({\mathcal{C}})\) and \(F(\partial D)\) have empty interior.
(H2) A dense subset of \({\mathcal{C}}\) consists of folds.
Instead of verifying such facts in the examples, we take the standard approach in numerical analysis: one proceeds with the inversion process and accepts an occasional breakdown.
Visualizable applications of the algorithm↩︎
The image of the function \[F: \mathbb{R}^2 \to \mathbb{R}^2 \;, \quad (x,y) \mapsto (\cos (x) - x^2\cos (x) + 2x \sin (x), y)\] is well represented by a piece of cloth pleated along vertical lines.
As indicated in Figure 4, the critical set \(\mathcal{C}=\{(k\pi,y), \; k \in {\mathbb{Z}}, \;y \in \mathbb{R}\}\) and its image \(F({\mathcal{C}})\) consist of vertical lines. All critical points
are folds. Points \(p\) in the image are covered a different number of times, the number of preimages of \(p\).
Figure 4: Pleats, the bifurcation diagram \(\mathcal{B}\) and many preimages \(P_i\) of \(g=F(P_0)\).
In Figure 4, the line \(r\) passes through \(P_0\), a preimage of \(g \in Y\). The image \(F(r)\) oscillates among images of critical lines. Inversion of \(F(r)\) yields \(\mathcal{B}\), the
bifurcation diagram associated with \(r\) from \(P_0\), the connected component of \(F^{-1}(F(r))\) containing \(P_0\). Bifurcation points at the intersection of \(r\) and \(\mathcal{C}\) in turn
gives rise to other preimages \(P_i\) of \(g=F(P_0)\). The line \(r\) is chosen so as to intersect \(\mathcal{C}\) abundantly.
Consider now the smooth (not analytic) function \[F:{\mathbb{C}}\to {\mathbb{C}}\;, \quad z \mapsto z^3 +\frac{5}{2}\;{\overline{z}}^2+z \;.\] The critical set \({\mathcal{C}}\) consists of the two
curves \(\mathcal{C} _1\) and \(\mathcal{C} _2\) in Figure 5, which roughly bound the three different regimes of the function: \(z \sim 0, 1\) and \(\infty\), where \(F\) behaves like \(z\), \({\
overline{z}}^2\) and \(z^3\) respectively. The three regimes already suggest that a line through the origin must hit the critical set at least four times.
We count preimages with Theorem 3. From its behavior at infinity, \(F\) is proper. Points in the unbounded tile in the image have three preimages, as for \(z \sim \infty\), the function is cubic. The
unbounded tile in the domain covers the unbounded tile in the image of the right hand side three times. Each of the five spiked tiles has five preimages, and the annulus surrounding the small
triangle, seven. Finally, the interior of the small triangle has nine preimages: \(F\) has nine zeros. The flower, in Figure 6, illustrates these facts.
Figure 5: The critical set \({\mathcal{C}}\), the line \(r\) and their images.
Figure 6: On the left, \({\mathcal{C}}, r, \mathcal{F}\) and \(\mathcal{B}\). On the right, \(r, {\mathcal{C}}, \mathcal{B}\) and an extension yielding a zero \(P_8\).
Let \(r\) be the vertical axis, \(P_0=(0,0) \in r\). Figure 5 shows \(r\) and \(F(r)\) and Figure 6 the flower \(\mathcal{F} = F^{-1}(F(\mathcal{C}_1)) \cup F^{-1}(F(\mathcal{C}_2))\): dotted black
lines and \({\mathcal{C}}_1\) form \(F^{-1}(F(\mathcal{C}_1))\), while \(F^{-1}(F(\mathcal{C}_2))\) consists of continuous black lines. Amplification of \(\mathcal{F}\) shows five thin triangles, one
in each ‘petal’ (one is visible in the petal on the left), each a full preimage of the triangle \(XYZ\) in the image (the enlarged detail in Figure 5). Add the other four preimages at tiles bounded
by \(F^{-1}(F(\mathcal{C}_1))\) to spot the nine zeros of \(F\). The flower is computationally expensive and is not computed in the algorithm we present.
On the left of Figure 6 are shown \({\mathcal{C}}\), \(\mathcal{F}\), \(r\) and \(\mathcal{B}\), the bifurcation diagram associated with \(r\) from \(P_0=(0,0)\). To emphasize \(r\) and \(\mathcal{B}
\), we removed \(\mathcal{F}\) on the right of Figure 6. The sets \(r\) and \(\mathcal{C}\) meet at four folds. The set \(\mathcal{B}\) contains the line \(r\) and eight of the nine zeros of \(F\).
The missing zero, \(P_8\), is on the petal on the left of Figure 6.
The branches originating from the four critical points in \(r\) lead by continuation to additional zeros \(P_i\) of \(F\). What about the missing zero? For the half-line \(s\) joining \(P_2\) to
infinity in Figure 6, bifurcation at \(u_*\in F^{-1}(F(s))\) obtains \(P_8\). For completeness, \[\begin{array}{c} P_1=( 0.2141, 0.3313) \;, \quad P_2 = (-0.5367,0.0000)\;, \quad P_3=( -0.7893,
2.5802)\;, \\ \quad P_4=(1.7752,1.3903)\;, \quad P_5=( 0.2141,-0.3313)\;, \quad P_6=(-0.7893, -2.5802)\;,\\ P_7=(1.7752,-1.3903)\;, \quad P_8=(-1.8633,0.0000) \;. \end{array}\]
Singularities and global properties of \(F\)↩︎
The example above shows that curves \(r\) with different intersection with the same critical component of \({\mathcal{C}}\) may yield different zeros. This will happen in other examples in the text:
curves will be parallel lines, each generating a set of zeros. We provide an explanation for this fact based on the geometric model.
This is already visible for the simpler function \(F\) in Figure 1. Informally, \(F\) at a fold point \(p\) of \({\mathcal{C}}\) looks like a mirror, in the sense that points on both sides of the arc
\({\mathcal{C}}_i\) of \({\mathcal{C}}\) near \(p\) are taken to the same side of the arc \(F({\mathcal{C}}_i)\) near \(F(p)\), as shown in Figure 2. In the example in Figure 1, at the three points
which are not folds, one might think of adjacent broken mirrors. Close to the image \(X = F(X_1)\) of such points, say \(X_1\), there are points with three preimages close to \(X_1\).
In higher dimensions (in particular, infinite dimensional spaces), there are critical points at which arbitrarily many splintered mirrors coalesce. At such higher order singularities, there are
points with clusters of preimages. H. McKean conjectured the existence of arbitrarily deep singularities for the operator \(F(u) = - u'' + u^2\), where \(u\) satisfies Dirichlet conditions in \([0,1]
\), and the result was proved in [28].
The original algorithm, \(2 \times 2\), identifies and explore cusps, a higher order singularity, in the two dimensional context. The lack of information about higher order singularities forces the
current algorithm to perform more searches, by essentially choosing curves \(r\) intercepting the critical set along different mirrors.
The region surrounded by \({\mathcal{C}}\) in Figure 1 has three mirrored images. The presence of higher singularities is related to the fact that there may be tiles on which the covering map induced
by the restriction \(F\) is not injective, as in the case of the annulus tile in the domain of \(F\) of this section. This in turn requires the image tile to have nontrivial topology (more precisely,
a nontrivial fundamental group), as simply connected domains are covered only by homeomorphisms.
Discretized nonlinear Sturm-Liouville maps↩︎
We consider the nonlinear Sturm-Liouville operator between Sobolev spaces, \[F:X=H^2([0,\pi]) \cap H^1_0([0,\pi]) \longrightarrow Y=L^2([0,\pi]) \;, \quad F(u) = -u''-f(u) \;.\] We compare two
theorems which count solutions of \(F(u) = g\) for special functions \(g\) for the operator and its discretizations. Recall that the linear operator \(u \mapsto - u''\) acting on functions satisfying
Dirichlet boundary conditions has eigenvalues equal to \(\lambda_k = k^2, k=1, 2, \ldots\) and associated eigenvectors \(\phi_k = \sin(kx)\).
Theorem 4 (Costa-Figueiredo-Srikanth [29]). Let \(f:\mathbb{R} \to \mathbb{R}\) be a smooth, convex function which is asymptotically linear with parameters \(\ell_-\), \(\ell_+\), \[\lim_{x \to - \
infty} f'(x) = \ell_-, \quad \lim_{x \to \infty} f'(x) = \ell_+ \;,\] \[\ell_- < \lambda_1 = 1, \;\;\lambda_k = k^2<\ell_+<(k+1)^2 = \lambda_{k+1}, \quad \ell_-, \ell_+ \notin \{k^2, k \in \mathbb{N}
\} \;.\] Then, for large \(t > 0\), \(F(u) = -t\sin(x)\) has exactly \(2k\) solutions.
The standard discretization of \(F\) is defined over the regular mesh \[\overline{I_h}=\left\{x_i = ih \;, \;i=0,...,n+1 \;, h=\frac{\pi}{n+1} \right\} \;.\] As usual, for points \(x_i \in I_h=\
overline{I_h} \setminus \{x_0,x_{n+1}\}\), approximate \[-u''(x_i) \sim \frac{1}{h^2} \left(-u(x_{i+1})+2u(x_i)-u(x_{i-1})\right)\] and consider the tridiagonal, \(n \times n\) symmetric matrix \(A^h
\) with diagonal entries \(2/h^2\) and remaining nonzero entries equal to \(-1/h^2\). Its (simple) eigenvalues are \[\lambda_k^h \;= \frac{2}{h^2} - \frac{2}{h^2}\cos(\pi-kh) , \;k = 1, \ldots, n\]
with associated eigenvectors \(\phi_k^h = \sin(k I_h)= (\sin(kx_i))\) for \(x_i \in I_h\). Similarly, set \(u^h = u(I_h)\) and \(f(u^h)=(f(u(I_h))^T\). Finally, the discretized operator is \[F^h: \
mathbb{R}^n \to \mathbb{R}^n , u^h \mapsto A^h u^h - f(u^h) .\]
Theorem 5 (Teles-Tomei [10]). Let \(f\) be smooth, convex, asymptotically linear function with parameters \(\ell_-\) and \(\ell_+\) satisfying \(\ell_-< \lambda_1^h \;, \lambda_k^h < \ell_+ < \
lambda_{k+1}^h\). Let \(y,p \in \mathbb{R}^n\), \(p_i > 0\). Then, for large parameters \(t >0\) and \(\ell_+\), the equation \[F^h(u^h)= A^hu^h-f(u^h)=y -tp,\] has exactly \(2^n\) solutions.
As \(h \to 0\), most solutions of \(F(u) = g\) disappear. The discretized problem is an interesting test case for our algorithm: \(F^h(u)=g\) has abundant solutions.
The critical set of \(F\) in both contexts has been studied in [9] and [8]. The property below is all it takes to implement the algorithm. Notice that Jacobians in both cases consist of linear
operators with eigenvalues labeled in increasing order.
Theorem 6. Lines in the domain with direction given by a positive function (vector, in the discrete case) intercept the critical set at points abundantly. Essentially, there is one intersection
associated with one eigenvalue.
This follows from min-max arguments on the Jacobians \(DF(u)\) along such lines. The argument is given in a more general context — for Jacobians given by semilinear elliptic operators — in
Proposition 4.
Notice the affinity beween such lines and rays from the origin in the examples of Section 3. As in Section 3.1, different curves might lead to a different set of solutions obtained by bifurcation and
Piecewise linear geometry and a data base of solutions↩︎
We follow [29] and [10], and consider a piecewise linear map \[\label{piecewise} f(x) = \begin{cases} \ell_- x\;, \;x < 0 \\ \ell_+ x\;, \;x > 0 \end{cases} \;.\tag{1}\] The associated function \(F^h
\) is globally continuous and linear when restricted to each orthant of \(\mathbb{R}^n\). The lack of differentiability leads us to be careful with the identification of the critical set \({\mathcal
{C}}\), an issue we address in Section 4.2.
For the case of the piecewise linear function \(f\), an explicit solution for the continuous problem was given by Lazer and McKenna [30]. For \(\ell_-<\lambda_1 = 1 <\ell_+\) and \(t>0\), two
solutions of \(-u''-\ell_+ u^{+}+\ell_- u^{-}=-t\sin(x)\) are \[\frac{ t\sin(x)}{\ell_+ -1} > 0 \;, \quad \frac{t\sin(x)}{\ell_- -1} < 0 \;.\] For the discretized operator, it is easy to verify that
two solutions are given by \[\label{LazerMcKenna} \frac{t\sin(I_h)}{\ell_+ -\lambda_1^h} \;\;\; \text{and} \;\;\; \frac{t\sin(I_h)}{\ell_- - \lambda_1^h}, \quad \lambda_1^h=\frac{2}{h^2} \left(1-\cos
(h) \right) \;.\tag{2}\] Again, the entries of the first (resp. second) solution are positive (resp. negative).
Following [10], we consider lines \(r\) aligned to the (normalized, positive) eigenvector \(\phi_1^h\) associated to the smallest eigenvalue \(\lambda_1^h\) of \(A^h\). Set \(V = \langle \phi_1^h \
rangle\), \(H = V^\perp\) and split \(\mathbb{R}^n = H \oplus V\). The critical set \(\mathcal{C}\) of the map \(F^h: \mathbb{R}^n \to \mathbb{R}^n\) consists of \(n\) hypersurfaces \(\mathcal{C}_j,
j=1, \ldots,n\). Each \(\mathcal{C}_j\) is a graph of a function \(c_j : H \to V\), and in particular projects diffeomorphically from \(\mathbb{R}^n\) to \(H\). Thus, topologically, \(\mathcal{C}_j\)
is trivial. Each line \(r\) intercepts all the critical components \(\mathcal{C}_j\) of \(F^h\).
The images \(F(\mathcal{C}_j)\) are more complicated: for \(\ell_+ >> 0\), \(F(\mathcal{C}_j)\) wraps around the line \(\{t(1, 1, \ldots, 1), t \in \mathbb{R} \}\subset \mathbb{R}^n\) substantially:
\(\binom{n-1}{j-1}\) times! It is this geometric turbulence which gives rise to the abundance of preimages for appropriate right hand sides. Said differently, some tiles in the image have nontrivial
fundamental group and are abundantly covered. The reader should compare this situation with the second example in Section 3: cubic behavior at infinity leads winding of the image of the outermost
critical curve and to nine zeros of \(F\).
Set \(n = 15\), \(g^h = -1000 \sin(I_h)\), \(f\) as in (1 ) with asymptotic parameters
\[\ell_-=\frac{\lambda_{1}^h}{2} \;\;\text{and} \;\; \ell_{+}^k=\frac{\lambda_k^h+\lambda_{k+1}^h}{2}, \;\text{for} \;k=1,...,14 \;, \;\;\ell_{+}^{15}=\lambda_{15}^h+\frac{\lambda_{1}^h}{2}.\] The
number of solutions \(N\) for different \(\ell_+^k\) is given below [10].
\(N\) 2 4 6 8 12 12 22 24 32 100 286 634 972 1320 2058
For small \(k\), \(N(g^h)\) is of the form \(2k\), as for the continuous counterpart. At some point, \(N(g^h)\) becomes exponential, \(2^k\). The numerics in [10] is elementary: solve a linear system
in each orthant and check if the solution belongs to the orthant, generating a reliable data bank of solutions. Here, we compare solutions obtained by inducing bifurcations with those in the data
bank. Another computational alternative, the shooting method for the discretized problem, did not perform well: we intend to explore the issue in a forthcoming paper.
The critical set of \(F^h\) for a piecewise linear \(f\)↩︎
We now relate criticality and the piecewise linear nature of the function \(f\). An (open) orthant \({\mathcal{O}}\subset {\mathbb{R}}^n\) is defined by the sign of its vector entries. For \(v \in {\
mathcal{O}}\), \(f(v)\) is multiplication by a diagonal matrix \(D^{\mathcal{O}}\), with diagonal entries equal to \(\ell_-\) or \(\ell_+\), according to the sign of the associated entry of \(v\).
Thus, the continuous map \(F^h: {\mathbb{R}}^n \to {\mathbb{R}}^n\) is of the form \(F^h = A^h - D^{{\mathcal{O}}}\) when restricted to \({\mathcal{O}}\).
Generically (i.e., for an open, dense set of pairs \((\ell_-, \ell_+)\)), all the matrices \(A^h - D^{{\mathcal{O}}}\) are invertible. Assuming this, in each orthant \({\mathcal{O}}\) the (linear)
map is trivially a local diffeomorphism. The critical set of \(F^h\) makes (topological) sense: it is the set in which \(F^h\) is not a local homeomorphism. Say a vector \(v\) is in the boundary of
exactly two orthants, \({\mathcal{O}}_1\) and \({\mathcal{O}}_2\) (equivalently, \(v\) has a single entry equal to zero). If \({\operatorname{sign}}\det (A^h - D^{{\mathcal{O}}_1}) = {\operatorname
{sign}}\det (A^h - D^{{\mathcal{O}}_1})\), the map \(F^h\) is a local homeomorphism at \(v\). If instead \({\operatorname{sign}}\det (A^h - D^{{\mathcal{O}}_1}) = -{\operatorname{sign}}\det (A^h - D^
{{\mathcal{O}}_1})\), \(F^h\) near \(v\) is a topological fold. The critical set consists of pieces of coordinate planes between two orthants in which \(\det (A^h - D^{{\mathcal{O}}})\) changes sign.
In the nongeneric situation, whenever \(A^h - D^{{\mathcal{O}}}\) is not invertible, we include \({\mathcal{O}}\) in the critical set.
The critical set is connected. A (generic) line \(r\) parallel to the vector \(\phi_1^h\) intercepts \(\mathcal{C}\) in as many eigenvalues of \(A^h\) there are between \(\ell_-\) and \(\ell_+\).
The visualizable case, \(n=2\)↩︎
Due to the lack of differentiability of \(f\), we cannot use Newton’s method as a local solver. We present the alternative approach in a visualizable example. If \(n=2\), we have \(h = \pi/3\) and
the matrix \(A^h\) has eigenvalues \(\lambda_1 \approx 0.9119\) and \(\lambda_2 \approx 2.7357\). For the values \(\ell_-=-1\) and \(\ell_+=2\) or \(\ell_+=4\), \(F^h = A^h - D^{{\mathcal{O}}}\) in
the interior of each quadrant is invertible. Figure 7 describes the critical sets and their images for the two values of \(\ell_+\). In each quadrant of the domain we indicate \({\operatorname{sign}}
\det (A^h - D^{\mathcal{O}})\). In each connected component of \(\mathbb{R}^2 \setminus F^h(\mathcal{C})\) we specify instead the number of preimages. The fact that \({\mathcal{C}}\) for \(\ell_+ = 4
\) is not a (topological) manifold is innocuous.
Figure 7: \(\mathcal{C}\) and \(F^h(\mathcal{C})\) for \(\ell_-=-1, \ell_+=2\) and for \(\ell_-=-1, \ell_+=4\)
Figure 8: \(\mathcal{B}\) (left) and \(F^h(r)\) (right) for \(\ell_-=-1, \ell_+=4\)
We consider \(\ell_-=-1\), \(\ell_+=4\). To invert \(g=-1000\sin(I_h)\), we start with the positive solution \(P_0 \approx ( 280.4396, 280.4396)\) (as in Section 4.2) and draw the half-line \(r =\
{P_0+s(0.2\sin(2I_h)-0.8\sin(I_h)), \; s \geq 0\} \subset \mathbb{R}^{2}\) through \(P_0\). We add the term \(0.2\sin(2I_h)\) to minimize the possibility that \(r\) intercepts \({\mathcal{C}}\) at
points which are not (topological) folds, i.e., points with more than one entry equal to zero (this is especially relevant for \(n\) large). Figure 8 shows \(r\), \(F^h(r)\) and the bifurcation
diagram \(\mathcal{B}\), the connected component of \(F^{-1}(F(r))\) containing \(P_0=(0,0)\). The critical points of \(F^h\) in \(r\) are \(a\) and \(c\). At these points, \(F^h\) is a fold, and
parts of \(r\) get mirrored along the critical set. Its preimages are obtained by solving linear systems. They in turn may intercept \({\mathcal{C}}\) (at \(b\)) and, by continuation, yield the
missing three solutions \(P_i\).
Discretizing with \(n=15\) points↩︎
We consider two situations. In the first, the asymptotic values \(\ell_-=0.4984\) e \(\ell_+=19.1248\) enclose four eigenvalues of \(A^h\) (\(k=4\)). A positive solution of \(F^h(u) = g^h=-1000\sin
(I_h)\) is \(P_0=55.1633\sin(I_h)\), from Section 4.2. We use half-lines which are perturbations of the eigenvector \(\sin(I_h)\) associated with the smallest eigenvalue, \[r=\{P_0 + s(\sin(I_h)-0.1\
sin(2I_h) -0.1\sin(3I_h)- 0.1\sin(4I_h)), \;\text{com} \;s \geq 0\} \subset \mathbb{R}^{15} \;.\] to ensure simple, transversal intersections of \(r\) with the critical set \(\mathcal{C}\) of \(F^h\)
. Figure 9 represents five open components \(R_k, k=1, \ldots,5\) of the regular set, \({\mathbb{R}}^{15} \setminus \mathcal{C}\). In each \(R_k\), the sign of \(\det(A^h - D^O)\) is constant. The
bifurcation diagram \(\mathcal{B}\) associated with \(r\) from \(P_0\) contains all the preimages in the data bank.
Figure 9: \(\mathcal{B}\) for \(k=4\).
The geometry relates naturally with standard properties: \(k\) is the number of negative eigenvalues of \(A^h - D^O\), the Morse index in this context (the problem admits a variational formulation).
The location of the solutions \(P_i\) in terms of the Morse index is in accordance with the continuous case [29]. The parameter \(k\) also relates to oscillation theory [31] in the discretized
context [32], an approach which does not extend to, say, the higher dimensional Dirichlet Laplacian.
The fact that for small \(k\) one expects \(2k\) solutions should be compared with the example in Figure 4. In that case, \(k\) critical components give rise to \(k + 1\) solutions and the critical
set contains only folds. Additional solutions indicate critical points which are not folds, in the spirit of Section 3.1.
We now take parameters \(\ell_-=0.4984\) e \(\ell_+=56.9367\), for which \(k=8\) and use the same \(g\) (for the finer mesh), with 24 preimages from the table in Section 4.1.
We initialize the algorithm differently: choose orthants randomly, and search for a solution by solving the associated linear system. A first solution \(P_0\) (notice that this is not the solution
given in equation (2 )), with \(k= 5\), was obtained after drawing 127 orthants. Define the generic line \[ r_1=&\{ P_0+ s(0.8\sin(I_h)-0.1\sin(2I_h)-0.1\sin(3I_h)-0.1\sin(4I_h)\nonumber \\ &-0.1\sin
(5I_h)-0.1\sin(6I_h)-0.1\sin(7I_h)+0.1\sin(8I_h))\}\nonumber \] containing \(P_0\). The computation of the bifurcation diagram \(\mathcal{B}\) led to 17 new candidate preimages, of which only 9 had a
small relative error, of the order of \(10^{-15}\). The spurious solutions are related to inversion of unstable matrices of the form \((A^h-D^o)\) in orthants trespassed along the homotopy process.
Stretches of the bifurcation diagram are either continuous or dotted, depending if they were obtained from \(s=0\) by taking \(s\) positive or negative. Points \(P_i\) at which continuous and dotted
stretches intercept are indeed regular points of \(F\).
A new solution \(P_{10}\) is obtained by sampling additional orthants, as in the computation of \(P_0\). The bifurcation diagram associated with \(r_2=\{P_{10}+s(\sin(I_h)-\sin(8I_h))\}\) yielded
four more preimages, after filtering candidates by relative error. Figure [fig:twolines] shows the (truncated) bifurcation diagrams associated with \(r_1\) and \(r_2\).
Additional solutions were obtained by sampling, leading to \(24\) solutions. The table below counts solutions \(u\) by \(k\), the number of negative eigenvalues of \(DF(u)\).
\(Number of preimages\) 1 2 2 4 6 4 2 2 1
Semi-linear perturbations of the Laplacian↩︎
We present a version of the Ambrosetti-Prodi theorem, combining material in [11], [13]–[16]). On a bounded set \(\Omega \subset \mathbb{R}^n\) with smooth boundary, let \[-\Delta_D: X = H^2(\Omega) \
cap H^1_D(\Omega) \to Y=L^2(\Omega)\] be the Dirichlet Laplacian, and denote its smallest eigenvalues by \(\lambda_1 < \lambda_2\). A vertical line is a line in \(X\) or \(Y\) with direction given by
\(\phi_1>0\), a positive eigenvector associated with \(\lambda_1\). Horizontal subspaces \(H_X \subset X\) and \(H_Y\subset Y\) consist of vectors perpendicular to \(\phi_1\). Horizontal hyperplanes
are parallel to the horizontal subspaces.
Theorem 7. Consider the function \[\label{AP} F: X \to Y \;, \quad u \mapsto - \Delta_D u - f(u)\qquad{(1)}\] where \(f: {\mathbb{R}}\to {\mathbb{R}}\) is a strictly convex smooth function satisfying
\[\label{ASY} - \infty < \lim_{x \to - \infty} f'(x) < \lambda_1 < \lim_{x \to \infty} f'(x) < \lambda_2 \;.\qquad{(2)}\] The critical set \({\mathcal{C}}\) of \(F\) contains only folds. The
orthogonal projection \({\mathcal{C}}\to H_X\) is a diffeomorphism, as is the projection of the image of each horizontal hyperplane \(F(H_V + x) \to H_Y, x \in X\). The inverse under \(F\) of
vertical lines in \(Y\) are curves in \(X\) intercepting each horizontal hyperplane and \({\mathcal{C}}\) exactly once, transversally. In particular, \(F\) is a global fold and the equation \(F(u) =
g \in Y\) has 0, 1 or 2 solutions.
In the spirit of Section 2.1, \(F: X \to Y\) is a global fold if there are diffeomorphisms \(\Phi: {\mathbb{R}}\times Z \to X\) and \(\Psi: Y \to {\mathbb{R}}\times Z\) such that \[\tilde{F} = \Psi \
circ F \circ \Phi(t, z): {\mathbb{R}}\times Z \to {\mathbb{R}}\times Z \;, \quad \tilde{F} (t, z) = (t^2, z)\] for some real Banach space \(Z\). The statement implies that the flower of \(F\) equals
the critical set \({\mathcal{C}}\): compared to the examples in the previous sections, the global geometry of \(F\) is very simple. Both functions \(F\) and \(\tilde{F}\) trivially satisfy the
geometric model in the Introduction: domain and counterdomain split in two tiles, both topological half-spaces, and both tiles in the domain are sent to the same tile in the counterdomain. Said
differently, the flower equals the critical set, which is topologically a hyperplane.
The underlying geometry led to numerical approaches to solving \(F(u) = g\). Smiley ([16]) suggested an algorithm based upon one-dimensional searches, later implemented in [11]. Finite dimensional
reduction applies for (generic) asymptotically linear functions \(f\) for which the image of \(f'\) contains a finite number of eigenvalues of \(-\Delta_D\) ([11], [12]). Under hypothesis (?? ), the
nonconvexity of \(f\) implies that some right hand side \(g\) has four preimages ([26]). Up to technicalities, the algorithm yields all solutions for convex and nonconvex nonlinearities.
Following a different geometric inspiration, Breuer, McKenna and Plum ([33]) computed four solutions of \[\label{Plum} - \Delta u + u^2 \;= \; 800 \sin( \pi x) \sin(\pi y), \;(x,y ) \in \Omega =
(0,1) \times (0,1) \;, \;u|_\Omega = 0 \;.\tag{3}\] The hardest one, call it \(u\), was obtained by interpreting it as a saddle point of a functional associated with the variational formulation of
the equation. The authors present a computer assisted proof that \(u\) is reachable by Newton’s method from a computed initial condition \(\tilde{u}\). We do not operate on this level of detail. Such
four solutions were also obtained in [17].
Equation (3 ) may be treated with our methods, but we introduce a situation with additional difficulties
In preparation, we count intersections of vertical lines and \({\mathcal{C}}\). For a bounded, smooth domain \(\Omega \subset {\mathbb{R}}^n\), the spectrum of \(- \Delta_D: X \to Y\) is \[0 < \
lambda_1 \le \lambda_2 \le \ldots \le \lambda_k < \lambda_{k+1} \le \ldots \to \infty \;,\] with associated (normalized) eigenvectors \(\phi_k\), where \(\phi_1 >0\) in \(\Omega\). Let \(f: {\mathbb
{R}}\to {\mathbb{R}}\) be a smooth function for which \(\lim_{x \to \pm \infty} f'(x) = \ell_{\pm}\), where \(\lambda_- < \lambda_1\) and \(\lambda_k \in (\lambda_k, \lambda_{k+1})\). To simplify
some arguments, we also consider \[\tilde{F}: \tilde{X} \to \tilde{Y} \;, \quad \tilde{F}(u) = -\Delta u - f(u)\] where \(\tilde{X} = C^{2,\alpha}_D(\Omega)\), the Hölder space of functions equal to
zero on \(\partial \Omega\), and \(\tilde{Y} = C^{0,\alpha}(\Omega)\), \(\alpha \in (0,1)\). For \(w \in \tilde{X}\), consider the vertical line \(\{w + t \phi_1, t \in {\mathbb{R}}\}\).
We use a natural extension of the familiar Rayleigh-Ritz technique to semibounded operators, Lemma XIII.3 of [34], which we transcribe without proof.
Proposition 3. Let \(H\) be a complex Hilbert space, \(T: D \subset H \to H\) a self-adjoint operator bounded from below with bottom eigenvalues \(\mu_1 \le \mu_2 \le \ldots \le \mu_k\) counted with
multiplicity. Let \(V\) be an \(n\)-dimensional subspace \(V \subset D\), \(n \ge k\). Let \(P\) be the orthogonal projection \(P: H \to V\) and suppose that the restriction \(P T P^\ast: V \to V\)
has eigenvalues \(\nu_1\le \ldots \nu_n\). Then \(\mu_i \le \nu_i, i=1, \ldots, k\).
Proposition 4. Vertical lines intercept the critical set \({\mathcal{C}}\) of \(\tilde{F}\) at at least \(k\) points (counted with multiplicity).
Proof. For the smooth function \(\tilde{F}\), the Jacobian \(D\tilde{F}(u): \tilde{X} \to \tilde{Y}\) is given by \(D\tilde{F}(u) v = - \Delta v - f'(u) v\). From standard arguments in linear
elliptic theory, since \(q = f' \circ u\) is a bounded (continuous) function, the operator \[T(u): X \to Y , \quad v \mapsto - \Delta_D v - f'(u) v\;\] is self-adjoint with spectrum consisting of
eigenvalues \[\mu_1(u) < \mu_2(u) \le \mu_3(u) \le \ldots \to \infty \;.\] Moreover, \(\sigma(D\tilde{F}(u)) = \sigma(T(u))\). We first show that for \(t <<0\), \(T(t)\) is a positive operator (i.e.,
all its eigenvalues are strictly positive). Indeed, by dominated convergence, since \(w + t \phi_1 \to \ell_-\) pointwise as \(t \to -\infty\), \[\mu_1(u(t)) = \min_{\| v \| = 1} \langle T(u(t)) v, v
\rangle = \min_{\| v \| = 1} \langle -\Delta_D v , v \rangle - \langle f'(w + t \phi_1) v, v \rangle \ge \lambda_1 - \ell_- > 0 \;.\] We now obtain estimates for \(\sigma(T(u(t))\) for \(t >> 0\).
Let \(V = \operatorname{span}\{\phi_i, i=1, \ldots,k\}\), the vector space generated by the first \(k\) eigenfunctions of \(-\Delta_D\). Then the matrix \(M(t)\) associated with \(P H P^T\) in this
(orthonormal) basis for \(V\) has entries \(M_{ij} = M_{ji} = \delta_{ij} \lambda_i - \langle v_i, f'(w + t \phi) v_j \rangle\). Again, as \(t \to \infty\), \(\langle v_i, f'(w + t \phi) v_j \to \
delta_{ij} \ell_+\), so that \(M(t)\) converge to the diagonal matrix with diagonal entries \(\lambda_i - \ell_+ < 0\). From Proposition 3, the first \(k\) eigenvalues of \(T(u(t))\) are strictly
negative for large \(t\). ◻
Figure 10: The domain \(\Omega\) and its mesh
In [35], Lazer and McKenna conjectured that, for an asymptotically linear \(f\) with parameters \(\ell_-<\lambda_1\) and \(\lambda_k<\ell_+<\lambda_{k+1}, k \in \mathbb{N}\), there should be at least
\(2k\) solutions of (4 ) for \(g = - t \phi_1, t >> 0\). A counterexample was provided by Dancer ([30]). The example we handle follows a positive result of Solimini [18].
Let \(\Omega\) be the annulus \[\Omega=\{x \in \mathbb{R}^2 | \;|x|<1 \;, \;|x-(-0.3,-0.3)|>0.2 \} \;.\]
We discretize \(- \Delta_D\) by piecewise linear finite elements on a mesh on \(\Omega\) with 274 triangles. The four smallest eigenvalues of the discretized operator \(-\Delta_D^h\) are simple, \[\
lambda_1^h \approx 9.0988, \quad \lambda_2^h \approx 16.3218, \quad \lambda_3^h \approx 22.9346, \quad \lambda_4^h \approx 30.4949 \;.\]
We consider \[\label{eq:PVCnaolinear} F(u)=-\Delta_D u-f(u)=g, \;\;\; \;u|_{\partial \Omega}=0 .\tag{4}\]
According to Solimini, for some \(\epsilon>0\) and parameters \(\ell_-\) and \(\ell_+\) satisfying \(\ell_-<\lambda_1 < \lambda_3< \ell_+< \lambda_3+\epsilon\) the equation (4 ) has exactly \(6\)
solutions for \(g = - t \phi_1\) for large, positive \(t\). For concreteness, we take \(f\) such that \(f'(x)=\alpha \;\text{arctan}(x)+\beta\), where \(\alpha\) and \(\beta\) are adjusted so that \
(\ell_- = -1\), \(\ell_+ = 25.3397\). Finally, set \(g^h=-1000 \;\phi_1^h\).
We first obtain a solution \(P_0\) by a continuation method. Set \(r=\{P_0+s(0.8\phi_1^h - 0.1\phi_2^h - 0.1\phi_3^h)\), a stretch of which (\(s \in [-1000, 1000]\)) we traverse with an increment \
(h_s=0.1\). As in the previous section, small terms are inserted so as to increase the possibilities that intersections with the critical set are transversal.
Along \(r\), the four smallest eigenvalues \(\mu_i^h\) of the Jacobian \(DF(u)\) are given in Figure 11 (for the underlying numerics, we used [36]). A point \(u \in r\) is a critical point of \(F\)
if and only if some such eigenvalue is zero.
Figure 11: The four smallest eigenvalues of \(DF(u)\) for \(u(s) \in r\).
In Figure [fig:morse], horizontal lines represent parts of the critical set \(\mathcal{C}\). The value \(k\) counts the number of negative eigenvalues of \(DF(u)\) at each of the regular components.
The point \(P_0 \in r\) belongs to a component for which \(k=2\). The bifurcation diagram \(\mathcal{B}\), containing the six solutions, is described in Figure [fig:morse] and the solutions are given
in Figure 12. Continuation to \(P_5\) required finer jumps along \(r\).
Figure 12: The six solutions.
No relative residue \(\epsilon({u^h})=\frac{||F^h(u^h)-g^h||_{Y^h}}{||g^h||_{Y^h}}\) is larger than \(10^{-12}\).
Predictor-corrector methods at regular points are well described in the literature ([1], [2], [37], [38]). Here we provide details about the inversion algorithm we employ in the examples in Section 5
in the neighborhood of a fold.
The algorithm must identify critical points \(u\) of \(F:X \to Y\). Due to the nature of the examples, this is accomplished by checking if some eigenvalue of the Jacobian \(DF(u)\) is zero. We assume
that the original problem \(F(u) = g\) admits a variational formulation, so that \(DF(u)\) is a self-adjoint map, and the task is simpler. The general case may also be handled, but we give no
We modify the prediction phase of the usual continuation method and perform correction in a standard fashion. Following ([5], [11], [12]), we use spectral data: by continuity, for \(u\) close to a
fold \(u_c\), the Jacobian \(DF(u)\) has an eigenvalue \(\lambda\) close to a zero eigenvalue \(\lambda_c=0\) of \(DF(u_c)\), and a normalized eigenvector \(\phi\) close to \(\phi_c\), a normalized
generator of \(\ker DF(u_c)\). The eigenvalue \(\lambda\) plays the role of arc length in familiar algorithms.
For a smooth function \(F: X \to Y\) between real Banach spaces, we search for the preimage \(u(t)\) of a smooth curve \(\gamma(t) \subset Y\) such that, at \(t = t_c\), \(\gamma(t_c) = F(u_c)\) is
the image of a fold \(u_c\). As usual, we consider the homotopy \[\begin{array}{lrll} H:&X \times \mathbb{R} &\longrightarrow Y, \;\;(u,t)&\longmapsto F(u)- \gamma(t) \\ \end{array}\] and assume the
hypothesis of the implicit function theorem: \[\begin{array}{lrll} DH(u,t): X \times \mathbb{R} \to Y \;, \quad (\hat{u}, \hat{t}) \;\mapsto DF(u) \;\hat{u} - \gamma' (t) \;\hat{t} \end{array}\] is
surjective at \((u_c, t_c)\). Clearly, \(\gamma'(t) \in Y\).
Proposition 5. \(DH(u_c, t_c)\) is surjective if and only if \(\gamma'(t_c) \notin \operatorname{Ran}DF(u_c)\).
Geometrically, the curve \(\gamma(t) \in Y\) crosses the image of the critical set \(F(\mathcal{C})\) transversally at the point \(\gamma(t_c) = F(u_c)\). Since \(\gamma\) is chosen by the
programmer, this is no real restriction.
Proof. As \(u_c\) is a fold, \(DF(u_c)\) is a Fredholm operator of index 0 with one dimensional kernel, and image given by a closed subspace of codimension one. Surjectivity of \(DH(u_c, t_c)(\hat{u}
, \hat{t}) = DF(u_c) \hat{u} - \gamma'(t_c) \hat{t}\) holds exactly if \(\gamma'(t)\) generates a complementary subspace to \(\operatorname{Ran}DF(u_c)\). ◻
The next proposition ensures that the inversion of appropriate operators may be performed as in the finite dimensional case. If \(X = Y = {\mathbb{R}}^n\), \(DH(z)\) is an \(n \times (n+1)\) matrix
of rank \(n\).
Proposition 6. For \((u,t)\) close to \((u_c, t_c)\), the Jacobian \(DH(u,t): X \times {\mathbb{R}}\to Y\) is a Fredholm operator of index 1. If \(\gamma'(t_c) \ne 0\), \(\dim \ker DH(u,t)= 1\) if
and only if \(u = u_c\), otherwise it is zero.
Proof. To show that \(DH(z_c)\) is a Fredholm operator of index 1 with one dimensional kernel, set \({\mathbb{R}}\sim \{ \gamma'(t_c) \hat{t} , \hat{t} \in {\mathbb{R}}\}\) and write \(DH(z_c)\) as
the composition \[( \hat{u} , \hat{t}) \in X \times {\mathbb{R}}\mapsto (DF(u_c) \hat{u} , \gamma'(t_c) \hat{t}) \in Y \times {\mathbb{R}}\ni (y, s) \mapsto y - s \in Y \;,\] easily seen to consist
of Fredholm operators of indices \(0\) and \(1\) respectively. Recall that the composition of Fredholm operators yields another Fredholm operator and indices add. We then have that \(DH(u_c, t_c)\)
is a Fredholm operator of index 1. If \(\gamma'(t_c) \ne 0\), then \(\dim \ker DH(u_c, t_c) \le 1\) and \(\dim \ker DF(u_c) = 1\) if and only if \(\dim \ker DH(u_c, t_c)=1\). Let \(L\) be Fredholm.
By standard perturbation properties of Fredholm operators, \(L+P\) is also Fredholm, \(\operatorname{ind}(L+P) = \operatorname{ind}L\) and \(DH(u,t)\) is Fredholm of index 1, for \((u,t)\) near \
((u_c, t_c)\), as \(H\) is of class \(C^1\). Smoothness of eigenvalues and eigenvectors proves the claims for \(\ker DH(u,t)\). ◻
To obtain a prediction from point \((u,t) = (u(t),t) \in H^{-1}(0)\), we must find a nonzero tangent vector \((\hat{u}, \hat{t}) \in T_{(u,t)} H^{-1}(0)\), so that \[\begin{array}{lrll} DH(u,t)(\hat
{u}, \hat{t})=DF(u) \hat{u} - \gamma'(t) \hat{t} =0 \;, \quad (\hat{u}, \hat{t}) \ne 0 . \end{array}\]
At points \(u= u(t)\) for which \(DF(u)\) is invertible, this is easy: set \(\hat{t} = 1\) and get \(\hat{u}\) by solving a linear system. Instead, we assume \(u\) close to a fold \(u_c \in \mathcal
{C}\). By the smoothness of simple eigenvalues and associated (normalized) eigenvectors, \(DF(u)\) has an eigenvalue \(\lambda\) and associated eigenvector \(\phi\) near an eigenvalue \(\lambda_c = 0
\) and eigenvector \(\phi_c\) of \(DF(u_c)\). We must compute a nonzero solution \((\hat{u} ,\hat{t})\) of \(DH(u,t)(\hat{u} ,\hat{t}) = 0\), or equivalently \(DF(u)\hat{u} = - \gamma'(t) \hat{t}\),
with a procedure which is continuous in \(t \sim t_c\).
For \(\hat{u} \in X\), split \(\hat{u}= \hat{v} + \hat{r} \phi\) for \(\langle \hat{v}, \phi \rangle =0\) and \(\langle \phi, \phi \rangle =1\). Clearly, \(\hat{v}\) and \(\hat{r}\) are continuous in
\(u\), since \(\phi= \phi(u)\) is. The tensor product \(\phi \otimes \phi\) denotes the rank one linear map \((\phi \otimes \phi) v = \langle \phi, v \rangle \phi\). In particular, as \(\| \phi \|=1
\), \((\phi \otimes \phi) \phi = \phi\).
Proposition 7. Let \(u\) be sufficiently close to a fold \(u_c \in X\) of \(F\), with the eigenvalue \(\lambda \sim 0\) such that \(| \lambda|< 1\) and associated normalized eigenvector \(\phi\). For
\(\alpha \ge 1\) the operator \(S(u)=DF(u)+\alpha \;\phi \otimes \phi : X \to Y\) is invertible.
Notice that \(S = DF(u)\) when restricted to \(\phi^\perp\).
Proof. The operator \(S(u)\) is a rank one perturbation of \(DF(u)\), and thus it is also a Fredholm operator of index zero: invertibility is equivalent to injectivity. For \(\hat{u}= \hat{v} + \hat
{r} \phi\), with \(\hat{v} \in \phi^\perp\) and \(\langle \phi, \phi \rangle =1\), \[S(u) \hat{u}= DF(u) (\hat{v} + \hat{r} \phi) + \alpha(\phi \otimes \phi) (\hat{v} + \hat{r} \phi) = 0 \;,\]
implies \[S(u) \hat{u}= DF(u) \;\hat{v} + \hat{r} \lambda \phi + \alpha \;\hat{r} \;\phi = 0 \;.\] Since \(DF(u)\) is self-adjoint, \(\phi^\perp\) is an invariant subspace and both terms are zero, \
[DF(u) \hat{v} =0 ,\; (\alpha + \lambda) \hat{r} \;\phi = 0 \;.\] As the restriction \(DF(u)\) to \(\ker DF(u)^\perp\) is an isomorphism, \(\hat{v} = 0\), \(\hat{r} = 0\). ◻
Proposition 8. Under the hypotheses of the proposition above, the solution of \[S(u) \hat{u} = - \lambda \gamma'(t) - \alpha \langle \phi, \gamma'(t) \rangle \phi\] is of the form \(\hat{u} = \hat{v}
- \langle \phi, \gamma'(t) \rangle \phi\) for some \(\hat{v} \in \phi^\perp\), \(\| \phi \| = 1\). Moreover, \[DF(u) \hat{u} = - \lambda \gamma'(t) \;and \;(\hat{u} , \lambda) \in T_{(u,t)} H^{-1}(0)
In other words, \((\hat{u} , \lambda)\) is the tangent vector required in the prediction phase.
Proof. As \(S(u)\) is invertible, \(\hat{u}\) is well defined for the given right hand side. For \(\hat{u} = \hat{v} + \hat{r} \phi\) with \(\hat{v} \in \phi^\perp\) and \(\langle \phi, \phi \rangle
=1\), take the inner product with \(\phi\) of \[S(u) \hat{u} = DF(u) \hat{u} + \alpha \hat{r} \phi = - \lambda \gamma'(t) - \alpha \langle \phi, \gamma'(t) \rangle \phi \;.\] to obtain \(\hat{r} = -
\langle \phi, \gamma'(t) \rangle\) and then \(DF(u) \hat{u} = - \lambda \gamma'(t)\) follows. ◻
In the application of Section 5 finite element methods applied to the Laplacian yields the usual sparse matrices. The term \(\alpha \phi \otimes \phi\) spoils sparseness, and one has to proceed by
inversion through standard techniques associated with rank one perturbations. The numerical inversion worked well with \(\alpha=1\).
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C.T. Kelley, Numerical methods for nonlinear equations, Acta Numer. 27, 2018, 207–287. | {"url":"https://academ.us/article/2311.10494/","timestamp":"2024-11-09T10:33:09Z","content_type":"text/html","content_length":"132632","record_id":"<urn:uuid:da1fdc32-d15e-48c6-80a3-57b008a9ae39>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00039.warc.gz"} |
Specific Examples
The use of SimuLase™ for the typical example of an edge-emitter and a VECSEL are described in the examples guide.
Below are some comparisons between experimental results and those of the fully microscopic calculations done with SimuLase™ for an:
A detailed explanation on how to use SimuLase™ for the examples of an edge emitting ridge waveguide structure and a VECSEL structure, including theory-experiment comparisons for all important
aspects, ranging from the PL to the input-output characteristics, can be found here.The edge emitting example is based on Ref. [30], the VECSEL example is based on Ref. [41].
Screen grabs from SimuLase in the analysis of an edge-emitting structure
Experimentally measured (solid lines) and calculated (dashed lines) temperature-dependent PL spectra for a mid-infrared type-II ‘W’ diode lase structure operating near 3.4µm at room temperature. The
active region consists of a short superlattice of five InAs/GaInSb/InAs/AlGaSb wells.
Temperature-dependent threshold current density for the mid-infrared type-II ‘W’ diode lase structure. Dots: Experimental data. Solid line: Total theoretical loss current. Dashed line: Calculated
Auger loss current. Dotted line: Calculated radiative loss current.
For more details on this example and calculations for radiative and Auger losses see Ref. [40].
Experimentally measured (black signs) and calculated (thick lines) temperature-dependent loss currents at threshold for a structure containing eight 4.4nm wide In0.83Ga0.17As0.67P0.33/InP quantum
wells lasing near 1.3µm. The experimental data was extracted from A.F. Phillips, et al., IEEE J. Sel. Topics Quantum Electron., vol. 5, 401 (1999). Black: total loss current; Red: radiative loss
current due to spontaneous emission; Blue: Auger loss current.
For more details on this example and calculations for radiative and Auger losses see Ref. [28].
Experimentally measured (black signs) and calculated (thick lines) temperature-dependent loss currents at the threshold for a structure containing four 2.5nm wide In0.65Ga0.35As/InP quantum wells
lasing near 1.5µm. The experimental data was extracted from A.F. Phillips et al., IEEE J. Sel. Topics Quantum Electron., vol. 5, 401 (1999). Black: total loss current; Red: radiative loss current due
to spontaneous emission; Blue: Auger loss current.
For more details on this example and calculations for radiative and Auger losses, see Ref. [28].
Experimentally measured (thin lines and signs) and calculated (thick lines) temperature-dependent loss currents at threshold for a structure with a 6.4nm wide Ga0.66In0.34N0.018As0.982/GaAs quantum
well lasing near 1.3µm. The experimental data was extracted from R. Fehse, et al., IEEE J. Sel. Topics Quantum Electron., vol. 8, 801 (2002). Black: total loss current; Red: radiative loss current
due to spontaneous emission; Blue: Auger loss current. Losses due to defect recombination have been excluded from the shown experimental data. The deviations between theory and experiment at high
temperatures are probably due to internal heating beyond the heat sink temperature.
Experimentally measured (black signs) and calculated (red lines) gain spectra for a structure containing of 2nm wide In0.1Ga0.9N quantum wells between 6nm wide GaN barriers. The experimental data was
extracted from B. Witzigmann, et al., Appl. Phys. Lett., 88, 021104 (2006).
For more details on this example and calculations for wide bandgap nitride systems see Ref. [31] and Ref.[32].
Experimentally measured (blue signs) and calculated (red lines) modal gain spectra for a structure consisting of three 6nm wide In0.32Ga0.68N0.01As0.99 quantum wells between In0.05Ga0.95N0.015As0.985
barriers. The experimental data were obtained with the Hakki-Paoli method.
The sheet carrier densities in the calculations ranged from 3.75 to 8.25×10¹²/cm² with increments of 0.75×10¹²/cm², the injection currents in the experiment were 7, 12, 14, 16 and 18mA.
Gain spectra for the same structure as above but for a wider density range. It demonstrates the heating of the device under high excitation conditions, leading to an otherwise unusual red shift of
the spectra. The comparison can be used to determine internal temperatures and heating. The experimental data was obtained with a different method than in the example above, which allows to examine a
wider density range, but is less accurate. Blue: experimental results for 0, 5, 10, 15, 20, 30, 40, 50, 60 and 70mA. Red: theoretical results for the assumed temperature of 300K and densities of 0,
1.5, 2.5, 3.0, 3.5 and 6.0×10¹²/cm². Green: theoretical results for 312K and 4.5 and 5.0×10¹²/cm². Magenta: 325K, 7.0*10¹²/cm². Black: 337K and 9.0×10¹²/cm².
Experimentally measured (blue signs) and calculated (red lines) modal gain spectra for a structure containing a 10nm wide In0.2Ga0.8As quantum well between 85nm wide AlxGa1-xAs graded index barriers
where the Aluminium concentration rises linearly from x=0.1 to x=0.6 with distance from the well.
Theory-experiment comparison for a structure consisting of a 8nm wide In0.05Ga0.95As quantum well between Al0.2Ga0.8 barriers. Blue signs show the experimental results, red lines the theoretical
ones. The carrier densities in the theory ranged from 0.6 to 3.0×10¹²/cm². The experimental currents are from 0 to 20mA.
Theory-experiment comparison for the excitonic absorption in a system of electronically coupled InGaAs quantum wells that are separated by thin InP barriers (‘superlattice’) under different external
electric fields.
For the case of a field of 44 kV/cm the red dashed line shows the theoretical result if the conduction band nonparabolicity is neglected. The red solid lines have been calculated with this
nonparabolicity. The lines for different fields have been shifted along the absorption axis for better clearity.
Determining Material Properties under Lasing Conditions from Low Excitation Luminescence | {"url":"https://www.nlcstr.com/real_life_example/specific-examples/","timestamp":"2024-11-02T09:03:41Z","content_type":"text/html","content_length":"152228","record_id":"<urn:uuid:0bd24d5b-b65c-4f4f-9204-b4427748adad>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00851.warc.gz"} |
Solution to Problem 693 | Beam Deflection by Method of Superposition
Problem 693
Determine the value of EIδ at the left end of the overhanging beam in Fig. P-693.
Solution 693
Click here to expand or collapse this section
The rotation at the left support is the combination of
Case No. 7
Case No. 12
$\theta_L = \dfrac{Pb(L^2 - b^2)}{6EIL}$
$\theta_L = \dfrac{ML}{3EI}$
The overhang beam is transformed into a simple beam and the end moment at the free end of the overhang is carried to the left support of the transformed beam.
$\theta = \dfrac{Pb(L^2 - b^2)}{6EIL} - \dfrac{ML}{3EI}$
$\theta = \dfrac{800(1.5)(4.5^2 - 1.5^2)}{6EI(4.5)} - \dfrac{600(4.5)}{3EI}$
$\theta = \dfrac{800}{EI} - \dfrac{900}{EI}$
$\theta = -\dfrac{100}{EI}$
The negative sign indicates that the rotation at the left end contributed by the end moment (taken as negative) is greater than the rotation at the left end contributed by the concentrated load
(taken as positive).
Case No. 5, the end deflection is
$\delta = \dfrac{ML^2}{2EI}$
The deflection at the overhang due to moment load alone is
$\delta_M = \dfrac{600(2^2)}{2EI}$
$\delta_M = \dfrac{1200}{EI}$
Total deflection at the left end of the given beam is
$\delta = 2\theta + \delta_M$
$\delta = 2\left( \dfrac{100}{EI} \right) + \dfrac{1200}{EI}$
$\delta = \dfrac{1400}{EI}$ answer
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Ch. 19 Problems - Physics | OpenStax
19.1 Ohm's law
What voltage is needed to make 6 C of charge traverse a 100-Ω resistor in 1 min?
a. The required voltage is 1 × 10^−3 V.
b. The required voltage is 10 V.
c. The required voltage is 1,000 V.
d. The required voltage is 10,000 V.
Resistors typically obey Ohm’s law at low currents, but show deviations at higher currents because of heating. Suppose you were to conduct an experiment measuring the voltage, V, across a resistor as
a function of current, I, including currents whose deviations from Ohm’s law start to become apparent. For a data plot of V versus I, which of the following functions would be best to fit the data?
Assume that a, b, and c are nonzero constants adjusted to fit the data.
a. $V=aI V=aI$
b. $V=aI+b V=aI+b$
c. $V=aI+b I 2 V=aI+b I 2$
d. $V=aI+b I 2 +c V=aI+b I 2 +c$
19.2 Series Circuits
What is the voltage drop across two 80-Ω resistors connected in series with 0.15 A flowing through them?
a. 12 V
b. 24 V
c. 36 V
d. 48 V
In this circuit, the voltage drop across the upper resistor is 4.5 V. What is the battery voltage?
a. 4.5V
b. 7.5V
c. 12V
d. 18V
19.3 Parallel Circuits
What is the equivalent resistance of this circuit?
a. The equivalent resistance of the circuit is 32.7 Ω.
b. The equivalent resistance of the circuit is 100 Ω.
c. The equivalent resistance of the circuit is 327 Ω.
d. The equivalent resistance of the circuit is 450 Ω.
What is the equivalent resistance of the circuit shown below?
a. The equivalent resistance is 25 Ω.
b. The equivalent resistance is 50 Ω.
c. The equivalent resistance is 75 Ω.
d. The equivalent resistance is 100 Ω.
19.4 Electric Power
When 12 V are applied across a resistor, it dissipates 120 W of power. What is the current through the resistor?
a. The current is 1,440 A.
b. The current is 10 A.
c. The current is 0.1 A.
d. The current is 0.01 A.
Warming 1 g of water requires 4.18 J of energy per °C. How long would it take to warm 1 L of water from 20 to 40 °C if you immerse in the water a 1-kW resistor connected across a 9.0-V battery?
a. 10 min
b. 20 min
c. 30 min
d. 40 min | {"url":"https://openstax.org/books/physics/pages/19-problems","timestamp":"2024-11-03T05:45:52Z","content_type":"text/html","content_length":"502030","record_id":"<urn:uuid:107585e2-1380-4c4a-a6c7-36c5e6f6a2bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00874.warc.gz"} |
Misocp-based solution approaches to the unit commitment problem with AC power flows
Tuncer, Deniz (2021) Misocp-based solution approaches to the unit commitment problem with AC power flows. [Thesis]
Unit Commitment (UC) and Optimal Power Flow (OPF) are fundamental problems in short-term electrical power systems planning. Generally, the UC problem is solved to determine the commitment status of
generators. Then, the OPF problem is solved to determine the power generation levels of committed generators. Instead of solving these problems in a serial manner, solving the UC problem and the OPF
problem with AC power flows simultaneously as a mixed-integer nonlinear program (MINLP) can yield better results, but there is only a limited number of studies in the literature utilizing such an
approach. Adopting this approach, we develop a base algorithm, in which we solve a mixedinteger second order conic program (MISOCP) relaxation of the UC problem with AC power flows. Then, we solve
the multiperiod OPF (MOPF) problem by a local solver, where commitment statuses from the first step are fixed to find a feasible solution to the original MINLP. The second step yields a feasible
solution to the original problem. We then assess the quality of the feasible solution by using the lower bound obtained from the first step. In order to obtain better lower bounds, we add some valid
inequalities that are originally developed for the OPF problem to the base algorithm, which we call enhanced algorithm. For the problem instances with small number of buses, the base and the enhanced
algorithms are able to provide small optimality gaps for the problem. However, it takes a long time to solve the MISOCP problem in larger instances. In order to solve the larger instances, we adopt a
Lagrangian decomposition method. With the addition of the mentioned valid inequalities, the quality of the lower bound of the Lagrangian subproblems are improved. Thanks to this decomposition method,
we obtain feasible solutions to the problem instances that the other algorithms are not able to provide feasible solutions within a reasonable time limit.
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Calculator.LR.FNs: Calculator for LR Fuzzy Numbers
Arithmetic operations scalar multiplication, addition, subtraction, multiplication and division of LR fuzzy numbers (which are on the basis of extension principle) have a complicate form for using in
fuzzy Statistics, fuzzy Mathematics, machine learning, fuzzy data analysis and etc. Calculator for LR Fuzzy Numbers package relieve and aid applied users to achieve a simple and closed form for some
complicated operator based on LR fuzzy numbers and also the user can easily draw the membership function of the obtained result by this package.
Version: 1.3
Published: 2018-05-02
DOI: 10.32614/CRAN.package.Calculator.LR.FNs
Author: Abbas Parchami (Department of Statistics, Faculty of Mathematics and Computer, Shahid Bahonar University of Kerman, Kerman, Iran)
Maintainer: Abbas Parchami <parchami at uk.ac.ir>
License: LGPL (≥ 3)
NeedsCompilation: no
CRAN checks: Calculator.LR.FNs results
Reference manual: Calculator.LR.FNs.pdf
Package source: Calculator.LR.FNs_1.3.tar.gz
Windows binaries: r-devel: Calculator.LR.FNs_1.3.zip, r-release: Calculator.LR.FNs_1.3.zip, r-oldrel: Calculator.LR.FNs_1.3.zip
macOS binaries: r-release (arm64): Calculator.LR.FNs_1.3.tgz, r-oldrel (arm64): Calculator.LR.FNs_1.3.tgz, r-release (x86_64): Calculator.LR.FNs_1.3.tgz, r-oldrel (x86_64):
Old sources: Calculator.LR.FNs archive
Please use the canonical form https://CRAN.R-project.org/package=Calculator.LR.FNs to link to this page. | {"url":"https://cran.rstudio.org/web/packages/Calculator.LR.FNs/index.html","timestamp":"2024-11-08T03:58:21Z","content_type":"text/html","content_length":"5635","record_id":"<urn:uuid:f8a0457b-d981-4692-96ac-944768431a46>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00135.warc.gz"} |
CSCI 5280 Project 3
Contrast Preserving Decolorization
Xiong Yuanjun
SID: 1155018814
1 Project Files
2 Project Description
Decolorization - the process to transform a color image to a grayscale one - is a basic tool in digital printing, stylized black-and-white photography, and in many single channel
image processing applications
The decolorization function is defined as g = f(c), which is a mapping from color image c to gray image g.
The mapping function is further expressed as
where {m} is the set of monomial basis and {w} is the corresponding coefficients.
With the weak color order assumption, the energy function w.r.t parameter {w} can be finally written as
The overall algorithm is like this
3 Implementation
The implementation is using MATLAB with the skeleton code provided. The main acceleration skill is to use 2D filter [1,-1] and [1;-1] to get the difference between neighbors.
4 Experiments
Test Images
Here we got 10 test images, with different contents. They are
Image 1 Image 2 Image 3 Image 4 Image 5
Image 6 Image 7 Image 8 Image 9 Image 10
In this program, there is one only one parameter we need to set. That is the variation of Gaussian functions, .
Here I use the
Experiment Results
Then we get 10 output image, listed below. Their original image are shown next to each image.
Numeric Evaluation
Using the function ccpr(), we get the following ccpr value:
We can set that the CCPR significantly improved the contrast of decolorized picture. | {"url":"https://yjxiong.me/CSCI5280/CCPR/CCPR.htm","timestamp":"2024-11-10T21:12:22Z","content_type":"text/html","content_length":"77545","record_id":"<urn:uuid:19aa9a1f-dbb0-40e3-bc2b-1fe2255886c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00185.warc.gz"} |
This research was designed to identify the problems teachers and students encountered when teaching and learning algebra at the secondary school level.
A total of seventeen (17) mathematics teachers and one hundred (100) students randomly selected served as the subjects of this study.
The instrument used for this research were Algebra Diagnostic Test (ADT) for students and Questionnaire for teachers and students. The data collected were analyzed using the statistical package for
social sciences (SPSS) which include simple percentages, the mean, standard deviation and analysis of variance (ANOVA).
The research questions were investigated and two null hypothesis were tested. The findings revealed that;.
i. There are significant problems encountered by teachers while teaching algebraic concepts that adversely affect students’ appreciation and achievement in algebra classes.
ii. There are significant problems encountered by students when learning algebraic concepts that lead to poor performance in mathematics and application to other subjects.
Based on these findings, appropriate recommendations were made and it was concluded that problems of teaching and learning algebraic concepts affect students achievement and performance in algebra
test and mathematics in general.
TABLE OF CONTENTS
Title page Pages
Certification ii
Dedication iii
Acknowledgement iv
Table of content v
Abstract vi
1.1 Background to the study. 1
1.2 Statement of the problem. 9
1.3 Purpose of the study. 10
1.4 Research Questions. 10
1.5 Research Hypothesis. 10
1.6 The Significance of the study. 11
1.7 Scope and Limitation of the study. 11
2.1 Introduction. 14
2.2 Problems of Teaching Algebra. 15
2.2.1 Professional Development and Mathematics 15
instruction of Teachers.
2.2.2 Limited Pedagogical Knowledge of Teachers. 16
2.2.3 Lack of Realization of the Effects of Students
Algebraic Thinking in effective Teaching. 17
2.2.4 Complexities in Coding Students Responses. 18
2.2.5 Limited content Knowledge base of Teacher. 18
2.2.6 Absence of Teaching Algebra as Modelling Tool. 19
2.3 Problems Students Encounter in Learning Algebra. 20
2.3.1 Isolation of Mathematical Ideas from Elementary
Algebra. 21
2.3.2 Nature of Learning Environment. 22
2.3.3 Misconception about the Use of
Letters and Symbols. 22
2.3.4 Nature of Algebraic Concepts. 23
2.3.5 Academic Procrastination among Students. 24
2.4 Research Finding on the Solution Strategies to
the problems of Teaching Algebra . 25
2.4.1 Improved Condition for professional and
Non-professional Mathematics Teacher. 25
2.4.2 Adequate Pedagogical Knowledge of Teachers. 26
2.4.3 Preference to the Effects of Students Algebraic
Thinking to Effective Teaching. 27
2.4.4 Devising projects for Coding Students Response. 28
2.4.5 Adequate Content Knowledge of Teachers. 29
2.4.6 Teaching of Algebra as a Modelling Tool . 30
2.5 Solution to the Problems of Learning
Algebra among Students. 31
2.5.1 Integrating Mathematical Ideas to Elementary
Algebra. 32
2.5.2 Fostering Positive Learning Environment. 32
2.5.3 Positive Concepts about Use of Letter and
Symbol In Algebra. 33
2.5.4 Reducing Abstraction Level in Algebra. 34
2.5.5 Counselling Students with Procrastinating Attitude. 35
2.6 The Dynamics of Problems of Teaching and
learning Algebra in the Secondary Schools. 36
3.1 Introduction. 43
3.2 Research Design. 44
3.3 Area of Study. 44
3.4 Population of the Study. 44
3.5 Sample and Sampling Technique. 45
3.6 Research Instruments. 46
3.7 Method of Data Collection. 47
3.8 Method of Data Analysis. 47
3.8.1 Scoring Algebra Diagnostic Test. 48
3.8.2 Statistical Treatment of the Data Analysis. 48
4.0 Introduction. 49
4.1 Analysis of the Demographic Information of the
Respondents (Teachers). 49
4.2 Analysis of the Items in the Teacher’s
Questionnaire . 51
4.5 Representation and analysis of Data According to
Research Hypothesis and Questions. 62
4.6 Analysis of the Demographic Information of the Respondents Students. 64
4.9 Representation and Analysis of Data According
to Research Hypothesis and Questions. 77
5.1 Discussion of Results. 80
5.2 Limitations. 82
5.3 Implications and Recommendation. 83
5.3.1 Science Teaching and Learning. 83
5.3 2 Education Policy Makers and Administrators . 86
5.3.3 Parents. 87
5.3.4 Future Research. 87
5.4 Conclusion. 88
CHAPTER ONE
1.1 Background to the Study
The place of mathematics in any enterprise is all encompassing and divergent in thought processes and logical reasoning. (Oshin, 1995). It’s usefulness and applications to other subject disciplines
in the education system is undeniably feasible; as most of the algorithms, systematization, organization, interpretation and analysis in presentation of facts, points, generalizations and argument
has helped to order the sequence of synthesized knowledge to producing comprehensive volumes of materials to be taught and learnt in these disciplines; some of which are sciences, commerce,
economics, medicine, maritime, defence and vocations. (Olayinka & Omoegun 2006;Butter& Wren 1951 cited in Udeinya & Okobiah, 1991 and Sule, n.d).
Olayinka & Omoegun (2006) and Usman & Umeano (2006), submitted that mathematics is an indispensable tool for national development it helps to build the computational, manipulative, deductive and
inductive thinking as well as problem solving skills of prospective individuals to function effectively within his ever dynamic world, through self discovery, development, worth and actualization.
Mathematics is unique in diversities. It is resourceful in scientific industrial, technological social and vocational progress of any society. (Asikhia, 2010).
The teaching and learning of mathematics concepts has been ranked as one of the most important activities in the education system. It’s distinctive nature as enunciated by Piaget (1972) and Piaget &
Garcia (1998) cited in Cooley, Martin, Vidakovic &Loch (2007) and Mashooque (2010) attest to the crucial placement of concepts learning and appreciation as well as utilization of mathematics ideas to
solve problems and analyze concepts in other school subjects. They highlighted visualization, abstractness, hierarchy of concepts, problem solving and discovery nature of mathematics as the
tremendous implication to the teaching and learning processes. In view of these, stakeholders (teachers, parents, educational planners and government) in educational planning and implementation have
helped to structure the mathematics curriculum into concept sequence to be taught and learnt at all levels of Nigerian education system. (primary secondary and tertiary).
At the secondary schools, topics that make up the branches of mathematics, such as measuration, geometry, inequalities, statistics, functions, algebra among others have been structurally arranged and
discussed in sequence, content, teaching activities and aids with respect to sub concepts, concept hierarchy, simplicity, difficult, technicality and applications (Macre, Kalejaiye, Chima, Garba,
Ademosu, chairman, Smith and Head, 2001) so that mathematics idea would be taught in such a way that repetition for re-inforcement, understanding and appreciation of topics are made at each level but
with elements of advancements in technicality and structure at higher classes in the secondary schools.
Algebra, as a branch of mathematics has been reckoned with as an important concept in mathematics. It is a generalized arithmetic which require the use of known and unknown quantities. (Osta and
Laban, 2007). They further define algebra as that branch of mathematics in which situation of life are represented with a first degree equations where the unknown appear in both sides of the equal
sign. In the same vein, Mashooque, (2010) disclosed that algebra uses symbols, letters and signs for generalizing arithmetic which have different meanings and interpretations in different situations.
It’s usage and applications to effective acquisition of knowledge, skills and understanding the titbits of other concepts in mathematics (measuration, geometry, inequalities, indices, statistics etc)
sciences, social sciences, maritime, medicine, defence and vocations among others: (Mashooque , 2010), attest to the crucial ordinal placement of this multidimensioned branch of mathematics in
solving problems.
Diophantus and Al-Khawarzmi Mohammed Ibn Musa; founder and consolidator of algebra cited in Oshin, (1995) revealed that algebra is one of the earliest mathematics inventions that transited from
arithmetic’s and got separated from it when equations and methods for reducing them were introduced. They proffered algebra as the science of transposition and cancellation. It is the branch of
mathematics that involves the solution of equation by such device as transposition and cancellation. In the same vein, Oshin (1995) revealed that as time passes, Al-Khawarizmi name was distorted into
“algorism” meaning “ the art of calculating” now referred to as Arithmetic, which has helped in the revolutionized mathematical manipulation, thereby making long division rather simple for children
and served as a model for later writings in their applications of arithmetic and algebra to the distribution of inheritance and astronomical inventions.
Deductively, algebra is wide in concept, forms, structure and applications, since it is studied virtually at all levels of education. (NCTM, 1989 cited in Cooley, Martins, Vidakovic and Loch 2007).
It is popularly known as Arithmetic in the primary schools, where pupils are taught the titbits and rudiments of counting, simple equations on sum, difference, product, divisions and word problems.
Algebra with it’s subdivisions retain it’s name at the secondary and tertiary schools with dissimilarities in concept sequence, classifications, technicality and application. For instance, at the
senior secondary schools, algebra was divided into equations (simple simultaneous and quadratic), set theory, inequality and variation with distinctive techniques for solving their problems (Macrae,
Kalejaiye, Chima, Garba, Ademosu, Channon 2001). All these concepts are taught and learnt at the senior secondary schools to facilitating proficiency of students in applying the techniques therein to
solve problems in other subjects.
Inspite the utilitarianism of algebra to the material world in developing skills in computation, manipulation, balancing and analyzing equations, logical reasoning, deductive thinking and problem
solving in individual to adapting and functioning effectively in the technologically dynamic world, students in the secondary schools; especially those in the senior secondary Two (ss2) have not
explored the resourcefulness of algebraic concepts, evident in their poor performance in not only algebraic sections of mathematics examination but also in mathematics in entirety and subjects like
physics, chemistry, Economics etc. (Kucheman, 1981 cited in Mashooque, 2010). This development is traceably linked to the problems associated with teaching and learning algebraic concepts at the
senior secondary schools.
Teaching and learning are tools for implementing educational policies and programmes. They constitute the basis for drawing out and developing the innate potentials of an individual to aiding
usefulness to oneself and the society. Teaching; a process of making someone attend, observe, reason and think. (Akande, 2004) and learning; a relative permanent change in behaviour due to experience
(Nwadinigwe, 2001) are two sides of a coin; as one complements the other in terms of effectiveness. The teaching and learning of mathematics concepts with algebra inclusive is beset with many
problems; most of which have adversely affected the performance of students in mathematics examinations despite several viable efforts put in by stakeholders to ameliorate the appreciation
achievement and utilization of mathematics concepts. (Okereke, 2005). Chimere, (2007) submitted that this ;pose grave danger to national development.
Teaching algebraic concepts has become evolving, multidimensional and divergent (Catherine, 2005). With the adventures in technological discoveries and mathematical inventions in algebraic
simplification, expression, manipulation and problem solving, concepts in this encompassing branch of mathematics absorb complexities in forms, structures and algorithms that directly affect
proficiency farewell of students in mathematics as whole and other subjects, many of which have advance the course of effective teaching. To this end, Catherine (2005) observed that these
complexities have implications on the content and pedagogical knowledge of teachers in effectively communicating algebraic ideas to the understanding and appreciation of the students, and submitted
that the adequacy or otherwise of content and pedagogical knowledge of teachers affect the achievement level of students in algebra classes.
But in most teaching and leaning encounters, it is revealed that teachers do the greater works by exposing titbits, algorithms and structures of algebra to students. Thereby, leaving him or her to
nothing for self discovery, attitudes of appreciation, which invariably discourage curiosity for further learning. This affects the true essence of self development (Catherine and Vistro, 2005).
Moreso, Udeinya & Okobiah (1991) highlighted poor methods of teaching, unqualified mathematics teachers, poor condition of service for teachers and apathy towards mathematics by the general public as
some of the threat hampering mathematics growth and development. They and Okereke, (2005) opined that it is a common knowledge that many mathematics teachers, mathematicians and the general public
are distressed about the state of general mathematics instruction in the country, inspite the pivotal role mathematics play as a key subject in the school curriculum.
On the part of students, Michele and Assude in their project work opined that algebra is a crucial domain as regard the relationship students develop with mathematics. This shows that algebra is
fundamental to students competence in mathematics ideas, structures and problems solving. Thus, it is an important concept that serves as a tool for analyzing other concepts in mathematics Algebra
also play a crucial role in facilitating students proficiencies in other subjects like physics, chemistry, economics etc because calculations, use of equation and expressions that are algebraic in
nature are absorbed by authors of the subjects to expound, analyse and solve concepts in these subjects. (Sule, n.d).
But, the state of affairs in the performance scale of student in mathematics and other subjects like physics, chemistry, economics etc is at a low ebb. Mashooque (2010) identified poor understanding
of algebra fundamental of students in the use of symbols, letter and signs, misconceptions in algebraic processes and poor attitude towards problems solving as some of the factors inhibiting students
appreciation and proficiencies in algebraic processes. Asikhia (2010) submitted that these anomalies emanated from the learning difficulties and challenges students encounter in the teaching and
learning process, which have implication to students functionality and adaptation in other subject as well as their future career.
In Sule, (n.d) project work on the relationship between attitude and problem solving in mathematics of secondary students in Kogi state stated that algebra concepts can be used to solve a multitude
of problems arising form diverse academic field, such as physics, chemistry, Economics, Sociology, Astronomy and statistics. By implication, poor academic performance in these field could be linked
to deficiency of students in interpreting, analyzing, balancing and solving algebraic expressions arising from numerous problems they encountered in learning algebra concepts.
1.2 Statement of the Problem
Inspite the utilitarian nature of algebra in facilitating potency in students to understand and apply basic concepts like equations, inequalities, set theories, variation, problem solving techniques
to interpret, analyse and solve problems in other branch of mathematics as well as other school subjects like physics, chemistry, economics etc for better functionality and adaptation of their
rudiments and structural patterns for improved performance, students have not been able to explore the usefulness of algebra to better their proficiency in these subjects due to the difficulties
encountered by teachers and students in the teaching and learning of algebra concepts. The students incessant poor performance in examination is posing threat to their educational and teacher’s
professional growth and development.
1.3 Purpose of the Study
The main purpose of this study is to identify the problems of teaching and learning algebraic concepts at the senior secondary school Two (SS 2) by undertaking a diagnostic test analysis of selected
mathematics teachers and students in Apapa Local Government Area of Lagos State Secondary Schools.
1.4 Research Question
In this study, the following research questions were raised:
i. What are the problems teachers encountered while teaching algebraic concepts, that adversely affect students’ appreciation and achievements in algebra classes?
ii. What are the problems students encountered while learning algebraic concepts that lead to poor performance in mathematics and application to other subjects.
1.5 Research Hypothesis
The following hypothesis stated below would be tested in this study.
HoI: There are no significant problems encountered by teachers while teaching algebraic concepts that adversely affects student’s appreciation and achievement in algebra classes.
HoII: There are no significant problems encountered by student’s while learning algebraic concepts that lead to poor performance in mathematics and applications to other subjects.
1.6 The Significant of the Study
This study is significant because it would help to provide valuable information to acquaint.
i. Teachers on problems of teaching algebra at the senior secondary schools and model for improved instruction.
ii. Students on problem of learning algebraic expressions and model for improved appreciation and performance in mathematics as well as application to other subjects.
1.7 Scope and Limitation of the Study
Considering the broad nature of algebraic concepts taught and learnt at different levels of education with their antecedent problems, massiveness in the number of mathematics teachers and students at
these schools, limited time frame, space and resources, this research study intend to focus on studying the problems of teaching and learning algebraic concepts among selected mathematics teachers
and SSS2 students in Apapa Local Government Area of Lagos State Secondary Schools.
Click “DOWNLOAD NOW” below to get the complete Projects
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NP-Problem -- from Wolfram MathWorld
A problem is assigned to the NP (nondeterministic polynomial time) class if it is solvable in polynomial time by a nondeterministic Turing machine.
A P-problem (whose solution time is bounded by a polynomial) is always also NP. If a problem is known to be NP, and a solution to the problem is somehow known, then demonstrating the correctness of
the solution can always be reduced to a single P (polynomial time) verification. If P and NP are not equivalent, then the solution of NP-problems requires (in the worst case) an exhaustive search.
Linear programming, long known to be NP and thought not to be P, was shown to be P by L. Khachian in 1979. It is an important unsolved problem to determine if all apparently NP problems are actually
A problem is said to be NP-hard if an algorithm for solving it can be translated into one for solving any other NP-problem. It is much easier to show that a problem is NP than to show that it is
NP-hard. A problem which is both NP and NP-hard is called an NP-complete problem. | {"url":"https://mathworld.wolfram.com/NP-Problem.html","timestamp":"2024-11-13T15:37:32Z","content_type":"text/html","content_length":"56056","record_id":"<urn:uuid:b611d168-6019-48eb-a8c8-c2732c6831b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00249.warc.gz"} |
KSEEB Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1
Students can Download Chapter 9 Data Handling Ex 9.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more
marks in your examinations.
Karnataka State Syllabus Class 6 Maths Chapter 9 Data Handling Ex 9.1
Question 1.
In a mathematics test, the following marks were obtained by 40 students, Arrange these marks in a table using tally marks.
a) Find how many students obtained marks equal to or more than 7.
b) How many students obtained marks below 4?
By observing the marks scored by 40 students in the test we can construct the Table as follows :
a) The students who obtained their marks equal to or more than 7 are the students who obtained their marks as either of 7, 8 and 9. Hence number of these students = 5 + 4 + 3 = 12
b) The students who obtained their marks below 4 are the students who obtained their marks below 4 are the students who obtained their marks as either of 1, 2 and 3
Hence, number of these students = 2 + 3 + 3 = 8
Question 2.
Following is the choice of sweets of 30 students of class VI.
Laddoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi,Ladoo, Barfi, Rasgulla, Ladoo, Jalbebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla,
Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
By observing the choices the choices of sweets of 30 students, we can construct the table as bellow.
a) Arrange the names of sweets in a table using tally marks
b) Which sweet is preferred by most of the students?
Ladoo is the most perferred sweet as the most number of students (i,e, 11) prefer ladoo
Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below.
Make a table and enter the data using tally marks. Find the number that appeared.
a) The minimum number of times
The number which appeared the minimum number of times (i,e, 4 times) is 4
b) The maximum number of times
The number which appeared the maximum number of times (i,e, 11 times) is 5
c) Find those numbers that appear an equal number of times.
1 and 6 are the number which appears for the same number of times (i,e, 7 times)
Question 4.
Following pictograph shows the number of tractors in five villages.
Observe the pictograph and answer the following questions.
i) Which village has the minimum number of tractors?
Village D has the minimum number of tractors, i.e. only 3 tractors
ii) Which village has the maximum number of tractors?
Village C has the maximum number of tractors i, e, 8 tractors.
iii) How many more tractors village C has compared to village B.
Number of more tractors that village C has = 8 – 5 = 3
iv) What is the total number of tractors in all the five village?
Total number of tractors in all these villages = 6 + 5 + 8 + 3 + 6 = 28.
Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
Observe this pictograph and answer the following questions:
From the above table, it can be concluded that in classes I, II, III, IV, V, VI, VII, VIII, there are 24, 18,20, 14, 10, 16, 12, 6 girls respectively.
a) Which class has the minimum number of girl student?
Class VIII has the minimum number of girls i.e. only 6 girls.
b) Is the number of girls in class VI less than the number of girls in Class V?
No, in class V and VI, there are 10 and 16 girls respectively clearly the number of girls is more in class VI than that in V
c) How many girls are there in class VII?
There are 12 girls in class VII.
Question 6.
The sale of electric bulbs on different days of a week is shown below.
Observe the pictograph and answer the following questions:
a) How many bulbs were sold on friday?
14 bulbs were sold on friday.
b) On Which day were the maximum number of bulbs sold?
The maximum numbers of bulbs (i.e., 18) were sold on Sunday.
c) On Which of the days same number of bulbs were sold?
Equal numbers of bulbs (i.e., 8) were sold on Wednesday and Saturday.
d) On Which of the days minimum number of bulbs were sold?
The minimum numbers of bulbs (i.e., 8) were so Id on Wednesday and Saturday.
e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Total bulbs sold in the week = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86
Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:
Observe this pictograph and answer the following questions:
a) Which merchant slid the maximum number of baskets?
b) How many fruit baskets were sold by answer?
c) The merchants who have sold 600 pr more number of baskets are planning to buy a godown for the next season, can you name them?
From the above pictograph, it can be observed that the number of fruit basket sold by Rahim, Lakhanpai Anwar, Martin, Ranjit sigh, and Joseph are 400, 550, 700, 950, 800 and 450 respectively.
a) Martin sold the maximum number of baskets i.e, 950
b) Anwar sold 700 baskets.
c) Anwar, martin, and Ranjith singh are the 3 merchants who have sold more than 600 baskets.
Therefore they are planning to by a godown for the next season. | {"url":"https://www.kseebsolutions.com/kseeb-solutions-for-class-6-maths-chapter-9-ex-9-1/","timestamp":"2024-11-13T01:26:07Z","content_type":"text/html","content_length":"78776","record_id":"<urn:uuid:7e7c1855-c439-4f3c-8a72-98c2999d928c>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00595.warc.gz"} |
A request for math help
An acquaintance, let's say a friend but we don't get together often, contacted Becky (my wife). She is probably in her 40s and wants to get some math help. She asked what Becky would charge for
tutoring. Becky said no charge but asked what sort of thing she had in mind, thinking maybe this was for work related issues. It wasn't. She just wants to be better at math, fairly basic math.
More power to her and Becky is very willing to help, but it seems that the first effort should be to set some sort of goals, choose some text or online resource, that sort of thing.
By contrast, Becky helped a teen-age girl with math some years ago. The girl's mother was homeschooling to protect her daughter from learning about evolution but the school system insisted that she
learn some math and the mother didn't know any. So there was a clear objective. The girl had to get through the math exam and there was a syllabus.
The woman who now has contacted Baecky mentions fractions, basic geometric concepts such as area, that sort of thing.
Any thoughts? I know of the Khan Academy but that seems to be organized for a person going from start to finish for a full year in, say geometry. Something different is wanted. This person doesn't
need to know where (or if) the three medians of a triangle intersect. But understanding that if the side lengths are all doubled then the area is quadrupled might be worthwhile.
It is stunning how often a normally intelligent person is floored by simple mathematical ideas. I am all for an adult wanting to do something about it but setting some realistic goals seems sensible.
Any ideas? | {"url":"https://www1.dal12.sl.bridgebase.com/forums/topic/70222-a-request-for-math-help/","timestamp":"2024-11-12T10:32:48Z","content_type":"application/xhtml+xml","content_length":"229579","record_id":"<urn:uuid:9d2324a3-7700-46ad-9fb3-e39fc1522dcc>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00861.warc.gz"} |
Maximum Likelihood Estimation (MLE)
Maximum Likelihood Estimation is a technique that we use to estimate the parameters for some statistical model. For example, let’s say that we have a truckload of oatmeal raisin cookies, and we
think that we can predict the number of raisins in each cookie based on features about the cookies, such as size, weight, etc. Further, we think that the sizes and weights can all be modeled by the
normal distribution. To define our cookie model, we would need to know the parameters, the mean(s) and the variance(s), which means that we need to count the raisins in each cookie, measure them,
weigh them, right? That sure would take a long time! But we have a group of hungry students, and each student is willing to count raisins, measure and weigh one cookie, ultimately producing these
features for a sample of the cookie population. Could that be enough information to at least guess the mean(s) and standard deviation(s) for the entire truckload? Yup!
What do we need to estimate parameters with MLE?
The big picture idea behind this method is that we are going to choose the parameters based on maximizing the likelihood of our sample data, based on the assumption that the number of raisins,
height, weight, etc. is normally distributed. Let’s review what we need in order to make this work:
• A hypothesis about the distribution. Check! We think that our raisins and other features each are normally distributed, defined by a mean and standard deviation
• Sample data. Check! Our hungry students gave us a nice list of counts and features for a sample of the cookies.
• A likelihood function: This is a function that takes, as input, a guess for the parameters of our model (the mean and standard deviation), our observed data, and plops out a value that represents
the likelihood (probability) of the parameters given the data. So a higher value means that our observed data is more likely, and we did a better job at guessing the parameters. If you think
back to your painful statistics courses, we model the probability of a distribution with the pdf, (probability density function). This is fantastic, because if we just maximize this function
then we can find the best parameters (the mean and standard deviation) to model our raisin cookies. How do we do that?
What are the steps to estimate parameters with MLE?
I am going to talk about this generally so that it could be applied to any problem for which we have the above components. I will define:
• θ: our parameters, likely a vector
• X: our observed data matrix (vectors of features) clumped together, with a row = one observation
• y: out outcome variable (in this case, raisin counts, which I will call continuous since you can easily have 1/2 a raisin, etc.)
This is clearly a supervised machine learning method, because we have both an outcome variable (y) and our features (X). We start with our likelihood function:
L(θ) = L(θ;X, ~y) = p(~y X; θ)
Oh no, statistics and symbols! What does it mean! We are saying that the Likelihood (L) of the parameters (θ) is a function of X and y, equal to the probability of y given X, parameterized by θ. In
this case, the “probability” is referring to the PDF (probability density function). To extend this equation to encompass multiple features (each normally distributed) we would write:
I apologize for the image - writing complex equations in these posts is beyond the prowess of my current wordpress plugins. This equation says that the likelihood of our parameters is obtained by
multiplying, for each feature 1 through m, the probability of y given the feature x, still parameterized by θ. Since we have said that our features are normally distributed, we plug in the function
for the Gaussian PDF. If you remember standard linear regression, the term in the numerator is the difference between the actual value and our predicted value (the error). So, we now have the
equation that we want to maximize! Do you see any issues?
I see a major issue - multiplication makes this really hard to work with. The trick that we are going to use is to take the log of the entire thing so that it becomes a sum. And the log of the
exponential function cancels out… do you see how this is getting more feasible? We also use a lowercase, italisized script “l” to represent “the log of the likelihood” :
The first line says that we are defining that stuff on the left as the log of the Likelihood. The second line is just writing “log” before the equation that we defined previously, the third line
moves the log into the summation (note that when you take the log of things multiplied together, that simplifies to summing them). We then distribute the log and simplify down to get a much more
feasible equation to maximize, however we are going to take a step further. If you look at the last line above, the first two terms do not depend on our variables of interest, so we can just look at
the last term.
This is equivalent to the cost function that we saw with linear regression! So we could solve for the parameters by maximizing or minimizing this function. And does it make sense that you could
apply this method to other distributions as well? You would start with their PDFs, take the log of the Likelihood, and then minimize to find the best parameters (and remember to do this we are
interested in finding where the derivative of the function with respect to our parameters is zero, because this means the rate of change of the function is zero because we are at a maximum or a
minimum point, where the slope is changing direction from positive to negative, or vice versa). And of course you can get R or Matlab to do all of this for you. Super cool, no?
Suggested Citation:
Sochat, Vanessa. "Maximum Likelihood Estimation (MLE)." @vsoch (blog), 17 Jun 2013, https://vsoch.github.io/2013/maximum-likelihood-estimation-mle/ (accessed 24 Oct 24). | {"url":"https://vsoch.github.io/2013/maximum-likelihood-estimation-mle/","timestamp":"2024-11-06T18:44:57Z","content_type":"text/html","content_length":"21875","record_id":"<urn:uuid:63f1790a-ba13-4ae5-b808-9cada0fa68a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00097.warc.gz"} |
Pi Symbols Copy and Paste πΏπΟπ«πΉε
Pi (Ο ) is a mathematical constant that represents the ratio of a circle's circumference to its diameter. It is an irrational number, meaning it cannot be expressed as a finite decimal or a fraction.
The value of Pi is approximately 3.14159, although it extends infinitely without repeating.
Pi is a fundamental constant in mathematics and appears in numerous mathematical formulas and equations. It has applications in various fields, including geometry, trigonometry, calculus, physics,
and engineering. | {"url":"https://www.getcoolsymbols.com/pi-symbols","timestamp":"2024-11-04T11:30:19Z","content_type":"text/html","content_length":"14828","record_id":"<urn:uuid:cb376138-851d-4d90-825c-bf50302c7e43>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00236.warc.gz"} |
Multivariate Linear Regression From Scratch | Towards AI
Multivariate Linear Regression From Scratch
Last Updated on July 25, 2023 by Editorial Team
Originally published on Towards AI.
With Code in Python
Photo by Hitesh Choudhary on Unsplash
In this new episode of the Machine Learning Basics series, we will create a model for a multivariate linear regression task and validate our model using Mathplotlib, Pandas, and NumPy libraries in
Python. While creating our model, we will use a Kaggle dataset as a training and validation resource. For this purpose, we will use and explain the essential parts of the following Kaggle Notebook:
Multivariate Linear Regression From Scratch
Explore and run machine learning code with Kaggle Notebooks U+007C Using data from Graduate Admission 2
Our previous episode explained basic machine learning concepts and the notions of hypothesis, cost, and gradient descent. In addition, using Python and some other libraries, we have written functions
for those algorithms and created and validated a model for a univariate linear regression task. If you are new to Machine Learning or want to refresh theΒ basic Machine Learning concepts, reading our
previous episode will be a good idea since we will not explain those subjects in this episode.
Univariate Linear Regression From Scratch
With Code in Python
We will use the gradient descent approach for the model creation task and will not discuss the Normal Equation since it will be the subject of another episode in this series. Therefore, we will write
code for the hypothesis, cost, and gradient descent functions before training our model.
We will also introduce two new concepts: Vectorization and Feature Scaling. So, letβ s start with vectorization.
Vectorization is the speed perspective in the implementation [1]. One way to implement Machine Learning algorithms is by using software loops in our code. Since we need to add or substruct values for
each data sample for implementing Machine Learning algorithms, it seems obvious to use software loops. On the other hand, various numerical linear algebra libraries we can utilize are either built-in
or quickly accessible in most programming languages. Those libraries are more efficient and can use parallel hardware in our computers if available. Therefore, instead of less efficient loops, we can
use vectors and matrices in those libraries to speed up. Moreover, we will write less code resulting in a cleaner implementation [2]. This concept is known as vectorization.
This blog post will use Pythonβ s NumPy library for vectorization.
Feature Scaling
We can speed up gradient descent by having each input value in roughly the same range [2]. This approach is called feature scaling. Applying feature scaling may be necessary for objective functions
to work precisely in some machine learning algorithms or penalize the coefficients appropriately in regularization [3].
Still, for our specific case, the gradient descent may converge more quickly if we apply feature scaling since ΞΈ will descend more rapidly in small ranges than large ranges, which causes
oscillations while finding the optimum ΞΈ values [2].
There are various techniques for scaling; we will present three of them:
For each sample value x of a feature, if we subtract the average value of that feature from x and divide the result by the standard deviation for that feature, we get a standardized solution. In
mathematical terms:
Mathematical Formulation for Standardization β Image by Author
Mean Normalization
For each sample value x of a feature, if we subtract the average value of that feature from x and divide the result by the difference between the maximum and minimum values of that feature, we get a
mean normalized solution. In mathematical terms:
Mathematical Formulation for Mean Normalization β Image by Author
Min-Max Scaling
For each sample value x of a feature, if we subtract the minimum value of that feature from x and divide the result by the difference between the maximum and minimum values of that feature, we get a
min-max scaled solution. In mathematical terms:
Mathematical Formulation for Min-Max Scaling β Image by Author
The Normalize Function
Below you will find the code we have used for feature scaling. You can find all three approaches we have discussed in the comments. We have chosen min-max scaling for feature scaling, which you can
see at the bottom of the code block.
The Python Code for Feature Scaling
We use the Pandas Libraryβ s min() and max() methods inside the code block. You can refer to the Imports section for importing Pandas Library. Interested readers may also refer to the Notebook
Multivariate Linear Regression From Scratch to see which columns are selected for scaling.
As stated previously, we need additional libraries to fulfill our requirements while writing the code for a multivariate regression task. These imports can be found below.
The Python Code for Imports
We will use the NumPy library for Matrix operations and vectorization purposes and the Pandas library for data processing tasks. The Matplotlib library will be used for visualization purposes.
Scikit-learn (or sklearn) is a prevalent library for Machine Learning. train_test_split from sklearn is used for splitting the dataset into two portions for training and validation. Although we will
not use train_test_split in our code in this blog post, we need to use it for splitting our data. Interested readers may refer to the Notebook Multivariate Linear Regression From Scratch for how it
is used.
The Hypothesis
The hypothesis will compute the predicted values for a given set of inputs and weights.
The hypothesis for a linear regression task with multiple variables is of the form:
The Hypothesis Formula β Image by Author
The equation above is not suitable for a vectorized solution. So letβ s define ΞΈ and x column vectors such that;
Column vectors of ΞΈ and x β Image by Author
Now we can rewrite the equation above as follows:
The Vectorized Hypothesis Formula β Image by Author
For most of the datasets available, each row represents one training (or test) sample such that:
A Common Dataset of m Samples and n Features β Image by Author
However, as described above, we need xβ = 1 for a vectorized solution. So, we must adjust our data set with a new column of xβ with all ones.
Modified Dataset of (m X n+1) Dimensionsβ Image by Author
Therefore, for a dataset of m samples and n features, we can insert a new column xβ of ones and then compute a column vector of (m x 1) dimensions such that every element of the column vector is a
predicted target for one dataset sample. The formula is as follows:
A Vectorized Hypothesis for an m Sample Dataset β Image by Author
Note that we have assigned a new column xβ to the x matrix, which is one. This is necessary for a vectorized solution, and we will see how to do it in the Model Training section.
As can be seen, we need two parameters to write the hypothesis function. These are an x matrix of (m X n+1) dimensions for training (or test) data and a ΞΈ column vector of (n+1) elements for
weights. m is the number of samples, and n is the number of features. So, letβ s write the Python function for the hypothesis:
The Python Code for the Hypothesis Function
Inside the hypothesis function, we return the dot product of the parameters using the NumPy library dot() method.
The Cost Function
We will use the squared error cost function as in the univariate linear regression case. Therefore, the cost function is:
Squared Error Cost Function β Image by Author
To achieve a vectorized solution, we can use the following vector:
The Squared Error Vector β Image by Author
This (m X 1) vector has some resemblance to the Hypothesis vector explained above, and we will exploit this similarity. The sum of this vectorβ s elements is the Ξ£ part of the cost function, and
that is:
The Sum of Squared Errors β Image by Author
We get the vector below if we factor the squares in each vector item.
Factorizationβ Image by Author
This equation can be rewritten as the dot product of the following row and column vectors.
The Dot Product β Image by Author
We get the following vectors if we change the ordering of the variables in the equation.
Reordering β Image by Author
And by replacing the vectorized hypothesis with XΞΈ, we get the following equation.
Vectorized Form β Image by Author
Finally, we get the vectorized cost function by multiplying this equation with one over two times m.
Vectorized Cost Function β Image by Author
We need three parameters for the cost function: the X, y, and ΞΈ vector. The X parameter is an (m X n+1) dimensions matrix where m is the number of examples and n is the number of features, and it is
the training (or test) dataset. The y parameter is an (m X 1) dimensions vector that holds the actual outputs for each sample. The ΞΈ parameter is an (n+1 X 1) dimensions vector for weights.
Note that we are using the hypothesis function inside the cost function, so to use the cost function, we must also write the hypothesis function.
The Python Code for the Cost Function
We can factor the cost function formula into three parts. A constant part, a vector of (mX1) dimensions, and the transpose of the same vector with (1Xm) dimensions.
So, inside the cost function, we first calculate the constant part. Then we calculate the (mX1) vector using the hypothesis function and the y vector. We can then use NumPyβ s transpose function to
figure the transpose of the same vector. Finally, we can dot product the two vectors employing the Numpy library, multiply by the constant and return the result.
Gradient Descent
The gradient descent computation is identical to the univariate version, but the equation must be expanded for n + 1 weights [2].
The Gradient Descent β Image by Author
In the formula above, n is the number of features, ΞΈβ±Ό (for j=0, 1, β ¦, n) is the corresponding weights of the hypothesis function, Ξ± is the learning rate, m is the number of examples, h(xβ ½β ±β
Ύ) is the result of the hypothesis function for the iα΅ Κ° training example, yβ ½β ±β Ύ is the actual value for the iα΅ Κ° training example and xβ β ½β ±β Ύ (for k=0, 1, β ¦, n; i=1, 2, β ¦, m) is
the value of kα΅ Κ° feature for the iα΅ Κ° training example.
Similar to what we have done in the cost function, we can rewrite the Ξ£ part of these equations as the dot product of two vectors for j=0, 1, β ¦, n:
Vectorized Form β Image by Author
So the vectorized solution for gradient descent becomes:
Vectorized Gradient Descentβ Image by Author
As can be seen, we need four parameters for the gradient descent function: the X, y, ΞΈ vector and the learning rate Ξ±.
Note that we are using the hypothesis function inside the gradient function, so to utilize the gradient function, we must also write the hypothesis function.
The Python Code for the Gradient Function
Inside the function, we first compute the constant part of the equation and the transpose of X. Then, we figure out the error vector by subtracting the actual output vector from the hypothesis
vector. Next, we compute the dot product of two previously calculated vectors and multiply it by the constant. As a final step, we subtract the outcome from the previous ΞΈ and return the result.
To create and validate our model, we will use the following dataset [4] from Kaggle:
Graduate Admission 2
Predicting admission from important parameters
The dataset includes several parameters critical for the application of Masterβ s Programs [5]:
GRE Scores ( out of 340 )
TOEFL Scores ( out of 120 )
University Rating ( out of 5 )
Statement of Purpose and Letter of Recommendation Strength ( out of 5 )
Undergraduate GPA ( out of 10 )
Research Experience ( either 0 or 1 )
Chance of Admit ( ranging from 0 to 1 )
The target variable is the β Chance of Admitβ parameter. We will use 80 % of the data for training and 20 % for testing. In addition, we will use a percentage estimation instead of finding the
probability for the target variable. We will not explain how to load and process data. Interested readers may refer to the Notebook Multivariate Linear Regression From Scratch for loading and
splitting the data.
There are two versions of CSV files in the dataset [5]. We will use version 1.1, which has 500 examples.
Model Training
Since we have written all the necessary functions, it is time to train the model.
To apply a vectorized solution, we must define an Xβ feature equal to 1 for all training examples. So our first task is to insert the Xβ column into the training and validation data. Next, we
need to initialize the ΞΈ vector, and we will set weights to all zero. As the last step before gradient descent, we will assign the learning rate and threshold values.
The Python Code for the Model Training Code Block
Inside the code block, the first thing to do is compute the m and n values for training and test data. Next, we will create a list of ones to insert into the DataFrames as the first column. This is
because we need to define the column Xβ =1 for a vectorized solution. Next, we set the ΞΈ vector as all zeros. After initializing the learning rate and threshold values, we can calculate the initial
cost value and initialize some essential variables for display purposes.
The threshold value will be used to check whether the gradient descent converges. We will subtract consecutive cost values inside the while loop to do that. If the difference is smaller than a
certain threshold, we will conclude that the gradient descent converges. Otherwise, we will keep taking the gradient and recalculating the cost value.
The Result
The result below is for the initial conditions in the model training code block. The training took 24 iterations. The final cost value was computed at approximately 66, and the ΞΈ vector is
calculated as:
ΞΈ: [2.03692557 1.13926025 1.14289218 6.02902784 6.60181418 6.82324225 1.20521232 1.25048269]
The Training Result β Image by Author
We can also utilize the learning curve to ensure that gradient descent works correctly [2]. The learning curve is a plot that shows us the change in the cost value in each iteration during the
training, and the cost value should decrease after every iteration [2]. Interested readers may refer to the Notebook Multivariate Linear Regression From Scratch for how to plot the learning curve.
The Learning Curveβ Image by Author
Model Validation
As explained above, we can use learning curves to debug our model. Our learning curve seems all right, and its shape is just as desired [2]. But we still require to validate our model. To do that, we
can use the validation dataset, which we have split before the training starts since, for validation, we must use data that has never been used in the training process. We will use the ΞΈ vector
calculated previously for validation.
Validation Result β Image by Author
The cost value of the validation dataset seems slightly bigger (worse) than the training dataset, which is expected.
In this blog post, we have written the hypothesis, cost, and gradient descent functions in Python with a vectorization method for a multivariate Linear Regression task. We have also scaled a portion
of our data, trained a Linear Regression model, and validated it by splitting our data. We have used the Kaggle dataset Graduate Admission 2 for this purpose.
We set the learning rate high and the convergence threshold low for demonstration purposes, resulting in a learning cost of approximately 66 and a validation cost of roughly 76. This high cost is
easily seen by checking the predicted and actual Chance of Admit values. By playing with the learning rate and threshold values, we can get a training cost value smaller than 20, resulting in better
But we must be aware that the dataset has been chosen only for demonstration purposes for a multivariate linear regression task, and there may be far better algorithms for this dataset.
Interested readers can find the code used in this blog post in the Kaggle notebook Multivariate Linear Regression from Scratch.
Thank you for reading!
[1] Ng, A. and Ma, T. (2022) CS229 Lecture Notes, CS229: Machine Learning. Stanford University. Available at: https://cs229.stanford.edu/notes2022fall/main_notes.pdf (Accessed: December 24, 2022).
[2] Ng, A. (2012) Machine Learning Specialization, Coursera.org. Stanford University. Available at: https://www.coursera.org/specializations/machine-learning-introduction (Accessed: January 9, 2023).
[3] Feature scaling (2022) Wikipedia. Wikimedia Foundation. Available at: https://en.wikipedia.org/wiki/Feature_scaling (Accessed: March 6, 2023).
[4] Mohan S Acharya, Asfia Armaan, Aneeta S Antony: A Comparison of Regression Models for Prediction of Graduate Admissions, IEEE International Conference on Computational Intelligence in Data
Science 2019
[5] Acharya, M.S. (2018) Graduate admission 2, Kaggle. Available at: https://www.kaggle.com/datasets/mohansacharya/graduate-admissions (Accessed: March 14, 2023).
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32nd M. Smoluchowski Symposium on Statistical Physics
Dr Maciej Majka (Jagiellonian University)
Binary mixtures, i.e. the systems composed of two different species of particles, exhibit a huge variety of dynamical modes and phase transitions. This includes demixing effects and several
combinations of mobility and arrest, e.g. the collective critical slow-down of the bigger particles mediated by the presence of the smaller molecules. It was recently realized that the interactions
in such system could be effectively mapped on the Spatially Correlated Noise (SCN), i.e. the thermal noise that affects neighboring particles in a similar manner[1,2]. Following this idea, the
over-damped SCN-driven Langevin dynamics was investigated as an effective, one-component model of dynamics in a dense binary mixture. It was found that the thermodynamically consistent
Fluctuation-Dissipation Relation for such system provides a novel insight into the arrest effects [3]. I will show that the mechanism of singular dissipation is embedded in the dissipation matrix,
accompanying SCN. The characteristic length of collective dissipation, which diverges at the critical packing is also identified. This is a new quantity, which conveniently grasps the difference
between the ergodic and non-ergodic dynamics and is a measure of cooperativity in the system. The model is fully analytically solvable, one-dimensional and admits arbitrary interactions between
particles. The transition is controlled by the interplay between the packing fraction and the noise correlation length, representing the packing fraction of smaller particles. As a practical example,
both the hard spheres and the system of ultra-soft particles were studied. The framework of this model makes it is possible to discuss e.g. the re-entrant arrest.
[1] M. Majka, P.F. Góra, Thermodynamically consistent Langevin dynamics with spatially correlated noise predicting frictionless regime and transient attraction effect, Phys. Rev. E, 94, 4, 042110
[2] M. Majka, P.F. Góra, Collectivity in diffusion of colloidal particles: from effective interactions to spatially correlated noise, J. Phys. A: Math. Theor., 50, 5, 054004 (2017)
[3] M. Majka, P.F. Góra, Molecular arrest induced by spatially correlated stochastic dynamics https://arxiv.org/abs/1707.07076 | {"url":"https://zakopane.if.uj.edu.pl/event/9/contributions/339/","timestamp":"2024-11-14T17:24:58Z","content_type":"text/html","content_length":"60012","record_id":"<urn:uuid:6e58f944-9fd3-42f2-bc86-89bf62cc8dc5>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00333.warc.gz"} |
Influence of Parameters in the Design of a Faceted Structure for Incoherent Beam Shaping
Abstract: A reflective faceted structure is proposed to reshaping an incoherent light beam into two focalized spots. To obtain the desired irradiance distribution on a detector, custom optimization
function is written, and the two dimensional tilt angles of each facet are optimized automatically in a pure non-sequential mode in Zemax OpticStudio 16. The result is also confirmed inside
LightTools 8.2 from Synopsys. For measuring the quality of the optimization result in the case of two spots focalization, four factors including efficiency on the detector, uniformity, the root mean
square error and the correlation coefficient are calculated. These four factors are used to evaluate the influence of several parameters on the irradiance distribution. These parameters include the
incidence angle, the divergence angle, the facet size, the source type and the resolution of the facet angular positions. Finally, an analysis of those parameters is made and the performance of this
type of component is demonstrated.
Keywords: Incoherent beam shaping; micro lens array; custom optimization | {"url":"http://www.jommpublish.org/p/79/","timestamp":"2024-11-08T22:12:45Z","content_type":"text/html","content_length":"113858","record_id":"<urn:uuid:e16edfd8-1ae1-463d-96af-341441a83ee6>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00187.warc.gz"} |
Saving HP EVs
After reading through the Smogon dex and Introduction to Competitive Pokémon, you embark on a mission to create a team, and after finally finding the members that suit you most, you went ahead and
opened Shoddy Battle, joyously inputting every single piece of information. As you select Scizor as a member, you hover ahead the Attack EVs section, maximizing it. Following that, you wish to give
Scizor as much overall bulk as possible, so you go ahead and prepare your fingers to input 252 in HP...
Halt! I warn you, this is an unwise action.
Take a moment to step back and consider the following. When Scizor’s HP is maximized, it produces the number 344. That number is unfortunately divisible by 8, which signals indirect damage hazard.
Stealth Rock, burn, Leech Seed, normal poisoning and even Darkrai’s Bad Dreams take away 1/8ths of the victim’s maximum HP. This means a maximum HP Scizor will faint in 8 times of exposure to one of
these (though it can not be poisoned). When your Scizor spams U-turn in order to weaken enemy counters, you might find your precious mantis dying upon its 8th switch in due to Stealth Rock.
Max HP Divisible by 8 but not by 16
Scizor alone is not the only one that suffers from this disorder when HP is maxed. In fact, there’s quite a list of Pokémon whose maximum HP is divisible by 8, yet not divisible by 16.
Base 70
Aggron*, Ariados, Cacturne, Castform, Cherrim, Darkrai, Delcatty, Drapion, Froslass+, Girafarig, Lucario*, Lunatone, Magnezone*, Manectric, Masquerain++, Mightyena, Mothim++, Omastar, Sceptile,
Scizor, Scyther++, Sharpedo, Solrock, Sudowoodo, Torkoal+, Venomoth+, Vespiquen++, Weavile+.
Base 78
Charizard++, Linoone, Typhlosion+.
Base 86
Cradily, Yanmega++.
Base 110
Lickilicky, Mamoswine, Regigigas, Walrein+, Whiscash*.
Base 150
Giratina, Giratina-O, Slaking, Drifblim+.
Base 190
The asterisk (*) indicates reduced Stealth Rock damage due to type. The plus (+) indicates increased Stealth Rock damage due to type. Similarly, a (++) indicates a 4x increase in standard Stealth
Rock damage.
Not only are the numbers divisible by 8, two turns of Leftovers recovery can not make up for the indirect damage. Furthermore, you’ll realize that you are wasting a point by maximizing HP on any of
these (Aggron is perhaps an exception since it takes reduced Stealth Rock damage, is immune to Poison, and Burn renders it useless anyways). To demonstrate what I mean, let’s put it in a simple
example we all can understand.
There are two Scizors; we will refer to them by Scizor 1, who has maximum HP EVs, and Scizor 2, who has 248 HP EVs. Both of them are sent out into the battlefield where Stealth Rock awaits to inflict
injury. The results should be as followed:
Scizor 1:
Scizor 2:
As you can see, they will enter the trap-laden battlefield with the same HP. However, should you want to switch Scizor out more than once…
Scizor 1:
Scizor 2:
What do we have here? The Scizor that started out with one point lower in HP ends up having higher HP upon its second time switching into the field.
Max HP Divisible by 4 or 2, but not by 16
But let’s not stop there shall we. We shall discuss the Pokémon that take increased damage from Stealth Rock as well. These Pokémon, when their HP is maxed, maximizes the pain that Stealth Rock
brings upon them
Adding to the previous list of Pokémon marked with the plus, here are some more that share their shame. The HP of all of these Pokémon are not divisible by 16
Base 20
Base 45
Base 52
Base 55
Base 60
Beautifly++, Butterfree++, Dodrio, Dustox, Parasect, Pelipper, Swellow, Wormadam
Base 61
Base 80
Aerodactyl, Glalie, Regice
Base 83
Base 100
Noctowl, Shaymin-S, Honchkrow
Base 120
Arceus (Only if it is one of the following types: Bug, Flying, Fire, Ice). Arceus is currently illegal.
Like previously, the double plus (++) indicates 50% damage when the Pokémon is switched in while Stealth Rock is active against you.
It may be tempting to maximize the HP of these Pokémon (especially Shuckle, who sports the lowest non Shedinja HP, yet the highest defenses in the game), but in reality, having one point less in HP
is better. The 100 Base HP receives a special note as maximizing HP allows the Pokémon to create 101 HP Substitutes, withstanding a Seismic Toss from Blissey. However, aside from possibly Shaymin-S,
the rest should not bother with such a task.
Now, with this outlook on entry hazards in mind, you can save one or even more points. But how much is a single point worth? It can give you a greater chance of survival, a greater chance of knocking
something out, or more importantly in some cases, outrunning the ignorant opposition. Did I say one point? You can save more when you plan to switch the Pokémon in more than once. This one point you
remove from HP can save your Pokémon from its deathbed during critical moments.
Max HP Leftovers Numbers
Now, there’s one more thing to discuss that falls in a similar issue. Known since the dawn of Ruby and Sapphire, it is the famed “Leftovers numbers”. These numbers allow for maximum recovery by the
hold item Leftovers as they are divisible by 16. Sounds great? Not always. Being divisible by 16 also means divisible by 8. Should these numbers be avoided? It depends on the situation you plan to
encounter. But before we go to that, we shall note the Pokémon whose HP is divisible by 16 when it is maxed.
Base 50
Cloyster, Deoxys, Deoxys-A, Deoxys-D, Deoxys-S, Hitmonchan, Hitmonlee, Hitmontop, Magcargo, Mawile, Rotom, Rotom-A, Sableye, Spiritomb.
Base 90
Abomasnow*, Ampharos, Arcanine+, Articuno++, Dewgong+, Donphan, Granbull, Machamp, Moltres++, Nidoqueen, Palkia, Politoed, Poliwrath, Raikou, Shiftry, Ursaring, Zapdos+.
Base 106
Mewtwo, Lugia+, Ho-oh++.
Base 130
Lapras+, Vaporeon.
Base 170
++ indicates 50% Stealth Rock damage
+ indicates 25% Stealth Rock damage
* indicates Abomasnow creates hail.
As stated before, to maximize or to not maximize depends almost entirely on your situation. For many of these Pokémon, maximizing HP also maximizes defensive capabilities, while having one point less
can sometimes produce the same type of problems as mentioned when you maximize a base 70 Pokémon’s HP (For instance, a Deoxys-D with 303 HP takes 37 damage from Stealth Rock, and two rounds of
leftovers recovery won’t completely heal it). However, as appealing as maximizing Leftovers recovery sounds, sandstorm and hail negate this advantage, and leaves your Pokémon suffering from increased
indirect damage. If you have one of these weathers in your team, it is generally a bad idea to maximize the HP EVs of these Pokémon unless they are immune to it. However, when you do not have
damaging weather on your side, it can go both ways. Do you wish to maximize the HP EVs of one of these Pokémon due to your own team, or do you maximize HP based on the possibility of sandstorm or
hail? You should consider the frequency of these weathers in different metagames. For example, sandstorm is common in OU due to Hippowdon’s and Tyranitar’s presence, but it is quite rare in Ubers due
to Kyogre and Groudon being everywhere, canceling out the sandstorm Tyranitar creates, or the hail Abomasnow brings. Therefore, it is safer to maximize HP of any of these Pokémon in the Uber metagame
compared to the OU metagame. Also consider other factors such as Rapid Spin or Taunt that could prevent the set up on entry hazards. It would probably be a more preferable option to max HP when you
have these and no damaging weather on your side (that hurts it obviously).
6 EVs in HP
Before we go, if you are going to go with a classic 252/252/6 Spread on some Pokémon, putting 6 EVs into HP automatically turns these Pokémon “Stealth Rock weak”, meaning that these Pokémon will fall
in 4 turns of switching when Stealth Rock is present.
Gyarados, Salamence, Togekiss, Dragonite, Ninjask++.
Crobat, Altaria, Armaldo, Houndoom, Jumpluff, Magmortar, Ninetales, Jynx.
Mantine, Pinsir, Rapidash, Entei.
Now that you’ve bled your eyes out upon reading all of these, you now know that EVs aren’t as simple as it would seem. Saving only one or two points may not be a big deal, but remember that there are
virtually no downsides to doing so, considering the prevalence of Stealth Rock, and the occasional Leech Seed and Will-O-Wisp.
Of course, there's yet another final word to this. It's not always maximizing HP, but make sure your HP isn't divisible by 8 (unless it is a lefties number and you decide to go by that due to
circumstance). This means Scizor with let's say...188 HP EVs is asking to die in 8 switch ins. | {"url":"https://www.smogon.com/smog/issue2/saving_hp2","timestamp":"2024-11-10T01:26:01Z","content_type":"application/xhtml+xml","content_length":"12647","record_id":"<urn:uuid:5877beb4-2d01-4b3d-9413-22074172c5de>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00606.warc.gz"} |
Browsing Codes
SAOImage DS9 is an astronomical imaging and data visualization application. DS9 supports FITS images and binary tables, multiple frame buffers, region manipulation, and many scale algorithms and
colormaps. It provides for easy communication with external analysis tasks and is highly configurable and extensible via XPA and SAMP. DS9 is a stand-alone application. It requires no installation or
support files. Versions of DS9 currently exist for Solaris, Linux, MacOSX, and Windows. All versions and platforms support a consistent set of GUI and functional capabilities. DS9 supports advanced
features such as multiple frame buffers, mosaic images, tiling, blinking, geometric markers, colormap manipulation, scaling, arbitrary zoom, rotation, pan, and a variety of coordinate systems. DS9
also supports FTP and HTTP access. The GUI for DS9 is user configurable. GUI elements such as the coordinate display, panner, magnifier, horizontal and vertical graphs, button bar, and colorbar can
be configured via menus or the command line. DS9 is a Tk/Tcl application which utilizes the SAOTk widget set. It also incorporates the X Public Access (XPA) mechanism to allow external processes to
access and control its data, GUI functions, and algorithms. | {"url":"https://ascl.net/code/all/page/57/limit/50/order/title/listmode/full/dir/asc","timestamp":"2024-11-02T20:38:20Z","content_type":"text/html","content_length":"75161","record_id":"<urn:uuid:a134e28b-6950-4fb3-9752-91c6af0bd71b>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00389.warc.gz"} |
Dear readers,
Designing the world’s largest amateur rocket is not an easy task. It’s an iterative process with a lot of dependencies. One of the very important dependencies relates to the rocket’s « propellant
delivery system », so how do you get fuel into the engine. There are several possibilities where a turbopump is clearly the sexiest, but also the hardest to master. On Nexø II, the fuel was delivered
via the DPR system, which basically consisted of a tank of 20 liters of helium at 300 bar and a set of control valves. Let’s look at how such a system will look like in Spica size.
Let’s assume that Spica without fuel has a weight of 1700 kg and that its BPM100 engine is running at a chamber pressure of 15 bar. To bring this up to a height of 100 km, we need about 2000 kg fuel.
At an OF ratio of 1.3, the LOX/ethanol has a medium density of approximately 1000 kg/m^3, so the size of the fuel tanks must be approximately 2000 liters. If we are to fly with 100% thrust all the
way (we most likely won’t), then we have a volume of 2000 liters, which at MECO has to be pressurized to around 20 bars. It will require 40,000 standard liter (SL) of compressed gas (nitrogen) to
achieve. It must of course be brought along in a compressed form, but it also has its price. During expansion from the high-pressure tank and down to 20 bar the gas is cooled and « loses » volume
according to the ideal gas laws. The temperature drop is calculated easily as the final temperature is given as
T2 = T1 * (P2/P1)^((gamma-1)/gamma)
Where T1 is the starting temperature, P1 and P2 are start and end pressures, respectively, and gamma is the ratio of specific heats, which for nitrogen is 1.40. Below, I have plotted the end
temperature as a function of starting pressure, assuming we expand to 20 bars starting from 295 K.
If we initially have 300 bars in the tank, then the expanded gas will have a final temperature of approx. 136 K, it will have « lost » more than half of its volume. Of course, it gets better the
closer we get to P2, but overall, we lose almost 40% of the gas volume due to this cooling. If we need 40,000 SL, we must bring about 67,000 SL. If we use the same tank as for Nexø II, then we need
11 of these. They each weigh 12 kg, so we have gas tanks for a total of 132 kg. Then additionally we have tubes, fittings and valves …
It is therefore a heavy system. And an expensive system. The high-pressure tank on Nexø II was one of the most expensive single components of the rocket, with a price tag of around 10,000 kroner,
imported from China. If anyone out there knows about large volume lightweight tanks at a reasonable price, we’ll be happy to hear from you.
Considering this, we’re looking at something else. We plan to introduce another cryogenic liquid on Spica. Besides, of course, having LOX on board, we are strongly considering taking the necessary
nitrogen for pressurization in liquid form. If we bring only 40 liters of liquid nitrogen which you then evaporate and heat to 200-300 degrees, then you have all the gas you need.
Heat is available in ample amounts in the BPM100 engine, but we are a little nervous about having to implement a nitrogen heat exchanger in the engine, so we consider bringing a small burner that
just has to evaporate the liquid nitrogen and heat it up. An important advantage of using a separate burner is the ability to test it without having to do a whole BPM100 test.
Nothing is certain at this point. We need to do some more calculations, but I find it very exciting. Though I’m not much for the extra complexity of having two cryogenic fluids on board. But in
spaceflight, things rarely come without these kind of costs.
15 Comments | {"url":"https://copenhagensuborbitals.com/bpm100-dpr/?lang=fr","timestamp":"2024-11-11T23:12:03Z","content_type":"text/html","content_length":"49295","record_id":"<urn:uuid:a982d9bc-d3de-4429-85f8-7fc77dc0abaa>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00860.warc.gz"} |
Towards the F-theorem: N = 2 field theories on the three-sphere
For 3-dimensional field theories with N = 2 supersymmetry the Euclidean path integrals on the three-sphere can be calculated using the method of localization; they reduce to certain matrix integrals
that depend on the R-charges of the matter fields. We solve a number of such large N matrix models and calculate the free energy F as a function of the trial R-charges consistent with the marginality
of the superpotential. In all our N = 2 superconformal examples, the local maximization of F yields answers that scale as N ^3/2 and agree with the dual M-theory backgrounds AdS[4] × Y , where Y are
7-dimensional Sasaki- Einstein spaces. We also find in toric examples that local F-maximization is equivalent to the minimization of the volume of Y over the space of Sasakian metrics, a procedure
also referred to as Z-minimization. Moreover, we find that the functions F and Z are related for any trial R-charges. In the models we study F is positive and decreases along RG flows.We therefore
propose the " F-theorem" that we hope applies to all 3-d field theories: the finite part of the free energy on the three-sphere decreases along RG trajectories and is stationary at RG fixed points.
We also show that in an infinite class of Chern-Simonsmatter gauge theories where the Chern-Simons levels do not sum to zero, the free energy grows as N5/3 at large N. This non-trivial scaling
matches that of the free energy of the gravity duals in type IIA string theory with Romans mass.
All Science Journal Classification (ASJC) codes
• Nuclear and High Energy Physics
• AdS-CFT correspondence
• Matrix models
• Renormalization group
• Strong coupling expansion
Dive into the research topics of 'Towards the F-theorem: N = 2 field theories on the three-sphere'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/towards-the-f-theorem-n-2-field-theories-on-the-three-sphere","timestamp":"2024-11-04T14:42:16Z","content_type":"text/html","content_length":"53578","record_id":"<urn:uuid:f9ad0a69-d784-49e5-806a-7850b8a61670>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00421.warc.gz"} |
Dreaming in numbers
I don’t dream in numbers, but if I did, I’m pretty sure it’d look a lot like this. In Nature by Numbers, a short movie by Cristobal Vila, inspired by, well, numbers and nature, Vila animates the
natural existence of Fibonacci sequences, the golden ratio, and Delaunay triangulation. Watch it. Even if you don’t know what those three things are, the video will rock your socks off.
Don’t forget to catch shots from the process too:
[via infosthetics]
28 Comments
• One time, I was sleeping and my wife decided to ask me some questions. She asked me if I loved her. I replied, “Of course I love you; it’s because of the numbers 2 1 5 6 3 4 0 3 2 4 0 8 4.” Or
some other string of random numbers. Perhaps it was the golden ratio.
• wow this is fantastic!!!
It’s really beautiful!
I will surely use this video in my classroom, soon. Thanks for sharing Nathan. Your blog always rocks!!
• Fantastic !!!
• Pingback: Eau de Golden Ratio « Layman's layout
• Pingback: Beautiful Math Dream « Barron of Blog
• Wonderful video.
• Amazing, fantastic, mind blowing !!!! loved it !! as i wish to be good artist and also inspired by nature now can clearly see the connection between the two sides of the brain…emotions and
calculations, maths !! Answering all “why” questions !!
I just loved it !!
count me one of your fan Cristobal Vila
• Beautiful. For some reason it reminds me of your math functions in real life post some time ago. https://flowingdata.com/2010/02/12/math-functions-in-the-real-world/
□ i was thinking the same exact thing when i first saw this.
• my boyfriend and i stumbled upon this video and all i can say is…wow. it was really beautiful. just amazing to see how much of an artform it is. i have such an amazement with math this just made
it ten times better.
• Pingback: BiofusionDesign | Nature by numbers
• This really makes one wonder – Nature, is it random chance or design?
• Wow! I was stunned by this video!
Marvelous! Stumbled. This made my day. ^^
• Fantático cheio de vida!
• Life based figures
• Wow..! i realy into math and was stunned by this video. GJ!
• Does anyone know the song in the background? I know I have heard it before.
□ In the end credits, it says the music is ‘Often a Bird’ by Wim Mertens.
• LOL…i’m tired but I should probably pay attention more often.
• Wow, probably the single most amazing thing I have ever seen.
• Great video! Although for the image, a+b is not equal to (a+b)/a
• You blog is really amazing and i thoroughly enjoyed this numbers logic. I am a bit inclined towards numerology and always get fascinated by the logic behind numbers and what all juice you can
extract from them.
Keep it up.
• one word….amaaaaaaazing.
this vid made believe even more that all the nature came from one creative source. (Allah)
Thank you for the post, keep the good work.
• I think it’s funny how the golden ratio is so apparent in this video, but the music is in a minor key. should have definitely enforced the Fibonacci sequence with a repeating pattern made of
major arpeggios. :D
• Absolutely beautiful video – seen it 3 times now!
I can’t explain why, but I’ve been seeing the number 137 for numerous years without even thinking about it. I understand that it could be my subconscious at work trying to hunt down the number
but hows this: often times just before i sleep i remove clothes from my alarm clock only to see the bright green LEDs write 1:37 (at night). I’ve gotten so used to seeing that number, I can spot
it a mile off….. I don’t know if it’s correlated to that 137.5 number in the video, and if it is, I’m off by 0.5 :P | {"url":"https://flowingdata.com/2010/05/10/dreaming-in-numbers/","timestamp":"2024-11-06T15:25:31Z","content_type":"text/html","content_length":"96181","record_id":"<urn:uuid:38f4d21c-7c33-48d2-8639-f3577ec751b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00209.warc.gz"} |
Pressure drop consistently lower than measured with SimpleFoam
The other thing to keep in mind is that you are being quite ambitious in trying to use CFD for this ... bear in mind that the pressure loss is all down to flow turbulence, and that the reason why the
90deg bend generates a larger pressure drop than a straight duct is because of flow separation and enhanced turbulence generation (which sucks energy out of the mean flow, ie the pressure field).
So, to get the correct pressure drop, you need to be able to model the smooth body separation and secondary flow patterns in the duct to
high accuracy
. This is not trivial! You will need a good mesh and a good turbulence model to get the answer spot on ... I don't want to dissuade you from trying, but do realise that this is not a trivial problem.
Lastly - if you want to be ultra accurate - remember also that for simplicity, RANS solvers often combine the following two terms
terms, into a single pressure gradient term,
(the turbulence term is the contribution from the normal Reynolds stresses). The solver then solves for
rather than for the static pressure
, so for a true comparison you also ned to subtract off the turbulence energy part ... this is typically negligible though, which is why the approximation is made!
Good luck. | {"url":"https://www.cfd-online.com/Forums/openfoam-solving/234316-pressure-drop-consistently-lower-than-measured-simplefoam.html","timestamp":"2024-11-05T14:04:22Z","content_type":"application/xhtml+xml","content_length":"109679","record_id":"<urn:uuid:99142ecc-815e-4e17-9681-aec6ff577503>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00061.warc.gz"} |
Data types - Perun2
Data types
The type system of Perun2 is very primitive. It consists of 9 internally defined data types and there are no more. Perun2 performs automatic casting between higher and lower data types when it is
possible as shown in the image below. For example, bool casted into string becomes a text value 0 or 1. It works only one direction.
Perun2 is statically typed. New variables are declared implicitly, with no need to specify their type.
Several expressions can be applied to multiple data types.
structure returns
[bool] ? [value] : [value] value
Ternary conditional operator. If the first argument is true, then the second argument is returned. Otherwise the result is the third argument. This structure works on every data type.
structure returns
[bool] ? [value] value
This is a simplified ternary operator. If the first argument equals false, then the whole expression returns empty value of certain data type. This value is empty text for string, integer zero for
number, empty period for period and empty collection for collections. This structure works on every data type.
Filters work only with collections. See
structure returns
[value] , [value] value collection
[value] , [value collection] value collection
[value collection] , [value] value collection
[value collection] , [value collection] value collection
Multiple elements can be joined together by commas in order to create a merged collection. If one of joined elements is a lazy evaluated collection (also known as definition), then the whole chain of
elements becomes lazy evaluated.
structure returns
[value collection variable] [ [number] ] value
To reach a certain element of a collection variable, use square brackets with an index written inside them. Indexing is zero-based. If index is out of range, then an empty value is returned.
The boolean data type can store only two values: true and false. When casted into number or string, true is treated as 1 and false is treated as 0.
structure returns
true bool
false bool
New boolean constants can be called by keywords true and false.
structure returns
[value] = [value] bool
[value] == [value] bool
[value] != [value] bool
[value] < [value] bool
[value] <= [value] bool
[value] > [value] bool
[value] >= [value] bool
Any two valid data type instances can be compared. Comparisons are carried out according to several rules. Boolean value true is considered greater than false. Two times are equal, if they share any
common moment in time. Strings are compared by their alphabetic order and they are sensitive to case size. Two collections are equal, if they contain the same elements in the same places. A
collection that contains more elements is considered greater.
structure returns
not [bool] bool
[bool] and [bool] bool
[bool] or [bool] bool
[bool] xor [bool] bool
Unlike other programming languages, Perun2 uses keywords instead of symbols for boolean operators. Elements can be grouped by brackets to determine the order of operations. All binary operators (and,
or, xor) have the same priority.
structure returns
[value] in [value collection] bool
[value] not in [value collection] bool
Checks if a specified value can be found inside a collection. Inclusion of the not keyword reverses the result.
structure returns
[string] like [string] bool
[string] not like [string] bool
This expression can be used to compare a string with a pattern. It uses several wildcart characters explained
. If the pattern is not valid and for example contains not closed brackets, then this expression always returns
structure returns
[string] resembles [string] bool
[string] not resembles [string] bool
This operator works similar to the Like operator. Go
for more info.
structure returns
[value] between [value] and [value] bool
[value] not between [value] and [value] bool
The Between operator works with any data type that is not a collection. Bounding values are included, so this expression is an equivalent to >= and <=. The order of bounds does not matter. A
non-existent value (NaN or never) as any argument makes the Between operator always return false provided there are no casts.
structure returns
[string] regexp [string] bool
[string] not regexp [string] bool
Perun2 uses the ECMAScript convention for regular expression matches. This operator is case sensitive.
Numbers in Perun2 can appear in three forms: as integers, in double-precision format or as NaN (not a number). The type of a number is assigned dynamically after every performed operation. For
example, division of two integers can result in an integer (8/4), a fraction (8/5) or NaN (8/0). In order to avoid integer overflows, the greatest effective amount of bits is used for integer
representation. This value depends on the operating system and equals at least 64 bits. The dot is always the decimal separator.
Numberic literals are consecutive digits with one optionary dot sign inside as a decimal separator. Preceding sign - makes the number negative.
suffix multiplier
kb 1024
mb 1024²
gb 1024³
tb 1024⁴
pb 1024⁵
Numbers can be followed by a suffix. This suffix multiplies the number, so it can be treated as a file size unit. Make sure that there is no space between the number and the suffix. For example, 100
megabytes would be expressed as 100mb.
suffix multiplier
k 1000
m 1000²
We can use these two decimal suffixes to express thousands and millions. For example, 3k means three thousand.
An integer literal can contain one K infix. It multiplies the preceding part by one thousand. This feature can be used to express years in a shorter way. For example, instead of 2023, we can write
structure returns
- [number] number
[number] + [number] number
[number] - [number] number
[number] * [number] number
[number] / [number] number
[number] % [number] number
Operators for multiplication, division and modulo have higher priority than operators for addition and subtraction. Expression elements can be grouped by brackets to enforce a desired order of
structure returns
[time variable].[time variable number] number
Numeric values can be obtained from time variables.
time variable number
year years
month months
weekday -
day days
hour hours
minute minutes
second seconds
Values are subject to two rules. Month take value from 1 (January) to 12 (December). Week days start from 1 (Monday) to 7 (Sunday). After all, these values do not have to be used by the language user
at all, as convenient time constants are more readable and not ambiguous.
Time expresses one moment in time. Time can point either a certain month, a certain day, a certain minute or a certain second. When casted into a string, time is written the same way as an equivalent
time constant or clock constant would be expressed. Month name is written in lowercase except for the first letter that is in uppercase.
structure returns
[month name] [nc] time
[nc] [month name] [nc] time
[nc] [month name] [nc] , [nc] : [nc] time
[nc] [month name] [nc] , [nc] : [nc] : [nc] time
Time constants can be built in four ways. All of them require an English month name and several numeric constants (shown above as [nc]). In the final presented form, numeric constants are in
sequence: days, years, hours, minutes and seconds.
month name
january february march april
may june july august
september october november december
You probably know English month names, but they are presented above anyway.
structure returns
[nc] : [nc] time
[nc] : [nc] : [nc] time
Clock constants need hours, minutes and (optionary) seconds.
structure returns
[time] + [period] time
[time] - [period] time
Time can be increased or decreased by a period.
structure returns
[time variable].date time
This expression returns a value of a time variable excluding its clock part (hours, minutes and seconds).
Period describes a period of time expressed in years, months, days, hours, minutes and seconds. Each of these units is an integer and can be negative. The ambiguity of years and months may introduce
a bit of confusion, as in reality each of them consists of different amount of days. Perun2 tries to represent them as realistically as possible. When conversion into days in inevitable, each
ambiguous month is treated as 30 days and each ambiguous year as 365 days. Period remembers several details such as amount of leap years it contains or days in months and uses them behind the curtain
for comparisons.
structure returns
1 [period singular] period
[number] [period plural] period
A new period unit can be defined by a number and a period keyword.
period singular period plural
year years
month months
week weeks
day days
hour hours
minute minutes
second seconds
Period keywords can take either singular or plural form.
structure returns
- [period] period
[period] + [period] period
[period] - [period] period
Negation, addition and subtraction can be performed on periods.
structure returns
[time] - [time] period
Subtraction performed on two times returns a period as a result.
String represents a sequence of Unicode characters.
Characters placed between two apostrophes form a new string literal. There is an exception: it cannot contain asterisks nor apostrophes. An asterisk turns this structure into an Asterisk Pattern. For
the sake of simplicity, string literals in Perun2 do not involve any escape characters nor escape sequences. Strings mean exactly what they show, so backslashes can be used safely for filesystem
This is the way to omit the limitations of apostrophe-defined string literals. A string literal formed with backtick characters can additionally contain asterisks and apostrophes. Just like before,
there are no escape sequences.
structure returns
[string variable] [ [number] ] string
Reach certain character of a string variable and return it as a new string. Indexing is zero-based. Negative indexes are allowed and enable reversed access from end to start. For example, character
at index -1 is the last one and at -2 is the penultimate. If index is out of range, then an empty string is returned.
structure returns
[string] + [string] string
Multiple strings can be concatenated by pluses. You should pay attention to adjacent elements. If two adjacent elements are numbers, they are summed. The same rule goes for periods and combinations
of times and periods.
Definition is a lazy evaluated collection of strings. Unlike all other data types, definition generates values on demand instead of generating all of them at once. This data type is crucial for
Perun2, as it enables efficient iteration over filesystem elements. It appears through important built-in variables such as files and directories.
Asterisk Patterns are apostrophe-defined string literals that contain at least one asterisk. They are explained deeply
Lists are just vectors of values. The easiest way to initialize them is by writing several values with commas between them. List of strings is the lowest data type in Perun2. | {"url":"https://perun2.org/docs/datatypes","timestamp":"2024-11-07T10:45:33Z","content_type":"text/html","content_length":"31489","record_id":"<urn:uuid:bfde07c7-a381-4ae9-bd19-d6a5b41704e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00800.warc.gz"} |
10 15 In Simplest Form
In the busy digital age, where screens dominate our every day lives, there's a long-lasting appeal in the simpleness of published puzzles. Amongst the myriad of ageless word games, the Printable Word
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View Question Write 10 15 In Simplest Form
10 15 In Simplest Form
Web Step 1 Enter the fraction you want to simplify The Fraction Calculator will reduce a fraction to its simplest form You can also add subtract multiply and divide fractions as well
Printable Word Searches offer a wonderful retreat from the consistent buzz of modern technology, permitting people to immerse themselves in a globe of letters and words. With a pencil in hand and a
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Write In Simplest Form Brainly
Write In Simplest Form Brainly
Web Answer Fraction 10 15 simplified to lowest terms is 2 3 10 15 2 3 2 3 is the simplified fraction of 10 15 Simplifying Fraction 10 15 using GCF The first way to simplify the fraction 10 15 is to
use the Greatest
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Simplest Form 5 5 Why Is Simplest Form 5 5 So Famous Simplest Form
Simplest Form 5 5 Why Is Simplest Form 5 5 So Famous Simplest Form
Web What is the Simplified Form of 10 15 A simplified fraction is a fraction that has been reduced to its lowest terms In other words it s a fraction where the numerator the top
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Find Each Of The Following Ratios In The Simplest Form Ratio In
Find Each Of The Following Ratios In The Simplest Form Ratio In
Web 3 sept 2022 nbsp 0183 32 9 3K views 10 months ago In this video we will simplify reduce the fraction 10 15 into its simplest form The key to simplifying fractions is to find a number that goes
into both the numerator
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Download 10 15 In Simplest Form
Fraction Calculator Mathway
Web Step 1 Enter the fraction you want to simplify The Fraction Calculator will reduce a fraction to its simplest form You can also add subtract multiply and divide fractions as well
What Is 10 15 Simplified To Simplest Form Calculatio
Web Answer Fraction 10 15 simplified to lowest terms is 2 3 10 15 2 3 2 3 is the simplified fraction of 10 15 Simplifying Fraction 10 15 using GCF The first way to simplify the fraction 10 15 is to
use the Greatest
Web Step 1 Enter the fraction you want to simplify The Fraction Calculator will reduce a fraction to its simplest form You can also add subtract multiply and divide fractions as well
Web Answer Fraction 10 15 simplified to lowest terms is 2 3 10 15 2 3 2 3 is the simplified fraction of 10 15 Simplifying Fraction 10 15 using GCF The first way to simplify the fraction 10 15 is to
use the Greatest
How To Simplify A Ratio To Its Simplest Form YouTube
Rections Perform The Operations As Indicated Write Each Answer In
The Area Of The Triangle Below Is frac 2 CameraMath
What Is 15 9 30 In Simplest Form Brainly
Fraction In Simplest Form Answers
Simplest Form Math Fractions
Simplest Form Math Fractions
14 9th Grade Fraction Worksheets Worksheeto | {"url":"https://reimbursementform.com/en/10-15-in-simplest-form.html","timestamp":"2024-11-10T01:02:03Z","content_type":"text/html","content_length":"20890","record_id":"<urn:uuid:5455c3f3-0e82-4d79-86ae-12acc31ddf37>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00154.warc.gz"} |
Mathematical Modeling of Real-World Problems
Insights from mathematical models make the headlines daily, though the accompanying numbers and graphs do not always make sense to the uninitiated. What makes it difficult for models to agree on
climate change predictions? Why is it so hard to predict and mitigate the impacts of a pandemic?
Exposing students to authentic problems equips them with sophisticated problem-solving skills that are essential to succeed in future job markets and to thrive as global citizens. The Computing with
R for Mathematical Modeling project (CodeR4MATH) is developing mathematical modeling activities within an interactive online environment that help high school students apply mathematical concepts and
skills such as writing equations and reading and understanding charts to solve real-world problems. Students explore open-ended questions using real-world data and a powerful and popular programming
Mathematical modeling activities
Mathematical modeling activities are a common component of computer science (CS) courses. However, not every secondary school offers CS courses, and when they do, the courses are typically electives.
By bringing real-world mathematical modeling experiences to high school math classrooms we aim to provide students and teachers with the opportunity to exercise computational thinking through math
modeling using the R programming language, an environment for statistical computing and graphics that is widely used by academics and data science professionals worldwide.
CodeR4MATH activities are designed to help students explore how mathematics and computation can be combined to build models that represent and help make sense of the world around them. While
CodeR4MATH activities mimic the types of challenges and learning processes that professionals face when solving problems, various levels of scaffolding are available to students through pre-populated
code snippets, hints, prompts, and quizzes.
Figure 1. New CodeR4Math activities for secondary students.
The CodeR4MATH activities follow a common pattern. First, they present a real-world situation and pose an open-ended question, then they guide students through the process of brainstorming the
problem, making assumptions, and defining the essential variables. This culminates in a simplified description or model of the problem. Students then create algorithms to solve the simplified model,
testing their algorithms and assumptions by coding in R. By the end of each activity, students have a mathematical model in the form of computer code, representing a solution to the simplified
version of the original, open-ended problem. Students revisit their list of initial assumptions, with the goal of improving their models.
In the “Lifehacking with R” activity, each student assumes the role of a college advisor, responsible for giving financial advice to first year college students who need to decide whether to purchase
a meal plan or pay for food on a per meal basis, called simply “meal plan” and “pay as you go.” Students quickly realize that one size does not fit all. In addition to the financial aspect, the
answer may depend on other factors, some of which are hard to quantify, such as personal preferences and values.
Brainstorming bare necessities
The first step is to brainstorm the problem. Students are given the cost of a meal plan at a specific college as well as hypothetical costs for each meal. They are told that the college has many
restaurants at its campus center, and receive a list of several. Students are free to use any search engine to browse these restaurants, the kinds of food they serve, prices, and more. They are asked
to jot down anything they think would influence the decision. To start, they consider the financial aspect. What is the cheapest option? What factors control the costs?
Creating algorithms and the first steps of coding
Keeping the initial assumptions in mind, students experiment with several code snippets, already populated with a full or partial R code (Figure 2). The activity offers high scaffolding when the
student runs existing code to check the output, medium scaffolding when the student can modify a small portion of the code and run it to check the new output, or minimal scaffolding when the student
follows examples to add new elements such as new variables or mathematical expressions. If code changes prevent it from running, a “Start Over” button allows the student to reset to the original
code. Students also have access to hints with templates that help them modify their code with new equations and graph outputs. After students submit their work, they can see the solution.
Figure 2. Example of an exercise from a CodeR4MATH activity with an R code snippet for students to complete and run. They can start over or check hints at any time, and the solution becomes available
after they submit their work.
Data-driven models and the impact of visualizations
After exploring and working with some code snippets, students dig into real datasets. They are able to explore tables (which R calls “data frames”), extract statistical information from different
variables, explore relationships and dependencies, work with data visualizations, and compare results. Figure 3 shows some of the elements that can appear in the CodeR4MATH activities, such as code
snippets, outputs like statistics, graphs, and tables, and multiple-choice and open-ended questions.
Figure 3. R code snippets and their outputs, including graphs, statistical information tables, and open-ended and multiple-choice questions.
Almost 140 students have participated in classroom implementations of CodeR4MATH activities. Our results have shown that even students who have never done any prior computer programming are able to
understand mathematical modeling and how professionals use computer programming to create, test, and validate models.
Through CodeR4MATH activities, students learn to apply the iterative process of mathematical modeling: defining the problem, making assumptions, simplifying the problem, defining variables, getting
solutions, validating the model, and reporting results. Throughout the process, students engage in computational thinking while learning the basics of coding in a widely used programming language.
The teacher’s role is to spark and facilitate brainstorming, control the pace of the activity, provide feedback to students, coordinate group or individual projects, and evaluate student work.
With a new set of activities, we hope even more students will have the opportunity to make sense of real-world problems through math modeling (Figure 1). Teachers and students can access free
web-based CodeR4MATH activities without downloading software or registering for accounts using any computer with a current Web browser.
Kenia T. Wiedemann (kwiedemann@concord.org) is a postdoctoral researcher.
This material is based upon work supported by the National Science Foundation under grant DRL-1742083. Any opinions, findings, and conclusions or recommendations expressed in this material are those
of the author(s) and do not necessarily reflect the views of the National Science Foundation. | {"url":"https://concord.org/newsletter/2020-fall/mathematical-modeling-real-world-problems/","timestamp":"2024-11-04T05:30:08Z","content_type":"text/html","content_length":"65719","record_id":"<urn:uuid:bf08831a-6a67-4014-af88-958265d2d9a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00208.warc.gz"} |
Low Temperature and Strong Magnetic Field Laboratory
Low Temperature and Strong Magnetic Field Laboratory is situated in the sub terrain floor (room 011) of the Department of Physics. Research is conducted using high magnetic fileds (up to 18 T), and
low temperatures (down to 0.3 K). We investigate magnetotransport properties (resistance, magnetoresistance, Hall effect) of various materials (low dimensional conductors, oxide
Our laboratory is equiped with:
• superconducting magnets: 18 T and 12 T
• variable temperature inserts: 1.5 K to 300 K
• He^3 temperature insert: down to 0.3 K
• equipment for AC and DC resistance measurement: Lock-in amplifiers, AC and DC current sources (e.g. Keithley 6221), nanovoltmenters (e.g. Keithley 181)
Current research projects:
Recent topics of our research:
• low dimensional conductors We have investigated magnetotransport properties of quasi-one-dimensional conductors: Bechgaard-Fabre salts, TTF-TCNQ, quasi-one-dimensional cuprates Sr[14-][x]Ca[x]Cu
[24]O[41], and quasi-two-dimensional organics based on BEDT-TTF, in order to examine the influence of dimensionality on the magnetotransport properties, and to prove the importance of the correct
choice of geometry for measurements of low dimensional conductors.
□ M. Čulo, E. Tafra, M.Basletić, S. Tomić, A.Hamzić, B.Korin-Hamzić, M.Dressel, J.A.Schlueter, Two-dimensional variable range hopping in the spin-liquid candidate κ-(BEDT-TTF)[2]Cu[2](CN)[3],
Physica B: Physics of Condensed Matter 460, 208 (2015)
□ M. Pinterić, M. Čulo, O. Milat, M. Basletić, B. Korin-Hamzić, E. Tafra, A. Hamzić, T. Ivek, T. Peterseim, K. Miyagawa, K. Kanoda, J. A. Schlueter, M. Dressel, S. Tomić, Anisotropic charge
dynamics in the quantum spin-liquid candidate κ-(BEDT-TTF)[2]Cu[2](CN)[3], Phys. Rev. B 90, 195139 (2014)
□ E. Tafra, M. Čulo, B. Korin-Hamzić, M. Basletić, A. Hamzić, C. S. Jacobsen, The Hall effect in the organic conductor TTF-TCNQ: choice of geometry for accurate measurements of a highly
anisotropic system, J. Phys.: Condens. Matter 24, 045602 (2012)
□ E. Tafra, B. Korin-Hamzić, M. Basletić, A. Hamzić, M. Dressel, J. Akimitsu, Influence of doping on the Hall coefficient in Sr[14-x]Ca[x]Cu[24]O[41], Phys. Rev. B 78, 155122 (2008)
• metallic heterostructures We have investigated Extrinsic Inverse Spin Hall Effect in copper-iridium alloy, and we have shown that alloys combining side-jump and skew scattering are promising for
eficient transformation of charge current into spin current in devices without magnetic components
• oxide heterostructures We have investigated magnetotransport properties of oxide heterostuctures based on SrTiO[3](STO), which are being considered as a substitution for classical ferromagnets in
spintronics. For Co-LSTO((La,Sr)Ti[0.98]Co[0.02]O[3])/STO, LaAlO[3](LAO)/STO and ion-irradiated STO, we examined the dimensionality, the width, and the origin of the high-mobility electron layer.
In photo from left to right: A. Hamzić, M. Basletić, A. Fert (visiting our lab) and E. Tafra
People working in the Low Temperature and Strong Magnetic Field Laboratory:
We are always open for collaboration and please don't hesitate to contact us. | {"url":"https://www.chem.pmf.hr/phy/en/research_groups/ltsmfl","timestamp":"2024-11-04T04:37:57Z","content_type":"text/html","content_length":"86529","record_id":"<urn:uuid:93d8efb1-f11d-4505-995c-e99d93d771c9>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00613.warc.gz"} |
Mastering the Fundamentals of Statics with R.C. Hibbeler's Engineering Mechanics: Statics - A Comprehensive Book Synthesis - Knowledge Hub: Your learning resources
Mastering the Fundamentals of Statics with R.C. Hibbeler’s Engineering Mechanics: Statics – A Comprehensive Book Synthesis
December 14, 2023 University Book synthesis
Engineering mechanics is a branch of science that deals with the behavior of physical bodies when subjected to forces or displacements. It is a fundamental discipline in the field of engineering,
providing the necessary tools and principles for analyzing and designing structures, machines, and systems. One of the key components of engineering mechanics is statics, which focuses on the study
of objects at rest or in equilibrium.
Statics is concerned with understanding and predicting the behavior of stationary objects under the influence of various forces. It provides engineers with the necessary tools to analyze and design
structures, machines, and systems that can withstand external loads and maintain stability. Without a solid understanding of statics, engineers would not be able to accurately predict how structures
will behave under different conditions, leading to potential failures and safety hazards.
Understanding the Basic Concepts of Statics
Equilibrium is a fundamental concept in statics, referring to a state in which an object is at rest or moving at a constant velocity. There are two types of equilibrium: static equilibrium and
dynamic equilibrium. Static equilibrium occurs when an object is at rest, with no net force or torque acting on it. Dynamic equilibrium occurs when an object is moving at a constant velocity, with no
net force or torque acting on it.
Forces are another important concept in statics. A force is a push or pull that can cause an object to accelerate or deform. There are several types of forces, including gravitational forces, normal
forces, frictional forces, tension forces, and shear forces. Each type of force has its own characteristics and effects on objects.
Moments are rotational forces that cause an object to rotate around a fixed point or axis. They are also known as torques and are measured in units of force multiplied by distance. Moments can be
classified into two types: clockwise moments and counterclockwise moments. The sum of all moments acting on an object determines its overall rotational equilibrium.
Analyzing Structures and Machines
Free-body diagrams are a crucial tool in statics for analyzing the forces acting on an object. They represent the object as a single isolated body and show all the external forces acting on it. By
drawing a free-body diagram, engineers can determine the magnitude and direction of each force and analyze how they interact with each other.
Force analysis is another important aspect of statics. It involves determining the internal forces within a structure or machine and analyzing their effects on its overall stability. There are
various methods and techniques for force analysis, including the method of joints and the method of sections.
The method of joints is used to analyze truss structures by considering the equilibrium of forces at each joint. The method involves drawing a free-body diagram of each joint and applying the
equations of equilibrium to solve for the unknown forces. The method of sections, on the other hand, involves cutting through a structure along a section and analyzing the forces acting on that
Solving Statics Problems
Trusses are rigid structures composed of interconnected members that are subjected to external loads. They are commonly used in bridges, roofs, and other structures that require high
strength-to-weight ratios. Analyzing trusses involves determining the internal forces in each member and ensuring that they can withstand the applied loads.
The method of joints is often used to analyze trusses by considering the equilibrium of forces at each joint. The method involves drawing a free-body diagram of each joint and applying the equations
of equilibrium to solve for the unknown forces. By repeating this process for all joints in the truss, engineers can determine the internal forces in each member.
The method of sections is another technique used to analyze trusses. It involves cutting through a truss along a section and analyzing the forces acting on that section. By applying the equations of
equilibrium to the cut section, engineers can determine the internal forces in each member.
Friction and its Effects on Statics
Friction is a force that opposes the relative motion or tendency of motion between two surfaces in contact. It plays a significant role in statics, as it can affect the stability and equilibrium of
objects. There are two types of friction: static friction and kinetic friction.
Static friction occurs when two surfaces are in contact but not moving relative to each other. It prevents the surfaces from sliding against each other and increases as the applied force increases.
Kinetic friction, on the other hand, occurs when two surfaces are in motion relative to each other. It opposes the motion and is generally less than static friction.
The angle of friction is a measure of the maximum angle at which an object can rest on an inclined plane without sliding. It is determined by the ratio of the maximum static friction force to the
normal force acting on the object. The coefficient of friction is a dimensionless quantity that represents the ratio of the frictional force between two surfaces to the normal force pressing them
Center of Gravity and Moment of Inertia
The center of gravity is the point at which the entire weight of an object can be considered to act. It is an important concept in statics, as it determines the stability and balance of objects.
Calculating the center of gravity involves determining the weighted average position of all the individual particles that make up an object.
The moment of inertia is a measure of an object’s resistance to changes in its rotational motion. It depends on both the mass distribution and shape of an object. Calculating the moment of inertia
involves integrating the mass distribution over the entire object and summing up all the individual moments.
Various shapes have different moments of inertia. For example, a solid cylinder has a moment of inertia that depends on its mass, radius, and axis of rotation. A hollow cylinder, on the other hand,
has a moment of inertia that depends on its mass, inner radius, outer radius, and axis of rotation. By calculating the moments of inertia for different shapes, engineers can determine their
rotational behavior.
Beams and Frames
Beams and frames are common structural elements used in buildings, bridges, and other structures. They are subjected to various external loads and must be designed to withstand these loads while
maintaining their stability. Shear and bending moment diagrams are important tools for analyzing the internal forces in beams and frames.
Shear diagrams represent the variation of shear forces along the length of a beam or frame. They show the magnitude and direction of the shear forces at different points along the structure. Bending
moment diagrams, on the other hand, represent the variation of bending moments along the length of a beam or frame. They show the magnitude and direction of the bending moments at different points
along the structure.
Deflection and slope are two important parameters that engineers consider when designing beams and frames. Deflection refers to the displacement of a point on a beam or frame from its original
position under an applied load. Slope refers to the change in angle of a beam or frame at a particular point under an applied load. By calculating the deflection and slope, engineers can ensure that
a structure will not deform excessively under load.
Virtual Work and Energy Methods
Virtual work and energy methods are powerful tools in statics for analyzing structures, machines, and systems. They involve considering the work done by external forces on an object or system and
using energy principles to determine its equilibrium.
Work done by a force is defined as the product of the force applied to an object and the displacement of that object in the direction of the force. It represents the energy transferred to or from an
object by an external force. By calculating the work done by all external forces on an object or system, engineers can determine its equilibrium.
The principle of virtual work states that if an object is in equilibrium, the virtual work done by all external forces on the object is zero. This principle allows engineers to analyze the
equilibrium of complex structures and systems by considering the work done by individual forces.
Conservation of energy is another important principle in statics. It states that the total energy of an isolated system remains constant over time. By applying the principle of conservation of
energy, engineers can analyze the equilibrium of systems and determine their stability.
Introduction to Dynamics
Dynamics is the branch of engineering mechanics that deals with the motion of objects under the influence of forces. It is concerned with understanding and predicting how objects move and accelerate
in response to external forces. There are two main branches of dynamics: kinematics and kinetics.
Kinematics is the study of motion without considering the forces that cause it. It involves analyzing the position, velocity, and acceleration of objects as they move through space. Kinematics
provides engineers with the necessary tools to describe and predict the motion of objects in a variety of applications.
Kinetics, on the other hand, is the study of motion taking into account the forces that cause it. It involves analyzing how forces affect the motion and acceleration of objects. Kinetics provides
engineers with the necessary tools to understand and predict how objects will respond to external forces.
Real-World Applications of Statics
Statics has numerous real-world applications in various fields of engineering. It is used in designing and analyzing structures, machines, and systems to ensure their stability, safety, and
In structural engineering, statics is used to design buildings, bridges, and other structures that can withstand external loads such as wind, earthquakes, and snow. By analyzing the forces acting on
a structure and ensuring its equilibrium, engineers can design structures that are safe and structurally sound.
In mechanical engineering, statics is used to design machines and mechanical systems that can perform their intended functions without failure or excessive deformation. By analyzing the forces and
moments acting on different components of a machine, engineers can ensure that they are properly designed and can withstand the applied loads.
In systems engineering, statics is used to analyze and design complex systems that involve multiple components and interactions. By considering the forces and moments acting on each component of a
system, engineers can ensure that it will function as intended and maintain its stability.
Statics is a fundamental discipline in the field of engineering, providing the necessary tools and principles for analyzing and designing structures, machines, and systems. It is essential for
ensuring the stability, safety, and efficiency of various engineering applications. By understanding the basic concepts of statics, analyzing structures and machines, solving statics problems,
considering frictional effects, calculating the center of gravity and moment of inertia, analyzing beams and frames, applying virtual work and energy methods, understanding dynamics, and applying
statics to real-world applications, engineers can effectively design and analyze a wide range of engineering systems. | {"url":"https://free-info-site.com/mastering-the-fundamentals-of-statics-with-r-c-hibbelers-engineering-mechanics-statics-a-comprehensive-book-synthesis/","timestamp":"2024-11-07T18:49:35Z","content_type":"text/html","content_length":"84724","record_id":"<urn:uuid:408d0821-c6fd-47ce-88b4-582a285ce9e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00462.warc.gz"} |
Theory Homotopy
(* Title: HOL/Analysis/Path_Connected.thy
Authors: LC Paulson and Robert Himmelmann (TU Muenchen), based on material from HOL Light
section ‹Homotopy of Maps›
theory Homotopy
imports Path_Connected Product_Topology Uncountable_Sets
definition✐‹tag important› homotopic_with
"homotopic_with P X Y f g ≡
(∃h. continuous_map (prod_topology (top_of_set {0..1::real}) X) Y h ∧
(∀x. h(0, x) = f x) ∧
(∀x. h(1, x) = g x) ∧
(∀t ∈ {0..1}. P(λx. h(t,x))))"
text‹‹p›, ‹q› are functions ‹X → Y›, and the property ‹P› restricts all intermediate maps.
We often just want to require that ‹P› fixes some subset, but to include the case of a loop homotopy,
it is convenient to have a general property ‹P›.›
abbreviation homotopic_with_canon ::
"[('a::topological_space ⇒ 'b::topological_space) ⇒ bool, 'a set, 'b set, 'a ⇒ 'b, 'a ⇒ 'b] ⇒ bool"
"homotopic_with_canon P S T p q ≡ homotopic_with P (top_of_set S) (top_of_set T) p q"
lemma split_01: "{0..1::real} = {0..1/2} ∪ {1/2..1}"
by force
lemma split_01_prod: "{0..1::real} × X = ({0..1/2} × X) ∪ ({1/2..1} × X)"
by force
lemma image_Pair_const: "(λx. (x, c)) ` A = A × {c}"
by auto
lemma fst_o_paired [simp]: "fst ∘ (λ(x,y). (f x y, g x y)) = (λ(x,y). f x y)"
by auto
lemma snd_o_paired [simp]: "snd ∘ (λ(x,y). (f x y, g x y)) = (λ(x,y). g x y)"
by auto
lemma continuous_on_o_Pair: "⟦continuous_on (T × X) h; t ∈ T⟧ ⟹ continuous_on X (h ∘ Pair t)"
by (fast intro: continuous_intros elim!: continuous_on_subset)
lemma continuous_map_o_Pair:
assumes h: "continuous_map (prod_topology X Y) Z h" and t: "t ∈ topspace X"
shows "continuous_map Y Z (h ∘ Pair t)"
by (intro continuous_map_compose [OF _ h] continuous_intros; simp add: t)
subsection✐‹tag unimportant›‹Trivial properties›
text ‹We often want to just localize the ending function equality or whatever.›
text✐‹tag important› ‹%whitespace›
proposition homotopic_with:
assumes "⋀h k. (⋀x. x ∈ topspace X ⟹ h x = k x) ⟹ (P h ⟷ P k)"
shows "homotopic_with P X Y p q ⟷
(∃h. continuous_map (prod_topology (subtopology euclideanreal {0..1}) X) Y h ∧
(∀x ∈ topspace X. h(0,x) = p x) ∧
(∀x ∈ topspace X. h(1,x) = q x) ∧
(∀t ∈ {0..1}. P(λx. h(t, x))))"
unfolding homotopic_with_def
apply (rule iffI, blast, clarify)
apply (rule_tac x="λ(u,v). if v ∈ topspace X then h(u,v) else if u = 0 then p v else q v" in exI)
apply simp
by (smt (verit, best) SigmaE assms case_prod_conv continuous_map_eq topspace_prod_topology)
lemma homotopic_with_mono:
assumes hom: "homotopic_with P X Y f g"
and Q: "⋀h. ⟦continuous_map X Y h; P h⟧ ⟹ Q h"
shows "homotopic_with Q X Y f g"
using hom unfolding homotopic_with_def
by (force simp: o_def dest: continuous_map_o_Pair intro: Q)
lemma homotopic_with_imp_continuous_maps:
assumes "homotopic_with P X Y f g"
shows "continuous_map X Y f ∧ continuous_map X Y g"
proof -
obtain h :: "real × 'a ⇒ 'b"
where conth: "continuous_map (prod_topology (top_of_set {0..1}) X) Y h"
and h: "∀x. h (0, x) = f x" "∀x. h (1, x) = g x"
using assms by (auto simp: homotopic_with_def)
have *: "t ∈ {0..1} ⟹ continuous_map X Y (h ∘ (λx. (t,x)))" for t
by (rule continuous_map_compose [OF _ conth]) (simp add: o_def continuous_map_pairwise)
show ?thesis
using h *[of 0] *[of 1] by (simp add: continuous_map_eq)
lemma homotopic_with_imp_continuous:
assumes "homotopic_with_canon P X Y f g"
shows "continuous_on X f ∧ continuous_on X g"
by (meson assms continuous_map_subtopology_eu homotopic_with_imp_continuous_maps)
lemma homotopic_with_imp_property:
assumes "homotopic_with P X Y f g"
shows "P f ∧ P g"
obtain h where h: "⋀x. h(0, x) = f x" "⋀x. h(1, x) = g x" and P: "⋀t. t ∈ {0..1::real} ⟹ P(λx. h(t,x))"
using assms by (force simp: homotopic_with_def)
show "P f" "P g"
using P [of 0] P [of 1] by (force simp: h)+
lemma homotopic_with_equal:
assumes "P f" "P g" and contf: "continuous_map X Y f" and fg: "⋀x. x ∈ topspace X ⟹ f x = g x"
shows "homotopic_with P X Y f g"
unfolding homotopic_with_def
proof (intro exI conjI allI ballI)
let ?h = "λ(t::real,x). if t = 1 then g x else f x"
show "continuous_map (prod_topology (top_of_set {0..1}) X) Y ?h"
proof (rule continuous_map_eq)
show "continuous_map (prod_topology (top_of_set {0..1}) X) Y (f ∘ snd)"
by (simp add: contf continuous_map_of_snd)
qed (auto simp: fg)
show "P (λx. ?h (t, x))" if "t ∈ {0..1}" for t
by (cases "t = 1") (simp_all add: assms)
qed auto
lemma homotopic_with_imp_subset1:
"homotopic_with_canon P X Y f g ⟹ f ` X ⊆ Y"
by (meson continuous_map_subtopology_eu homotopic_with_imp_continuous_maps)
lemma homotopic_with_imp_subset2:
"homotopic_with_canon P X Y f g ⟹ g ` X ⊆ Y"
by (meson continuous_map_subtopology_eu homotopic_with_imp_continuous_maps)
lemma homotopic_with_imp_funspace1:
"homotopic_with_canon P X Y f g ⟹ f ∈ X → Y"
using homotopic_with_imp_subset1 by blast
lemma homotopic_with_imp_funspace2:
"homotopic_with_canon P X Y f g ⟹ g ∈ X → Y"
using homotopic_with_imp_subset2 by blast
lemma homotopic_with_subset_left:
"⟦homotopic_with_canon P X Y f g; Z ⊆ X⟧ ⟹ homotopic_with_canon P Z Y f g"
unfolding homotopic_with_def by (auto elim!: continuous_on_subset ex_forward)
lemma homotopic_with_subset_right:
"⟦homotopic_with_canon P X Y f g; Y ⊆ Z⟧ ⟹ homotopic_with_canon P X Z f g"
unfolding homotopic_with_def by (auto elim!: continuous_on_subset ex_forward)
subsection‹Homotopy with P is an equivalence relation›
text ‹(on continuous functions mapping X into Y that satisfy P, though this only affects reflexivity)›
lemma homotopic_with_refl [simp]: "homotopic_with P X Y f f ⟷ continuous_map X Y f ∧ P f"
by (metis homotopic_with_equal homotopic_with_imp_continuous_maps homotopic_with_imp_property)
lemma homotopic_with_symD:
assumes "homotopic_with P X Y f g"
shows "homotopic_with P X Y g f"
proof -
let ?I01 = "subtopology euclideanreal {0..1}"
let ?j = "λy. (1 - fst y, snd y)"
have 1: "continuous_map (prod_topology ?I01 X) (prod_topology euclideanreal X) ?j"
by (intro continuous_intros; simp add: continuous_map_subtopology_fst prod_topology_subtopology)
have *: "continuous_map (prod_topology ?I01 X) (prod_topology ?I01 X) ?j"
proof -
have "continuous_map (prod_topology ?I01 X) (subtopology (prod_topology euclideanreal X) ({0..1} × topspace X)) ?j"
by (simp add: continuous_map_into_subtopology [OF 1] image_subset_iff flip: image_subset_iff_funcset)
then show ?thesis
by (simp add: prod_topology_subtopology(1))
show ?thesis
using assms
apply (clarsimp simp: homotopic_with_def)
subgoal for h
by (rule_tac x="h ∘ (λy. (1 - fst y, snd y))" in exI) (simp add: continuous_map_compose [OF *])
lemma homotopic_with_sym:
"homotopic_with P X Y f g ⟷ homotopic_with P X Y g f"
by (metis homotopic_with_symD)
proposition homotopic_with_trans:
assumes "homotopic_with P X Y f g" "homotopic_with P X Y g h"
shows "homotopic_with P X Y f h"
proof -
let ?X01 = "prod_topology (subtopology euclideanreal {0..1}) X"
obtain k1 k2
where contk1: "continuous_map ?X01 Y k1" and contk2: "continuous_map ?X01 Y k2"
and k12: "∀x. k1 (1, x) = g x" "∀x. k2 (0, x) = g x"
"∀x. k1 (0, x) = f x" "∀x. k2 (1, x) = h x"
and P: "∀t∈{0..1}. P (λx. k1 (t, x))" "∀t∈{0..1}. P (λx. k2 (t, x))"
using assms by (auto simp: homotopic_with_def)
define k where "k ≡ λy. if fst y ≤ 1/2
then (k1 ∘ (λx. (2 *⇩[R] fst x, snd x))) y
else (k2 ∘ (λx. (2 *⇩[R] fst x -1, snd x))) y"
have keq: "k1 (2 * u, v) = k2 (2 * u -1, v)" if "u = 1/2" for u v
by (simp add: k12 that)
show ?thesis
unfolding homotopic_with_def
proof (intro exI conjI)
show "continuous_map ?X01 Y k"
unfolding k_def
proof (rule continuous_map_cases_le)
show fst: "continuous_map ?X01 euclideanreal fst"
using continuous_map_fst continuous_map_in_subtopology by blast
show "continuous_map ?X01 euclideanreal (λx. 1/2)"
by simp
show "continuous_map (subtopology ?X01 {y ∈ topspace ?X01. fst y ≤ 1/2}) Y
(k1 ∘ (λx. (2 *⇩[R] fst x, snd x)))"
apply (intro fst continuous_map_compose [OF _ contk1] continuous_intros continuous_map_into_subtopology continuous_map_from_subtopology | simp)+
by (force simp: prod_topology_subtopology)
show "continuous_map (subtopology ?X01 {y ∈ topspace ?X01. 1/2 ≤ fst y}) Y
(k2 ∘ (λx. (2 *⇩[R] fst x -1, snd x)))"
apply (intro fst continuous_map_compose [OF _ contk2] continuous_intros continuous_map_into_subtopology continuous_map_from_subtopology | simp)+
by (force simp: prod_topology_subtopology)
show "(k1 ∘ (λx. (2 *⇩[R] fst x, snd x))) y = (k2 ∘ (λx. (2 *⇩[R] fst x -1, snd x))) y"
if "y ∈ topspace ?X01" and "fst y = 1/2" for y
using that by (simp add: keq)
show "∀x. k (0, x) = f x"
by (simp add: k12 k_def)
show "∀x. k (1, x) = h x"
by (simp add: k12 k_def)
show "∀t∈{0..1}. P (λx. k (t, x))"
fix t show "t∈{0..1} ⟹ P (λx. k (t, x))"
by (cases "t ≤ 1/2") (auto simp: k_def P)
lemma homotopic_with_id2:
"(⋀x. x ∈ topspace X ⟹ g (f x) = x) ⟹ homotopic_with (λx. True) X X (g ∘ f) id"
by (metis comp_apply continuous_map_id eq_id_iff homotopic_with_equal homotopic_with_symD)
subsection‹Continuity lemmas›
lemma homotopic_with_compose_continuous_map_left:
"⟦homotopic_with p X1 X2 f g; continuous_map X2 X3 h; ⋀j. p j ⟹ q(h ∘ j)⟧
⟹ homotopic_with q X1 X3 (h ∘ f) (h ∘ g)"
unfolding homotopic_with_def
apply clarify
subgoal for k
by (rule_tac x="h ∘ k" in exI) (rule conjI continuous_map_compose | simp add: o_def)+
lemma homotopic_with_compose_continuous_map_right:
assumes hom: "homotopic_with p X2 X3 f g" and conth: "continuous_map X1 X2 h"
and q: "⋀j. p j ⟹ q(j ∘ h)"
shows "homotopic_with q X1 X3 (f ∘ h) (g ∘ h)"
proof -
obtain k
where contk: "continuous_map (prod_topology (subtopology euclideanreal {0..1}) X2) X3 k"
and k: "∀x. k (0, x) = f x" "∀x. k (1, x) = g x" and p: "⋀t. t∈{0..1} ⟹ p (λx. k (t, x))"
using hom unfolding homotopic_with_def by blast
have hsnd: "continuous_map (prod_topology (subtopology euclideanreal {0..1}) X1) X2 (h ∘ snd)"
by (rule continuous_map_compose [OF continuous_map_snd conth])
let ?h = "k ∘ (λ(t,x). (t,h x))"
show ?thesis
unfolding homotopic_with_def
proof (intro exI conjI allI ballI)
have "continuous_map (prod_topology (top_of_set {0..1}) X1)
(prod_topology (top_of_set {0..1::real}) X2) (λ(t, x). (t, h x))"
by (metis (mono_tags, lifting) case_prod_beta' comp_def continuous_map_eq continuous_map_fst continuous_map_pairedI hsnd)
then show "continuous_map (prod_topology (subtopology euclideanreal {0..1}) X1) X3 ?h"
by (intro conjI continuous_map_compose [OF _ contk])
show "q (λx. ?h (t, x))" if "t ∈ {0..1}" for t
using q [OF p [OF that]] by (simp add: o_def)
qed (auto simp: k)
corollary homotopic_compose:
assumes "homotopic_with (λx. True) X Y f f'" "homotopic_with (λx. True) Y Z g g'"
shows "homotopic_with (λx. True) X Z (g ∘ f) (g' ∘ f')"
by (metis assms homotopic_with_compose_continuous_map_left homotopic_with_compose_continuous_map_right homotopic_with_imp_continuous_maps homotopic_with_trans)
proposition homotopic_with_compose_continuous_right:
"⟦homotopic_with_canon (λf. p (f ∘ h)) X Y f g; continuous_on W h; h ∈ W → X⟧
⟹ homotopic_with_canon p W Y (f ∘ h) (g ∘ h)"
by (simp add: homotopic_with_compose_continuous_map_right image_subset_iff_funcset)
proposition homotopic_with_compose_continuous_left:
"⟦homotopic_with_canon (λf. p (h ∘ f)) X Y f g; continuous_on Y h; h ∈ Y → Z⟧
⟹ homotopic_with_canon p X Z (h ∘ f) (h ∘ g)"
by (simp add: homotopic_with_compose_continuous_map_left image_subset_iff_funcset)
lemma homotopic_from_subtopology:
"homotopic_with P X X' f g ⟹ homotopic_with P (subtopology X S) X' f g"
by (metis continuous_map_id_subt homotopic_with_compose_continuous_map_right o_id)
lemma homotopic_on_emptyI:
assumes "P f" "P g"
shows "homotopic_with P trivial_topology X f g"
by (metis assms continuous_map_on_empty empty_iff homotopic_with_equal topspace_discrete_topology)
lemma homotopic_on_empty:
"(homotopic_with P trivial_topology X f g ⟷ P f ∧ P g)"
using homotopic_on_emptyI homotopic_with_imp_property by metis
lemma homotopic_with_canon_on_empty: "homotopic_with_canon (λx. True) {} t f g"
by (auto intro: homotopic_with_equal)
lemma homotopic_constant_maps:
"homotopic_with (λx. True) X X' (λx. a) (λx. b) ⟷
X = trivial_topology ∨ path_component_of X' a b" (is "?lhs = ?rhs")
proof (cases "X = trivial_topology")
case False
then obtain c where c: "c ∈ topspace X"
by fastforce
have "∃g. continuous_map (top_of_set {0..1::real}) X' g ∧ g 0 = a ∧ g 1 = b"
if "x ∈ topspace X" and hom: "homotopic_with (λx. True) X X' (λx. a) (λx. b)" for x
proof -
obtain h :: "real × 'a ⇒ 'b"
where conth: "continuous_map (prod_topology (top_of_set {0..1}) X) X' h"
and h: "⋀x. h (0, x) = a" "⋀x. h (1, x) = b"
using hom by (auto simp: homotopic_with_def)
have cont: "continuous_map (top_of_set {0..1}) X' (h ∘ (λt. (t, c)))"
by (rule continuous_map_compose [OF _ conth] continuous_intros | simp add: c)+
then show ?thesis
by (force simp: h)
moreover have "homotopic_with (λx. True) X X' (λx. g 0) (λx. g 1)"
if "x ∈ topspace X" "a = g 0" "b = g 1" "continuous_map (top_of_set {0..1}) X' g"
for x and g :: "real ⇒ 'b"
unfolding homotopic_with_def
by (force intro!: continuous_map_compose continuous_intros c that)
ultimately show ?thesis
using False
by (metis c path_component_of_set pathin_def)
qed (simp add: homotopic_on_empty)
proposition homotopic_with_eq:
assumes h: "homotopic_with P X Y f g"
and f': "⋀x. x ∈ topspace X ⟹ f' x = f x"
and g': "⋀x. x ∈ topspace X ⟹ g' x = g x"
and P: "(⋀h k. (⋀x. x ∈ topspace X ⟹ h x = k x) ⟹ P h ⟷ P k)"
shows "homotopic_with P X Y f' g'"
by (smt (verit, ccfv_SIG) assms homotopic_with)
lemma homotopic_with_prod_topology:
assumes "homotopic_with p X1 Y1 f f'" and "homotopic_with q X2 Y2 g g'"
and r: "⋀i j. ⟦p i; q j⟧ ⟹ r(λ(x,y). (i x, j y))"
shows "homotopic_with r (prod_topology X1 X2) (prod_topology Y1 Y2)
(λz. (f(fst z),g(snd z))) (λz. (f'(fst z), g'(snd z)))"
proof -
obtain h
where h: "continuous_map (prod_topology (subtopology euclideanreal {0..1}) X1) Y1 h"
and h0: "⋀x. h (0, x) = f x"
and h1: "⋀x. h (1, x) = f' x"
and p: "⋀t. ⟦0 ≤ t; t ≤ 1⟧ ⟹ p (λx. h (t,x))"
using assms unfolding homotopic_with_def by auto
obtain k
where k: "continuous_map (prod_topology (subtopology euclideanreal {0..1}) X2) Y2 k"
and k0: "⋀x. k (0, x) = g x"
and k1: "⋀x. k (1, x) = g' x"
and q: "⋀t. ⟦0 ≤ t; t ≤ 1⟧ ⟹ q (λx. k (t,x))"
using assms unfolding homotopic_with_def by auto
let ?hk = "λ(t,x,y). (h(t,x), k(t,y))"
show ?thesis
unfolding homotopic_with_def
proof (intro conjI allI exI)
show "continuous_map (prod_topology (subtopology euclideanreal {0..1}) (prod_topology X1 X2))
(prod_topology Y1 Y2) ?hk"
unfolding continuous_map_pairwise case_prod_unfold
by (rule conjI continuous_map_pairedI continuous_intros continuous_map_id [unfolded id_def]
continuous_map_fst_of [unfolded o_def] continuous_map_snd_of [unfolded o_def]
continuous_map_compose [OF _ h, unfolded o_def]
continuous_map_compose [OF _ k, unfolded o_def])+
fix x
show "?hk (0, x) = (f (fst x), g (snd x))" "?hk (1, x) = (f' (fst x), g' (snd x))"
by (auto simp: case_prod_beta h0 k0 h1 k1)
qed (auto simp: p q r)
lemma homotopic_with_product_topology:
assumes ht: "⋀i. i ∈ I ⟹ homotopic_with (p i) (X i) (Y i) (f i) (g i)"
and pq: "⋀h. (⋀i. i ∈ I ⟹ p i (h i)) ⟹ q(λx. (λi∈I. h i (x i)))"
shows "homotopic_with q (product_topology X I) (product_topology Y I)
(λz. (λi∈I. (f i) (z i))) (λz. (λi∈I. (g i) (z i)))"
proof -
obtain h
where h: "⋀i. i ∈ I ⟹ continuous_map (prod_topology (subtopology euclideanreal {0..1}) (X i)) (Y i) (h i)"
and h0: "⋀i x. i ∈ I ⟹ h i (0, x) = f i x"
and h1: "⋀i x. i ∈ I ⟹ h i (1, x) = g i x"
and p: "⋀i t. ⟦i ∈ I; t ∈ {0..1}⟧ ⟹ p i (λx. h i (t,x))"
using ht unfolding homotopic_with_def by metis
show ?thesis
unfolding homotopic_with_def
proof (intro conjI allI exI)
let ?h = "λ(t,z). λi∈I. h i (t,z i)"
have "continuous_map (prod_topology (subtopology euclideanreal {0..1}) (product_topology X I))
(Y i) (λx. h i (fst x, snd x i))" if "i ∈ I" for i
proof -
have §: "continuous_map (prod_topology (top_of_set {0..1}) (product_topology X I)) (X i) (λx. snd x i)"
using continuous_map_componentwise continuous_map_snd that by fastforce
show ?thesis
unfolding continuous_map_pairwise case_prod_unfold
by (intro conjI that § continuous_intros continuous_map_compose [OF _ h, unfolded o_def])
then show "continuous_map (prod_topology (subtopology euclideanreal {0..1}) (product_topology X I))
(product_topology Y I) ?h"
by (auto simp: continuous_map_componentwise case_prod_beta)
show "?h (0, x) = (λi∈I. f i (x i))" "?h (1, x) = (λi∈I. g i (x i))" for x
by (auto simp: case_prod_beta h0 h1)
show "∀t∈{0..1}. q (λx. ?h (t, x))"
by (force intro: p pq)
text‹Homotopic triviality implicitly incorporates path-connectedness.›
lemma homotopic_triviality:
shows "(∀f g. continuous_on S f ∧ f ∈ S → T ∧
continuous_on S g ∧ g ∈ S → T
⟶ homotopic_with_canon (λx. True) S T f g) ⟷
(S = {} ∨ path_connected T) ∧
(∀f. continuous_on S f ∧ f ∈ S → T ⟶ (∃c. homotopic_with_canon (λx. True) S T f (λx. c)))"
(is "?lhs = ?rhs")
proof (cases "S = {} ∨ T = {}")
case True then show ?thesis
by (auto simp: homotopic_on_emptyI simp flip: image_subset_iff_funcset)
case False show ?thesis
assume LHS [rule_format]: ?lhs
have pab: "path_component T a b" if "a ∈ T" "b ∈ T" for a b
proof -
have "homotopic_with_canon (λx. True) S T (λx. a) (λx. b)"
by (simp add: LHS image_subset_iff that)
then show ?thesis
using False homotopic_constant_maps [of "top_of_set S" "top_of_set T" a b]
by (metis path_component_of_canon_iff topspace_discrete_topology topspace_euclidean_subtopology)
have "∃c. homotopic_with_canon (λx. True) S T f (λx. c)" if "continuous_on S f" "f ∈ S → T" for f
using False LHS continuous_on_const that by blast
ultimately show ?rhs
by (simp add: path_connected_component)
assume RHS: ?rhs
with False have T: "path_connected T"
by blast
show ?lhs
proof clarify
fix f g
assume "continuous_on S f" "f ∈ S → T" "continuous_on S g" "g ∈ S → T"
obtain c d where c: "homotopic_with_canon (λx. True) S T f (λx. c)" and d: "homotopic_with_canon (λx. True) S T g (λx. d)"
using RHS ‹continuous_on S f› ‹continuous_on S g› ‹f ∈ S → T› ‹g ∈ S → T› by presburger
with T have "path_component T c d"
by (metis False ex_in_conv homotopic_with_imp_subset2 image_subset_iff path_connected_component)
then have "homotopic_with_canon (λx. True) S T (λx. c) (λx. d)"
by (simp add: homotopic_constant_maps)
with c d show "homotopic_with_canon (λx. True) S T f g"
by (meson homotopic_with_symD homotopic_with_trans)
subsection‹Homotopy of paths, maintaining the same endpoints›
definition✐‹tag important› homotopic_paths :: "['a set, real ⇒ 'a, real ⇒ 'a::topological_space] ⇒ bool"
"homotopic_paths S p q ≡
homotopic_with_canon (λr. pathstart r = pathstart p ∧ pathfinish r = pathfinish p) {0..1} S p q"
lemma homotopic_paths:
"homotopic_paths S p q ⟷
(∃h. continuous_on ({0..1} × {0..1}) h ∧
h ∈ ({0..1} × {0..1}) → S ∧
(∀x ∈ {0..1}. h(0,x) = p x) ∧
(∀x ∈ {0..1}. h(1,x) = q x) ∧
(∀t ∈ {0..1::real}. pathstart(h ∘ Pair t) = pathstart p ∧
pathfinish(h ∘ Pair t) = pathfinish p))"
by (auto simp: homotopic_paths_def homotopic_with pathstart_def pathfinish_def)
proposition homotopic_paths_imp_pathstart:
"homotopic_paths S p q ⟹ pathstart p = pathstart q"
by (metis (mono_tags, lifting) homotopic_paths_def homotopic_with_imp_property)
proposition homotopic_paths_imp_pathfinish:
"homotopic_paths S p q ⟹ pathfinish p = pathfinish q"
by (metis (mono_tags, lifting) homotopic_paths_def homotopic_with_imp_property)
lemma homotopic_paths_imp_path:
"homotopic_paths S p q ⟹ path p ∧ path q"
using homotopic_paths_def homotopic_with_imp_continuous_maps path_def continuous_map_subtopology_eu by blast
lemma homotopic_paths_imp_subset:
"homotopic_paths S p q ⟹ path_image p ⊆ S ∧ path_image q ⊆ S"
by (metis (mono_tags) continuous_map_subtopology_eu homotopic_paths_def homotopic_with_imp_continuous_maps path_image_def)
proposition homotopic_paths_refl [simp]: "homotopic_paths S p p ⟷ path p ∧ path_image p ⊆ S"
by (simp add: homotopic_paths_def path_def path_image_def)
proposition homotopic_paths_sym: "homotopic_paths S p q ⟹ homotopic_paths S q p"
by (metis (mono_tags) homotopic_paths_def homotopic_paths_imp_pathfinish homotopic_paths_imp_pathstart homotopic_with_symD)
proposition homotopic_paths_sym_eq: "homotopic_paths S p q ⟷ homotopic_paths S q p"
by (metis homotopic_paths_sym)
proposition homotopic_paths_trans [trans]:
assumes "homotopic_paths S p q" "homotopic_paths S q r"
shows "homotopic_paths S p r"
using assms homotopic_paths_imp_pathfinish homotopic_paths_imp_pathstart unfolding homotopic_paths_def
by (smt (verit, ccfv_SIG) homotopic_with_mono homotopic_with_trans)
proposition homotopic_paths_eq:
"⟦path p; path_image p ⊆ S; ⋀t. t ∈ {0..1} ⟹ p t = q t⟧ ⟹ homotopic_paths S p q"
by (smt (verit, best) homotopic_paths homotopic_paths_refl)
proposition homotopic_paths_reparametrize:
assumes "path p"
and pips: "path_image p ⊆ S"
and contf: "continuous_on {0..1} f"
and f01 :"f ∈ {0..1} → {0..1}"
and [simp]: "f(0) = 0" "f(1) = 1"
and q: "⋀t. t ∈ {0..1} ⟹ q(t) = p(f t)"
shows "homotopic_paths S p q"
proof -
have contp: "continuous_on {0..1} p"
by (metis ‹path p› path_def)
then have "continuous_on {0..1} (p ∘ f)"
by (meson assms(4) contf continuous_on_compose continuous_on_subset image_subset_iff_funcset)
then have "path q"
by (simp add: path_def) (metis q continuous_on_cong)
have piqs: "path_image q ⊆ S"
by (smt (verit, ccfv_threshold) Pi_iff assms(2) assms(4) assms(7) image_subset_iff path_defs(4))
have fb0: "⋀a b. ⟦0 ≤ a; a ≤ 1; 0 ≤ b; b ≤ 1⟧ ⟹ 0 ≤ (1 - a) * f b + a * b"
using f01 by force
have fb1: "⟦0 ≤ a; a ≤ 1; 0 ≤ b; b ≤ 1⟧ ⟹ (1 - a) * f b + a * b ≤ 1" for a b
by (intro convex_bound_le) (use f01 in auto)
have "homotopic_paths S q p"
proof (rule homotopic_paths_trans)
show "homotopic_paths S q (p ∘ f)"
using q by (force intro: homotopic_paths_eq [OF ‹path q› piqs])
show "homotopic_paths S (p ∘ f) p"
using pips [unfolded path_image_def]
apply (simp add: homotopic_paths_def homotopic_with_def)
apply (rule_tac x="p ∘ (λy. (1 - (fst y)) *⇩[R] ((f ∘ snd) y) + (fst y) *⇩[R] snd y)" in exI)
apply (rule conjI contf continuous_intros continuous_on_subset [OF contp] | simp)+
by (auto simp: fb0 fb1 pathstart_def pathfinish_def)
then show ?thesis
by (simp add: homotopic_paths_sym)
lemma homotopic_paths_subset: "⟦homotopic_paths S p q; S ⊆ t⟧ ⟹ homotopic_paths t p q"
unfolding homotopic_paths by fast
text‹ A slightly ad-hoc but useful lemma in constructing homotopies.›
lemma continuous_on_homotopic_join_lemma:
fixes q :: "[real,real] ⇒ 'a::topological_space"
assumes p: "continuous_on ({0..1} × {0..1}) (λy. p (fst y) (snd y))" (is "continuous_on ?A ?p")
and q: "continuous_on ({0..1} × {0..1}) (λy. q (fst y) (snd y))" (is "continuous_on ?A ?q")
and pf: "⋀t. t ∈ {0..1} ⟹ pathfinish(p t) = pathstart(q t)"
shows "continuous_on ({0..1} × {0..1}) (λy. (p(fst y) +++ q(fst y)) (snd y))"
proof -
have §: "(λt. p (fst t) (2 * snd t)) = ?p ∘ (λy. (fst y, 2 * snd y))"
"(λt. q (fst t) (2 * snd t - 1)) = ?q ∘ (λy. (fst y, 2 * snd y - 1))"
by force+
show ?thesis
unfolding joinpaths_def
proof (rule continuous_on_cases_le)
show "continuous_on {y ∈ ?A. snd y ≤ 1/2} (λt. p (fst t) (2 * snd t))"
"continuous_on {y ∈ ?A. 1/2 ≤ snd y} (λt. q (fst t) (2 * snd t - 1))"
"continuous_on ?A snd"
unfolding §
by (rule continuous_intros continuous_on_subset [OF p] continuous_on_subset [OF q] | force)+
qed (use pf in ‹auto simp: mult.commute pathstart_def pathfinish_def›)
text‹ Congruence properties of homotopy w.r.t. path-combining operations.›
lemma homotopic_paths_reversepath_D:
assumes "homotopic_paths S p q"
shows "homotopic_paths S (reversepath p) (reversepath q)"
using assms
apply (simp add: homotopic_paths_def homotopic_with_def, clarify)
apply (rule_tac x="h ∘ (λx. (fst x, 1 - snd x))" in exI)
apply (rule conjI continuous_intros)+
apply (auto simp: reversepath_def pathstart_def pathfinish_def elim!: continuous_on_subset)
proposition homotopic_paths_reversepath:
"homotopic_paths S (reversepath p) (reversepath q) ⟷ homotopic_paths S p q"
using homotopic_paths_reversepath_D by force
proposition homotopic_paths_join:
"⟦homotopic_paths S p p'; homotopic_paths S q q'; pathfinish p = pathstart q⟧ ⟹ homotopic_paths S (p +++ q) (p' +++ q')"
apply (clarsimp simp: homotopic_paths_def homotopic_with_def)
apply (rename_tac k1 k2)
apply (rule_tac x="(λy. ((k1 ∘ Pair (fst y)) +++ (k2 ∘ Pair (fst y))) (snd y))" in exI)
apply (intro conjI continuous_intros continuous_on_homotopic_join_lemma; force simp: joinpaths_def pathstart_def pathfinish_def path_image_def)
proposition homotopic_paths_continuous_image:
"⟦homotopic_paths S f g; continuous_on S h; h ∈ S → t⟧ ⟹ homotopic_paths t (h ∘ f) (h ∘ g)"
unfolding homotopic_paths_def
by (simp add: homotopic_with_compose_continuous_map_left pathfinish_compose pathstart_compose image_subset_iff_funcset)
subsection‹Group properties for homotopy of paths›
text✐‹tag important›‹So taking equivalence classes under homotopy would give the fundamental group›
proposition homotopic_paths_rid:
assumes "path p" "path_image p ⊆ S"
shows "homotopic_paths S (p +++ linepath (pathfinish p) (pathfinish p)) p"
proof -
have §: "continuous_on {0..1} (λt::real. if t ≤ 1/2 then 2 *⇩[R] t else 1)"
unfolding split_01
by (rule continuous_on_cases continuous_intros | force simp: pathfinish_def joinpaths_def)+
show ?thesis
apply (rule homotopic_paths_sym)
using assms unfolding pathfinish_def joinpaths_def
by (intro § continuous_on_cases continuous_intros homotopic_paths_reparametrize [where f = "λt. if t ≤ 1 /2 then 2 *⇩[R] t else 1"]; force)
proposition homotopic_paths_lid:
"⟦path p; path_image p ⊆ S⟧ ⟹ homotopic_paths S (linepath (pathstart p) (pathstart p) +++ p) p"
using homotopic_paths_rid [of "reversepath p" S]
by (metis homotopic_paths_reversepath path_image_reversepath path_reversepath pathfinish_linepath
pathfinish_reversepath reversepath_joinpaths reversepath_linepath)
lemma homotopic_paths_rid':
assumes "path p" "path_image p ⊆ s" "x = pathfinish p"
shows "homotopic_paths s (p +++ linepath x x) p"
using homotopic_paths_rid[of p s] assms by simp
lemma homotopic_paths_lid':
"⟦path p; path_image p ⊆ s; x = pathstart p⟧ ⟹ homotopic_paths s (linepath x x +++ p) p"
using homotopic_paths_lid[of p s] by simp
proposition homotopic_paths_assoc:
"⟦path p; path_image p ⊆ S; path q; path_image q ⊆ S; path r; path_image r ⊆ S; pathfinish p = pathstart q;
pathfinish q = pathstart r⟧
⟹ homotopic_paths S (p +++ (q +++ r)) ((p +++ q) +++ r)"
apply (subst homotopic_paths_sym)
apply (rule homotopic_paths_reparametrize
[where f = "λt. if t ≤ 1 /2 then inverse 2 *⇩[R] t
else if t ≤ 3 / 4 then t - ( 1 / 4)
else 2 *⇩[R] t - 1"])
apply (simp_all del: le_divide_eq_numeral1 add: subset_path_image_join)
apply (rule continuous_on_cases_1 continuous_intros | auto simp: joinpaths_def)+
proposition homotopic_paths_rinv:
assumes "path p" "path_image p ⊆ S"
shows "homotopic_paths S (p +++ reversepath p) (linepath (pathstart p) (pathstart p))"
proof -
have p: "continuous_on {0..1} p"
using assms by (auto simp: path_def)
let ?A = "{0..1} × {0..1}"
have "continuous_on ?A (λx. (subpath 0 (fst x) p +++ reversepath (subpath 0 (fst x) p)) (snd x))"
unfolding joinpaths_def subpath_def reversepath_def path_def add_0_right diff_0_right
proof (rule continuous_on_cases_le)
show "continuous_on {x ∈ ?A. snd x ≤ 1/2} (λt. p (fst t * (2 * snd t)))"
"continuous_on {x ∈ ?A. 1/2 ≤ snd x} (λt. p (fst t * (1 - (2 * snd t - 1))))"
"continuous_on ?A snd"
by (intro continuous_on_compose2 [OF p] continuous_intros; auto simp: mult_le_one)+
qed (auto simp: algebra_simps)
then show ?thesis
using assms
apply (subst homotopic_paths_sym_eq)
unfolding homotopic_paths_def homotopic_with_def
apply (rule_tac x="(λy. (subpath 0 (fst y) p +++ reversepath(subpath 0 (fst y) p)) (snd y))" in exI)
apply (force simp: mult_le_one path_defs joinpaths_def subpath_def reversepath_def)
proposition homotopic_paths_linv:
assumes "path p" "path_image p ⊆ S"
shows "homotopic_paths S (reversepath p +++ p) (linepath (pathfinish p) (pathfinish p))"
using homotopic_paths_rinv [of "reversepath p" S] assms by simp
subsection‹Homotopy of loops without requiring preservation of endpoints›
definition✐‹tag important› homotopic_loops :: "'a::topological_space set ⇒ (real ⇒ 'a) ⇒ (real ⇒ 'a) ⇒ bool" where
"homotopic_loops S p q ≡
homotopic_with_canon (λr. pathfinish r = pathstart r) {0..1} S p q"
lemma homotopic_loops:
"homotopic_loops S p q ⟷
(∃h. continuous_on ({0..1::real} × {0..1}) h ∧
image h ({0..1} × {0..1}) ⊆ S ∧
(∀x ∈ {0..1}. h(0,x) = p x) ∧
(∀x ∈ {0..1}. h(1,x) = q x) ∧
(∀t ∈ {0..1}. pathfinish(h ∘ Pair t) = pathstart(h ∘ Pair t)))"
by (simp add: homotopic_loops_def pathstart_def pathfinish_def homotopic_with)
proposition homotopic_loops_imp_loop:
"homotopic_loops S p q ⟹ pathfinish p = pathstart p ∧ pathfinish q = pathstart q"
using homotopic_with_imp_property homotopic_loops_def by blast
proposition homotopic_loops_imp_path:
"homotopic_loops S p q ⟹ path p ∧ path q"
unfolding homotopic_loops_def path_def
using homotopic_with_imp_continuous_maps continuous_map_subtopology_eu by blast
proposition homotopic_loops_imp_subset:
"homotopic_loops S p q ⟹ path_image p ⊆ S ∧ path_image q ⊆ S"
unfolding homotopic_loops_def path_image_def
by (meson continuous_map_subtopology_eu homotopic_with_imp_continuous_maps)
proposition homotopic_loops_refl:
"homotopic_loops S p p ⟷
path p ∧ path_image p ⊆ S ∧ pathfinish p = pathstart p"
by (simp add: homotopic_loops_def path_image_def path_def)
proposition homotopic_loops_sym: "homotopic_loops S p q ⟹ homotopic_loops S q p"
by (simp add: homotopic_loops_def homotopic_with_sym)
proposition homotopic_loops_sym_eq: "homotopic_loops S p q ⟷ homotopic_loops S q p"
by (metis homotopic_loops_sym)
proposition homotopic_loops_trans:
"⟦homotopic_loops S p q; homotopic_loops S q r⟧ ⟹ homotopic_loops S p r"
unfolding homotopic_loops_def by (blast intro: homotopic_with_trans)
proposition homotopic_loops_subset:
"⟦homotopic_loops S p q; S ⊆ t⟧ ⟹ homotopic_loops t p q"
by (fastforce simp: homotopic_loops)
proposition homotopic_loops_eq:
"⟦path p; path_image p ⊆ S; pathfinish p = pathstart p; ⋀t. t ∈ {0..1} ⟹ p(t) = q(t)⟧
⟹ homotopic_loops S p q"
unfolding homotopic_loops_def path_image_def path_def pathstart_def pathfinish_def
by (auto intro: homotopic_with_eq [OF homotopic_with_refl [where f = p, THEN iffD2]])
proposition homotopic_loops_continuous_image:
"⟦homotopic_loops S f g; continuous_on S h; h ∈ S → t⟧ ⟹ homotopic_loops t (h ∘ f) (h ∘ g)"
unfolding homotopic_loops_def
by (simp add: homotopic_with_compose_continuous_map_left pathfinish_def pathstart_def image_subset_iff_funcset)
subsection‹Relations between the two variants of homotopy›
proposition homotopic_paths_imp_homotopic_loops:
"⟦homotopic_paths S p q; pathfinish p = pathstart p; pathfinish q = pathstart p⟧ ⟹ homotopic_loops S p q"
by (auto simp: homotopic_with_def homotopic_paths_def homotopic_loops_def)
proposition homotopic_loops_imp_homotopic_paths_null:
assumes "homotopic_loops S p (linepath a a)"
shows "homotopic_paths S p (linepath (pathstart p) (pathstart p))"
proof -
have "path p" by (metis assms homotopic_loops_imp_path)
have ploop: "pathfinish p = pathstart p" by (metis assms homotopic_loops_imp_loop)
have pip: "path_image p ⊆ S" by (metis assms homotopic_loops_imp_subset)
let ?A = "{0..1::real} × {0..1::real}"
obtain h where conth: "continuous_on ?A h"
and hs: "h ∈ ?A → S"
and h0[simp]: "⋀x. x ∈ {0..1} ⟹ h(0,x) = p x"
and h1[simp]: "⋀x. x ∈ {0..1} ⟹ h(1,x) = a"
and ends: "⋀t. t ∈ {0..1} ⟹ pathfinish (h ∘ Pair t) = pathstart (h ∘ Pair t)"
using assms by (auto simp: homotopic_loops homotopic_with image_subset_iff_funcset)
have conth0: "path (λu. h (u, 0))"
unfolding path_def
proof (rule continuous_on_compose [of _ _ h, unfolded o_def])
show "continuous_on ((λx. (x, 0)) ` {0..1}) h"
by (force intro: continuous_on_subset [OF conth])
qed (force intro: continuous_intros)
have pih0: "path_image (λu. h (u, 0)) ⊆ S"
using hs by (force simp: path_image_def)
have c1: "continuous_on ?A (λx. h (fst x * snd x, 0))"
proof (rule continuous_on_compose [of _ _ h, unfolded o_def])
show "continuous_on ((λx. (fst x * snd x, 0)) ` ?A) h"
by (force simp: mult_le_one intro: continuous_on_subset [OF conth])
qed (force intro: continuous_intros)+
have c2: "continuous_on ?A (λx. h (fst x - fst x * snd x, 0))"
proof (rule continuous_on_compose [of _ _ h, unfolded o_def])
show "continuous_on ((λx. (fst x - fst x * snd x, 0)) ` ?A) h"
by (auto simp: algebra_simps add_increasing2 mult_left_le intro: continuous_on_subset [OF conth])
qed (force intro: continuous_intros)
have [simp]: "⋀t. ⟦0 ≤ t ∧ t ≤ 1⟧ ⟹ h (t, 1) = h (t, 0)"
using ends by (simp add: pathfinish_def pathstart_def)
have adhoc_le: "c * 4 ≤ 1 + c * (d * 4)" if "¬ d * 4 ≤ 3" "0 ≤ c" "c ≤ 1" for c d::real
proof -
have "c * 3 ≤ c * (d * 4)" using that less_eq_real_def by auto
with ‹c ≤ 1› show ?thesis by fastforce
have *: "⋀p x. ⟦path p ∧ path(reversepath p);
path_image p ⊆ S ∧ path_image(reversepath p) ⊆ S;
pathfinish p = pathstart(linepath a a +++ reversepath p) ∧
pathstart(reversepath p) = a ∧ pathstart p = x⟧
⟹ homotopic_paths S (p +++ linepath a a +++ reversepath p) (linepath x x)"
by (metis homotopic_paths_lid homotopic_paths_join
homotopic_paths_trans homotopic_paths_sym homotopic_paths_rinv)
have 1: "homotopic_paths S p (p +++ linepath (pathfinish p) (pathfinish p))"
using ‹path p› homotopic_paths_rid homotopic_paths_sym pip by blast
moreover have "homotopic_paths S (p +++ linepath (pathfinish p) (pathfinish p))
(linepath (pathstart p) (pathstart p) +++ p +++ linepath (pathfinish p) (pathfinish p))"
using homotopic_paths_lid [of "p +++ linepath (pathfinish p) (pathfinish p)" S]
by (metis 1 homotopic_paths_imp_path homotopic_paths_imp_subset homotopic_paths_sym pathstart_join)
have "homotopic_paths S (linepath (pathstart p) (pathstart p) +++ p +++ linepath (pathfinish p) (pathfinish p))
((λu. h (u, 0)) +++ linepath a a +++ reversepath (λu. h (u, 0)))"
unfolding homotopic_paths_def homotopic_with_def
proof (intro exI strip conjI)
let ?h = "λy. (subpath 0 (fst y) (λu. h (u, 0)) +++ (λu. h (Pair (fst y) u))
+++ subpath (fst y) 0 (λu. h (u, 0))) (snd y)"
have "continuous_on ?A ?h"
by (intro continuous_on_homotopic_join_lemma; simp add: path_defs joinpaths_def subpath_def conth c1 c2)
moreover have "?h ∈ ?A → S"
using hs
unfolding joinpaths_def subpath_def
by (force simp: algebra_simps mult_le_one mult_left_le intro: adhoc_le)
ultimately show "continuous_map (prod_topology (top_of_set {0..1}) (top_of_set {0..1}))
(top_of_set S) ?h"
by (simp add: subpath_reversepath image_subset_iff_funcset)
qed (use ploop in ‹simp_all add: reversepath_def path_defs joinpaths_def o_def subpath_def conth c1 c2›)
moreover have "homotopic_paths S ((λu. h (u, 0)) +++ linepath a a +++ reversepath (λu. h (u, 0)))
(linepath (pathstart p) (pathstart p))"
by (rule *; simp add: pih0 pathstart_def pathfinish_def conth0; simp add: reversepath_def joinpaths_def)
ultimately show ?thesis
by (blast intro: homotopic_paths_trans)
proposition homotopic_loops_conjugate:
fixes S :: "'a::real_normed_vector set"
assumes "path p" "path q" and pip: "path_image p ⊆ S" and piq: "path_image q ⊆ S"
and pq: "pathfinish p = pathstart q" and qloop: "pathfinish q = pathstart q"
shows "homotopic_loops S (p +++ q +++ reversepath p) q"
proof -
have contp: "continuous_on {0..1} p" using ‹path p› [unfolded path_def] by blast
have contq: "continuous_on {0..1} q" using ‹path q› [unfolded path_def] by blast
let ?A = "{0..1::real} × {0..1::real}"
have c1: "continuous_on ?A (λx. p ((1 - fst x) * snd x + fst x))"
proof (rule continuous_on_compose [of _ _ p, unfolded o_def])
show "continuous_on ((λx. (1 - fst x) * snd x + fst x) ` ?A) p"
by (auto intro: continuous_on_subset [OF contp] simp: algebra_simps add_increasing2 mult_right_le_one_le sum_le_prod1)
qed (force intro: continuous_intros)
have c2: "continuous_on ?A (λx. p ((fst x - 1) * snd x + 1))"
proof (rule continuous_on_compose [of _ _ p, unfolded o_def])
show "continuous_on ((λx. (fst x - 1) * snd x + 1) ` ?A) p"
by (auto intro: continuous_on_subset [OF contp] simp: algebra_simps add_increasing2 mult_left_le_one_le)
qed (force intro: continuous_intros)
have ps1: "⋀a b. ⟦b * 2 ≤ 1; 0 ≤ b; 0 ≤ a; a ≤ 1⟧ ⟹ p ((1 - a) * (2 * b) + a) ∈ S"
using sum_le_prod1
by (force simp: algebra_simps add_increasing2 mult_left_le intro: pip [unfolded path_image_def, THEN subsetD])
have ps2: "⋀a b. ⟦¬ 4 * b ≤ 3; b ≤ 1; 0 ≤ a; a ≤ 1⟧ ⟹ p ((a - 1) * (4 * b - 3) + 1) ∈ S"
apply (rule pip [unfolded path_image_def, THEN subsetD])
apply (rule image_eqI, blast)
apply (simp add: algebra_simps)
by (metis add_mono affine_ineq linear mult.commute mult.left_neutral mult_right_mono
add.commute zero_le_numeral)
have qs: "⋀a b. ⟦4 * b ≤ 3; ¬ b * 2 ≤ 1⟧ ⟹ q (4 * b - 2) ∈ S"
using path_image_def piq by fastforce
have "homotopic_loops S (p +++ q +++ reversepath p)
(linepath (pathstart q) (pathstart q) +++ q +++ linepath (pathstart q) (pathstart q))"
unfolding homotopic_loops_def homotopic_with_def
proof (intro exI strip conjI)
let ?h = "(λy. (subpath (fst y) 1 p +++ q +++ subpath 1 (fst y) p) (snd y))"
have "continuous_on ?A (λy. q (snd y))"
by (force simp: contq intro: continuous_on_compose [of _ _ q, unfolded o_def] continuous_on_id continuous_on_snd)
then have "continuous_on ?A ?h"
using pq qloop
by (intro continuous_on_homotopic_join_lemma) (auto simp: path_defs joinpaths_def subpath_def c1 c2)
then show "continuous_map (prod_topology (top_of_set {0..1}) (top_of_set {0..1})) (top_of_set S) ?h"
by (auto simp: joinpaths_def subpath_def ps1 ps2 qs)
show "?h (1,x) = (linepath (pathstart q) (pathstart q) +++ q +++ linepath (pathstart q) (pathstart q)) x" for x
using pq by (simp add: pathfinish_def subpath_refl)
qed (auto simp: subpath_reversepath)
moreover have "homotopic_loops S (linepath (pathstart q) (pathstart q) +++ q +++ linepath (pathstart q) (pathstart q)) q"
proof -
have "homotopic_paths S (linepath (pathfinish q) (pathfinish q) +++ q) q"
using ‹path q› homotopic_paths_lid qloop piq by auto
hence 1: "⋀f. homotopic_paths S f q ∨ ¬ homotopic_paths S f (linepath (pathfinish q) (pathfinish q) +++ q)"
using homotopic_paths_trans by blast
hence "homotopic_paths S (linepath (pathfinish q) (pathfinish q) +++ q +++ linepath (pathfinish q) (pathfinish q)) q"
by (smt (verit, best) ‹path q› homotopic_paths_imp_path homotopic_paths_imp_subset homotopic_paths_lid
homotopic_paths_rid homotopic_paths_trans pathstart_join piq qloop)
thus ?thesis
by (metis (no_types) qloop homotopic_loops_sym homotopic_paths_imp_homotopic_loops homotopic_paths_imp_pathfinish homotopic_paths_sym)
ultimately show ?thesis
by (blast intro: homotopic_loops_trans)
lemma homotopic_paths_loop_parts:
assumes loops: "homotopic_loops S (p +++ reversepath q) (linepath a a)" and "path q"
shows "homotopic_paths S p q"
proof -
have paths: "homotopic_paths S (p +++ reversepath q) (linepath (pathstart p) (pathstart p))"
using homotopic_loops_imp_homotopic_paths_null [OF loops] by simp
then have "path p"
using ‹path q› homotopic_loops_imp_path loops path_join path_join_path_ends path_reversepath by blast
show ?thesis
proof (cases "pathfinish p = pathfinish q")
case True
obtain pipq: "path_image p ⊆ S" "path_image q ⊆ S"
by (metis Un_subset_iff paths ‹path p› ‹path q› homotopic_loops_imp_subset homotopic_paths_imp_path loops
path_image_join path_image_reversepath path_imp_reversepath path_join_eq)
have "homotopic_paths S p (p +++ (linepath (pathfinish p) (pathfinish p)))"
using ‹path p› ‹path_image p ⊆ S› homotopic_paths_rid homotopic_paths_sym by blast
moreover have "homotopic_paths S (p +++ (linepath (pathfinish p) (pathfinish p))) (p +++ (reversepath q +++ q))"
by (simp add: True ‹path p› ‹path q› pipq homotopic_paths_join homotopic_paths_linv homotopic_paths_sym)
moreover have "homotopic_paths S (p +++ (reversepath q +++ q)) ((p +++ reversepath q) +++ q)"
by (simp add: True ‹path p› ‹path q› homotopic_paths_assoc pipq)
moreover have "homotopic_paths S ((p +++ reversepath q) +++ q) (linepath (pathstart p) (pathstart p) +++ q)"
by (simp add: ‹path q› homotopic_paths_join paths pipq)
ultimately show ?thesis
by (metis ‹path q› homotopic_paths_imp_path homotopic_paths_lid homotopic_paths_trans path_join_path_ends pathfinish_linepath pipq(2))
case False
then show ?thesis
using ‹path q› homotopic_loops_imp_path loops path_join_path_ends by fastforce
subsection✐‹tag unimportant›‹Homotopy of "nearby" function, paths and loops›
lemma homotopic_with_linear:
fixes f g :: "_ ⇒ 'b::real_normed_vector"
assumes contf: "continuous_on S f"
and contg:"continuous_on S g"
and sub: "⋀x. x ∈ S ⟹ closed_segment (f x) (g x) ⊆ t"
shows "homotopic_with_canon (λz. True) S t f g"
unfolding homotopic_with_def
apply (rule_tac x="λy. ((1 - (fst y)) *⇩[R] f(snd y) + (fst y) *⇩[R] g(snd y))" in exI)
using sub closed_segment_def
by (fastforce intro: continuous_intros continuous_on_subset [OF contf] continuous_on_compose2 [where g=f]
continuous_on_subset [OF contg] continuous_on_compose2 [where g=g])
lemma homotopic_paths_linear:
fixes g h :: "real ⇒ 'a::real_normed_vector"
assumes "path g" "path h" "pathstart h = pathstart g" "pathfinish h = pathfinish g"
"⋀t. t ∈ {0..1} ⟹ closed_segment (g t) (h t) ⊆ S"
shows "homotopic_paths S g h"
using assms
unfolding path_def
apply (simp add: closed_segment_def pathstart_def pathfinish_def homotopic_paths_def homotopic_with_def)
apply (rule_tac x="λy. ((1 - (fst y)) *⇩[R] (g ∘ snd) y + (fst y) *⇩[R] (h ∘ snd) y)" in exI)
apply (intro conjI subsetI continuous_intros; force)
lemma homotopic_loops_linear:
fixes g h :: "real ⇒ 'a::real_normed_vector"
assumes "path g" "path h" "pathfinish g = pathstart g" "pathfinish h = pathstart h"
"⋀t x. t ∈ {0..1} ⟹ closed_segment (g t) (h t) ⊆ S"
shows "homotopic_loops S g h"
using assms
unfolding path_defs homotopic_loops_def homotopic_with_def
apply (rule_tac x="λy. ((1 - (fst y)) *⇩[R] g(snd y) + (fst y) *⇩[R] h(snd y))" in exI)
by (force simp: closed_segment_def intro!: continuous_intros intro: continuous_on_compose2 [where g=g] continuous_on_compose2 [where g=h])
lemma homotopic_paths_nearby_explicit:
assumes §: "path g" "path h" "pathstart h = pathstart g" "pathfinish h = pathfinish g"
and no: "⋀t x. ⟦t ∈ {0..1}; x ∉ S⟧ ⟹ norm(h t - g t) < norm(g t - x)"
shows "homotopic_paths S g h"
using homotopic_paths_linear [OF §] by (metis linorder_not_le no norm_minus_commute segment_bound1 subsetI)
lemma homotopic_loops_nearby_explicit:
assumes §: "path g" "path h" "pathfinish g = pathstart g" "pathfinish h = pathstart h"
and no: "⋀t x. ⟦t ∈ {0..1}; x ∉ S⟧ ⟹ norm(h t - g t) < norm(g t - x)"
shows "homotopic_loops S g h"
using homotopic_loops_linear [OF §] by (metis linorder_not_le no norm_minus_commute segment_bound1 subsetI)
lemma homotopic_nearby_paths:
fixes g h :: "real ⇒ 'a::euclidean_space"
assumes "path g" "open S" "path_image g ⊆ S"
shows "∃e. 0 < e ∧
(∀h. path h ∧
pathstart h = pathstart g ∧ pathfinish h = pathfinish g ∧
(∀t ∈ {0..1}. norm(h t - g t) < e) ⟶ homotopic_paths S g h)"
proof -
obtain e where "e > 0" and e: "⋀x y. x ∈ path_image g ⟹ y ∈ - S ⟹ e ≤ dist x y"
using separate_compact_closed [of "path_image g" "-S"] assms by force
show ?thesis
using e [unfolded dist_norm] ‹e > 0›
by (fastforce simp: path_image_def intro!: homotopic_paths_nearby_explicit assms exI)
lemma homotopic_nearby_loops:
fixes g h :: "real ⇒ 'a::euclidean_space"
assumes "path g" "open S" "path_image g ⊆ S" "pathfinish g = pathstart g"
shows "∃e. 0 < e ∧
(∀h. path h ∧ pathfinish h = pathstart h ∧
(∀t ∈ {0..1}. norm(h t - g t) < e) ⟶ homotopic_loops S g h)"
proof -
obtain e where "e > 0" and e: "⋀x y. x ∈ path_image g ⟹ y ∈ - S ⟹ e ≤ dist x y"
using separate_compact_closed [of "path_image g" "-S"] assms by force
show ?thesis
using e [unfolded dist_norm] ‹e > 0›
by (fastforce simp: path_image_def intro!: homotopic_loops_nearby_explicit assms exI)
subsection‹ Homotopy and subpaths›
lemma homotopic_join_subpaths1:
assumes "path g" and pag: "path_image g ⊆ S"
and u: "u ∈ {0..1}" and v: "v ∈ {0..1}" and w: "w ∈ {0..1}" "u ≤ v" "v ≤ w"
shows "homotopic_paths S (subpath u v g +++ subpath v w g) (subpath u w g)"
proof -
have 1: "t * 2 ≤ 1 ⟹ u + t * (v * 2) ≤ v + t * (u * 2)" for t
using affine_ineq ‹u ≤ v› by fastforce
have 2: "t * 2 > 1 ⟹ u + (2*t - 1) * v ≤ v + (2*t - 1) * w" for t
by (metis add_mono_thms_linordered_semiring(1) diff_gt_0_iff_gt less_eq_real_def mult.commute mult_right_mono ‹u ≤ v› ‹v ≤ w›)
have t2: "⋀t::real. t*2 = 1 ⟹ t = 1/2" by auto
have "homotopic_paths (path_image g) (subpath u v g +++ subpath v w g) (subpath u w g)"
proof (cases "w = u")
case True
then show ?thesis
by (metis ‹path g› homotopic_paths_rinv path_image_subpath_subset path_subpath pathstart_subpath reversepath_subpath subpath_refl u v)
case False
let ?f = "λt. if t ≤ 1/2 then inverse((w - u)) *⇩[R] (2 * (v - u)) *⇩[R] t
else inverse((w - u)) *⇩[R] ((v - u) + (w - v) *⇩[R] (2 *⇩[R] t - 1))"
show ?thesis
proof (rule homotopic_paths_sym [OF homotopic_paths_reparametrize [where f = ?f]])
show "path (subpath u w g)"
using assms(1) path_subpath u w(1) by blast
show "path_image (subpath u w g) ⊆ path_image g"
by (meson path_image_subpath_subset u w(1))
show "continuous_on {0..1} ?f"
unfolding split_01
by (rule continuous_on_cases continuous_intros | force simp: pathfinish_def joinpaths_def dest!: t2)+
show "?f ∈ {0..1} → {0..1}"
using False assms
by (force simp: field_simps not_le mult_left_mono affine_ineq dest!: 1 2)
show "(subpath u v g +++ subpath v w g) t = subpath u w g (?f t)" if "t ∈ {0..1}" for t
using assms
unfolding joinpaths_def subpath_def by (auto simp: divide_simps add.commute mult.commute mult.left_commute)
qed (use False in auto)
then show ?thesis
by (rule homotopic_paths_subset [OF _ pag])
lemma homotopic_join_subpaths2:
assumes "homotopic_paths S (subpath u v g +++ subpath v w g) (subpath u w g)"
shows "homotopic_paths S (subpath w v g +++ subpath v u g) (subpath w u g)"
by (metis assms homotopic_paths_reversepath_D pathfinish_subpath pathstart_subpath reversepath_joinpaths reversepath_subpath)
lemma homotopic_join_subpaths3:
assumes hom: "homotopic_paths S (subpath u v g +++ subpath v w g) (subpath u w g)"
and "path g" and pag: "path_image g ⊆ S"
and u: "u ∈ {0..1}" and v: "v ∈ {0..1}" and w: "w ∈ {0..1}"
shows "homotopic_paths S (subpath v w g +++ subpath w u g) (subpath v u g)"
proof -
obtain wvg: "path (subpath w v g)" "path_image (subpath w v g) ⊆ S"
and wug: "path (subpath w u g)" "path_image (subpath w u g) ⊆ S"
and vug: "path (subpath v u g)" "path_image (subpath v u g) ⊆ S"
by (meson ‹path g› pag path_image_subpath_subset path_subpath subset_trans u v w)
have "homotopic_paths S (subpath u w g +++ subpath w v g)
((subpath u v g +++ subpath v w g) +++ subpath w v g)"
by (simp add: hom homotopic_paths_join homotopic_paths_sym wvg)
also have "homotopic_paths S … (subpath u v g +++ subpath v w g +++ subpath w v g)"
using wvg vug ‹path g›
by (metis homotopic_paths_assoc homotopic_paths_sym path_image_subpath_commute path_subpath
pathfinish_subpath pathstart_subpath u v w)
also have "homotopic_paths S … (subpath u v g +++ linepath (pathfinish (subpath u v g)) (pathfinish (subpath u v g)))"
using wvg vug ‹path g›
by (metis homotopic_paths_join homotopic_paths_linv homotopic_paths_refl path_image_subpath_commute
path_subpath pathfinish_subpath pathstart_join pathstart_subpath reversepath_subpath u v)
also have "homotopic_paths S … (subpath u v g)"
using vug ‹path g› by (metis homotopic_paths_rid path_image_subpath_commute path_subpath u v)
finally have "homotopic_paths S (subpath u w g +++ subpath w v g) (subpath u v g)" .
then show ?thesis
using homotopic_join_subpaths2 by blast
proposition homotopic_join_subpaths:
"⟦path g; path_image g ⊆ S; u ∈ {0..1}; v ∈ {0..1}; w ∈ {0..1}⟧
⟹ homotopic_paths S (subpath u v g +++ subpath v w g) (subpath u w g)"
by (smt (verit, del_insts) homotopic_join_subpaths1 homotopic_join_subpaths2 homotopic_join_subpaths3)
text‹Relating homotopy of trivial loops to path-connectedness.›
lemma path_component_imp_homotopic_points:
assumes "path_component S a b"
shows "homotopic_loops S (linepath a a) (linepath b b)"
proof -
obtain g :: "real ⇒ 'a" where g: "continuous_on {0..1} g" "g ∈ {0..1} → S" "g 0 = a" "g 1 = b"
using assms by (auto simp: path_defs)
then have "continuous_on ({0..1} × {0..1}) (g ∘ fst)"
by (fastforce intro!: continuous_intros)+
with g show ?thesis
by (auto simp: homotopic_loops_def homotopic_with_def path_defs Pi_iff)
lemma homotopic_loops_imp_path_component_value:
"⟦homotopic_loops S p q; 0 ≤ t; t ≤ 1⟧ ⟹ path_component S (p t) (q t)"
apply (clarsimp simp: homotopic_loops_def homotopic_with_def path_defs)
apply (rule_tac x="h ∘ (λu. (u, t))" in exI)
apply (fastforce elim!: continuous_on_subset intro!: continuous_intros)
lemma homotopic_points_eq_path_component:
"homotopic_loops S (linepath a a) (linepath b b) ⟷ path_component S a b"
using homotopic_loops_imp_path_component_value path_component_imp_homotopic_points by fastforce
lemma path_connected_eq_homotopic_points:
"path_connected S ⟷
(∀a b. a ∈ S ∧ b ∈ S ⟶ homotopic_loops S (linepath a a) (linepath b b))"
by (auto simp: path_connected_def path_component_def homotopic_points_eq_path_component)
subsection‹Simply connected sets›
text✐‹tag important›‹defined as "all loops are homotopic (as loops)›
definition✐‹tag important› simply_connected where
"simply_connected S ≡
∀p q. path p ∧ pathfinish p = pathstart p ∧ path_image p ⊆ S ∧
path q ∧ pathfinish q = pathstart q ∧ path_image q ⊆ S
⟶ homotopic_loops S p q"
lemma simply_connected_empty [iff]: "simply_connected {}"
by (simp add: simply_connected_def)
lemma simply_connected_imp_path_connected:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟹ path_connected S"
by (simp add: simply_connected_def path_connected_eq_homotopic_points)
lemma simply_connected_imp_connected:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟹ connected S"
by (simp add: path_connected_imp_connected simply_connected_imp_path_connected)
lemma simply_connected_eq_contractible_loop_any:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟷
(∀p a. path p ∧ path_image p ⊆ S ∧ pathfinish p = pathstart p ∧ a ∈ S
⟶ homotopic_loops S p (linepath a a))"
(is "?lhs = ?rhs")
assume ?rhs then show ?lhs
unfolding simply_connected_def
by (metis pathfinish_in_path_image subsetD homotopic_loops_trans homotopic_loops_sym)
qed (force simp: simply_connected_def)
lemma simply_connected_eq_contractible_loop_some:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟷
path_connected S ∧
(∀p. path p ∧ path_image p ⊆ S ∧ pathfinish p = pathstart p
⟶ (∃a. a ∈ S ∧ homotopic_loops S p (linepath a a)))"
(is "?lhs = ?rhs")
assume ?lhs
then show ?rhs
using simply_connected_eq_contractible_loop_any by (blast intro: simply_connected_imp_path_connected)
assume ?rhs
then show ?lhs
by (meson homotopic_loops_trans path_connected_eq_homotopic_points simply_connected_eq_contractible_loop_any)
lemma simply_connected_eq_contractible_loop_all:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟷
S = {} ∨
(∃a ∈ S. ∀p. path p ∧ path_image p ⊆ S ∧ pathfinish p = pathstart p
⟶ homotopic_loops S p (linepath a a))"
by (meson ex_in_conv homotopic_loops_sym homotopic_loops_trans simply_connected_def simply_connected_eq_contractible_loop_any)
lemma simply_connected_eq_contractible_path:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟷
path_connected S ∧
(∀p. path p ∧ path_image p ⊆ S ∧ pathfinish p = pathstart p
⟶ homotopic_paths S p (linepath (pathstart p) (pathstart p)))"
(is "?lhs = ?rhs")
assume ?lhs
then show ?rhs
unfolding simply_connected_imp_path_connected
by (metis simply_connected_eq_contractible_loop_some homotopic_loops_imp_homotopic_paths_null)
assume ?rhs
then show ?lhs
using homotopic_paths_imp_homotopic_loops simply_connected_eq_contractible_loop_some by fastforce
lemma simply_connected_eq_homotopic_paths:
fixes S :: "_::real_normed_vector set"
shows "simply_connected S ⟷
path_connected S ∧
(∀p q. path p ∧ path_image p ⊆ S ∧
path q ∧ path_image q ⊆ S ∧
pathstart q = pathstart p ∧ pathfinish q = pathfinish p
⟶ homotopic_paths S p q)"
(is "?lhs = ?rhs")
assume ?lhs
then have pc: "path_connected S"
and *: "⋀p. ⟦path p; path_image p ⊆ S;
pathfinish p = pathstart p⟧
⟹ homotopic_paths S p (linepath (pathstart p) (pathstart p))"
by (auto simp: simply_connected_eq_contractible_path)
have "homotopic_paths S p q"
if "path p" "path_image p ⊆ S" "path q"
"path_image q ⊆ S" "pathstart q = pathstart p"
"pathfinish q = pathfinish p" for p q
proof -
have "homotopic_paths S p (p +++ reversepath q +++ q)"
using that
by (smt (verit, best) homotopic_paths_join homotopic_paths_linv homotopic_paths_rid homotopic_paths_sym
homotopic_paths_trans pathstart_linepath)
also have "homotopic_paths S … ((p +++ reversepath q) +++ q)"
by (simp add: that homotopic_paths_assoc)
also have "homotopic_paths S … (linepath (pathstart q) (pathstart q) +++ q)"
using * [of "p +++ reversepath q"] that
by (simp add: homotopic_paths_assoc homotopic_paths_join path_image_join)
also have "homotopic_paths S … q"
using that homotopic_paths_lid by blast
finally show ?thesis .
then show ?rhs
by (blast intro: pc *)
assume ?rhs
then show ?lhs
by (force simp: simply_connected_eq_contractible_path)
proposition simply_connected_Times:
fixes S :: "'a::real_normed_vector set" and T :: "'b::real_normed_vector set"
assumes S: "simply_connected S" and T: "simply_connected T"
shows "simply_connected(S × T)"
proof -
have "homotopic_loops (S × T) p (linepath (a, b) (a, b))"
if "path p" "path_image p ⊆ S × T" "p 1 = p 0" "a ∈ S" "b ∈ T"
for p a b
proof -
have "path (fst ∘ p)"
by (simp add: continuous_on_fst | {"url":"https://www.isa-afp.org/browser_info/current/HOL/HOL-Analysis/Homotopy.html","timestamp":"2024-11-04T07:22:13Z","content_type":"application/xhtml+xml","content_length":"1049321","record_id":"<urn:uuid:d3bd075b-98d1-4932-8436-19df496d2720>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00376.warc.gz"} |
Karl Petersen (UNC-CH), Ergodic Theory and Dynamical Systems Seminar - Department of Mathematics
Karl Petersen (UNC-CH), Ergodic Theory and Dynamical Systems Seminar
November 8, 2016 @ 3:30 pm - 4:30 pm
Title: A well-known but still fascinating example in ergodic theory: The Gauss Map
Abstract: Defining Tx to be the fractional part of 1/x for x in the unit interval produces a map that is isomorphic to the shift map on continued fraction expansions. We review some interesting
(long known) properties of this map. We will see that this map preserves a measure equivalent to Lebesgue measure, called the Gauss measure. The map is ergodic with respect to this measure, so using
the Ergodic Theorem and properties of continued fractions one can determine average rate of growth of continued fraction denominators and digits and find that the typical speed of approximation of
irrationals by rationals is the entropy of the map. We probably will not have time to discuss the related ideas of Farey (properly called Haros) fractions and denominators, the Minkowski ? function,
and associated dynamical systems and C* algebras, mention of which can be found in the notes on the speaker’s web page. | {"url":"https://math.unc.edu/event/karl-petersen-unc-ch-ergodic-theory-and-dynamical-systems-seminar/","timestamp":"2024-11-09T07:53:55Z","content_type":"text/html","content_length":"112543","record_id":"<urn:uuid:60d8081c-59cd-41e3-9067-bee77f1c47e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00631.warc.gz"} |
Swept-frequency cosine
y = chirp(t,f0,t1,f1) generates samples of a linear swept-frequency cosine signal at the time instances defined in array t. The instantaneous frequency at time 0 is f0 and the instantaneous frequency
at time t1 is f1.
y = chirp(t,f0,t1,f1,method) specifies an alternative sweep method option.
y = chirp(t,f0,t1,f1,"quadratic",phi,shape) specifies the shape of the spectrogram of a quadratic swept-frequency signal.
y = chirp(___,cplx) returns a real chirp if cplx is specified as "real" and returns a complex chirp if cplx is specified as "complex".
Linear Chirp
Generate a chirp with linear instantaneous frequency deviation. The chirp is sampled at 1 kHz for 2 seconds. The instantaneous frequency is 0 at t = 0 and crosses 250 Hz at t = 1 second.
t = 0:1/1e3:2;
y = chirp(t,0,1,250);
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.1 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,1e3,"spectrogram",TimeResolution=0.1, ...
Quadratic Chirp
Generate a chirp with quadratic instantaneous frequency deviation. The chirp is sampled at 1 kHz for 2 seconds. The instantaneous frequency is 100 Hz at t = 0 and crosses 200 Hz at t = 1 second.
t = 0:1/1e3:2;
y = chirp(t,100,1,200,"quadratic");
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.1 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,1e3,"spectrogram",TimeResolution=0.1, ...
Convex Quadratic Chirp
Generate a convex quadratic chirp sampled at 1 kHz for 2 seconds. The instantaneous frequency is 400 Hz at t = 0 and crosses 300 Hz at t = 1 second.
t = 0:1/1e3:2;
fo = 400;
f1 = 300;
y = chirp(t,fo,1,f1,"quadratic",[],"convex");
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.1 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,1e3,"spectrogram",TimeResolution=0.1, ...
Symmetric Concave Quadratic Chirp
Generate a concave quadratic chirp sampled at 1 kHz for 4 seconds. Specify the time vector so that the instantaneous frequency is symmetric about the halfway point of the sampling interval, with a
minimum frequency of 100 Hz and a maximum frequency of 500 Hz.
t = -2:1/1e3:2;
fo = 100;
t1 = max(t);
f1 = 500;
y = chirp(t,fo,t1,f1,"quadratic",[],"concave");
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.1 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,t,"spectrogram",TimeResolution=0.1, ...
Logarithmic Chirp
Generate a logarithmic chirp sampled at 1 kHz for 10 seconds. The instantaneous frequency is 10 Hz initially and 400 Hz at the end.
t = 0:1/1e3:10;
fo = 10;
f1 = 400;
y = chirp(t,fo,10,f1,"logarithmic");
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.2 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,t,"spectrogram",TimeResolution=0.2, ...
Use a logarithmic scale for the frequency axis. The spectrogram becomes a line, with high uncertainty at low frequencies.
ax = gca;
ax.YScale = "log";
Complex Chirp
Generate a complex linear chirp sampled at 1 kHz for 10 seconds using single support precision. The instantaneous frequency is –200 Hz initially and 300 Hz at the end. The initial phase is zero.
fs = 1e3;
t = single(0:1/fs:10);
fo = -200;
f1 = 300;
ph0 = 0;
y = chirp(t,fo,t(end),f1,"linear",ph0,"complex");
Compute and plot the spectrogram of the chirp. Divide the signal into segments such that the time resolution is 0.2 second. Specify 99% of overlap between adjoining segments and a spectral leakage of
pspectrum(y,t,"spectrogram",TimeResolution=0.2, ...
Verify that a complex chirp has real and imaginary parts that are equal but with ${90}^{\circ }$ phase difference.
x = chirp(t,fo,t(end),f1,"linear",0)...
+ 1j*chirp(t,fo,t(end),f1,"linear",-90);
pspectrum(x,t,"spectrogram",TimeResolution=0.2, ...
Input Arguments
t — Time array
vector | matrix | N-D array
Time array, specified as a vector, matrix, or N-D array.
If you specify t using single-precision data, the chirp function generates a single-precision signal y.
Data Types: single | double
f0 — Instantaneous frequency at time 0
0 (default) | real scalar in Hz
Initial instantaneous frequency at time 0, specified as a real scalar expressed in Hz.
Data Types: single | double
t1 — Reference time
1 (default) | positive scalar in seconds
Reference time, specified as a positive scalar expressed in seconds.
Data Types: single | double
f1 — Instantaneous frequency at time t1
100 (default) | real scalar in Hz
Instantaneous frequency at time t1, specified as a real scalar expressed in Hz.
Data Types: single | double
method — Sweep method
"linear" (default) | "quadratic" | "logarithmic"
Sweep method, specified as "linear", "quadratic", or "logarithmic".
• "linear" — Specifies an instantaneous frequency sweep f[i](t) given by
${f}_{i}\left(t\right)={f}_{0}+\beta t,$
$\beta =\left({f}_{1}-{f}_{0}\right)/{t}_{1}$
and the default value for f[0] is 0. The coefficient β ensures that the desired frequency breakpoint f[1] at time t[1] is maintained.
• "quadratic" — Specifies an instantaneous frequency sweep f[i](t) given by
${f}_{i}\left(t\right)={f}_{0}+\beta {t}^{2},$
$\beta =\left({f}_{1}-{f}_{0}\right)/{t}_{1}{}^{2}$
and the default value for f[0] is 0. If f[0] > f[1] (downsweep), the default shape is convex. If f[0 ]< f[1] (upsweep), the default shape is concave.
• "logarithmic" — Specifies an instantaneous frequency sweep f[i](t) given by
${f}_{i}\left(t\right)={f}_{0}×{\beta }^{t},$
$\beta ={\left(\frac{{f}_{1}}{{f}_{0}}\right)}^{\frac{1}{{t}_{1}}}$
and the default value for f[0] is 10^–6.
Data Types: char | string
phi — Initial phase
0 (default) | positive scalar in degrees
Initial phase, specified as a positive scalar expressed in degrees.
Data Types: single | double
shape — Spectrogram shape of quadratic chirp
"convex" | "concave"
Spectrogram shape of quadratic chirp, specified as "convex" or "concave". shape describes the shape of the parabola with respect to the positive frequency axis. If not specified, shape is "convex"
for the downsweep case with f[0] > f[1], and "concave" for the upsweep case with f[0] < f[1].
Data Types: char | string
cplx — Output complexity
"real" (default) | "complex"
Output complexity, specified as "real" or "complex".
Data Types: char | string
Output Arguments
y — Swept-frequency cosine signal
Swept-frequency cosine signal, returned as a vector.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Thread-Based Environment
Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool.
This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.
Version History
Introduced before R2006a
R2024b: Generate Single-Precision Outputs
The chirp function supports single-precision outputs. | {"url":"https://de.mathworks.com/help/signal/ref/chirp.html","timestamp":"2024-11-12T10:04:21Z","content_type":"text/html","content_length":"127018","record_id":"<urn:uuid:8d541141-ad1a-4664-b221-5cb05ddc29c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00147.warc.gz"} |
The fixed-point property and piecewise-projective transformations of the line
Printable PDF
Department of Mathematics,
University of California San Diego
Math 211A: Seminar in Algebra
Professor Nicolas Monod
The fixed-point property and piecewise-projective transformations of the line
We describe a new and elementary proof of the fact that many groups of piecewise-projective transformation of the line are non-amenable by constructing an explicit action without fixed points. One
the one hand, such groups provide explicit counter-examples to the Day-von Neumann problem. On the other hand, they illustrate that we can distinguish many "layers" of relative non-amenability
between nested subgroups.
Adrian Ioana, Alireza Golsefidy
November 18, 2024
3:00 PM
APM 7321
Research Areas
Algebra Representation Theory | {"url":"https://math.ucsd.edu/seminar/fixed-point-property-and-piecewise-projective-transformations-line","timestamp":"2024-11-03T00:38:10Z","content_type":"text/html","content_length":"34965","record_id":"<urn:uuid:698ca59c-3292-4aeb-ab9b-0628fc54ab3f>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00718.warc.gz"} |
Math activites; high school distance formula
Algebra Tutorials!
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Rational Expressions for multiple variables | algerbraic. formulas | graph hyberbola | adding & subtracting integers word document | intermediate algebra worksheets | evaluating
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Functions ti 83
Solve Two-Step Equations
Multiply, Dividing;
Exponents; Square Roots; Author Message
and Solving Equations
LinearEquations N0che Posted: Friday 29th of Dec 12:10
Solving a Quadratic I am going to high school now. As math has always been my weakness, I purchased the course material in advance. I am plan on learning a
Equation couple of chapters before the classes start. Any kind of help would be highly appreciated that could help me to start studying math
Systems of Linear activites; high school distance formula myself.
Equations Introduction
Equations and Registered:
Inequalities 24.08.2002
Solving 2nd Degree From: Missouri
Review Solving Quadratic
System of Equations AllejHat Posted: Sunday 31st of Dec 10:59
Solving Equations & Sounds like your concepts are not clear . Excelling in math activites; high school distance formula requires that your concepts be strong .
Inequalities I know students who actually start tutoring juniors in their first year. Why don’t you try Algebrator? I am quite sure, this program will
Linear Equations help you.
Functions Zeros, and
Applications Registered:
Rational Expressions and 16.07.2003
Functions From: Odense, Denmark
Linear equations in two
Lesson Plan for
Comparing and Ordering DoniilT Posted: Sunday 31st of Dec 15:26
Rational Numbers Yeah, I think so too . Algebrator explains everything in such great detail that even a beginner can learn the tricks of the trade, and
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Solving Equations solution with such finesse that you’ll learn a lot from it.
Radicals and Rational
Exponents Registered:
Solving Linear Equations 27.08.2002
Systems of Linear From:
Solving Exponential and
Logarithmic Equations
Solving Systems of SanG Posted: Sunday 31st of Dec 21:47
Linear Equations Algebrator is the program that I have used through several algebra classes - College Algebra, Pre Algebra and Basic Math. It is a really a
DISTANCE,CIRCLES,AND great piece of algebra software. I remember of going through difficulties with lcf, function range and radical inequalities. I would simply
QUADRATIC EQUATIONS type in a problem homework, click on Solve – and step by step solution to my algebra homework. I highly recommend the program.
Solving Quadratic
Equations Registered:
Quadratic and Rational 31.08.2001
Inequalit From: Beautiful
Applications of Systems Northwest Lower
of Linear Equations in Michigan
Two Variables
Systems of Linear
Test Description for
RATIONAL EX
Exponential and
Logarithmic Equations
Systems of Linear
Equations: Cramer's Rule
Introduction to Systems
of Linear Equations
Literal Equations &
Equations and
Inequalities with
Absolute Value
Rational Expressions
SOLVING LINEAR AND
Steepest Descent for
Solving Linear Equations
The Quadratic Equation
Linear equations in two
math activites; high school distance formula
Related topics:
Home restrictions on hyperbolas | slope worksheets middle school | answers for prentice hall algebra 1 book | free fraction worksheets elementary | equation solver
Rational Expressions for multiple variables | algerbraic. formulas | graph hyberbola | adding & subtracting integers word document | intermediate algebra worksheets | evaluating
Graphs of Rational expressions worksheets | free online gre papers | simplifying by factoring square roots calculator | solving simultaneous equations calculator | change log base
Functions ti 83
Solve Two-Step Equations
Multiply, Dividing;
Exponents; Square Roots; Author Message
and Solving Equations
LinearEquations N0che Posted: Friday 29th of Dec 12:10
Solving a Quadratic I am going to high school now. As math has always been my weakness, I purchased the course material in advance. I am plan on learning a
Equation couple of chapters before the classes start. Any kind of help would be highly appreciated that could help me to start studying math
Systems of Linear activites; high school distance formula myself.
Equations Introduction
Equations and Registered:
Inequalities 24.08.2002
Solving 2nd Degree From: Missouri
Review Solving Quadratic
System of Equations AllejHat Posted: Sunday 31st of Dec 10:59
Solving Equations & Sounds like your concepts are not clear . Excelling in math activites; high school distance formula requires that your concepts be strong .
Inequalities I know students who actually start tutoring juniors in their first year. Why don’t you try Algebrator? I am quite sure, this program will
Linear Equations help you.
Functions Zeros, and
Applications Registered:
Rational Expressions and 16.07.2003
Functions From: Odense, Denmark
Linear equations in two
Lesson Plan for
Comparing and Ordering DoniilT Posted: Sunday 31st of Dec 15:26
Rational Numbers Yeah, I think so too . Algebrator explains everything in such great detail that even a beginner can learn the tricks of the trade, and
LinearEquations crack some of the most difficult mathematical problems. It elaborates on each and every intermediate step that it took to reach a certain
Solving Equations solution with such finesse that you’ll learn a lot from it.
Radicals and Rational
Exponents Registered:
Solving Linear Equations 27.08.2002
Systems of Linear From:
Solving Exponential and
Logarithmic Equations
Solving Systems of SanG Posted: Sunday 31st of Dec 21:47
Linear Equations Algebrator is the program that I have used through several algebra classes - College Algebra, Pre Algebra and Basic Math. It is a really a
DISTANCE,CIRCLES,AND great piece of algebra software. I remember of going through difficulties with lcf, function range and radical inequalities. I would simply
QUADRATIC EQUATIONS type in a problem homework, click on Solve – and step by step solution to my algebra homework. I highly recommend the program.
Solving Quadratic
Equations Registered:
Quadratic and Rational 31.08.2001
Inequalit From: Beautiful
Applications of Systems Northwest Lower
of Linear Equations in Michigan
Two Variables
Systems of Linear
Test Description for
RATIONAL EX
Exponential and
Logarithmic Equations
Systems of Linear
Equations: Cramer's Rule
Introduction to Systems
of Linear Equations
Literal Equations &
Equations and
Inequalities with
Absolute Value
Rational Expressions
SOLVING LINEAR AND
Steepest Descent for
Solving Linear Equations
The Quadratic Equation
Linear equations in two
Rational Expressions
Graphs of Rational
Solve Two-Step Equations
Multiply, Dividing;
Exponents; Square Roots;
and Solving Equations
Solving a Quadratic
Systems of Linear
Equations Introduction
Equations and
Solving 2nd Degree
Review Solving Quadratic
System of Equations
Solving Equations &
Linear Equations
Functions Zeros, and
Rational Expressions and
Linear equations in two
Lesson Plan for
Comparing and Ordering
Rational Numbers
Solving Equations
Radicals and Rational
Solving Linear Equations
Systems of Linear
Solving Exponential and
Logarithmic Equations
Solving Systems of
Linear Equations
Solving Quadratic
Quadratic and Rational
Applications of Systems
of Linear Equations in
Two Variables
Systems of Linear
Test Description for
RATIONAL EX
Exponential and
Logarithmic Equations
Systems of Linear
Equations: Cramer's Rule
Introduction to Systems
of Linear Equations
Literal Equations &
Equations and
Inequalities with
Absolute Value
Rational Expressions
SOLVING LINEAR AND
Steepest Descent for
Solving Linear Equations
The Quadratic Equation
Linear equations in two
math activites; high school distance formula
Related topics:
restrictions on hyperbolas | slope worksheets middle school | answers for prentice hall algebra 1 book | free fraction worksheets elementary | equation solver for multiple variables |
algerbraic. formulas | graph hyberbola | adding & subtracting integers word document | intermediate algebra worksheets | evaluating expressions worksheets | free online gre papers |
simplifying by factoring square roots calculator | solving simultaneous equations calculator | change log base ti 83
Author Message
N0che Posted: Friday 29th of Dec 12:10
I am going to high school now. As math has always been my weakness, I purchased the course material in advance. I am plan on learning a couple of chapters before the
classes start. Any kind of help would be highly appreciated that could help me to start studying math activites; high school distance formula myself.
From: Missouri
AllejHat Posted: Sunday 31st of Dec 10:59
Sounds like your concepts are not clear . Excelling in math activites; high school distance formula requires that your concepts be strong . I know students who
actually start tutoring juniors in their first year. Why don’t you try Algebrator? I am quite sure, this program will help you.
From: Odense, Denmark
DoniilT Posted: Sunday 31st of Dec 15:26
Yeah, I think so too . Algebrator explains everything in such great detail that even a beginner can learn the tricks of the trade, and crack some of the most difficult
mathematical problems. It elaborates on each and every intermediate step that it took to reach a certain solution with such finesse that you’ll learn a lot from it.
SanG Posted: Sunday 31st of Dec 21:47
Algebrator is the program that I have used through several algebra classes - College Algebra, Pre Algebra and Basic Math. It is a really a great piece of algebra
software. I remember of going through difficulties with lcf, function range and radical inequalities. I would simply type in a problem homework, click on Solve – and
step by step solution to my algebra homework. I highly recommend the program.
From: Beautiful
Northwest Lower
Author Message
N0che Posted: Friday 29th of Dec 12:10
I am going to high school now. As math has always been my weakness, I purchased the course material in advance. I am plan on learning a couple of chapters before the classes
start. Any kind of help would be highly appreciated that could help me to start studying math activites; high school distance formula myself.
From: Missouri
AllejHat Posted: Sunday 31st of Dec 10:59
Sounds like your concepts are not clear . Excelling in math activites; high school distance formula requires that your concepts be strong . I know students who actually start
tutoring juniors in their first year. Why don’t you try Algebrator? I am quite sure, this program will help you.
From: Odense, Denmark
DoniilT Posted: Sunday 31st of Dec 15:26
Yeah, I think so too . Algebrator explains everything in such great detail that even a beginner can learn the tricks of the trade, and crack some of the most difficult
mathematical problems. It elaborates on each and every intermediate step that it took to reach a certain solution with such finesse that you’ll learn a lot from it.
SanG Posted: Sunday 31st of Dec 21:47
Algebrator is the program that I have used through several algebra classes - College Algebra, Pre Algebra and Basic Math. It is a really a great piece of algebra software. I
remember of going through difficulties with lcf, function range and radical inequalities. I would simply type in a problem homework, click on Solve – and step by step solution
to my algebra homework. I highly recommend the program.
From: Beautiful
Northwest Lower
Posted: Friday 29th of Dec 12:10
I am going to high school now. As math has always been my weakness, I purchased the course material in advance. I am plan on learning a couple of chapters before the classes start. Any kind of help
would be highly appreciated that could help me to start studying math activites; high school distance formula myself.
Posted: Sunday 31st of Dec 10:59
Sounds like your concepts are not clear . Excelling in math activites; high school distance formula requires that your concepts be strong . I know students who actually start tutoring juniors in
their first year. Why don’t you try Algebrator? I am quite sure, this program will help you.
Posted: Sunday 31st of Dec 15:26
Yeah, I think so too . Algebrator explains everything in such great detail that even a beginner can learn the tricks of the trade, and crack some of the most difficult mathematical problems. It
elaborates on each and every intermediate step that it took to reach a certain solution with such finesse that you’ll learn a lot from it.
Posted: Sunday 31st of Dec 21:47
Algebrator is the program that I have used through several algebra classes - College Algebra, Pre Algebra and Basic Math. It is a really a great piece of algebra software. I remember of going through
difficulties with lcf, function range and radical inequalities. I would simply type in a problem homework, click on Solve – and step by step solution to my algebra homework. I highly recommend the | {"url":"https://rational-equations.com/in-rational-equations/graphing-equations/math-activites-high-school.html","timestamp":"2024-11-06T04:09:55Z","content_type":"text/html","content_length":"96914","record_id":"<urn:uuid:5834bd0a-a48c-4475-9483-c52821ce1579>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00470.warc.gz"} |
Excel Statistical Functions - Free Excel Tutorial
This section will learn how to use Excel’s Statistical Functions such as: Avedev, Average, AverageA, AverageIF, AverageIFS, etc.
AVEDEV – returns the average of the absolute deviations of the numbers that you provided.
AVERAGE – returns the average of the numbers that you provided
AVERAGEA – returns the average of its arguments, including numbers, text, and logical values.
AVERAGEIF – returns the average of all numbers in a range of cells that meet a given criteria.
AVERAGEIFS – returns the average of all numbers in a range of cells that meet multiple criteria.
BETA.DIST – calculate the cumulative beta distribution or beta probability density function.
BETADIST – returns the cumulative beta probability density function.
BETA.INV – returns the inverse of the beta cumulative probability density function.
BETAINV – returns the inverse of the beta cumulative probability density function.
BINOM.DIST – returns the individual term binomial distribution probability.
BINOM.INV – returns the inverse of the Cumulative Binomial Distribution that is greater than or equal to a criterion value.
BINOMDIST – returns the individual term binomial distribution probability.
CHIDIST – returns the right-tailed probability of the chi-squared distribution.
CHIINV – returns the inverse of the right-tailed probability of the chi-squared distribution.
CHITEST – returns the value from the chi-squared distribution for the statistic and the appropriate degrees of freedom.
COUNT – counts the number of cells that contain numbers, and counts numbers within the list of arguments.
COUNTA – counts the number of cells that are not empty in a range.
COUNTBLANK – use to count the number of empty cells in a range of cells.
COUNTIF – count the number of cells in a range that meet a given criteria.
COUNTIFS -returns the count of cells in a range that meet one or more criteria.
COVAR – returns covariance, the average of the products of deviations for each data point in two given sets of values.
FORECAST – used to calculate or predict a future value by using existing values.
FREQUENCY – calculates how often values occur within a range of values.
GROWTH – calculates the predicted exponential growth based on existing data.
INTERCEPT – calculates the point at which a line will intersect the y-axis by using existing x-values and y-values.
LARGE -returns the largest numeric value from the numbers that you provided. Or returns the largest value in the array.
LINEST – calculates the statistics for a line by using the “least squares” method to calculate a straight line that best fits your data, and then returns an array that describes the line.
MAX – returns the largest numeric value from the numbers that you provided.
MAXA – returns the largest numeric value from a range of values.
MEDIAN – returns the median of the given numbers.
MIN – returns the smallest numeric value from the numbers that you provided.
MINA -returns the smallest numeric value from the numbers that you provided, while counting text and the logical values.
MODE – returns the most frequently occurring number found in an array or range of numbers.
MODE.MULT – returns a vertical array of the most frequently occurring number found in an array or range of numbers.
MODE.SNGL -returns the most frequently occurring number found in an array or range of numbers.
PERCENTILE -returns the kth percentile from a supplied range of values.
PERCENTRANK – returns the rank of a value in a set of values as a percentage of the set.
PERMUT – returns the number of permutations for a given number of items.
QUARTILE – returns the quartile from a supplied range of values.
RANK – returns the rank of a given number in a supplied range of cells.
SLOPE – returns the slope of the linear regression line through data points in known_y’s and know_x’s.
SMALL -returns the nth smallest numeric value from the numbers that you provided.
STDEV – returns the standard deviation of a population based on a sample of numbers.
STDEVA – returns the standard deviation of a population based on a sample of numbers, text, and logical values.
STDEVP – returns the standard deviation of a population based on an entire population of numbers.
STDEVPA – returns the standard deviation of a population based on an entire population of numbers, text or logical values.
VAR – returns the variance of a population based on a sample of numbers.
VARA – returns the variance of a population based on a sample of numbers, text, or logical values.
VARP – returns the variance of a population based on an entire population of numbers.
VARPA – returns the variance of a population based on an entire population of numbers, text, or logical values. | {"url":"https://www.excelhow.net/excel-statistical-functions.html","timestamp":"2024-11-12T23:35:16Z","content_type":"text/html","content_length":"90607","record_id":"<urn:uuid:02f26619-9b7d-4fb1-ba2b-a978ff20e4f9>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00734.warc.gz"} |
Linear Algebra
Numerisk analys 5 sv February 2004.
Linjär Algebra: Fast utan att vara så JOBBIGT - Amazon UK
Additionally, the book presents special topics from applied linear algebra including matrix functions, the singular value decomposition, the Kronecker product and linear matrix equations. The
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Elementary Linear Algebra, International Metric Edition – Bokab
Wellesley-Cambridge Press : Reading a book can help a student grasp the concepts of linear algebra with relative ease. So, we have listed some of the best books on linear algebra for your
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Algebra or Linear algebra . 132: This is a good contemporary book on linear algebra.
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Dozens of applications connect key concepts in linear algebra to real-world examples in Physics, Chemistry, Circuits, and more. Linear Algebra for Everyone Gilbert Strang. ISBN 978-1-7331466-3-0
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This book is meant to provide an introduction to vectors, matrices, and least squares methods, basic topics in applied linear algebra. Our goal is to give the beginning student, with little or no
prior exposure to linear algebra, a good ground-ing in the basic ideas, as well as an appreciation for how they are used in … Linear algebra Linear Algebra and Its Applications Linear Algebra and Its
Applications, 5th Edition Linear Algebra and Its Applications, 5th Edition 5th Edition | ISBN: 9780321982384 / 032198238X. 2,010. expert-verified solutions in this book. Buy on Amazon.com 5th Edition
| ISBN: 9780321982384 / … A standard book for a first course in linear algebra is Gilbert Strang's Linear Algebra and Its Applications. After getting an initial exposure, Sheldon Axler's Linear
Algebra Done Right is a good book for getting a more abstract view of linear algebra (at Carnegie Mellon, this is … Linear Algebra by Jim Hefferon, along with its answers to exercises, is a text for
a first undergraduate course.It is Free.
Elementary Linear Algebra, International Metric Edition – Bokab
Linear Algebra Done Right 3rd ed. 2015 Edition by Sheldon Axler (errata | videos) Linear Algebra 2nd Edition by Kenneth M Hoffman, Ray Kunze (see solutions here) Good Linear Algebra textbooks (not
complete) Introduction to Linear Algebra, Fifth Edition by Gilbert Strang, Solution Manual Se hela listan på machinelearningmastery.com 2017-09-04 · The title of the book sounds a bit mysterious. Why
should anyone read this book if it presents the subject in a wrong way?
The book covers less mathematics than a typical text on applied linear algebra. We use only one theoretical concept from linear algebra, linear independence, and only one computational tool, the QR
factorization; our approach to most applica-tions relies on only one method, least squares (or some extension). In this sense Welcome to Linear Algebra for Beginners: Open Doors to Great Careers! | {"url":"https://kopavguldgplcuku.netlify.app/10764/2483.html","timestamp":"2024-11-07T19:33:51Z","content_type":"text/html","content_length":"12334","record_id":"<urn:uuid:fdf1691e-3a86-4f32-abd0-5285ad37c729>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00374.warc.gz"} |
Logarithms problems and solutions
Logarithms Questions with solutions
Once you learn the logarithmic rules clearly, you are ready to use them as formulas in log questions. The logarithm practice problems worksheet is given here with answers and the variety of examples
for students who study the logarithms and also the solutions to learn how to find the logarithms by the log properties in the problems.
$(1).\,\,$ Evaluate $\log_{5}{7^{\displaystyle -3\log_{7}{5}}}$
$(2).\,\,$ $\log{\Big(\dfrac{x^2}{yz}\Big)}$ $+$ $\log{\Big(\dfrac{y^2}{zx}\Big)}$ $+$ $\log{\Big(\dfrac{z^2}{xy}\Big)}$
$(3).\,\,$ Evaluate $\log_{\sqrt{2}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$
List of the logarithmic expressions problems with solutions to learn how to evaluate them by simplification.
List of the logarithmic equations problems with solutions to learn how to solve the logarithm equations.
$(1).\,\,$ Solve $\log_{3}{\big(5+4\log_{3}{(x-1)}\big)}$ $\,=\,$ $2$
$(2).\,\,$ Evaluate $\dfrac{x}{y}+\dfrac{y}{x}$ if $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$
$(3).\,\,$ Solve $x$ $+$ $\log{(1+2^x)}$ $=$ $x\log{5}$ $+$ $\log{6}$ | {"url":"https://www.mathdoubts.com/log-problems-and-solutions/","timestamp":"2024-11-02T14:42:40Z","content_type":"text/html","content_length":"27624","record_id":"<urn:uuid:932b2223-830d-4039-bbb5-6cefe928a715>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00497.warc.gz"} |
How Many Hours is 3/4 of a Day?
Three-quarters of a day is equivalent to 18 hours. But what does that really mean in terms of everyday activities? Let’s look at some examples to put it into perspective.
Understanding the Calculation
To understand how we get to 18 hours from 3/4 of a day, here’s a simple breakdown:
• Hours in a Day: There are 24 hours in a day.
By multiplying the number of hours in a day by 3/4, we get: 24 hours/day × .75 = 18 hours
Examples of Activities That Last 18 Hours
1. Long-Distance Flights: Many long-distance international flights, such as those from New York to Sydney, can take around 18 hours. This includes flight time and possible layovers.
2. Extended Work Shifts: Some professions, such as healthcare and emergency services, may require extended work shifts that last up to 18 hours, including overtime.
3. Road Trips: Driving long distances, such as from Los Angeles to Denver, typically takes around 18 hours, including breaks for food and rest.
4. Marathon Events: Certain endurance events or charity marathons can span approximately 18 hours, providing participants with a challenging and rewarding experience.
Breaking Down 3/4 of a Day into Hours
Understanding 3/4 of a day in terms of hours helps visualize it better.
• Hours: 3/4 of a day is 18 hours.
So, 3/4 of a day is equal to 18 hours.
Real-Life Applications
Knowing how long 18 hours is can help in planning and managing time effectively. Here are some practical uses:
• Travel Planning: When planning long trips or flights, knowing that 3/4 of a day is 18 hours helps in scheduling rest breaks and activities upon arrival.
• Work and Study Sessions: For extended work or study sessions, understanding that 3/4 of a day is 18 hours helps in managing energy levels and planning productive breaks.
• Event Planning: If you have events or activities lasting around 18 hours, knowing it’s 3/4 of a day helps in organizing and scheduling efficiently.
Practical Uses of 18 Hours
1. Fitness Challenges: Some intense fitness challenges, such as multi-day hikes or cycling tours, can last up to 18 hours. Understanding this duration helps in planning rest and nutrition.
2. Entertainment Marathons: Planning a TV show or movie marathon for 18 hours provides a full day of entertainment, requiring careful planning for breaks and meals.
3. Long-Term Commitments: Engaging in projects or hobbies for 18 hours can lead to significant progress and achievement, whether it’s a work project, a creative endeavor, or learning a new skill. | {"url":"https://sitemap.hillhouse4design.com/18-hours-to-days/","timestamp":"2024-11-01T19:16:32Z","content_type":"text/html","content_length":"47405","record_id":"<urn:uuid:758e540f-988b-495a-b3f0-a1be79cb72a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00246.warc.gz"} |
What is L * W * H?
What is L * W * H?
Length, width, height, depth.
How do you find the length and width in algebra?
Perimeter let us call it P right you should actually write let statements let X be the length Y be the width and P be the parameter. The parameter P is 2x plus 2y right opposite sides are equal. So
What equation is length x width x height?
When we multiply the length, width, and height of a cuboid, we get its volume. This means, Length x Width x Height = Volume of Cuboid. In other words, the capacity or volume of a cuboid or any
rectangular box can be measured if we multiply these three dimensions together. Let us understand this with an example.
What formula is length x width?
The formula for the area, ‘A’ of a rectangle whose length and width are ‘l’ and ‘w’ respectively is the product of length and width, that is, “A = l × w”.
What is formula of L * B * H?
Volume Formula of Cuboid
The volume of cuboid = Base Area × Height cubic units. The base area for cuboid = l × b square units. Hence, the volume of a cuboid, V = l × b × h = lbh units3, where ‘l’ ‘b’ and ‘h’ represent the
length, breadth, and height of the cuboid.
How do you multiply length width and height?
Volume (LxWxH) – YouTube
What is the formula to find length?
If you have the area A and width w , its length w is determined as h = A/w . If you have the perimeter P and width w , its length can be found with h = P/2−w . If you have the diagonal d and width w
, it’s length is h = √(d²−w²) .
How do you find length width height and area?
To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself
to find the area. This is the same as saying length2 or length squared.
How do you find the volume of a LxWxH?
Multiply the length (L) times the width (W) times the height (H). The formula looks like this: LxWxH For this example, to calculate the volume of the object the formula would be 10 x 10 x 10 = 1,000
cubic inches.
What is the formula for area?
Area and Perimeter Formula Chart
Figures Area Formula Variables
Area of Square Area = a2 a = sides of the square
Area of a Triangle Area = 1/2 b×h b = base h = height
Area of a Circle Area = πr2 r = radius of the circle
Area of a Trapezoid Area = 1/2 (a + b)h a =base 1 b = base 2 h = vertical height
What is the formula to find width?
If you have the area A and length h , its width w is w = A/h . If you have the perimeter P and length h , its width is w = P/2−h . If you have the diagonal d and length h , it’s width can be found
with w = √(d²−h²) .
What are the 3 formulas for volume?
The volume of three-dimensional mathematical shapes like cube, cuboid, cylinder, prism and cone etc.
Volume Formulas of Various Geometric Figures.
Shapes Volume Formula Variables
Rectangular Solid or Cuboid V = l × w × h l = Length w = Width h = Height
What is L and B in rectangle?
Ans: The Area of a rectangle (A) = l x b, where l is the length and b are the breadths of the rectangle. Given, l = 500 m and b = 300 m. Thus, Area of the rectangle A = 500 m x 300 m = 1, 50,000
square meters. Now, let us calculate the cost of painting the land.
What is volume if we multiply length width and height?
Remember, the equation for volume is V = length x width x height, so simply multiply all three sides together to get your volume.
How do you find length width height and volume?
Divide the volume by the product of the length and width to calculate the height of a rectangular object. For this example, the rectangular object has a length of 20, a width of 10 and a volume of
6,000. The product of 20 and 10 is 200, and 6,000 divided by 200 results in 30. The height of the object is 30.
How do you find the width?
How do I calculate the width of a rectangle?
1. If you have the area A and length h , its width w is w = A/h .
2. If you have the perimeter P and length h , its width is w = P/2−h .
3. If you have the diagonal d and length h , it’s width can be found with w = √(d²−h²) .
Is width and breadth same?
Main Difference – Breadth vs Width
Breadth is used to talk about measurements of larger entities and figurative entities while width is used to talk about the measurements of smaller entities and physical objects. Breadth and width
are two words that are commonly used when discussing the measurements of an object.
What is area formula?
Area = length × lengthArea = l2. Area = length x breadthArea = l × b. Area = π × radius × radiusArea = π × r2(π = 3.14) Using these formulas, area for different quadrilaterals (a special case of
polygons with four sides and angles ≠ 90o) is also calculated. Application.
What is the formula for calculating area?
The area is measurement of the surface of a shape. To find the area of a rectangle or a square you need to multiply the length and the width of a rectangle or a square. Area, A, is x times y.
What is volume in math 4th grade?
VOLUME is the amount of space inside a 3D object.
How do you find volume in 5th grade math?
Volume of a Rectangular Prism | 5th Grade Math – YouTube
Why formula is area?
Area and Perimeter Formula are the two major formulas for any given two-dimensional shape in Mathematics.
Area and Perimeter Formula Chart.
Figures Area Formula Variables
Area of Square Area = a2 a = sides of the square
Area of a Triangle Area = 1/2 b×h b = base h = height
What is a formula of area of rectangle?
Area of a Rectangle. A = l × b. The area of any rectangle is calculated, once its length and width are known. By multiplying length and breadth, the rectangle’s area will obtain in a square-unit
How do you find the length width and height?
What Is the Length, Width & Height of a Cube? : Math Tutorials – YouTube
What are the 2 formulas for volume?
Volume Formulas of Various Geometric Figures
Shapes Volume Formula Variables
Rectangular Solid or Cuboid V = l × w × h l = Length w = Width h = Height
Cube V = a3 a = Length of edge or side
Cylinder V = πr2h r = Radius of the circular base h = Height
Prism V = B × h B = Area of base, (B = side2 or length.breadth) h = Height | {"url":"https://mattstillwell.net/what-is-l-w-h/","timestamp":"2024-11-11T07:43:20Z","content_type":"text/html","content_length":"42756","record_id":"<urn:uuid:d20ec039-d0fa-440a-b252-8647a0fce76e>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00290.warc.gz"} |
Negative Zero
My wife brings up the following story any time she wants to make the point that I’m pedantic: When one of my daughters was in second grade, her math teacher told the class that any number divided by
zero was one. I dashed off an impassioned email to the teacher, insisting that the result had to be undefined. Supposedly this is evidence that I’m sometimes difficult to be around.
Turns out the joke might be on me — although it’s still hard to support the second-grade teacher’s answer. I recently learned a bunch of things I didn’t know about floating point math:
• There is a value for negative zero, separate from regular (positive?) zero. These two zeroes are defined to be equal to each other and yet they are distinct values.
• x ÷ 0.0, for x ≠ ±0.0, is not an error. Instead, the result is either positive infinity or negative infinity, following the usual sign convention.
• The case of ±0.0 ÷ ±0.0 is an error (specifically it’s “not a number” or NaN).
• –0.0 + –0.0 = –0.0, –0.0 + 0.0 = 0.0, and –0.0 × 0.0 = –0.0
These rules stem from the IEEE 754 “Standard for Floating-Point Arithmetic,” which standardized floating point representations across platforms. The most recent version of the standard was completed
in 2008 but the original version was issued in 1985, so this behavior is not new. The rules above are true in both C (gcc) and Swift on my Mac, and also true in Swift on an iPhone. Python on the Mac
supports negative zero for floats, but throws an exception when you attempt to divide by zero of any sign.
There are a couple of surprising corollaries to these rules:
• Because 0.0 and -0.0 must compare as equal, the test (x < 0.0) does not return true for every negative number—it fails for negative zero. Therefore, to determine the sign of a zero value, you
need to use the platform’s built-in sign function, for instance Double.sign in Swift. Or I guess you could bit-manipulate the raw representation of the double, which is very much a C programmer’s
• If a = b ÷ c, it does not necessarily follow that b = a × c, because this also fails for the case where c is zero of either sign.
I’m not a number theorist, but I find the concepts above surprising.
One immediate problem: Infinity is not a number, like zero or 3.25 or π. Rather, infinity is a concept. It is true that the rational numbers are countably infinite—but infinity is not a member of the
set of rational numbers.
Furthermore, from a number theory perspective, division by zero is nonsensical. You can understand why if you get precise about what division means. Technically, “division” is “multiplication by a
number’s inverse,” where the inverse satisfies: a × a^-1 = 1. Zero is the only number in the set of real numbers that simply has no multiplicative inverse. And since this inverse doesn’t exist, we
can’t go around multiplying by it.
But surely the people who designed floating point numbers knew all this. So, I got wondering about why the behavior described came to be written into the IEEE standard.
To start, let’s consider the problem that floating-point math is trying to address. The real numbers are uncountably infinite, and yet we wish to represent this entire set within the bounds of finite
computer memory. With a 64-bit double, there are 2^64 possible symbols, and the designers of the IEEE standard were trying to map these symbols onto the set of real numbers in a way that was both
useful to real-world applications and also economically feasible given the constraints of early 80s silicon. Given the basic requirements, clearly approximations were going to be used.
The reasoning for negative zero appears to date to a 1987 paper^1 by William Kahan, a Berkeley professor who is considered the “father of floating point” and who later won the Turing Award for his
work in drafting IEEE 754. It turns out that the existence of negative zero is intimately tied to the ability to divide by zero.
Let’s start by discussing the usual reason that division by zero is not allowed. A naïve approach to division by zero is the observation that:
In other words, as x gets smaller, the result of 1/x gets larger. But this is only true when x approaches 0 from the positive side (which is why there’s that little plus sign above). Running the same
thought experiment from the negative side:
As a result, the generic limit of 1/x as x approaches 0 is undefined because there is a discontinuity (what Kahan calls a slit) in the function 1/x.
However, by introducing a signed zero, Kahan and the IEEE committee could work around the difficulty. Intuitively, the sign of a zero is taken to indicate the direction the limit is being approached
from. As Kahan states in his 1987 paper:
Rather than think of +0 and -0 as distinct numerical values, think of their sign bit as an auxiliary variable that conveys one bit of information (or misinformation) about any numerical variable that
takes on 0 as its value. Usually this information is irrelevant; the value of 3+x is no different for x := +0 than for x := -0…. However, a few extraordinary arithmetic operations are affected by
zero’s sign; for example 1/ (+0) = +∞ but 1/ (–0) = –∞.
I’ve made my peace with the concept by adopting a rationalization proposed by my partner Mike Perkins: The 2^64 available symbols are clearly inadequate to represent the entirety of the set of real
numbers. So, the IEEE designers set aside a few of those symbols for special meanings. In this sense, ∞ doesn’t really mean “infinity”—instead, it means “a real number that is larger than we can
otherwise represent in our floating-point symbol set.” And therefore +0 doesn’t really mean “zero,” but rather “a real number that is larger than true 0 but smaller than any positive number we can
Incidentally, while researching this issue, I discovered that even Kahan doesn’t love the idea of negative zero:
Signed zero “well, the signed zero was a pain in the ass that we could eliminate if we used the projective mode. If there was just one infinity and one zero you could do just fine; then you didn’t
care about the sign of zero and you didn’t care about the sign of infinity. But if, on the other hand, you insisted on what I would have regarded as the lesser choice of two infinities, then you are
going to end up with two signed zeros. There really wasn’t a way around that and you were stuck with it.” (From an interview of Kahan conducted in 2005.)
I’m not certain if writing a blog post ten years later makes up for railing against a poor second-grade teacher. For her part, my daughter, now in high school, just rolled her eyes when I started
talking about division by zero at dinner. So maybe that “difficult to be around” thing is hereditary.
Kahan, W., “Branch Cuts for Complex Elementary Functions, or Much Ado About Nothing’s Sign Bit,” The State of the Art in Numerical Analysis, (Eds. Iserles and Powell), Clarendon Press, Oxford, 1987,
available here.
Category: Signal Processing
By Howdy Pierce | {"url":"https://www.cardinalpeak.com/blog/negative-zero","timestamp":"2024-11-06T11:47:24Z","content_type":"text/html","content_length":"326473","record_id":"<urn:uuid:3cfd761a-d7c5-4b56-92a3-c46b16ea27b4>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00845.warc.gz"} |
How to Use the SUM Function using Query function
In Google Sheets, the SUM function is used to add up a series of numbers in a range of cells. The function can be used in combination with the QUERY function to calculate the sum of specific values
returned from a QUERY.
Here's how to use the SUM function with the QUERY function in Google Sheets:
=SUM(QUERY(data, query_expression, [headers]))
• data: The range of cells containing the data you want to perform the QUERY on.
• query_expression: A query expression in Google Visualization API Query Language that you'll use to filter the data.
• headers: (Optional) The number of header rows in the data range. If omitted, Google Sheets will try to guess the number of header rows automatically.
1. Open your Google Sheet containing the data you want to work with.
2. Click on an empty cell where you want to display the result of the SUM function with the QUERY function.
3. Type the formula, starting with an equal sign (=), followed by the SUM function, and then the QUERY function. For example: =SUM(QUERY(A1:C10, "SELECT B WHERE A = 'X'", 1))
□ Replace A1:C10 with the range of cells containing your data.
□ Replace the query expression "SELECT B WHERE A = 'X'" with the appropriate query to filter your data. In this example, the query will return all values in column B where the corresponding
value in column A is "X".
□ Replace 1 with the number of header rows in your data range. If your data has no header rows, you can omit this parameter or set it to 0.
4. Press Enter to execute the formula. The result of the SUM function with the QUERY function will be displayed in the selected cell.
Let's assume we have the following data in our Google Sheet:
Category Amount
A 10
B 20
A 30
C 40
B 50
C 60
Now, we want to calculate the sum of the Amount column for all rows where the Category is "B". Here's how we can use the SUM and QUERY functions together:
1. Click on an empty cell where you want to display the result (e.g., C8).
2. Type the formula: =SUM(QUERY(A1:B6, "SELECT B WHERE A = 'B'", 1))
3. Press Enter to execute the formula.
4. The result (70) will be displayed in cell C8, which is the sum of the Amount column for all rows with Category "B".
Did you find this useful? | {"url":"https://sheetscheat.com/google-sheets/how-to-use-the-sum-function-using-query-function","timestamp":"2024-11-11T13:22:04Z","content_type":"text/html","content_length":"15155","record_id":"<urn:uuid:b2775839-f998-4359-8a92-d52926c53d79>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00302.warc.gz"} |
What is Average Total Assets?
Average Total Assets
Average total assets is a financial metric that represents the average value of a company’s total assets during a specific period, usually a year or a quarter. This metric is often used in financial
analysis and accounting to calculate various performance ratios, such as return on assets (ROA).
To calculate the average total assets, you need to take the sum of the total assets at the beginning and the end of the period, and then divide by 2.
Here’s the formula for calculating average total assets:
\(\text{Average Total Assets} = \frac{\text{Beginning Total Assets + Ending Total Assets}}{2} \)
This calculation provides a more accurate representation of a company’s asset base during the period, which helps in analyzing the company’s efficiency and profitability more effectively.
Example of Average Total Assets
let’s say we have a company named XYZ Corp. We want to calculate the average total assets for the financial year 2022.
The company’s total assets at the beginning of the year (January 1, 2022) were $500,000, and at the end of the year (December 31, 2022), the total assets were $700,000.
Using the formula for average total assets, we can calculate as follows:
\(\text{Average Total Assets} = \frac{\text{Beginning Total Assets + Ending Total Assets}}{2} \)
\(\text{Average Total Assets} = \frac{500,000 + 700,000}{2} \)
\(\text{Average Total Assets} = \frac{1,200,000}{2} \)
\(\text{Average Total Assets} = 600,000 \)
So, the average total assets for XYZ Corp during the financial year 2022 were $600,000. This figure can be used in various financial analyses and ratios, such as return on assets (ROA) or asset
turnover, to assess the company’s efficiency and profitability during that period. | {"url":"https://www.superfastcpa.com/what-is-average-total-assets/","timestamp":"2024-11-09T07:01:44Z","content_type":"text/html","content_length":"394712","record_id":"<urn:uuid:f8cb6e61-190f-445a-ae81-3625530bf3af>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00833.warc.gz"} |
Price Per CWT CalculatorPrice Per CWT Calculator - Calculator Flares
Price Per Cwt Calculator
Basic Price Per Cwt Calculator
Enter the values to calculate the price per hundredweight (cwt):
Advanced Price Per Cwt Calculator
Enter the values to calculate the discounted price per hundredweight (cwt):
The Price Per CWT Calculator is an essential tool for anyone dealing with commodities, freight, and bulk goods. CWT, or hundredweight, is a common unit of measure used to standardize prices across
industries like agriculture, shipping, and wholesale markets. The Price Per CWT Calculator allows businesses and consumers to quickly determine the cost per hundredweight, helping to make accurate
and informed pricing decisions in various scenarios.
What is CWT?
CWT, or hundredweight, is a unit of measurement that equals 100 pounds in the United States and Canada. The term originates from an old English unit, which was used for weighing bulk goods like
grains and livestock. Today, CWT is widely used in many sectors, especially in commodities trading and shipping, to streamline pricing and ensure consistency.
Why CWT is Important in Pricing and Shipping
CWT simplifies the pricing process by offering a standard unit of measure. Instead of quoting prices based on total weight, businesses can quote per hundredweight, making it easier to compare costs
across different shipments or products. This method is particularly useful in LTL shipping (Less Than Truckload), where the cost efficiency of transporting smaller shipments is crucial.
How to Calculate Price Per CWT
To calculate the price per CWT, you need to know the total price and the total weight in pounds. The formula is straightforward:
For example, if a shipment costs $250 and weighs 150 pounds, the price per CWT would be calculated as follows:
Here’s the step-by-step solution:
1. Divide the total price by the weight: 250 ÷ 150 = 1.6667
2. Multiply by 100: 1.6667 × 100 = 166.67
So, the Price Per CWT is indeed $166.67.
Calculating CWT in Different Scenarios
Agricultural Commodities
In the agricultural sector, CWT is often used to price commodities like grain, livestock, and dairy products.
For example, if a farmer sells 1,000 pounds of wheat for $500, the price per CWT would be
Here’s the step-by-step solution:
1. Divide the total price by the weight: 500 ÷ 1000 = 0.5
2. Multiply by 100: 0.5 × 100 = 50
So, the Price Per CWT is indeed $50.
Freight and Shipping
For freight and LTL shipping, CWT helps determine the cost of shipping goods based on their weight. If you have a shipment weighing 2,000 pounds, and the shipping company charges $800, the price per
CWT would be:
Here’s the step-by-step solution:
1. Divide the total price by the weight: 800 ÷ 2000 = 0.4
2. Multiply by 100: 0.4 × 100 = 40
So, the Price Per CWT is indeed $40.
Wholesale Markets
In wholesale purchasing, CWT can standardize prices for bulk goods. Suppose a retailer buys 500 pounds of rice for $350. The price per CWT would be:
Here’s the step-by-step solution:
1. Divide the total price by the weight: 350 ÷ 500 = 0.7
2. Multiply by 100: 0.7 × 100 = 70
So, the Price Per CWT is indeed $70.
The Role of a CWT Calculator
Using a CWT calculator simplifies the process of determining the cost per hundredweight, especially when dealing with large quantities of goods. By inputting the total price and weight, the
calculator instantly provides the price per CWT, saving time and reducing the risk of errors.
Know the Different CWT Units
Short Hundredweight
In the United States, a short hundredweight equals 100 pounds. This is the standard unit used in most calculations involving CWT.
Long Hundredweight
In the United Kingdom, a long hundredweight equals 112 pounds. Although not as commonly used as the short hundredweight, it is still relevant in certain international trades.
Converting Between Units
To convert from one unit to another, you must adjust for the difference in weight. For instance, converting from a long hundredweight to a short hundredweight would involve multiplying by a factor of
0.892857 (since 100/112 = 0.892857).
Factors Affecting CWT Pricing
Market Conditions
The market plays a significant role in determining the price per CWT. Supply and demand, geopolitical factors, and trade policies can cause fluctuations in prices, especially for commodities like
grain and livestock.
Freight Costs
Shipping costs can vary based on factors such as distance, fuel prices, and the weight of the goods being transported. These variables directly influence the price per CWT in the freight industry.
Commodity Type
Different commodities have varying densities and values, affecting their price per CWT. For example, a high-value commodity like gold will have a much higher price per CWT than a low-value commodity
like coal.
Practical Applications of CWT in Shipping
LTL Shipping
LTL shipping is a cost-effective method for shipping goods that do not require a full truckload. The cost is often calculated using CWT to standardize prices across different shipments. This method
allows shippers to only pay for the portion of the truck they use, making it an efficient solution for small to medium-sized loads.
Full Truckload (FTL) Shipping
In FTL shipping, the entire truck is reserved for one shipment. Although CWT is less commonly used in FTL shipping, it can still provide valuable insights into the cost efficiency of transporting
goods, particularly when comparing different shipments.
How to Use a CWT Calculator for Accurate Pricing
A CWT calculator is a handy tool for quickly determining the price per CWT. To use it, simply enter the total weight and total price, and the calculator will do the rest. This tool is especially
useful for businesses that regularly deal with bulk shipments, as it helps standardize pricing and ensures accurate cost calculations.
Example Calculations Using the CWT Formula
Let’s look at a few more examples to reinforce the concept of CWT and its calculations:
Example 1: Shipping Agricultural Goods
A farmer ships 3,000 pounds of corn to a buyer for $1,200. To find the price per CWT, use the following calculation:
Here’s the step-by-step solution:
1. Divide the total price by the weight: 1200 ÷ 3000 = 0.4
2. Multiply by 100: 0.4 × 100 = 40
So, the Price Per CWT is indeed $40.
Example 2: Freight Shipping
A company ships 5,000 pounds of equipment, and the freight cost is $2,500. The price per CWT would be:
Here’s the step-by-step solution:
1. Divide the total price by the weight: 2500 ÷ 5000 = 0.5
2. Multiply by 100: 0.5 × 100 = 50
So, the Price Per CWT is indeed $50.
Example 3: Wholesale Purchasing
A retailer buys 2,000 pounds of sugar for $1,800. The price per CWT is calculated as follows:
Frequently Asked Questions About CWT
Q1: What does CWT stand for?
CWT stands for hundredweight, which is a unit of measurement equal to 100 pounds in the United States.
Q2: How do you calculate CWT?
To calculate CWT, divide the total weight of the goods in pounds by 100. This will give you the number of hundredweights. Then, to find the price per CWT, divide the total price by the number of
Q3: Why is CWT used in shipping?
CWT is used in shipping to standardize pricing, making it easier to compare costs across different shipments and commodities.
Q4: What is the difference between short hundredweight and long hundredweight?
A short hundredweight is 100 pounds, while a long hundredweight is 112 pounds. The short hundredweight is commonly used in the United States, while the long hundredweight is used in the United
Q5: How does CWT affect freight costs?
Freight costs are often calculated based on the CWT of the shipment. This allows for standardized pricing, especially in LTL shipping. | {"url":"https://calculatorflares.com/price-per-cwt-calculator/","timestamp":"2024-11-03T07:18:28Z","content_type":"text/html","content_length":"212210","record_id":"<urn:uuid:7d3defc0-bae0-42d5-81d0-5dcff017707d>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00515.warc.gz"} |
Which of the following points represents the x-intercepts of the quadratic equation y = x2 + 5x + 6?
Which of the following points represents the x-intercepts of the quadratic equation y = x2 + 5x + 6?
Solution 1
The x-intercepts of a quadratic equation are the values of x for which y = 0.
So, to find the x-intercepts of the equation y = x^2 + 5x + 6, we set y = 0 and solve for x:
0 = x^2 + 5x + 6
This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 5, and c = 6.
We can solve th Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem.
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Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions. | {"url":"https://knowee.ai/questions/81625652-which-of-the-following-points-represents-the-xintercepts-of-the","timestamp":"2024-11-14T07:34:14Z","content_type":"text/html","content_length":"365674","record_id":"<urn:uuid:bca94189-c0cd-4544-baa6-b85d67603613>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00645.warc.gz"} |
12. [Binomial Distribution (Bernoulli Trials)] | Probability | Educator.com
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while
watching the lecture.
Section 1: Probability by Counting
Experiments, Outcomes, Samples, Spaces, Events 59:30
Combining Events: Multiplication & Addition 1:02:47
Choices: Combinations & Permutations 56:03
Inclusion & Exclusion 43:40
Independence 46:09
Bayes' Rule 1:02:10
Section 2: Random Variables
Random Variables & Probability Distribution 38:21
Expected Value (Mean) 46:14
Variance & Standard Deviation 47:23
Markov's Inequality 26:45
Tchebysheff's Inequality 42:11
Section 3: Discrete Distributions
Binomial Distribution (Bernoulli Trials) 52:36
Geometric Distribution 52:50
Negative Binomial Distribution 51:39
Hypergeometric Distribution 36:27
Poisson Distribution 52:19
Section 4: Continuous Distributions
Density & Cumulative Distribution Functions 57:17
Mean & Variance for Continuous Distributions 36:18
Uniform Distribution 32:49
Normal (Gaussian) Distribution 1:03:54
Gamma Distribution (with Exponential & Chi-square) 1:08:27
Beta Distribution 52:45
Moment-Generating Functions 51:58
Section 5: Multivariate Distributions
Bivariate Density & Distribution Functions 50:52
Marginal Probability 42:38
Conditional Probability & Conditional Expectation 1:02:24
Independent Random Variables 51:39
Expected Value of a Function of Random Variables 37:07
Covariance, Correlation & Linear Functions 59:50
Section 6: Distributions of Functions of Random Variables
Distribution Functions 1:07:35
Transformations 1:00:16
Moment-Generating Functions 1:18:52
Order Statistics 1:04:56
Sampling from a Normal Distribution 1:00:07
The Central Limit Theorem 1:09:55
Hi and welcome back to the probability lectures here on www.educator.com, my name is Will Murray.0000
We are starting a chapter now of working through our discrete distributions.0006
The first we are going to study is the binomial distribution.0012
If you are studying Bernoulli trials in your probability course then you are using the binomial distribution.0020
These are synonymous terms for the same idea, let us learn what that idea is.0026
Before I give you the formulas for the binomial distribution, I want to tell you the general setting.0029
It is very important to be able to recognize this setting.0037
When you get some random problem, you have to figure out is this a binomial distribution, is this a geometric distribution?0040
Let me tell you the setting for the binomial distribution.0046
It describes a sequence of N independent tests, each one of which can have 2 outcomes.0049
You can think of running a test N times and each time you can either succeed or fail.0056
Every time there is 2 outcomes, that is why it is called binomial, success or failure.0061
It is also known as Bernoulli trials, as I mentioned.0066
The technical example of the binomial distribution or a Bernoulli trial is flipping a coin.0069
You want to think of flipping a coin exactly N times in a row.0075
By the way, N is always constant for the binomial distribution or for Bernoulli trials.0079
That is different from some of the distributions we are going to encounter later like the geometric distribution.0089
We want to think about it, like I said, flipping a coin is the prototypical example.0097
Although, that is somewhat limiting because people often think of the coin as being fair,0103
meaning it got a 50-50 chance of coming up heads or tails.0108
That certainly is a binomial distribution but you can also use a binomial distribution,0113
For example, if your coin is loaded, it is more likely to come up heads than tails, that is still a binomial distribution.0123
We will see how we adjust the formulas to account for that.0130
You can also think about any other kind of situation where you either have success or failure.0133
For example, one sports team is going to play another sports team and each time your home team will either win or lose.0139
If you say we are going to play the other team 15 × and each time we will win or lose.0149
At the end, we will have a string of wins and losses.0155
That is a binomial distribution, that is a set of Bernoulli trials.0158
You think of, wait a second, there are 6 different things that can happen when I roll a dice not just 2.0165
Suppose you are only interested in whether the dice comes up 6 or not.0171
If you roll a 6, it is a success and somebody pays you some money.0176
If you do not roll a 6, it is a failure.0180
1 through 5 essentially can sort of want those altogether and count those all as a single category of failure and rolling a 6 is a success.0182
Essentially, rolling a dice just boils down to you roll it, do I get a 6, do I not get a 6?0193
You can think of that as a set of Bernoulli trials.0201
The probability of success if it is a fair dice, there is just 1/6 and the probability of failure is 5/6.0204
There is all these different situations, they all come down to studying the binomial distribution.0213
Let us go ahead and take a look at the formulas that you get for the binomial distribution.0222
That is different from some of the other distributions that we are going to study later.0239
In particular, the geometric distribution that we will study in the very next video, that it is different from this,0243
and you keep flipping a coin until you get a head and that could take indefinitely long.0251
You say ahead of time that I’m guaranteed I'm going to flip this coin N times,0261
or we are going to play the other team N times.0265
It is fixed, it stays constant throughout the experiment and you know that ahead of time.0268
P is the probability of success on any given, looks like we got cutoff a little bit there.0276
If you are dealing with a fair coin then P would be ½.0288
If you are dealing with a sports team playing another sports team, it depends on the relative strength of the teams.0295
Maybe if your team is the underdog, maybe they only win 1/3 of the games then P would be 1/3.0302
That is your chance of winning any particular game with the other team.0311
Maybe, if you are rolling a dice and you are trying to get a 6 then your probability of getting a 6 would be 1 and 6.0316
Your probability of failure, now, we are going to call that Q but Q is not really very difficult to figure out0325
because that is the probability of failing on any given trial, that is just 1 – P.0334
The probability of failure is, we call it Q but sometimes we will swap back and forth and alternate between Q and 1 – P.0341
If you are flipping a coin and it is a fair coin then your probability of failing to get a head or getting a tail would be ½.0351
If you have a sports team and we said the probability of winning any given match is 1/3 because we are the underdogs here,0359
If we are rolling a dice and we are trying to get a 6, anything else is considered failure.0372
The probability of getting anything other than a 6 is 5 out of 6 there.0378
It is very easy to fill in Q, that is just 1 – P.0384
The formula for probability distribution, there are some terrible notation here and I do not like to use it0388
but it is kind of universal in the probability textbooks.0395
This P of Y here, what that represents is the probability of Y successes.0401
If you are flipping a coin N ×, this is the probability that you will get exactly Y heads.0424
If your sports team is going to play the other team N ×, this is the probability that you will win exactly Y games.0430
If you are going to roll a dice N ×, this is the probability that you will get exactly Y successes.0439
The formula that we have here is N choose Y, that is not a fraction there.0447
The other notation that we have for that is the binomial coefficient notation C of NY.0454
The actual way you calculate that is as N! ÷ Y! × N – Y!.0460
It is not just a fraction N/Y, that is really the formula for combinations.0470
The rest of this formula here, we have P ⁺Y and here is the really unpleasant part here.0483
This P right here is not the same as the P on the left hand side.0488
I said the P on the left hand side is the probability of exactly Y successes.0493
This P right here is this P right here, it is the probability of getting a success on any given trial.0499
It is this P up here, whatever the probability of successes on any given trial, that is what you fill in for this P.0510
That is the really unfortunate notation that you see with the binomial distribution,0518
is that they use P for 2 different things in the same formula.0523
I think that is really a high crime there but I was not given the choice to make up the notation myself.0527
I do not want to mislead you by using different notation from everybody else in the world.0536
Just be careful there that that P and that P are 2 different uses of the word P.0543
We end up using the variable P in many places.0552
Remember, we said Q is just 1 – P, that is easy to fill in and N – Y,0559
that is our formula, and then let us think about the range of values that Y could be.0564
If you are going to flip a coin N ×, how many heads could you get?0569
The fewest heads you can get would be 0, if you do not get any heads at all.0573
The most heads you can get would be N heads.0577
Our range of possibilities for Y is from 0 to N.0580
That is a formula we will be using over and over again.0585
There are another couple of issues we need to straighten out, before we jump into the examples.0589
The key properties of a binomial distribution and we will need to know these properties for every distribution we encounter.0596
The binomial distribution is just the first one, but we will be getting into the geometric distribution and the Poisson distribution,0603
For every single one, we want to know the mean, variance, and the standard deviation.0615
The mean, which is also known as the expected value means the exact same thing, mean and expected value.0622
We use the Greek letter μ for mean and we also say E of Y for expected value.0634
Remember, N is the number of trials and P is the probability of success on any given trial.0651
The variance, two different notations for that, V of Y and σ² is N × P × Q.0657
Sometimes you will see that written as N × P × 1 – P but they mean the same thing.0668
Standard deviation is always just the square root of the variance.0673
If you tell me the variance, I can always calculate the standard deviation very easily.0678
Just take the square root of the variance and that is the square root of N × P × Q.0685
We will be calculating those in some of our examples.0691
In our first example, we have the Los Angeles angels are going to play the Tasmanian devils in a 5 game series.0698
Maybe this is football or baseball, let us say baseball.0706
The angels have a 1/3 chance of winning any given game.0711
I guess the Tasmanian devils are bit stronger than the Angels.0714
What is the chance that the Angels will win exactly 3 games here?0718
Let me write down our general formula for the binomial distribution.0724
The general formula for binomial distribution is P of Y is equal to N choose Y, binomial coefficient there, combinations N choose Y.0730
The Y I’m interested in is 3 games because I want to find the chance that they are going to win exactly 3 games.0750
My Y will be 3, my P is the probability that the Angels will win any particular game and that is 1/3 there.0757
My N is the number of games that they are going to play in total.0769
It is a 5 game series, that is where I’m getting that from, N = 5 and Q is just 1 – P.0773
Q is 1 - P which is 2/3, 1 -1/3 there.0782
The probability of 3, probability of winning 3 games is 5 choose 3 N choose Y 5 choose 3 × 1/3³ × 2/3⁵ – Y.0788
Be careful here, it is rather seductive to get your binomial coefficients and your fractions mix up here0812
because we are mixing them both on the same formula.0820
The fractions are 1/3 and 2/3, that 5/3 is a binomial coefficient, it is not a fraction.0823
We want to expand that out as a binomial coefficient, 5!/3! × 2! × 1/3.0830
I got 1/3³ and then 1/3³ × 2/3⁵ – 3, that is 2/3².0843
5!/3! that means the 1, 2, 3, will cancel out.0856
Let us see, I’m going to have 3²/5 in the denominator here because I got 3 copies of 1/30866
and then 2 more here and 2² in the numerator.0874
5 × 4/2 that is 20/2 is 10 × 2²/3⁵.0880
10 × 2², 2² is 4, 10 × 4 is 40.0891
3⁵, 3 × 3 is 9 × 3 is 27 × 3 is 81 × 3 is 243.0896
That is my exact probability of winning exactly 3 games.0918
443 is just about 6 × 40 because 240 is 6 × 40.0923
This is very close to 1/6, if you want to make an approximation there.0929
The wraps that one up, let us just see how we solve that.0944
The probability of exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.0952
I’m going to fill in all the numbers that I know here.0961
My N came from the fact that it was a 5 game series.0963
My Y is the number of game that I want to win, that was the 3 here.0967
The 1/3 is the little P, that is the probability that I will win any particular game.0974
My Q is just 1 – P, that is 2/3.0983
I drop all those numbers in here, very careful, the 5 choose 3 is a binomial coefficient.0986
I simplify these fractions down while I’m simplifying down the binomial coefficient there.0999
Doing the arithmetic, it simplifies down to 40/243 which I noticed is approximately equal to 1/6.1007
That is my probability of winning exactly 3 games out of this 5 game series.1014
In example 2 here, we got a big exam coming up and we studied most of the material but not all of it.1023
In fact any given problem, we have a ¾ chance of getting that problem right.1031
Most likely, we will get a problem right but not guaranteed.1036
The exam that we are going to take is 10 problems and I guess we are really hoping to get an 80% score or better.1039
I would like to score 80% on this exam but I really only studied ¾ of the problems.1049
Remember that you use the binomial distribution, when you have a sequence of trials and each trial ends in success or failure.1060
We have 10 problems here, each problem we will do our best to solve it and will either succeed or fail.1071
It is exactly 10 problems, each one is N success or failure.1077
Let me go ahead and set up the generic formula for binomial distribution.1085
The probability of getting exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.1088
In this case, let me fill in here, while my N is the number of trials here.1102
That is the number of problems I will be struggling with, N is 10.1107
P is my probability of getting any particular problem right.1114
My Y is the number of successes that I would like to have.1124
In this case, I want to score 80% or better which means out of 10 problems, I got to get 8 of them, or 9 of them, or 10 of them right.1129
The Q, remember is always 1 – P, that is our chance of failure on any given problem.1148
1 – P, 1 -3/4 is ¼, if you give me a single problem, that is the chance I will not get it right.1155
Since, I need to find the probability of getting exactly 8, 9, or exactly 10 problems.1166
I will be adding up 3 different quantities here, P of 8.1173
I will give myself some space because I do not get a little bit messy.1177
+ P of 9 + P of getting exactly 10 problems right.1181
P of 8, just dropping Y=8 into this formula, is 10 choose 8, N was 10 × P ⁺Y.1190
P is ¾ ⁺Y is 8 × Q was ¼ ⁺N- Y.1200
N was 10 so N – 8 is 2 + P of 9 that is 10 choose 9 × ¾⁹ × ¼¹ + P of 10 is 10 choose 10 × 3/3 ⁺10 × ¼ ⁺10 – 10 which is 0.1208
I want to simplify that, these numbers are going to get a bit messy.1239
At some point, I’m going to throw out my hands in despair and just go to the calculator.1242
In particular, these binomial coefficients, I know how to simplify those.1250
Remember, you are not going to mix up the fractions.1254
This 10 choose 8, that is 10!/8!/10 -8! = 2!.1257
The 10! And 8! Cancel each other just leaving 10 × 9/2.1267
That is 45 there, 45 × 3⁸/4 ⁺10 because we have 8 factors of 3 and 8 factors of 4 and then 2 more factors of 4.1273
That is just 10!/9! which is all the factors are cancel except the 10 there.1301
There is only one way to choose 10 things out of 10 possibilities.1323
In case you want to confirm that to the formula, it is 10! ÷ 10! × 10 - 10 is 0!.1327
But 0! Is just 1, remember, so that is 1.1335
That is 1 × ¾ ⁺10, 3 ⁺10/4 ⁺10, that ¼⁰ is just 1, that does not do anything.1341
At this point, I do not think the numbers are going to get any nicer by trying to simplify them as fractions.1351
I’m going to go ahead and threw these numbers in, all these numbers to my calculator.1359
Let me show you what I got for each one of those.1363
For the first one, I got the 0.2816 +, in the second one I got 0.1877 +, in the last part I got 0.0563.1365
What these really represent these 3 numbers right now, represent your probabilities1383
of getting exactly 8 problems right, that is P of 8 right there.1388
The probability that you get exactly 8 problems right, you score exactly 80% on the test.1394
This is probability of getting 90% on the test so you got exactly 9 problems right.1399
This is your probability of getting 100%, getting all 10 problems right.1405
Not very likely, you got 5% chance every single test, if you are only ready with ¾ of the material going in.1408
If we add those up, 28 + 18 + 5 turns out to be, I did this on my calculator, 52.56 is approximately,1419
52.5%, I will round that up to 53%, that is your probability of getting 80% or more on this exam.1433
You studied ¾ of the material, your probability of getting 80% on the exam is 53%.1445
I started with the basic formula for the binomial distribution, here it is.1456
And then, I filled in all the quantities I know.1461
The N = 10 that come from the stem of the problem.1464
The P, the probability of getting any problem right is ¾, that also comes from the stem of the problem.1469
The Y that we are interested in, we want to get 80% or better, that means that we want to get 8 problems,1480
Because if you are shooting for 80% and if you end up getting 90 or 100, that certainly is acceptable.1491
The Q is always 1 – P, since P was ¾, Q was ¼.1500
We just drop those in for the different values of Y, the 8, 9, and 10.1506
We get these binomial coefficients and some fairly nasty fractions which I did not want to simplify by hand.1512
We sorted out the binomial coefficients into 45 and 10 and 1.1520
Each one of those multiplied by some fractions gave me some percentages,1525
the probabilities of getting 8 problems, 9 problems, 10 problems right.1530
We would put those all together and we get a total probability of 53%.1534
If you are shooting for an 80% on an exam and you study 3/4 of the material, your chance of getting that 80% is 53%.1540
You are likely get 80% but it is definitely not a sure thing.1552
Example 3, we are going to keep going with that same exam from example 2.1561
It is telling us that each problem is worth 10 points, what is your expected grade and your standard deviation?1565
As soon as you see the word expected in a probability problem, that does not mean the English meaning of the word expected.1575
That does not mean what grade you are going to get but that means on average, what your grade be if you take this exam many times.1586
It is asking for the expected value of your exam score.1599
We have learned the formula for the expected value of the binomial distribution.1605
That is the same as the mean of the binomial distribution.1610
The formula we learned this back on the third slide of this lecture, it is N × P.1613
In this case, the N was 10 and the P is the probability of you getting any particular problem right, that was ¾.1620
The expected value of Y, I’m using Y here to mean the number of problems that you get right.1629
We just figured out down below that that is 10 × ¾ is 7 ½.1651
That 7 ½ problems but each problem is 10 points.1658
That is 75 points on the exam, that is your expecting great.1667
Remember, I said that that is a technical term, that is you are expected grade.1675
In real life, there is no way you can get a 75 on the exam because all the problems are worth 10 points each.1681
In real life, when you take a single exam, you will have to get a multiple of 10.1690
You might get a 60% on the exam, a 70, 80, etc.1699
You will not get a 75 on the exam, I guarantee you because we are not talking about partial credit here.1707
Your actual score would be 60, 70, 80, or so on.1715
What I mean when I say that your expected grade is 75, what I mean is that if you take this exam many times,1721
or if you take many exams, your average over the long run will be 75 points.1729
Maybe, for example if you take 2 exams, your total on the 2 exams might be 150 which means you are averaging 75 points per exam,1752
even though you are not going to get exactly 75 points on any exams here.1761
That was your expecting grade, your standard deviation, a good steppingstone to calculating that is to find the variance first.1766
Let us find the variance of Y, variance of your score.1774
Variance, we learned the formula for that, it was also on the third slide of this video, NPQ.1778
Our N here is 10, our P is ¾, and our Q is ¼.1786
If we simplify that, we get 30/16 that is not extremely revealing at this point.1796
But remember, that was just the variance, that is not our standard deviation.1805
To get the standard deviation, you take the square root of the variance.1808
Our σ is the square root of the variance, that is always true.1813
It is √ 38/16 and I can simplify that a bit into √30 on top and √16 is just 4.1819
It does not really do anything good after that, I just threw it into a calculator.1833
What it came back with was that that is approximately equal to 1.369 problems.1839
Our unit here is the problem because Y was measured in the number of problems that we get right, 1.369 problems.1848
Our standard deviation in terms of points on the exam, that is equal to 13.69 points because each problem was worth 10 points.1857
Our σ there is 13, it is approximately equal to 13.69 points on the exam.1876
You can estimate your score on the exam, your expected grade would be 75 points.1890
Your standard deviation as you take many exams will be 13.69 points up above and below 75 points.1896
This really came back to remembering the formulas from the third slide that we had earlier on the videos.1909
If you do not remember those, just go back, check them out on the third slide of this video and you will see them.1915
N is 10, P is ¾, our Q is 1 - P is ¼.1931
That tells us the expected value and the variance, in terms of the problems on the exam1939
because we define our random variable in terms of the problems that we expect to get right.1946
To convert into actual points on the exam, we multiplied by 10 because each problem is worth 10 points.1953
7 1/2 problems I expect to get 7 1/2 problems right that means I expect on average to get 75 points on the exam.1960
I will never get exactly 75 because with 10 point problems, my score will definitely be some multiple of 10.1968
But on average, if I take many exams, I will get 75 points.1976
The variance, drop in those numbers I get 30/16, that is the variance not the standard deviation.1981
To get the standard deviation, you take the √ of that and that simplifies down into 1.369 problems.1988
Converting that to points gives me a standard deviation of 13.69 points on this exam.1996
In example 4, we are going watch the heralded Long Beach jack rabbits play and2008
they are going to be playing in the world pogo sticking championship.2015
Apparently, they are very good at this, as you expect jack rabbits to be.2019
Each year they have an 80% chance of winning the world championships.2025
They are obviously the dominant force in the world pogo sticking championships.2029
The question is we want to find the probability that they will win exactly 5 × in the next 7 years.2035
7 years of championship, they will play every year, each year they got 80% chance.2042
We also want to find the probability that they will win at least 5 × in the next 7 years.2048
I put an extra slide in here for us to work these out.2059
This is still the example of the long beach poly jack rabbits in their pogo sticking championship.2064
Why is this a Bernoulli trial, it is because we are playing 7 championships.2080
We have a probability of winning each year or losing each year.2087
Let me fill in the generic formula for Bernoulli trials.2091
P of getting exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.2095
The N here is the number of trials that we are doing here.2118
We are going to track this over 7 years so our N is 7.2122
P is our probability of success on any given trial.2129
That is the probability that the jack rabbits will win in any given year.2134
I will give that as a fraction, I will try to work this not using fractions, that is 4/5.2139
Q is always 1 – P so that is 1 - 4/5.2147
Finally, what is the Y value that we will be interested in here?2157
Our Y value, we want to win exactly 5 × for the first part of the problem.2162
We want to find the probability of exactly Y successes.2169
In the second part of the problem, we want to win at least 5 ×.2173
That means we could win 5 ×, we could win 6 ×, we could win 7 ×.2176
P of 5, when I plug in Y = 5 here, that is 7 choose 5 × P is 4/5 ⁺Y is 5 × Q is 1/5 ⁺N – Y, that 7 – 5 is 2.2185
Now, I just have to expand and simplify these fractions.2216
That is just 7 × 6 because the 5! takes care of all the other factors, ÷ 2!.2236
That is equal to, 7 × 6, 6/2 is 3, 7 × 3 is 21.2247
I still have 4⁵/5⁷ because there are five 5 in the first fraction and two 5 in the second fraction.2255
What will we get here is 21 × 4⁵ ÷ 5⁷.2274
That does not really turn out to be any particularly interesting numbers.2287
I just left that as a fraction, I did not bother to plug that into my calculator but that is our answer to part A.2291
That is the probability that the jack rabbits will come home with exactly 5 championships within the next 7 years.2298
In part B, we want to get at least 5 championships.2306
That means we really want to figure out the probability of getting 5 or 6, or7 championships.2311
We figured out the probability of 5 already, let us find the probability of 6.2319
We use a same formula except we put in Y equal 6, so 7 choose 6, 4/5⁶, 1/5⁷ -6 that is 1 there.2324
Let me go ahead and figure out P of 7, the probability of winning all 7 matches.2372
Although, it might be a little easier if you think about this directly but I want to practice the formula.2382
It is 7 choose 7 × 4/5⁷ × 1/5⁷ -7 which is 0.2386
We just have a 4/5⁷ because the 1/5⁰ is just 1.2407
It would have been easier to think about that as saying we have a 4/4 chance of winning.2421
In order to win 7 matches, we have to win all 7 × in a row.2428
That is probably an easier way to get there more directly but I just want to practice using the probability distribution formula.2438
The probability that we will win at least 5 matches, you just add up those 3 numbers.2447
The probability of 5 + the probability of 6 + the probability of 7.2455
The fractions are actually fun to workout here, I did work them out.2462
Also, in the numerator we got lots of 4 and we factor out 4⁵/5⁷.2488
The numerator, I still have a 21 + 7 × 4 because 4⁶ is 4⁵ × 4 + 4².2496
But 65 is 5 × 13, we cancel out one of those 5’s and we get 4⁵ × 13/5⁶2530
because one of the 5 was canceled with the 5 from the 65.2546
That is our probability of winning at least 5 games or 5 championships over the next 7 years.2556
This is kind of a classic binomial distribution problem, classic Bernoulli trials.2567
It does not have to be coin flipping even though people always talk about coin flipping.2572
In this case, it is a question of the Long Beach jack rabbits in the pogo sticking championships.2576
They will play for 7 years, that is why we have our N = 7.2593
The question was asking us what our chance of winning exactly 5 × is, that is where we get that 5.2599
And then later on, we want the probability of winning at least 5 ×.2607
We are going to calculate 6 and 7 as well.2611
What we are really doing here is dropping 5, and 6, and 7 in for Y into the binomial distribution formula.2614
I just dropped Y = 5, that 5 in the denominator came from the probability but that came from Y there and that 2 was 7 – Y.2623
And then, I simplified all the fractions and I got that is my probability of winning exactly 5 championships.2638
I found the probability of winning 6 championships the same way by running all the way through Y = 6, then I ran through Y=7.2646
There is an easier way to calculate that but I want to practice the binomial distribution formula.2655
Add all those together to find the probability of winning at least 5 × because it is at least 5 ×,2663
that is why you put a greater than or equal to.2669
We have a common denominator 5⁷ on all of these.2675
These numbers, you can do some nice factoring, factor out 4⁵ and simplify everything down2679
and you get still bit of a cumbersome number but that tell us our probability of winning at least 5 × in 7 years.2686
I want you to hang onto the numbers from this example because in example 5,2696
which we are about to do refers back to this example.2699
It is the same scenario with the long beach poly jack rabbits playing in the pogo sticking championship.2704
I think they are going to play for a different number of years but the probability will be the same.2711
Let us check that out but remember the numbers from this example.2716
In example 5, again, each year the Long Beach jackrabbit has an 80% chance of winning the world pogo sticking championship.2723
We want to find the expected number of championships that they will win in the next 5 years and the standard deviation in that total.2730
I’m going to use my generic formulas for expected value and variance, and standard deviation.2741
You can find those formulas on the third slide in this lecture.2748
If you do not remember those formulas, where they come from,2753
just check back in the third slide of this lecture and you will see the following formulas.2755
The expected value of Y is N × P, this is for the binomial distribution.2761
The standard deviation which is always just the square root of the variance.2774
That makes it in this case, the square root of NPQ.2781
We are going to calculate each one of those for this particular problem.2786
N, remember is the number of trials that we are running.2790
In the previous example, we are running this over 7 years but now we are just running it over 5 years.2793
I’m getting that from right here, that is my N, N = 5.2799
The P here is the probability of winning any particular year and we are given that that is 80%.2805
I’m not going to go ahead and figure out what Q is.2814
That is easy, that is 1 - 4/5 is 1/5.2821
I think that is all I need to know for this one.2827
The expected value, the expected number of championships that they will win over the next 5 years is N × P.2833
N is 5, P is 4/5, that is just 4 championships.2844
That of course should not be at all surprising because we are going to play for 5 years in a row.2853
We have an 80% chance of winning in any given year.2864
If we play for 5 years, we expect to win 4 out of 5 of those years on average.2871
That is very intuitive result there but it is good that it is backed up by the formulas because probability can sometimes be counterintuitive.2877
We are not being asked the variance in the problem but it is kind of a steppingstone to finding the standard deviation.2888
That is the variance, that is not our full answer yet.2910
To get our full answer, we want to find the standard deviation2913
which is the square root of the quantity that we found above, √NPQ, which is √4/5.2916
I could distribute that square root in the numerator and get 2/√5.2929
Not a very enlightening number, I did go ahead and plug that into my calculator.2934
What my calculator told me was that that is 0.894.2939
That is the standard deviation in the number of championships we expect to win over a 5 year span.2945
I will go ahead and box that up because that is our final answer there.2958
I got these formulas for expected value of variance and standard deviation straight off the third slide of this lecture series.2965
You can just go back and look at those formulas.2975
Make sure you are working with the binomial distribution before you involve those formulas.2980
This one is a binomial distribution because what is happening is that,2985
the Long Beach jackrabbits are playing the championship year after year.2990
You can think of that as being almost like flipping a coin,2997
except that it is not a 50-50 coin because the jackrabbits are dominant3000
that every year the coin has an 80% chance of coming up heads.3004
Every year, they have an 80% chance of winning and only 20% chance of losing.3008
The expected value for binomial distribution we said was NP.3015
Variance is NPQ, the standard deviation is always the square root of the variance, square root of NPQ.3020
I’m just dropping in the numbers and get the numbers from the stem of the problem.3027
N is the number of trials that you are going to run.3033
In this case, that is the number of possible championships.3036
We are going to play for 5 years, that N = 5 come from right there from the stem of the problem.3042
P = 4/5 that comes from their chance of winning in any given year.3049
The Q is always 1 – P, that is 1 – 4/5 is 1/5.3060
We just plot those numbers right into our formulas, the expected value is NP.3065
It simplifies down to 4 championships which makes perfect intuitive sense.3070
If you are going to play for 5 years and you got an 80% chance of winning each year,3075
The variance NPQ simplifies down to 4/5 but that is not what we want.3084
We want the standard deviation which is the square root of the variance, and that is 2 ÷ √5.3091
I just threw that number into my calculator and it spat out the number 0.8943100
is the standard deviation in the number of championships that we expect to win over any given 5 year span.3107
This is part of the probability lecture series here on www.educator.com.3119
I hope you will stick around and learn about that.3125
It looks a bit like the binomial distribution but there are certain key issues where it is different.3128
In particular, you are not running a fixed number of trials anymore, you are running the trials over and over until you get a success.3133
That turned out to change the probability distribution and it is going to change our mean and variance, and so on.3139
We will look at that out in the next lecture, I hope you will stick around for that.3145
In the meantime, as I said, these are the probability lectures here on www.educator.com.3149
My name is Will Murray, thank you very much for watching, bye.3154
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Did Singapore Airport Services Really Pay A High Price for Singapore Food? Don't be fooled!
TODAY published an article on 3 December 2008 on the sale of Singapore Food Industries to Singapore Airport Terminal Services ( SATS). . The Editor used the market capitalisation of Singapore Food to
state whether the sale price was at a discount or premium.So is market capitalisation a good measure of whether an acquisition is cheap or expensive?
Let's give an example, if 2 slaves each cost $4. ( just for simplicity sake lah...we hate slavery...don't be soooo uptight!). If one slave has a debt of $1 but another has a debt of $3. If a master
was to buy a slave and he has to pay back the debt owed by the slave, which is actually a more expensive buy? Of cos the slave who has a debt of $3 is the more expensive buy. On the same token, if
one slave has $1 dollar in his pocket and the other has $2 in the pocket. When you buy a slave, you get his dollar in the pocket. So in this instance, the slave who has $2 in the pocket is a cheaper
buy as you get to pocket the $2. So let's use this concept to measure whether the buy price of Singapore Food is cheap or not, shall we?It is actually a well used concept with a jargonic name called
Enterprice Value. [S:It is a measure of the theoretical takeover price that an investor would have to pay in order to acquire a particular firm.:S]
It is better intepreted as the
true cost of the acquistion in the market place
( See Opinions at the bottom of the article by CC for more insight. )Read
to find out more about Enterprise Value.
Enterprise Value ( EV) = Market Cap + Debt - Cash
Debt = $74,968,000 (ABOVE)
Cash = $17,428,000 (ABOVE)
Enterprise value = 516,261,702 X 0.89( Using the price given in the above TODAY article) + 74,968,000 - 17,428,000 = 517,012,915
Therefore, the theoretical cost of the acquisition if we based on market-determined price of 89 cents listed on the Singapore Stock exchange in TODAY's article should be $517,012,915. If we divide it
by the total number of shares = 517,012,915 / 516,261,702 = $1.0015 per share.
[S:Today stated that 93 cents is a 4.5% premium to the 89 cents of Singapore Food. At first glance, it seems Singapore Airport Terminal has paid a premium. But based on Enterprise value, it seemed
SATS did have a good deal after all, paying 93 cents instead of $1.0015, a 7% discount!:S]
The cost of the acquisition to SATs if they are able to acquire all the shares in the market place based on last done market- determined price of 89 cents on the Singapore Stock Exchange in the TODAY
paper above is therefore theoretically ( academically) actually $1.0015 per share. Mainstream media states 93 cents as the price paid per share.
But what is the true true true true true true true cost that it paid? Since SATs paid 93 cents. Based on the below calculation,
516,261,702 X 0.93+ 74,968,000 - 17,428,000 = 537663382.86
537663382.86/ 516,261,702 (shares)=$1.04 per share!!
(This article has been heavily edited after contributions from readers. SGDividends SUCKS!! )
Important: The objective of the articles in this blog is to set you thinking about the company before you invest your hard-earned money. Do not invest solely based on this article. Unlike House or
Instituitional Analysts who have to maintain relations with corporations due to investment banking relations, generating commissions,e.t.c, SGDividends say things as it is, factually. Unlike Analyst
who have to be "uptight" and "cheem", we make it simplified and cheapskate. -The Vigilante Investor, SGDividends Team
9 comments:
1. Does this makes sense or just the other way round?
Enterprise Value (EV) = Market Cap + Debt - Cash
So more debt, less cash means enterprise value increased?I think the term 'Enterprise Value' is misleading.
The equation is ofcos correct but its does not imply the 'value' of the firm. It merely represents the true cost of the acquisition.
SATS paid 93cts cash but after the net debt assumed the actual cost will be the $1.01 calculated.
Hence its incorrect to say that there is a discount of 7%. It should be the actual cost will be 8.6% higher.
2. Hi CC,
You know something. We thought about it and we think you make sense.
Enterprise value should be intepreted as the true cost of the acquisition and not as the value of the firm as for example:
Company A has $100 cash
Company B has $400 cash.
Ceteris Paribus.
It does not make sense that Company B is valued less than Company A since it has more cash. (Enterprise Value (EV) = Market Cap + Debt - Cash)
Therefore, taken in context, the cost of acquisition to SATs is actually $1.0015 instead of 93 cents.
Which means the cost to SATs is actually 0.0715/0.93 = 7.7% more than 93 cents.And they actually "paid" more.
Agreed. Will update the above.
Thanks CC =)
3. jus wondering......should EV be computed using the offer price 0.93 (actual acquisistion price) by SATS instead of 0.89 (mkt last done price) ??? if we are talking abt acquisition cost here ?
4. Hi Anonymous,
Enterprise Value (EV) = Market Cap + Debt - Cash
Are you refering to Market Cap?
Market Cap is using market price X number of oustanding share. This is considered what the general market values SingFood seen in the Singapore Exchange (SGX). This is based purely on academic
theory used by companies before they takeover a company to find the true cost of take over.So maybe SATS did this theoretical calculation first before. Maybe.
We know what you mean actually. Since SATS pay 0.93 per share and SATs still have to pay debt and after minusing off the cash, the real real cost to SATs is actually
516,261,702 X 0.93( Using the price given in the above TODAY article) + 74,968,000 - 17,428,000 = 537663382.86
537663382.86 / 516,261,702=1.04
This is the real real cost : $1.04.And the Economic value ( Cost of acquisition in the market place) = $1.0015.
5. This comment has been removed by the author.
6. Quote: "This article has been heavily edited after contributions from readers. SGDividends SUCKS!! "
Don't be that harsh lah... it's just different ways of looking at/interpreting the figures.
7. You are super funny la. :)
It's always enjoyable reading your posts.
Btw, I think SATS did make a good move in a strategic viewpoint. They have always been wanting to branch out for a few years already.
Paying a premium for a company which have tentacles extended outside is usually reasonable for acquisitions. Just only how much.
But as according to your post above, it seems that maybe they got a fair deal?
There are so many ways of valuating the business but what is more important is what SATS is going to do with the new acquisition in the long run.
8. hi Dancerene,
Its ok. We are used to self critism and we have low self esteem. Maybe cos we are small in built =)
Hi PassiveLifeIncome,
Thanks!Yeah, it just makes perfect sense for SATS to buy SingFood.
Anyway Singfood has been providing food services for Commercial ships. Now it will include Commercial planes!
9. Like to throw in the tale of BenQ buying Siemens mobile phone division for the grand price of -250 million euro (yes, the negative is not a typo) in 2005. The strategic goal of this deal is that
BenQ wants to compete with the big boys(eg. Nokia,Motorola) in the global market.
Siemens gave the division to BenQ along with that cash gift in exchange for the promise that BenQ keeps the factory and jobs in Germany. The total debts and liabilities nearly bankrupted BenQ 1
year later...and Siemens mobile phones became collectibles and a footnote in history.
Of course, providing packaged food and mobile phones are very different business with different cost structure. But still amusing to think about that deal in terms of enterprise value and price
per share.
Good article and thanks for highlighting this real real cost...in this era of media bombarding us with easily digestible bottom-lines and ratios.
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seislet transform
Signal and noise separation in prestack seismic data using velocity-dependent seislet transform
Next: Introduction Up: Reproducible Documents
Published as Geophysics, 80, no. 6, WD117-WD128, (2015)
Signal and noise separation in prestack seismic data using velocity-dependent seislet transform
Yang Liu^, Sergey Fomel^, Cai Liu^
^College of Geo-exploration Science and Technology,
Jilin University
No.938 Xi minzhu street,
Changchun, China, 130026
^Bureau of Economic Geology,
John A. and Katherine G. Jackson School of Geosciences
The University of Texas at Austin
University Station, Box X
Austin, TX, USA, 78713-8924
The seislet transform is a wavelet-like transform that analyzes seismic data by following varying slopes of seismic events across different scales and provides a multiscale orthogonal basis for
seismic data. It generalizes the discrete wavelet transform (DWT) in the sense that DWT in the lateral direction is simply the seislet transform with a zero slope. Our earlier work used plane-wave
destruction (PWD) to estimate smoothly varying slopes. However, PWD operator can be sensitive to strong noise interference, which makes the seislet transform based on PWD (PWD-seislet transform)
occasionally fail in providing a sparse multiscale representation for seismic field data. We adopt a new velocity-dependent (VD) formulation of the seislet transform, where the normal moveout
equation serves as a bridge between local slope patterns and conventional moveout parameters in the common-midpoint (CMP) domain. The velocity-dependent (VD) slope has better resistance to strong
random noise, which indicates the potential of VD seislets for random noise attenuation under 1D earth assumption. Different slope patterns for primaries and multiples further enable a VD-seislet
frame to separate primaries from multiples when the velocity models of primaries and multiples are well disjoint. Results of applying the method to synthetic and field-data examples demonstrate that
the VD-seislet transform can help in eliminating strong random noise. Synthetic and field-data tests also show the effectiveness of the VD-seislet frame for separation of primaries and pegleg
multiples of different orders.
Signal and noise separation in prestack seismic data using velocity-dependent seislet transform
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Mortgage, Loan & Basic, Interest Formula
A mortgage is a type of loan used to finance the purchase of real estate. When you take out a mortgage, you borrow money from a lender, typically a bank or a mortgage company, to buy a home or other
property. In exchange for lending you the money, the lender typically charges interest, which is a fee calculated as a percentage of the loan amount, and may also require you to provide collateral,
such as the property itself.
This is the total amount of money you borrow from the lender to purchase the property. This is the percentage charged by the lender for borrowing the money. It determines the cost of borrowing and
affects your monthly mortgage payments. The term of the mortgage is the length of time over which you agree to repay the loan. Common mortgage terms are 15, 20, or 30 years, though other terms are
This is the amount you pay each month to the lender, which typically includes both principal and interest. The monthly payment amount is determined based on the loan amount, interest rate, and term
of the loan. This is the initial upfront payment you make toward the purchase price of the property. It is usually expressed as a percentage of the total purchase price, with common down payment
amounts ranging from 3% to 20% or more.
If you fail to make your mortgage payments as agreed, the lender may take steps to foreclose on the property, which means they can seize the property and sell it to recover the remaining balance of
the loan. It’s important to carefully consider your financial situation and ability to repay before taking out a mortgage.
Mortgage loan basics:
A mortgage loan is a type of loan specifically used to purchase real estate, such as a home or land. It allows individuals or families to borrow money from a lender to buy property, with the property
itself serving as collateral for the loan.
Types of Mortgages:
1. Fixed-Rate Mortgage:
With a fixed-rate mortgage, the interest rate remains constant for the entire term of the loan. This means your monthly payments remain the same, providing predictability and stability.
2. Adjustable-Rate Mortgage (ARM):
An ARM typically offers a lower initial interest rate for a set period, after which the rate adjusts periodically based on market conditions. This can result in fluctuations in monthly payments.
3. Government-Backed Mortgages:
These are loans insured or guaranteed by government agencies such as the Federal Housing Administration (FHA) or the Department of Veterans Affairs (VA). They often have more lenient eligibility
4. Jumbo Loans:
These are loans that exceed the conforming loan limits set by government-sponsored entities like Fannie Mae and Freddie Mac. They’re typically used for high-priced properties and may have stricter
5. Loan Amount and Down Payment:
The loan amount is the total amount of money borrowed from the lender to purchase the property. The down payment is the initial payment made by the buyer towards the purchase price. It’s typically
expressed as a percentage of the total purchase price. A higher down payment can result in a lower loan amount and potentially lower monthly payments.
6. Interest Rate and Term:
The interest rate is the percentage charged by the lender for borrowing the money. It can be fixed or adjustable. The term of the mortgage is the length of time over which you agree to repay the
loan. Common terms include 15, 20, or 30 years.
7. Monthly Payments and Amortization:
Monthly payments typically include both principal (the amount borrowed) and interest (the cost of borrowing). They’re calculated based on the loan amount, interest rate, and term. An amortization
schedule outlines how your payments are applied to principal and interest over the life of the loan. Initially, a larger portion of each payment goes towards interest, but over time, more goes
towards principal.
8. Closing Costs:
These are fees and expenses associated with finalizing the mortgage loan and transferring ownership of the property. They include appraisal fees, loan origination fees, title insurance, and more.
अनुप्रास अलंकार के 50 उदाहरण विवरण सहित हिंदी में
Standard or conforming mortgages:
Standard or conforming mortgages refer to loans that meet the criteria set by government-sponsored enterprises (GSEs) such as Fannie Mae and Freddie Mac. These criteria include specific loan limits,
down payment requirements, and underwriting guidelines. Conforming mortgages are attractive to lenders because they can be sold to Fannie Mae or Freddie Mac in the secondary mortgage market, which
allows lenders to free up capital to make more loans.
Fannie Mae and Freddie Mac set loan limits each year, which vary by location. These limits dictate the maximum amount borrowers can borrow while still qualifying for conforming loan status. Loan
limits are typically higher in areas with higher housing costs. Conforming mortgages often require a down payment, which is a percentage of the home’s purchase price paid upfront by the borrower.
While down payment requirements can vary, they typically range from 3% to 20% of the home’s purchase price.
Borrowers must meet certain credit score and income requirements to qualify for a conforming mortgage. Lenders typically evaluate factors such as credit history, debt-to-income ratio, employment
history, and assets. Conforming mortgages often have competitive interest rates compared to other types of loans. These rates can be fixed or adjustable and may vary based on factors such as credit
score, down payment amount, and loan term.
Borrowers who make a down payment of less than 20% on a conforming mortgage may be required to pay private mortgage insurance (PMI). PMI protects the lender in case the borrower defaults on the loan
and typically adds an additional cost to the monthly mortgage payment. Conforming mortgages can have various features, including fixed-rate or adjustable-rate options, different loan terms (e.g., 15
years, 30 years), and options for government-backed mortgage insurance.
Overall, conforming mortgages offer borrowers a standardized and regulated way to finance their home purchases. By adhering to the guidelines set by Fannie Mae and Freddie Mac, lenders can provide
borrowers with competitive interest rates and terms, making homeownership more accessible to a broader range of individuals and families.
Mortgage Principal and interest:
Principal and interest are two components of your mortgage payment:
Principal: The principal is the amount of money you initially borrowed from the lender to purchase your home. Each month, a portion of your mortgage payment goes towards paying down the principal
balance. Over time, as you make payments, the amount of principal you owe decreases.
Interest: Interest is the cost of borrowing money from the lender. It’s calculated as a percentage of the remaining principal balance. In the early years of your mortgage, a significant portion of
your monthly payment goes towards paying interest. However, as you continue making payments, the amount of interest you pay decreases, while the amount going towards the principal increases.
An amortization schedule is typically worked out taking the principal left at the end of each month, multiplying by the monthly rate and then subtracting the monthly payment. This is typically
generated by an amortization calculator using the following formula:
is the periodic amortization payment
is the principal amount borrowed
is the rate of interest expressed as a fraction; for a monthly payment, take the (Annual Rate)/12
is the number of payments; for monthly payments over 30 years, 12 months x 30 years = 360 payments.
When you make a mortgage payment, the total amount is typically divided between principal and interest, with any additional amounts going towards taxes, insurance, and possibly private mortgage
insurance (PMI) or homeowners association (HOA) fees, depending on your loan terms and escrow arrangements.
The proportion of your payment that goes towards principal and interest changes over time, a concept known as amortization. Initially, more of your payment goes towards interest, but as you pay down
the principal balance, a greater portion of your payment goes towards reducing the amount you owe.
It’s important to note that the total monthly mortgage payment (including principal and interest) is determined by factors such as the loan amount, interest rate, and term of the loan. You can use a
mortgage calculator to estimate your monthly principal and interest payments based on these factors.
Read More: Mortgage, Loan & Basic, Interest Formula
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Computer Science
AQA Computer Science GCSE
Programming Concepts - Mathematical Operators
Basic mathematical operators are addition (+), subtraction (-), multiplication (*) and division (/). BIDMAS rules apply when you're using these, so take care, especially when dealing with
multiplication and division.
Basic Mathematical Operators - notes from class
You then need to add in MOD and DIV - two specialist forms of division.
All Mathematical Operators - a summary with DIV and MOD added
Some exam style questions dealing with operators and trace tables. It would be a good idea to code these questions in Python as well.
Question 0 - a simple introduction
Question 2 - a bit trickier (particularly the last part) | {"url":"http://bluesquarething.co.uk/aqacs/csunit02/prog5ops.htm","timestamp":"2024-11-15T03:22:49Z","content_type":"text/html","content_length":"7247","record_id":"<urn:uuid:dc32523d-4c81-49b5-ac2c-492811fad0c0>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00226.warc.gz"} |
Lesson 3
Patterns in Multiplication
Warm-up: Choral Count: $\frac{1}{4}$ and $\frac{1}{8}$ (10 minutes)
The purpose of this Choral Count is to invite students to practice counting by a unit fraction and notice patterns in the count. These understandings will be helpful later in this lesson when
students recognize every fraction can be written as the product of a whole number and unit fraction.
• “Count by \(\frac{1}{4}\), starting at 0.”
• Record as students count.
• Stop counting and recording at \(\frac{11}{4}\).
• Repeat with \(\frac{1}{8}\).
• Stop counting and recording at \(\frac{15}{8}\).
Activity Synthesis
• “What patterns do you notice?” (In both counts, the numerators go up by 1, and denominators stay the same.)
• “How many groups of \(\frac{1}{4}\) do we have?” (11)
• “Where do you see them?” (Each count represents a new group of \(\frac{1}{4}\).)
• “How might we represent 11 groups of \(\frac{1}{4}\) with an expression?” (\(11 \times \frac{1}{4}\))
• “How many groups of \(\frac{1}{8}\) do we have?” (15)
• “How might we represent 15 groups of \(\frac{1}{8}\) with an expression?” (\(15 \times \frac{1}{8}\))
• “How would our count change if we counted by \(\frac{2}{4}\) or \(\frac{2}{8}\)?” (Each numerator would be a multiple of 2 or an even number.)
Activity 1: Describe the Pattern (15 minutes)
Students may have previously noticed a connection between the whole number in a given multiplication expression and the numerator of the fraction that is the resulting product. In this activity, they
formalize that observation. Students reason repeatedly about the product of a whole number and a unit fraction, observe regularity in the value of the product, and generalize that the numerator in
the product is the same as the whole-number factor (MP8).
Representation: Develop Language and Symbols. Provide students with access to a chart that shows definitions and examples of the terms that will help them articulate the patterns they see, including
whole number, fraction, numerator, denominator, unit fraction, and product.
Supports accessibility for: Language, Memory
• Groups of 2
• “Work with your partner to complete the tables. One person should start with Set A and the other with Set B.”
• “Afterwards, analyze your completed tables together and look for patterns.”
• 5–7 minutes: partner work time on the first two problems
• Monitor for the language students use to explain patterns:
□ The whole number in each expression is only being multiplied by the numerator of each fraction.
□ Language describing patterns in the denominator of the product (The denominator in the product is the same as the unit fraction each time.)
□ “Groups of” language to justify or explain patterns (The number of groups of each unit fraction is going up each time because it is one more group.)
• “Pause after you've described the patterns in the second problem.”
• Select 1–2 students to share the patterns they observed.
• “Now let's apply the patterns you noticed to complete the last two problems.”
• 3 minutes: independent or partner work time
Student Facing
1. Here are two tables with expressions. Find the value of each expression. Use a diagram if you find it helpful.
Leave the last two rows of each table blank for now.
Set A
│ expression │value│
│\(1 \times \frac{1}{8}\) │ │
│\(2 \times \frac{1}{8}\) │ │
│\(3 \times \frac{1}{8}\) │ │
│\(4 \times \frac{1}{8}\) │ │
│\(5 \times \frac{1}{8}\) │ │
│\(6 \times \frac{1}{8}\) │ │
Set B
│ expression │value│
│\(2 \times \frac{1}{3}\) │ │
│\(2 \times \frac{1}{4}\) │ │
│\(2 \times \frac{1}{5}\) │ │
│\(2 \times \frac{1}{6}\) │ │
│\(2 \times \frac{1}{7}\) │ │
│\(2 \times \frac{1}{8}\) │ │
2. Study your completed tables. What patterns do you see in how the expressions and values are related?
3. In the last two rows of the table of Set A, write \(\frac{11}{8}\) and \(\frac{13}{8}\) in the “value” column. Write the expressions with that value.
4. In the last two rows of the table of Set B, write \(\frac{2}{12}\) and \(\frac{2}{15}\) in the “value” column. Write the expressions with that value.
Activity Synthesis
• “How did you use the patterns to write expressions for \(\frac{11}{8}, \frac{13}{8}, \frac{2}{12}\), and \(\frac{2}{15}\)?” (I knew that each expression had a whole number and a unit fraction.
The whole number is the same as the numerator of the product.)
• Select students to share their multiplication expressions for these four fractions.
• “Can you write any fraction as a multiplication expression using its unit fraction?” (Yes, because the numerator is the number of groups and the denominator represents the size of each group.)
• “What would it look like to write \(\frac{3}{10}\) as a multiplication expression using a whole number and a unit fraction?” (\(\frac{3}{10} = 3\times \frac{1}{10}\))
Activity 2: What's Missing? (20 minutes)
This activity serves two main purposes. The first is to allow students to apply their understanding that the result of \(a \times \frac{1}{b}\) is \(\frac{a}{b}\). The second is for students to
reinforce the idea that any non-unit fraction can be viewed in terms of equal groups of a unit fraction and expressed as a product of a whole number and a unit fraction.
The activity uses a “carousel” structure in which students complete a rotation of steps. Each student writes a non-unit fraction for their group mates to represent in terms of equal groups, using a
diagram, and as a multiplication expression. The author of each fraction then verifies that the representations by others indeed show the written fraction. As students discuss and justify their
decisions they create viable arguments and critiqe one another’s reasoning (MP3).
MLR8 Discussion Supports. Display sentence frames to support small-group discussion after checking their fraction diagram and equation: “I agree because . . .”, “I disagree because . . . .”
Advances: Conversing
• Groups of 3
• “Let's now use the patterns we saw earlier to write some true equations showing multiplication of a whole number and a fraction.”
• 3 minutes: independent work time on the first set of problems
• 2 minutes: group discussion
• Select students to explain how they reasoned about the missing numbers in the equations.
• If not mentioned in students' explanations, emphasize that: “We can interpret \(\frac{5}{10}\) as 5 groups of \(\frac{1}{10}\), \(\frac{8}{6}\) as 8 groups of \(\frac{1}{6}\), and so on.”
• “In an earlier activity, we found that we can write any fraction as a multiplication of a whole number and a unit fraction. You'll now show that this is the case using fractions written by your
group mates.”
• Demonstrate the 4 steps of the carousel using \(\frac{7}{4}\) for the first step.
• Read each step aloud and complete a practice round as a class.
• “What questions do you have about the task before you begin?”
• 5–7 minutes: group work time
Student Facing
1. Use the patterns you observed earlier to complete each equation so that it’s true.
1. \(5 \times \frac{1}{10} = \underline{\hspace{0.5in}}\)
2. \(8 \times \frac{1}{6} = \underline{\hspace{0.5in}}\)
3. \(4 \times \underline{\hspace{0.5in}} = \frac{4}{5}\)
4. \(6 \times \underline{\hspace{0.5in}} = \frac{6}{10}\)
5. \( \underline{\hspace{0.5in}} \times \frac{1}{4}= \frac{3}{4}\)
6. \( \underline{\hspace{0.5in}} \times \frac{1}{12}= \frac{7}{12}\)
2. Your teacher will give you a sheet of paper. Work with your group of 3 and complete these steps on the paper. After each step, pass your paper to your right.
□ Step 1: Write a fraction with a numerator other than 1 and a denominator no greater than 12.
□ Step 2: Write the fraction you received as a product of a whole number and a unit fraction.
□ Step 3: Draw a diagram to represent the expression you just received.
□ Step 4: Collect your original paper. If you think the work is correct, explain why the expression and the diagram both represent the fraction that you wrote. If not, discuss what revisions
are needed.
Lesson Synthesis
“Today we looked at two sets of multiplication expressions. In the first set, the number of groups changed while the unit fraction stayed the same. We found a pattern in their values.”
“Then we looked at expressions in which the unit fraction changed and the number of groups stayed the same. We found a pattern there as well.”
Display the two tables that students completed in the first activity.
“In the first table, why does it make sense that the numerator in the product is the same number as the whole-number factor?” (Because there are as many groups of \(\frac{1}{8}\) as the whole-number
“In the second table, why does it make sense that the numerator in the product is always 2?” (Because all the expressions represent 2 groups of a unit fraction.)
“We also discussed how we could write any fraction as a product of a whole number and unit fraction. Tell a partner about how we could write \(\frac{8}{3}\) as a product of a whole number and a
fraction.” (\(\frac{8}{3} =8 \times \frac{1}{3}\))
Cool-down: Fraction Multiplication (5 minutes) | {"url":"https://curriculum.illustrativemathematics.org/k5/teachers/grade-4/unit-3/lesson-3/lesson.html","timestamp":"2024-11-07T22:07:32Z","content_type":"text/html","content_length":"86487","record_id":"<urn:uuid:3228546e-55c2-4218-8de7-294ca75b7020>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00773.warc.gz"} |
Mean-Variance Portfolio Optimization - QuantStrategy.io - blog
An investor is offered different assets or securities in a given market, each with a varied rate of return depending on the underlying risk.
How to choose an asset, a set of various assets, or a security investment that best optimizes risk and return on investment while maximizing expected utility is a practical concern for rational
investors. But we can get around this problem by using portfolio optimization with the mean-variance method.
In this post, we’ll explore the concept of portfolio optimization and demonstrate how to create some standard portfolios for portfolio optimization using the mean-variance approach.
What is Portfolio Optimization?
Let’s first understand what portfolio optimization is. Portfolio optimization in investing refers to the process of choosing assets in a way that maximizes return while minimizing risk. For example,
to ensure they make the highest possible return, an investor might be interested in choosing five stocks from a list of twenty. Private equity investments can be managed and diversified with the help
of portfolio optimization techniques. Investing in Bitcoin and Ethereum, among other cryptocurrencies, has recently benefited from the application of portfolio optimization strategies.
The goal of asset optimization in each of these scenarios is to strike a balance between risk and return, with return on a stock being the profits realized over time and risk being the standard
deviation in the value of an asset. Numerous portfolio optimization techniques are simply advancements of asset diversification techniques used in investing. The key concept behind this is that
diversifying your portfolio o,f assets rather ,than keeping them in the same way reduces your risk.
What is Mean-Variance Optimization?
Investment banks and asset management companies put a lot of effort into finding the best techniques for portfolio optimization. One of the original techniques is known as mean-variance optimization,
often known as the Markowitz Method or the HM technique because it was created by Harry Markowitz. It entails assessing an asset’s risk to its expected return and making investments based on that
risk/return ratio. Here’s how it functions.
Investors can use a mean-variance analysis as a technique to help disperse risk across their portfolios. In it, the investor calculates the risk of an asset, which is represented by the “variance,”
and then compares it with the asset’s expected return. The objective of a mean-variance optimization is to maximize return based on the investment’s risk.
Investors employ mean-variance analysis to make investment decisions. They evaluate how much risk they are ready to accept in exchange for various degrees of profit. They can use mean-variance
analysis to determine which investment offers the best return at a particular level of risk or the lowest risk at a particular level of return. They can then investbased of that information. This
enables them to diversify their portfolios across various risk levels and choose investments depending on desired returns.
Understanding Mean-Variance Portfolio Optimization
The Mean-Variance Portfolio Theory (MPT)
Mean-Variance Portfolio Theory, often called The Modern Portfolio Theory (MPT), was developed by Harry Markowitz in 1952. Investors can use the key concepts presented in the theory as practical
guidelines for building investment portfolios to ensure the highest expected return given a specific amount of risk.
Assumptions of Modern Portfolio Theory
The following are some fundamental presumptions upon which the MPT is based.
1. Markets are open and transparent, and investors may obtain all information about the expected return, variances, and covariances of securities or other assets.
2. Investors tend to avoid unneeded risks since they are risk-averse. For example, investors like bank deposits over equities since the latter may offer higher returns but come with a high risk of
losing money. Bank deposits, however, pay lower returns but are guaranteed.
3. Investors are not satisfied, thus they would choose the security with the higher predicted return if presented with two options with the same standard deviation.
4. Any amount can be held as an asset.
5. A fixed single-time horizon exists.
6. There are no transaction fees or taxes.
7. Investors based onions exclusively on the basis of variance and expected returns.
8. There is a risk-free rate of return in the market, and unlimited cash can be borrowed or invested at this rate.
Now that we have learned our working hypothesis, we may now start to comprehend the intuition behind MPT. To get started, consider a portfolio that has two equities, A and B.
We can observe from the image above that Stock A has lower expected returns and lower risk than Stock B. Before MPT, it was supposed that these two stocks could be combined in any number of linear
Supposing that there’s a linear relationship among our stocks, all of our potential portfolio combinations should, in theory, fall somewhere along the dotted line in the accompanying image. We would
be mistaken, in Markowitz’s opinion. According to Markowitz’s theory, the relationship between our stocks has a curved form known as the efficient frontier rather than being linear. So, our potential
portfolio combinations are accurately represented by the efficient frontier.
You might now be wondering how this is feasible. All of this efficient frontier is possible by correlation. In order to better understand how correlation gives us the curved shape that represents the
portfolio configurations, we must first comprehend how our portfolio returns and portfolio risk are determined.
Portfolio Returns & Risk
In a nutshell, the expected return of the entire portfolio is equal to the proportional sum of the expected returns of the individual assets.
Wi stands for the ith asset of the portfolio
E(ri) denotes the expected return on i-th asset of the portfolio
This computation of the portfolio’s expected return is based on the same reasoning that leads us to conclude that the relationship between the two stocks is linear. This relationship becomes
non-linear when we calculate the risk of our portfolio.
Let us determine the portfolio’s risk, which also represents the variance of the portfolio.
ρ(A, B) denotes the correlation between the stocks
σ stands for the standard deviation of each stock
Now that we understand how the correlation between the two stocks affects our risk equation, we can comprehend why our potential portfolio combinations have an efficient frontier structure.
Our two assets’ correlation, which is expressed as a value between -1 and 1, reveals the relationship between our two stocks. The returns of the two assets often move in the same direction, and vice
versa, if our correlation is positive. This is how you would compute the correlation between Stock A and Stock B:
Where ρ(A , B) denotes the covariance among the A and B stocks.
In case,e of low or negative correlations they lower the overall portfolio risk while having no impact on the expected return of our portfolio, as can be shown from our calculation for portfolio risk
above. Let’s return to our example for a better understanding.
Three distinct situations are shown in the figure above. The highest decrease in the portfolio’s risk occurs when there is a -1 correlation between the two stocks. The risk of our portfolio also
rises as a result of the rising stock correlation.
Although this was meant to be a brief introduction to modern portfolio theory, we hope it has given you a basic grasp of the subject so you can learn more about it. The remainder of this blog article
will focus on using the PortfolioLab library to implement this portfolio optimization strategy into practice.
Mean-Variance Optimization with PortfolioLab
In this part, we’ll demonstrate how customers can employ a variety of mean-variance optimization (MVO) techniques offered by the PortfolioLab Python library to improve their portfolios. Download the
required library to carry out tthe optimization process, such as pyPortfolioOpt library or PortfolioLab. We will be using PortfolioLab for this purpose.
Strategies based on Harry Markowtiz’s methodologies for computing effective frontier solutions are used in PortfolioLab’s mean-variance optimization class. Users can create the best portfolio
solutions for a variety of objective functions using the PortfolioLab library, such as:
The Data
We’ll be using the previous closing prices for 17 assets in this lesson. The portfolio is made up of a varied range of assets, from bonds to commodities, and each asset has a unique risk-return
Producing a few Sample Portfolios
First, we will practice creating an optimal portfolio for the inverse-variance solution string. It’s one of the most straightforward yet effective allocation strategies, outperforming several
challenging optimization goals.
We need to make a new MeanVarianceOptimisation object and employ the allocate method to determine the asset weights of the inverse-variance portfolio.
Keep in mind that the allocate method needs the following three parameters to function:
1. asset_names (a collection of strings containing the names of the assets).
2. asset_prices (a data frame containing historical asset prices – daily close)
3. Solution (the kind of solution or algorithm to employ in the weight calculation)
Users can also give expected asset returns and a covariance matrix of asset returns in place of historical asset prices. The allocation mechanism also allows for much more customization. We’ll make
it simple for consumers by demonstrating how to build an optimal portfolio using the Inverse Variance portfolio solution. The only thing that needs to be modified to produce a different optimized
portfolio is the value of the “solution” parameter string.
Let’s construct one more portfolio — the highest Sharpe Ratio portfolio. The tangency portfolio may also be used to refer to this.
Custom Input for Users
While many of the calculations needed to construct optimal portfolios are provided by PortfolioLab, users also have the option to give input for their calculations. They can input a covariance matrix
of asset returns and the predicted asset returns in place of supplying the raw historical closing prices for their assets to create their optimal portfolio. You can consult the official documentation
if you want additional information regarding the customizability within PortfolioLab’s MVO implementation.
The allocate method uses the subsequent parameters to accomplish this:
• ‘covariance_matrix’: (pd.DataFrame/NumPy matrix) a covariance matrix of returns of an asset.
• ‘expected_asset_returns: (list) a lof average asset returns
We will use the ReturnsEstimators class offered by PortfolioLab to perform some of the essential calculations.
We also determine the anticipated returns or mean asset returns.
We compute all the necessary input matrices and use these custom inputs to build an effective risk portfolio. The fundamental goal ofan efficient risk portfolio is effective risk management for the
investor’s specified target return. In this case, the target return is set to 0.2.
Additionally, many traders and portfolio managers choose to give particular assets in their portfolio a defined weight limit. For example, one would desire to restrict the weights given to French-2Y
and US-30Y in the portfolio mentioned above. Let’s attempt to reduce their allocations and slightly diversify the total portfolio. We set a minimum bound on French-5Y and French-10Y and a maximum
bound on their weights. Keep in mind that the indexing begins at 0.
You can observe how the weight bounds assist us in creating a more diverse portfolio and prevent the weights from getting overly centered on a few assets.
The Capital Market Line
The tangent of the efficient frontier is the capital market line. The capital market line (CML) takes its shape from the Capital Asset Pricing Model, which is based on the Modern Portfolio Theory.
The capital market line represents the portfolio’s projected return as a linear of the risk-free rate, the market portfolio’s standard deviation and return, and the portfolio’s standard deviation.
On the efficient frontier, the market portfolio is situated where the line from the risk-free asset (where risk is zero) is tangent to the efficient frontier. Since all assets beneath the efficient
frontier offer the same level of return with a greater degree of risk or a lower rate of return with the same level of risk, they are all inefficient portfolios. Conversely, the places above the
CML are not reachable.
At the point when the capital market line meets the efficient frontier, all rational investors will hold the combination of risk-free assets and M and the portfolio of risky assets.
Since each investor’s portfolio of risky assets will always be the market portfolio, we can calculate the optimal mix of investments for a portfolio of investors without knowing their levels of risk
A two-asset portfolio made up of a risk-free asset and a risky asset can be used to calculate the risk-return characteristics provided by CML.
Let’s investigate the following CML equation:
• E(Rc) is the anticipated return on any portfolio
• Rf is the risk-free return rate
• E(Rm) is the anticipated return on the market portfolio
• σc is the standard deviation of the portfolio’s return
• σm is the standard deviation of the return
The equation may be written as:
On the capital market line, the factor (Em – r / σM) is equal to the line’s gradient and is commonly referred to as the market price of risk.
As a result, we can write the CML equation above as follows:
Expected return = Market price risk + risk-free rate x Portfolio risk
Investment Strategies
While developing an investment strategy every investor wants to build a portfolio of stocks with the best long-term returns without taking on a lot of risk. The premise of MPT, which incorporates
mean-variance analysis, is that investors are risk averse. As a result, they concentrate on developing a portfolio that maximizes the projected returnabouto a particular degree of risk. Investors are
aware that high-return investments always carry some risk. Diversifying the portfolio of investments is the answer to reducing risk.
Mutual funds, stocks, bonds, etc., can be included in a portfolio, and each of these financial instruments carries a different level of risk when combined. In an ideal world, a gain in another
security would offset a loss in one security’s value.
Comparing a portfolio consisting of more than one type of security to one consisting of only one type is thought to be a better strategic choice. Mean-variance analysis can be a key component of an
investment strategy.
Limitations of the Mean-Variance Framework in Portfolio Optimization
Despite being popular, this portfolio optimization strategy has certain drawbacks:
• It is sensitive to input expectations of expected variances, covariances, and expected returns, which are susceptible to uncertainty and minute changes.
• Inadequate portfolio allocations may result from inaccurate estimates. Furthermore, non-normal behavior, skewness, and kurtosis are ignored in the mean-variance optimizer, which assumes that
asset returns follow a normal distribution. This may lead toan insufficient assessment of tail risk. optimization assumption made by the optimized strategy is that asset returns would be linear,
which may not accurately reflect the complexity of non-linear connections, particularly in the case of options or derivative instruments.
• Additionally, as mean-variance optimization primarily depends on historical data, it may not reliably predict future market circumstances, particularly in the case of structural changes or
extreme occurrences.
With the help of mean-variance optimization by PortfolioLab, users can build several common portfolios right away. However, there is also plenty of freedom for those who want to design their
portfolio optimization problems with unique constraints, data, and objectives.
Besides, it is imperative to exercise utmost caution. This post is not intended to provide investment advice; rather, it provides a framework for creating your own optimized portfolio with standard
examples. To avoid any errors, make sure to update every relevant detail while making any changes to the portfolio optimization.
How would you describe your trading/investing activity?
How would you describe your trading/investing activity?
What aspect of trading do you find most challenging?
What aspect of trading do you find most challenging?
What is the biggest frustrating problem you have in your trading experience?
What is the biggest frustrating problem you have in your trading experience?
If you had a magic wand and could instantly improve one aspect of your trading or investing experience, what would it be?
If you had a magic wand and could instantly improve one aspect of your trading or investing experience, what would it be?
Imagine your ideal trading tool. If a magic wand could make it happen overnight, what would that tool look like?
Imagine your ideal trading tool. If a magic wand could make it happen overnight, what would that tool look like?
Do you incorporate Moving Averages (MA) into your trading strategies?
Do you incorporate Moving Averages (MA) into your trading strategies?
If yes, how do you primarily utilize Moving Averages?
If yes, how do you primarily utilize Moving Averages?
What factors influence your choice of Moving Average period (e.g., timeframe, asset class, market conditions)?
What factors influence your choice of Moving Average period (e.g., timeframe, asset class, market conditions)?
What challenges do you encounter when using Moving Averages in your trading?
What challenges do you encounter when using Moving Averages in your trading?
Do you utilize support and resistance levels in your trading strategies?
Do you utilize support and resistance levels in your trading strategies?
If yes, how do you primarily use support and resistance levels?
If yes, how do you primarily use support and resistance levels?
What methods do you employ to identify accurate support and resistance levels?
What methods do you employ to identify accurate support and resistance levels?
What challenges do you face when using support and resistance levels in your trading?
What challenges do you face when using support and resistance levels in your trading?
Do you incorporate the Relative Strength Index (RSI) into your trading strategies?
Do you incorporate the Relative Strength Index (RSI) into your trading strategies?
If yes, how do you primarily utilize RSI?
If yes, how do you primarily utilize RSI?
What factors influence your choice of RSI period?
What factors influence your choice of RSI period?
What challenges do you encounter when using RSI in your trading?
What challenges do you encounter when using RSI in your trading?
Do you incorporate the Moving Average Convergence Divergence (MACD) indicator into your trading strategies?
Do you incorporate the Moving Average Convergence Divergence (MACD) indicator into your trading strategies?
If yes, how do you primarily utilize MACD?
If yes, how do you primarily utilize MACD?
What factors influence your choice of MACD parameters (e.g., fast EMA, slow EMA, signal line)?
What factors influence your choice of MACD parameters (e.g., fast EMA, slow EMA, signal line)?
What challenges do you face when using MACD in your trading?
What challenges do you face when using MACD in your trading?
Which technical indicators do you use most frequently in your trading?
Which technical indicators do you use most frequently in your trading?
How do you combine different indicators to create a robust trading strategy?
How do you combine different indicators to create a robust trading strategy?
What are the most significant challenges you face when using technical indicators?
What are the most significant challenges you face when using technical indicators?
Would you be willing to participate in an online interview or answer follow-up questions?^*
Would you be willing to participate in an online interview or answer follow-up questions?^*
Please provide your Name.^*
Please provide your Name.^*
Please provide your email^*
Please provide your email^* | {"url":"https://quantstrategy.io/blog/mean-variance-portfolio-optimization/","timestamp":"2024-11-12T19:18:44Z","content_type":"text/html","content_length":"496725","record_id":"<urn:uuid:3c25abdc-f8d0-4e67-9f6b-9136d4952021>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00801.warc.gz"} |
Frontiers | Localization in Flow of Non-Newtonian Fluids Through Disordered Porous Media
• ^1Institute of Terrestrial Ecosystems, Department of Environmental Systems Science, ETH Zürich, Zurich, Switzerland
• ^2Departamento de Física, Universidade Federal do Ceará, Campus do Pici, Fortaleza, Brazil
• ^3Institute of Environmental Engineering, Department of Civil, Environmental and Geomatic Engineering, ETH Zürich, Zürich, Switzerland
• ^4Instituto Federal do Ceará, Campus Tianguá, Ceará, Brasil
• ^5EAWAG, Swiss Federal Institute of Aquatic Science and Technology, Dübendorf, Switzerland
• ^6Instituto de Física, Universidade Federal da Bahia, Salvador, Brazil
• ^7Swiss Federal Research Institute WSL, Birmensdorf, Switzerland
We combine results of high-resolution microfluidic experiments with extensive numerical simulations to show how the flow patterns inside a “swiss-cheese” type of pore geometry can be systematically
controlled through the intrinsic rheological properties of the fluid. Precisely, our analysis reveals that the velocity field in the interstitial pore space tends to display enhanced channeling under
certain flow conditions. This observed flow “localization”, quantified by the spatial distribution of kinetic energy, can then be explained in terms of the strong interplay between the disordered
geometry of the pore space and the nonlinear rheology of the fluid. Our results disclose the possibility that the constitutive properties of the fluid can enhance the performance of chemical reactors
and chromatographic devices through control of the channeling patterns inside disordered porous media.
Flow through porous media is of great interest in chemical engineering, physics, and biology [1–3]. Previous studies have shown that the disordered characteristics of the pore structure naturally
leads to heterogeneous flow patterns [4–8] and preferential channeling [9–11]. Understanding how to control and manipulate these flow patterns can help to optimize catalysts [12, 13] or
chromatographic devices [14, 15], and allows to steer chemical reactions inside the porous medium itself [16–18].
In order to understand the physics of important problems like, for example, blood flow through the kidney [19] or oil flow through porous rocks [20, 21], one must also consider the nonlinear
constitutive behavior of the fluids involved in these processes. Technological applications which make use of non-Newtonian fluids are ubiquitous nowadays [22–25]. It is, for instance, the case of
shear-thinning solvents that are present in dropless paints [26], shear-thickening fluids being used as active dampers [27] and hybrid fluids as components of enhanced body armors [28]. While
Newtonian flows in irregular media have been extensively investigated theoretically and confirmed by many experiments, the study of non-Newtonian fluids lack a generalized framework due to their
diverse constitutive nature. Non-Newtonian flows through porous media have mainly been studied theoretically [29, 30] and through numerical simulations [31, 32], where the main focus of interest was
to find non-Darcian models for the flow of generalized Newtonian fluids [30, 33–37]. In the particular case of power-law fluids, it has been shown that, in spite of the nonlinear nature of the
fluid’s rheology and the geometrical complexity of the pore volume, the general behavior of the system can still be quantified in terms of a universal permeability extending over a broad range of
Reynolds conditions and power-law exponents [38].
However, quantitative experiments with non-Newtonian materials which go beyond simple bulk measurements [39, 40] are scarce [41] because the design of the experimental pore geometry and the operating
conditions need to be adjusted in order to match the nonlinear constitutive regime of the fluid’s rheology. Here we combine the results of microfluidic experiments [42] with fluid dynamics
simulations to demonstrate how the nonlinear rheological properties of a fluid can be effectively exploited in order to control the macroscopic transport properties of a flow through the external
operational flow conditions. These results have important consequences for the design of chemical reactors and chromatographic systems as well as for the enhancement of oil recovery and transport in
porous media in general.
Under steady-state conditions the motion of an incompressible fluid through the interstitial space of a porous medium is described by mass and momentum conservation, respectively,
together with appropriate boundary conditions. The variables ϱ, $u$ and p are the fluid’s density, velocity and pressure, and $T$ is the deviatoric stress tensor which depends on the fluid’s
rheology. For many fluids, this constitutive relation is well described by a simple linear rheology $T=2μE$, where $Eij=1/2(∂jui+∂iuj)$ is the shear strain rate tensor [43] and the proportionality
constant μ defines the kinematic viscosity. Examples of these so-called Newtonian fluids are water, light oil and most diluted gases. However, many fluids present in industrial products, biology and
environmental flows obey much more complex nonlinear constitutive laws [23, 25, 44]. These fluids are called non-Newtonian fluids. The constitutive behavior of most non-Newtonian fluids can be
described by a generalization of the Newtonian relation, namely,
Here the apparent viscosity $μ(γ˙)$ is a nonlinear function of the second principal invariant $γ˙=2E:E$ of the shear strain rate tensor $E$ alone [43]. Examples of fluids—which are often called
generalized Newtonian fluids—are colloidal suspensions, protein or polymeric mixtures, heavy petroleum, blood or debris flows, only to mention a few [45–48].
Our analysis is based on experimental results from the setup presented recently in Eberhard et al. [42]. Their main purpose was to map the local viscosity of a non-Newtonian flow in a porous
microfluidic channel by means of a high-resolution technique of image velocimetry, namely, Ghost Particle Velocimetry (GVP) [49]. The geometry of the microfluidic chip is shown in Figure 1. As
non-Newtonian fluid we used a 0.5 wt% xanthan gum solution, which is a polysaccharide mainly found in food industry [50] and enhanced oil recovery [22, 24]. It has a shear-thinning rheology which
changes its apparent viscosity over several orders of magnitude. While polymeric solutions often show viscoelastic behavior [51], the concentration of xanthan gum in our experimental solution was so
low that no measurable elastic behavior could be observed during the experiment. The rheology of xanthan gum closely follows a Carreau model,
approaching the viscosity of the solvent (water) $μ∞=0.001$$Pa⋅s$ in the limit of very high shear [52]. Conversely, for low shear, Eq. 4 reduces to $μC(γ˙)=μ0$, corresponding to a constant viscosity
$μ0=24$$Pa⋅s$. At an intermediate range of shear rates, the fluid follows a power-law relation $μ∼γ˙n−1$ with $n=0.3$ in our specific case. The remaining Carreau parameter $λ=50 s$ was determined
using a nonlinear least-square fit to the experimentally measured values [42]. As shown in Figure 1, the experimental pore structure consisted of a microfluidic device containing a quasi-2D porous
medium of size 30 mm × 15 mm and depth of 100 µm. It contains pillars of radius 100 µm that are randomly allocated and can overlap, forming a “swiss-cheese” pore geometry with void fraction
approximately equal to 0.8.
FIGURE 1
FIGURE 1. Sketch of the experimental setup showing the whole pore geometry together with the simulated flow field in the mid plane for the non-Newtonian case at qin 5 µL/min. The porous region is L
30 mm long, B 15 mm wide and has a height of h 100 µm. The radius of the circular pillars is also 100 mµ. The gray shaded region marks the part of the porous device where the experimental flow
velocity measurement has been performed.
Figure 2 compares the velocimetry measurements obtained from ref. 42 in a section of the mid plane of the microfluidic chip with those obtained from numerical simulations calculated with exactly the
same pore geometry, fluid properties and flow conditions. More precisely, the flow rates for the presented cases are q[in] = 0.05 µL/min (Figure 2A) and q[in] = 5 µL/min (Figure 2B) for the xanthan
case, and q[in] = 5 µL/min for the measurement with water (Figure 2C). The color scale has been normalized to the 95% quantile of the velocity distribution to facilitate the comparison of the flow
fields at different flow rates. Although differences between the experimentally measured and simulated velocity fields can be visually detected, they are mostly local and could be explained by the
natural difficulties of exactly reproducing in the mathematical model the detailed features of the flow, the fluid rheology, and the flow operational conditions. To perform numerical simulations, the
computational mesh was generated by capturing the two-dimensional technical drawing of the device geometry into Ansys’ meshing module [53]. In the horizontal plane, we first created an unstructured
quadrilateral mesh with an average cell size of $≈2 μm2$. The three-dimensional structure was obtained by extruding ten vertical layers, resulting in a computational mesh composed of approximately
30 million unstructured hexahedral cells (corresponding to roughly 20 µm^3/cell), which was then imported into Ansys Fluent^TM [53]. Non-slip boundary conditions were applied on all solid walls of
the microfluidic chip, which is a reasonable assumption on surfaces without hyper-hydrophilic coatings and at scales much larger than the polymer coil size [54]. The fluid was injected via a constant
velocity inlet corresponding to the inflow rate reported in the experiment. The density of the non-Newtonian fluid used in the computational simulations matches exactly the experimental value of the
xanthan solution, namely, ρ[xan] = 1.0 g/cm^3. As for the rheology of the fluid, we used Eq. 4 to interpolate the local viscosity values obtained from the independent rheometer experiments [42].
Finally, the steady-state flow solution in terms of the velocity and pressure fields was calculated using a second-order integration scheme and convergence was achieved if the residuals reached a
threshold of $10−6$.
FIGURE 2
FIGURE 2. Comparison of the experimentally measured [42] and simulated velocity fields in the mid plane of the porous device for the two non-Newtonian cases at (A) qin 0.05 µL/min, (B) qin 5 µL/min,
and for the Newtonian case (C). Note the different velocity ranges between (A–C). For comparison, the color scales has been normalized to the 95% quantile of the velocity distribution.
In order to check the variability of our results with respect to the disorder level of the pore space, numerical simulations have also been performed with three additional realizations of the
swiss-cheese geometry, but keeping the same physico-chemical properties of the fluid, operational parameters of the flow, and boundary conditions. Considering the independence of the rheometry and
velocimetry measurements, the excellent agreement between results from the numerical model and experiments (Figures 2A–C) clearly demonstrates the global consistency of our methodological approach.
In order to highlight the differences between the Newtonian and non-Newtonian flows and the tendency for stronger localization in the non-Newtonian flow, we show in Figure 3A the contour plot of the
ratio between the local velocity magnitudes measured with the xanthan solution and water normalized by their respective mean velocities. In both cases, the applied flow rate was set to q[in] = 5 µL/
FIGURE 3
FIGURE 3. (A) Contour plot of the ratio between the local velocity magnitudes measured with the non-Newtonian (xanthan) and Newtonian (water) fluids, normalized by the corresponding mean velocities
in the observation mid planes, both obtained at qin 5 µL/min. (B) Participation ratio as a function of Reynolds number. Gray markers label the participation values obtained from the simulations of
four different realizations of the pore structure. The realization which corresponds to the experimental device is marked in darker gray. The black solid line was obtained by averaging the four
realizations of the “swiss-cheese” pore geometry. The participation ratio calculated from the experimentally measured velocity fields of the xanthan gum solution at q[in] 0.05 µL/min and q[in] 5 µL/
min are labeled with yellow and blue stars, respectively.
The channeling effect present in the flow fields shown in Figure 2 can be statistically quantified in terms of their spatial distributions of kinetic energy $e∝|u|2$. This is performed here in terms
of a measure utilized in previous studies on localization of vibrational modes in harmonic chains [55], namely, the participation number $Π$ defined as,
where the total volume of the fluid in a domain is given by $V=∫Ω1dω$. Thus the participation ratio varies between $Π=1$, corresponding to a limiting state of equal partition of kinetic energy $(e(x)
=const⋅∀x∈Ω)$ and the value $Π≈0$ for a sufficiently large system $(V→∞)$, indicating strong localization, namely, the presence of intense channeling effects in the flow field [56]. For reference,
the Reynolds number is defined here as,
where v is the mean velocity at the entrance of the pore zone and $dp$ is the diameter of the solid obstacles. Figure 3B shows how the participation ratio obtained with our computational model varies
as a function of Reynolds number. The thick black curve is obtained by averaging the results from the pore geometry used in the microfluidic experiment (dark gray) with four additional realizations
of the “swiss-cheese” pore geometry (light gray markers) having the same porosity. The spread of the markers therefore gives a good indication for the statistical variability of the participation
ratio for different pore geometries. The values of the participation ratios obtained from the two experimentally measured velocity fields are marked with yellow and blue stars, respectively, and are
in good agreement with the numerical calculations.
At low Reynolds numbers, the participation ratio is practically constant at $Π≈0.6$, similar to a Stokes flow with fixed viscosity $μ=μ0=24$$Pa⋅s$. By increasing the flow rate, the local shear within
the interstitial pore space also generally increases to eventually reach a point where its range of variability overlaps with the range in which the rheology of the xanthan gum solution follows a
power-law behavior. At this point, the local viscosity spans over a wide range of values, leading to a drop in the participation ratio by almost 20% at a Reynolds numbers around $Re=3×10−4$.
Interestingly, while the absolute value of the participation ratio varies slightly from realization to realization, the location of the participation ratio minimum is determined by the fluid’s
rheology and does not seem to be influenced by the details of the pore geometry. More precisely, we find for the two experimentally measured velocity fields a participation ratio of $Π=0.556$ for q
[in] = 0.05 µL/min and $Π=0.520$ for q[in] = 5 µL/min. For the simulations, the mean participation ratios averaged over four realizations yield $Π=0.582±0.002$ for q[in] = 0.05 µL/min and $Π=
0.525±0.002$ for q[in] = 5 µL/min. The participation ratios of the two simulations which share the same pore geometry as the experimental setup are $Π=0.582$ (q[in] = 0.05 µL/min) and $Π=0.528$ (q
[in] = 5 µL/min). After reaching the minimum, where the flow is most heterogeneously distributed in the porous medium, the participation ratio increases again as the flow rate pushes the fluid’s
rheology beyond the power-law regime, ultimately approaching the flow pattern of water. In this limiting case, the Newtonian behavior is recovered, but inertial effects on the flow should prevent the
participation ratio to reach the same value obtained for very low Reynolds numbers, namely, $Π≈0.6$.
A question that naturally arises is how rheology influences the flow’s heterogeneity in space. At low and high Reynolds number, the Carreau fluid has an almost constant viscosity equal to the low and
high shear limits $μ0=24$$Pa⋅s$ and $μ∞=0.001$$Pa⋅s$, respectively. In the intermediate regime, however, the local viscosity covers a broad spectrum of values [42]. In this case, both experimental
and simulation results reveal that the interplay between the disordered geometry of the pore space and the fluid rheology leads to a larger flow heterogeneity and therefore to a stronger localization
It is well accepted that flow and transport processes in porous media are fundamentally controlled by the complex interplay between the fluid and the pore space structure. Here we showed that these
processes can be tailored by tuning the rheology of the fluid. Precisely, the heterogeneity of the pore scale structure of the medium causes a high variability of shear rates that, when matched with
the nonlinear viscosity window of the non-Newtonian fluid, can substantially enhance macroscopic properties of the system like the participation ratio. These effects may be exploited to improve
filters and catalysts or to enhance chemical reactions by spreading the transported chemicals more uniformly throughout the pore space. In particular, localization should have a deleterious influence
on the effectiveness of catalysts subjected to flow, for example, in a packed bed chemical reaction. Precisely, the preferential channeling at the minimum of the participation number should be
avoided to maximize the activity of the surface area available for reaction in the system.
Data Availability Statement
The raw data supporting the conclusions of this article will be made available by the authors, without undue reservation.
Author Contributions
HS and JA designed the research. UE carried out the experiments with input from ES, MH, and JJ-M All authors discussed the results. HS and JA wrote the paper with input from all other authors.
Conflict of Interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
We acknowledge Prof. R. Stocker and E. Burmeister from the Department of Civil, Environmental and Geomatic Engineering at ETH Zurich for kindly providing access to laboratory equipment and material.
ES acknowledges SNSF PRIMA Grant 179834. JA acknowledges financial support from the Brazilian agencies CNPq, CAPES and FUNCAP. HS and JA acknowledge financial support from Petrobras (“Física do
Petróleo em Meios Porosos”, Project Number: F0185. RA acknowledges financial support from the Brazilian agency CNPq and the INCT-SC Project.
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Keywords: localization, microfluidics, particle velocimetry, non-newtonian fluids, porous media
Citation: Seybold HJ, Eberhard U, Secchi E, Cisne RLC, Jiménez-Martínez J, Andrade RFS, Araújo AD, Holzner M and Andrade J (2021) Localization in Flow of Non-Newtonian Fluids Through Disordered
Porous Media. Front. Phys. 9:635051. doi: 10.3389/fphy.2021.635051
Received: 29 November 2020; Accepted: 04 January 2021;
Published: 16 February 2021.
Edited by:
Ferenc Kun
, University of Debrecen, Hungary
Reviewed by:
Taotao Fu
, Tianjin University, China
Simon Haward
, Okinawa Institute of Science and Technology Graduate University, Japan
Sandro Longo
, University of Parma, Italy
Copyright © 2021 Seybold, Eberhard, Secchi, Cisne, Jiménez-Martínez, Andrade, Araújo, Holzner and Andrade. This is an open-access article distributed under the terms of the Creative Commons
Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original
publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
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Publications (Excerpt)
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms
Title: Item-based reliability-centred life-cycle costing using monte carlo simulation.
Written by: Reifferscheidt, Jan and Weigell, Jürgen and Jahn, Carlos
in: <em>Journal of Physics: Conference Series</em>. (2021).
Volume: <strong>2018</strong>. Number:
on pages:
how published:
DOI: 10.15480/882.3846
Abstract: This paper presents a time sequential probabilistic simulation model for the detailed design of maintenance strategies for turbine critical items. The term item shall refer to any part,
component, device, subsystem, or functional unit of a wind turbine that can be individually described and considered. The model enables wind farm operators and turbine manufactures to find the most
cost effective maintenance strategy for each turbine critical item. Cost optimizations are realized through a better adaptation of the maintenance strategy to the itemspecific failure modes,
degradation processes, failure detection capabilities and the given operational configuration of the wind farm. Based on a time sequential Monte Carlo simulation technique, the maintenance activities
at turbine level are simulated over the windfarm`s operational lifetime, considering correlations between the stochastic variables. The results of the Monte Carlo simulation are evaluated using
statistical means, thereby, determining the optimal maintenance strategy and associated parameters. The developed model is implemented as a Python application and equally applicable for onshore and
offshore windfarms | {"url":"https://www.tuhh.de/mls/en/research/publications-excerpt?showUid=182450&cHash=fc0c79724e361eae3e8c6f8dbae99a83","timestamp":"2024-11-12T03:12:02Z","content_type":"text/html","content_length":"52814","record_id":"<urn:uuid:530e64db-891d-4ecb-8dd3-41e14fb0351e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00845.warc.gz"} |
Defining a Distribution Model with the FCMP Procedure
A severity distribution model consists of a set of functions and subroutines that are defined using the FCMP procedure. The FCMP procedure is part of Base SAS software. Each function or subroutine
must be named as distribution-name_keyword, where distribution-name is the identifying short name of the distribution and keyword identifies one of the functions or subroutines. The total length of
the name should not exceed 32. Each function or subroutine must have a specific signature, which consists of the number of arguments, sequence and types of arguments, and return value type. The
summary of all the recognized function and subroutine names and their expected behavior is given in Table 23.4.
Consider following points when you define a distribution model:
• When you define a function or subroutine requiring parameter arguments, the names and order of those arguments must be the same. Arguments other than the parameter arguments can have any name,
but they must satisfy the requirements on their type and order.
• When the SEVERITY procedure invokes any function or subroutine, it provides the necessary input values according to the specified signature, and expects the function or subroutine to prepare the
output and return it according to the specification of the return values in the signature.
• You can typically use most of the SAS programming statements and SAS functions that you can use in a DATA step for defining the FCMP functions and subroutines. However, there are a few
differences in the capabilities of the DATA step and the FCMP procedure. Refer to the documentation of the FCMP procedure to learn more.
• You must specify either the PDF or the LOGPDF function. Similarly, you must specify either the CDF or the LOGCDF function. All other functions are optional, except when necessary for correct
definition of the distribution. It is strongly recommended that you define the PARMINIT subroutine to provide a good set of initial values for the parameters. The information provided by PROC
SEVERITY to the PARMINIT subroutine enables you to use popular initialization approaches based on the method of moments and the method of percentile matching, but you can implement any algorithm
to initialize the parameters by using the values of the response variable and the estimate of its empirical distribution function.
• The LOWERBOUNDS subroutines should be defined if the lower bound on at least one distribution parameter is different from the default lower bound of 0. If you define a LOWERBOUNDS subroutine but
do not set a lower bound for some parameter inside the subroutine, then that parameter is assumed to have no lower bound (or a lower bound of ). Hence, it is recommended that you explicitly
return the lower bound for each parameter when you define the LOWERBOUNDS subroutine.
• The UPPERBOUNDS subroutines should be defined if the upper bound on at least one distribution parameter is different from the default upper bound of . If you define an UPPERBOUNDS subroutine but
do not set an upper bound for some parameter inside the subroutine, then that parameter is assumed to have no upper bound (or a upper bound of ). Hence, it is recommended that you explicitly
return the upper bound for each parameter when you define the UPPERBOUNDS subroutine.
• If you want to use the distribution in a model with regression effects, then make sure that the first parameter of the distribution is the scale parameter itself or a log-transformed scale
parameter. If the first parameter is a log-transformed scale parameter, then you must define the SCALETRANSFORM function.
• In general, it is not necessary to define the gradient and Hessian functions, because PROC SEVERITY uses an internal system to evaluate the required derivatives. The internal system typically
computes the derivatives analytically. But it might not be able to do so if your function definitions use other functions that it cannot differentiate analytically. In such cases, derivatives are
approximated using a finite difference method and a note is written to the SAS log to indicate the components that are differentiated using such approximations. PROC SEVERITY does reasonably well
with these finite difference approximations. But, if you know of a way to compute the derivatives of such components analytically, then you should define the gradient and Hessian functions.
In order to use your distribution with PROC SEVERITY, you need to record the FCMP library that contains the functions and subroutines for your distribution and other FCMP libraries that contain FCMP
functions or subroutines used within your distribution’s functions and subroutines. Specify all those libraries in the CMPLIB= system option by using the OPTIONS global statement. For more
information about the OPTIONS statement, see the SAS Statements: Reference. For more information about the CMPLIB= system option, see the SAS System Options: Reference.
Each predefined distribution mentioned in the section Predefined Distributions has a distribution model associated with it. The functions and subroutines of all those models are available in the
Sashelp.Svrtdist library. The order of the parameters in the signatures of the functions and subroutines is the same as listed in Table 23.2. You do not need to use the CMPLIB= option in order to use
the predefined distributions with PROC SEVERITY. However, if you need to use the functions or subroutines of the predefined distributions in SAS statements other than the PROC SEVERITY step (such as
in a DATA step), then specify the Sashelp.Svrtdist library in the CMPLIB= system option by using the OPTIONS global statement prior to using them.
Table 23.4 shows functions and subroutines that define a distribution model, and subsections after the table provide more detail. The functions are listed in alphabetical order of the keyword suffix.
Table 23.4: List of Functions and Subroutines That Define a Distribution Model
Name Type Required Expected to Return
dist_CDF Function YES Cumulative distribution
function value
dist_CDFGRADIENT Subroutine NO Gradient of the CDF
dist_CDFHESSIAN Subroutine NO Hessian of the CDF
dist_CONSTANTPARM Subroutine NO Constant parameters
dist_DESCRIPTION Function NO Description of the distribution
dist_LOGCDF Function YES Log of cumulative distribution
function value
dist_LOGCDFGRADIENT Subroutine NO Gradient of the LOGCDF
dist_LOGCDFHESSIAN Subroutine NO Hessian of the LOGCDF
dist_LOGPDF Function YES Log of probability density
function value
dist_LOGPDFGRADIENT Subroutine NO Gradient of the LOGPDF
dist_LOGPDFHESSIAN Subroutine NO Hessian of the LOGPDF
dist_LOGSDF Function NO Log of survival
function value
dist_LOGSDFGRADIENT Subroutine NO Gradient of the LOGSDF
dist_LOGSDFHESSIAN Subroutine NO Hessian of the LOGSDF
dist_LOWERBOUNDS Subroutine NO Lower bounds on parameters
dist_PARMINIT Subroutine NO Initial values
for parameters
dist_PDF Function YES Probability density
function value
dist_PDFGRADIENT Subroutine NO Gradient of the PDF
dist_PDFHESSIAN Subroutine NO Hessian of the PDF
dist_QUANTILE Function NO Quantile for a given CDF value
dist_SCALETRANSFORM Function NO Type of relationship between
the first distribution parameter
and the scale parameter
dist_SDF Function NO Survival function value
dist_SDFGRADIENT Subroutine NO Gradient of the SDF
dist_SDFHESSIAN Subroutine NO Hessian of the SDF
dist_UPPERBOUNDS Subroutine NO Upper bounds on parameters
1. Either the dist_CDF or the dist_LOGCDF function must be defined.
2. Either the dist_PDF or the dist_LOGPDF function must be defined.
The signature syntax and semantics of each function or subroutine are as follows: | {"url":"http://support.sas.com/documentation/cdl/en/etsug/65545/HTML/default/etsug_severity_details25.htm","timestamp":"2024-11-05T03:32:48Z","content_type":"application/xhtml+xml","content_length":"122871","record_id":"<urn:uuid:730442de-0a08-4bc9-b21b-9c452030ef5c>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00722.warc.gz"} |
Level Order Traversal in Binary Tree | Explained with Code and Example
Traversal is an algorithm for visiting each node in a different fashion. Earlier we have seen see pre-order, in-order and post-order BT traversal.
Now we are interested in level order traversal.
What is level order traversal in Binary Tree?
In level order traversal, we visit (print) nodes of the binary tree at level zero, then at level one, two and so on…
For the above binary tree, the level order traversal is 4, 1, 5, 7, 3, 6.
• The node at level one is 4.
• The node at level two is 1 and 5.
• The node at level three is 7, 3, 6.
Note: The number of levels in the binary tree is one more than the height of the tree (or root node). In the above example, the height of the BT is two and the total number of levels is three (2+1).
1. Find the height of the binary tree
2. Repeat for each level (1 to height+1)
- Print the value of nodes at that level.
(Use recursion here to print the values.)
To implement level order traversal, it is required to find the height of a Binary Tree.
Python Program
Here is a simple Python program for level order traversal using recursion.
class node:
def __init__(self, val, left=None, right=None):
def height(root):
if root:
return 1+max(height(root.left), height(root.right))
return -1
def printAtLevel(root, level):
if root:
if level==1:
printAtLevel(root.left, level-1)
printAtLevel(root.right, level-1)
def levelOrderTraversal(root):
for i in range(1,height(root)+2):
printAtLevel(root, i)
If Python is not your primary language, try to solve this problem in C/C++, Java or any programming language you are comfortable with. You can use the same algorithm and logic.
The worst-case complexity of the algorithm is with a skewed binary tree where each node will be having either one or two nodes.
In the skewed tree, to print the element at level one (function printAtLevel()) take the time of O(1), to print the element at level two takes the time of O(2)… to print the element at level n takes
the time of O(n)…
(Where, n is the number of nodes in the binary tree.)
To print all the elements at each level is O(1)+O(2)+O(3)+…+O(n)=O(n^2).
So, in the worst case, the time complexity of level order traversal in binary tree is O(n^2). | {"url":"https://www.csestack.org/level-order-traversal-binary-tree/","timestamp":"2024-11-12T13:45:13Z","content_type":"text/html","content_length":"137350","record_id":"<urn:uuid:7dbb5eb6-1a5d-4626-a4e1-1e1511f3b72f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00359.warc.gz"} |
7 Roulette Terms You Should Know
7 Roulette Terms You Should Know
People have been playing roulette for hundreds of years, with the earliest versions dating back to 18th century France.
The basic premise of the game is to bet on a number and spin the roulette wheel. A ball will land on a colored number, and everyone who wagered on that outcome will win.
Since it has been around for so long, there are lots of terms and words that relate specifically to roulette. Many of them are French due to roulette’s origins.
In this article, we cover some of the words and terms you may find when playing a game of roulette.
American Roulette Vs European Roulette
When you are learning how to play roulette, you will hear the terms American roulette and European roulette be used a lot. These terms refer to the 2 different types of roulette wheels you will find.
The American wheel has 38 pockets, 2 of which are green and numbered 0 and 00. The rest are red and black and are numbered 1-36. This means the probability of winning is 1/38.
The European wheel has 37 pockets; it has the same numbers and colors as the American wheel but has no green 00 pockets. This means the probability of winning on this wheel is 1/37.
The order of the numbers also changes depending on the wheel.
Biased Numbers
This is a group of numbers that are statistically more likely to be landed on. This is done through analysis and observation and is often the result of a faulty wheel.
En Prison
This is French for ‘in prison’ and is an option that is sometimes given on European roulette wheels.
It only applies to even-money wages and only comes into play when the wheel lands on 0.
If this happens, the players don’t immediately lose their bet. Instead, it is locked (or imprisoned) for another spin.
If they win the next spin, they get their bet back; if not, it goes to the house. This is a favorable rule because it is essentially a free retry.
A similar rule is La Partage, where the player only loses half their bet but doesn’t get the option to leave for another spin.
House Edge
It is a common term that the house always wins in gambling. This is because they set the rules and make them in favor of the house.
The house edge is usually determined by a percentage of how likely they are to come away winning compared to the player. For example, with an American roulette wheel, the house edge is 5.36%, whereas
a European roulette wheel has a house edge of 2.7%.
Inside And Outside Bets
These are the 2 different types of bets you can make.
Inside bets refer to the numbers on the inside of the roulette table layout. They tend to have a larger payout. Below are some examples of inside bets that can be made.
• Corner - a bet on four numbers in a square configuration, also called carre
• Six-line - a 6-number wagering system, also called sixain
• Split - bet on 2 adjacent numbers, also called cheval
• Straight-up - a bet on a singular number and color, also called en plein. This is one of the largest payouts
• Street - a bet on three numbers in a horizontal line
• Trio - a bet of 2 sets of 3 number configurations, also called transversale
An outside bet is when you bet on one of the options of the outside of the roulette table layout. These bets cover a larger area of probability, so have a smaller payout. Below are some outside bets
that could be made.
• Red (rouge) or black (noir)
• Even (pair) or odd (impair)
• 1-18 (manque)
• 19-36 (passe)
• Column bets also called Colonne
• A dozen bets, also called douzaine- the first group is called premiere, the last group is called derniere
This is also known as a combo or accumulator bet. It is a strategy that involves doubling your bet after a win and returning to the initial bet if you lose.
This approach is high risk, with you either winning double your money or losing everything. It has great benefits if you are on a winning streak.
Parlay as a strategy can also be applied in blackjack.
This is the amount of money you win on a bet. The payout depends on the type of bet you made and how much money you bet.
More Terms
• Airball roulette - A roulette wheel powered by air pressure.
• Backtrack/balltrack - The outer rim where the ball is spun.
• Biased wheel - A faulty wheel that causes a certain number to be landed on more often.
• Bottom track - The bottom area of the wheel that the ball goes into before bouncing into a pocket.
• Cold table - A table where you and/or other players have been consistently losing.
• Dealer - A casino worker who deals out winnings and spins the roulette wheel, also called croupier which is exclusive to roulette.
• Electronic roulette - A roulette wheel powered by electricity.
• Even money - A bet that pays out 1 to 1.
• Gaffed wheel - A wheel rigged by either players or the casino.
• Hot table - A table where players have been regularly winning.
• Mini roulette - A smaller wheel with only 13 numbers labeled 1-12. The payout is adjusted accordingly.
• Orphelins - This is exclusive to European roulette. It refers to the digits that are close to each other on the roulette wheel but not on the board, such as 1-20-14-31-9 and 14-34-6.
• Positive progression - Any strategy that increases your bets after a win.
• Section slicing - Discover biases by dividing the board based on where the wheel has been hitting.
• Surrender - Some casinos only take half the player’s wager if it lands on 0 or 00 and they have bet even.
• Visual wheel tracking - Ability to tell where the ball will land by how it’s spinning.
Final Thoughts
With this expanded vocabulary, you will be able to follow along with any roulette game with ease and feel like a proper high roller. | {"url":"https://techheals.com/roulette-terms-you-should-know","timestamp":"2024-11-04T13:45:56Z","content_type":"text/html","content_length":"31360","record_id":"<urn:uuid:f639814f-3e8c-4567-bb27-e99703c9967a>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00559.warc.gz"} |
numpy.in1d(ar1, ar2, assume_unique=False, invert=False)[source]¶
Test whether each element of a 1-D array is also present in a second array.
Returns a boolean array the same length as ar1 that is True where an element of ar1 is in ar2 and False otherwise.
We recommend using isin instead of in1d for new code.
ar1 : (M,) array_like
Input array.
ar2 : array_like
The values against which to test each value of ar1.
assume_unique : bool, optional
If True, the input arrays are both assumed to be unique, which can speed up the calculation. Default is False.
invert : bool, optional
If True, the values in the returned array are inverted (that is, False where an element of ar1 is in ar2 and True otherwise). Default is False. np.in1d(a, b, invert=True) is
equivalent to (but is faster than) np.invert(in1d(a, b)).
New in version 1.8.0.
in1d : (M,) ndarray, bool
The values ar1[in1d] are in ar2.
See also
Version of this function that preserves the shape of ar1.
Module with a number of other functions for performing set operations on arrays.
in1d can be considered as an element-wise function version of the python keyword in, for 1-D sequences. in1d(a, b) is roughly equivalent to np.array([item in b for item in a]). However, this idea
fails if ar2 is a set, or similar (non-sequence) container: As ar2 is converted to an array, in those cases asarray(ar2) is an object array rather than the expected array of contained values.
>>> test = np.array([0, 1, 2, 5, 0])
>>> states = [0, 2]
>>> mask = np.in1d(test, states)
>>> mask
array([ True, False, True, False, True])
>>> test[mask]
array([0, 2, 0])
>>> mask = np.in1d(test, states, invert=True)
>>> mask
array([False, True, False, True, False])
>>> test[mask]
array([1, 5]) | {"url":"https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.in1d.html","timestamp":"2024-11-10T06:06:11Z","content_type":"text/html","content_length":"12529","record_id":"<urn:uuid:8033ad25-67d0-406b-b3e3-05d2a6d404fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00782.warc.gz"} |