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# Aaron will jog from home at X miles per hour and then walk
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Aaron will jog from home at X miles per hour and then walk [#permalink] 20 Apr 2008, 16:22
Aaron will jog from home at X miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours joggin and walking?
a) xt/y
b) x+t/xy
c)xyt/x+y
d)x+y+t/xy
e) y+t/x - t/y
Are there alternative ways of solving this question?? I understood the answer, but i just didnt like the way they did it?
thx
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Re: OG 11 Diagnostic #24 [#permalink] 20 Apr 2008, 17:45
Victor81 wrote:
Aaron will jog from home at X miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours joggin and walking?
a) xt/y
b) x+t/xy
c)xyt/x+y
d)x+y+t/xy
e) y+t/x - t/y
Are there alternative ways of solving this question?? I understood the answer, but i just didnt like the way they did it?
thx
Which way is the OE?
Set d = total distance, and this is what we are looking for.
time going there + time going tack = t
d/x + d/y = t
d = txy / (x+y)
Ans = C
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Re: OG 11 Diagnostic #24 [#permalink] 22 Apr 2008, 07:42
Thanks man, yours is a much better way of solving. I had the right idea, just couldnt fully put it together. The OG answer is as follows:
Let J be the number of hours Aaron spends jogging
Let T -J bet the total number of hours he spends walking
So Aaron jogs XJ Miles and walks y(t-J) miles
Because Aaron travels the same route, XJ = y(t-J)
J = yt/x+Y
the number of miles he can jog is xj, so substitute for J
xJ = Yt/x+Y
x(yt/x+Y) = xyt/x+y
Re: OG 11 Diagnostic #24 [#permalink] 22 Apr 2008, 07:42
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# If x, y, and z are integers, y + z = 13, and xz = 9, which of the foll
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If x, y, and z are integers, y + z = 13, and xz = 9, which of the foll [#permalink]
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31 Aug 2018, 00:07
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If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?
(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x
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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which of the foll [#permalink]
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31 Aug 2018, 00:18
XZ = 9.
Possible numbers are (1,9) (3,3) or (9,1)
Then Y+Z = 13.
Taking Z as 1,3 or 9.
Y becomes 12,10 or 4.
In all cases Y is greater than 3.
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Re: If x, y, and z are integers, y + z = 13, and xz = 9, which of the foll [#permalink]
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31 Aug 2018, 00:30
Bunuel wrote:
If x, y, and z are integers, y + z = 13, and xz = 9, which of the following must be true?
(A) x is even
(B) x = 3
(C) y is odd
(D) y > 3
(E) z < x
we have y + z = 13
This is possible when
odd + even = odd
which means
one of x or y is odd and the other is even.
xz = 9
This is possible when both x and z are odd
we finally have in terms of odd or even
x= odd
so answer choice A and C are out.
since x , y and z are integers
in xz=9
x can hold following values
1,3,9
and so can z
answer choice B and E are out
Finally , greatest possible value of z can be 9
thus the smallest possible value of x is 13-9 = 4 > 3
Re: If x, y, and z are integers, y + z = 13, and xz = 9, which of the foll [#permalink] 31 Aug 2018, 00:30
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Mathematics
Alternate titles: Gaussian elimination; pivoting
Gauss elimination, in linear and multilinear algebra, a process for finding the solutions of a system of simultaneous linear equations by first solving one of the equations for one variable (in terms of all the others) and then substituting this expression into the remaining equations. The result is a new system in which the number of equations and variables is one less than in the original system. The same procedure is applied to another variable and the process of reduction continued until there remains one equation, in which the only unknown quantity is the last variable. Solving this equation makes it possible to “back substitute” this value in an earlier equation that contains this variable and one other unknown in order to solve for another variable. This process is continued until all the original variables have been evaluated. The whole process is greatly simplified using matrix operations, which can be performed by computers.
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Or click Continue to submit anonymously: | 679 | 3,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-22 | latest | en | 0.845011 |
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Assignment Help Operation Management
##### Reference no: EM13251753
The Arctic Oil Company has recently drilled two new wells in a remote area of Alaska. The company is planning to install a pipeline to carry the oil from the two new wells to a transportation and refining (T&R) center. The locations of the oil wells and the T&R center are summarized in the following table. Assume a unit change in either coordinate represents 1 mile.
X Coordinate Y Coordinate
Oil Well 1 50 150
Oil Well 2 30 40
T&R Center 230 70
Installing the pipeline is a very expensive undertaking, and the company wants to minimize the amount of pipeline required. Because the shortest distance between two points is a straight line, one of the analysts assigned to the project believes that a separate pipe should be run from each well to the T&R Center. Another alternative is to run separate pipes from each well to some intermediate substation where the two lines are joined into a single pipeline that continues on to the T&R center. Arctic Oil's Management wants to determine which alternative is best. Furthermore, if using the intermediate substation is best, management wants to determine where this station should be located.
a. Create a spreadsheet model to determine how many miles of pipeline Arctic Oil must install if it runs separate pipelines from each oil well to the T&R center. How much pipe will be needed? Use Solver.
b.If Arctic Oil wants to build a substation, where should it be built? How much pipe is needed for this solution? Use Solver.
c. Which alternaive is best?
d. Suppose the substation cannot be built within a 10-mile radius of the coordinates X=80, Y=95. (Assume the pipeline can run through this area, but the substation cannot be built in this area.) What is the optimal location of the substation now, and how much pipe will be needed? Use Solver.
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$384 \pi$
Here, we have $\iiint_E\sqrt{x^2+y^2} dV=\int_0^{2\pi} \int_{-5}^{4}\int_0^{4} r^2 dr dz d\theta$ $=\int_0^{2\pi} d\theta \cdot \int_{-5}^{4} dz \cdot \int_0^{4} r^2 dr$ $=[\theta]_0^{2\pi} [z]_{-5}^{4} \cdot [\dfrac{r^3}{3}]_0^{4}$ $=18 \pi[\dfrac{1}{3}(4)^3-\dfrac{1}{3}(0)^3]$ $=384 \pi$ | 178 | 325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-51 | latest | en | 0.263644 |
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## MATLAB Basics
#### Introduction
In MATLAB basics and implementation, we will cover the basic concepts of the MATLAB programming language and its implementation. This includes installation, paradigm, MATLAB window, variable assignment, arrays in MATLAB and their indexing, etc in great detail. Although this series of the blog is intended to develop a deep understanding of developing MATLAB programs for Image Processing, this section ‘MATLAB Basics’ is equally beneficial for those who want to learn MATLAB programming in other fields.
#### MATLAB Installation
MATLAB is a proprietary programming language, so you have to purchase it. You can follow its official website for installation or you can follow the instructions provided here or here. GNU Octave is one of the major free and open-source alternative software to MATLAB. It resembles almost all of the features and programming structures of MATLAB. You can find the installation guide for Octave here.
The paradigm of a programming language is a way of classification on the basis of programming features. There are multiple ways in which one language differs from the other. For example, whether the language is High-level or Low-level? Similarly, whether it is compiled or interpreted? or it supports object-oriented programming or not? etc. The following figure illustrates the position of some of the well-known languages including MATLAB in the programming paradigm.
The arrow (A –> B) indicates that the compiler/interpreter of language B is written in language A. So the interpreter used in MATLAB (and part of its libraries) are written in C, C++, and Java.
Introduction to MATLAB
MATLAB is a high level, object-oriented and an interpreted language. It is developed by Mathworks primarily for numerical computing. However, it became popular for developing algorithms, creating models and Data Analysis.
MATLAB is the abbreviation for Matrix Laboratory. Therefore, it is preferable to have some basic knowledge of matrices before we discuss the MATLAB basics and implementation. In the previous section of the blog, I’ve discussed the concepts of Matrix and its usage in MATLAB. You can read the blog here.
#### MATLAB Window
Let’s have a look at how a MATLAB window looks like.
In the figure, you can see four sections or sub-windows in the MATLAB window. There are other sections and windows also that we will discuss later. So let’s discuss these windows briefly.
1. Command Window – This is the main window of MATLAB where you can declare a variable, run a command and get the results. Even for the longer codes written in script files, you’ll get the result in the command window only.
2. Workspace Window – It shows the information about variables currently stored in the MATLAB memory.
3. Editor Window – Here you can write and edit your programs and also save it as a script file. When you write and save a program in the editor, it generates a file with extension .m.
4. Current Folder Window – Here you’ll see the files in the current folder. You can see the path of the current folder in the address bar in Fig. 3.2. When you run a script file and this file is not present in the current folder it will ask for two options- ‘Change Folder’ or ‘Add to Path’. It is preferable to opt for the Change Folder as it will automatically take you to the folder where the file is stored.
There are other windows also, like Figure window for displaying graphical results, Help window for getting help regarding a function or commands, etc. We will discuss them later as we progress.
#### Working on Command Window
Working on the command window is the easiest way of understanding MATLAB syntax and commands. Once you start working in the command window, you’ll realize that you can perform a lot of operations without any prior knowledge of programming. Just to begin with, let’s start with some basic arithmetic operations.
Fig. 3.3 shows a snippet of the MATLAB command window with some of the basic arithmetic operations. Also, there are some variable assignments. For example, a = 5; which means a variable namely a is created and the number 5 is assigned to this variable.
#### Variable Assignment
The following equation shows a general form of variable assignment in MATLAB.
Variable\_name = Numerical\_value \textbf{or} Expression
For example,
>> x = 5;
>> y = 2*x + 3;
We use the semicolon to suppress the output. You can observe the missing output display where a semicolon is present at the end in Fig 3.3.
After the assignment of variables a and b (in Fig 3.3), there is an addition operation using these variables. So c = a + b indicates that the variable c stores the summation of values stored in a and b. Similar is the case for the subtraction d = b – a. Since the semicolon is absent in both the operations, you find the display of output values of c and d on the screen.
#### Some Arithmetic operations in MATLAB
The following table shows the list of MATLAB operations with their symbols and examples.
#### Some built-in functions in MATLAB
You can have a look at some of the built-in functions along with their descriptions and examples in the following table.
#### Managing Variables
You can manage the variables that you have created using certain commands. This includes erasing variables, getting information about the variables, etc.
>> clear
This command clears all the variables present in the workspace.
>> clear a b c
The clear command followed by one or more variables deletes only those variables from the workspace. In this example, the command clears the variables a, b and c.
>> who
This command displays the list of variables in the memory.
>> whos
‘whos’ command displays the variables with details including name, size, and a class of the variables. Adding variable name after whos will display information about the specified variables only.
#### Creating Arrays
Arrays are a group of elements associated with a single variable. Like all other programming languages, MATLAB has its own way of creating, addressing and handling arrays. In general, we use the following syntax for creating an array.
Variable\_name = \textbf{[} Array Elements \textbf{]}
For example,
>> A = [3 5 1 4 9 0]
A = % output
3 5 1 4 9 0
>> B = [2 5 9 ; 4 1 0 ; 6 2 -1]
B = % output
2 5 9
4 1 0
6 2 -1
Here A is a row vector of length 6 and B is a 3×3 matrix. If you wish to create a column vector, you just need to put a semicolon (;) between the elements.
>> a=[1;2;3;4]
a = % output
1
2
3
4
Therefore, you should keep the following points in mind while creating arrays.
Row Vector – First write the name of the vector following an equality sign. And on the right side of it, within the square brackets write the vector elements with spacing. You can insert commas (,) also between the consecutive elements.
Column Vector – Same as row vector except that we insert semicolon (;) between the consecutive elements.
Matrix – Withing the square brackets, elements of a row are written as in the case of a row vector, that is, with spacings. While each of the rows is separated by a semicolon.
Note that if you are not familiar with the terminologies or concept of vector or/and matrix, please refer to the previous section of the blog here before we move on to the MATLAB basics and implementation further. I’ve explained these concepts in detail.
#### Creating special vectors – with constant spacing
You can create a vector with constant spacings between the consecutive elements. There are two ways in which you can do this depending upon the requirements.
1. xi:d:xf – Here xi is the first element, d is the difference between the consecutive elements, and xf is the upper limit of the last element. This syntax is used when you know the first element, the common difference between the consecutive elements, and the upper limit of the last element. So the xf may or may not be the last element of the vector. For example,
>> x = 1:2:9
x = % output
1 3 5 7 9
>> x = 2:3:15
x = % output
2 5 8 11 14
>> x = 2:0.5:5
x = % output
2.0 2.5 3.0 3.5 4.0 4.5 5.0
If you skip the d parameter and use the syntax xi:xf, the default different is taken as 1. For example,
>> x = 1:5
x = % output
1 2 3 4 5
>> x = -3:2
x = % output
-3 -2 -1 0 1 2
You can use a negative common difference also.
>> x = 10:-2:1
x = % output
10 8 6 4 2
>> x = 3:-1:-2
x = % output
3 2 1 0 -1 -2
2. linspace(xi,xf,n)– This command is used when you know the first element, the last element and you want to create n number of equally spaced elements in the vector. For example,
>> x = linspace(1,10,5)
x = % output
1.00 3.25 5.50 7.75 10.00
>> x = linspace(-2,6,5)
x = % output
-2 0 2 4 6
>>x = linspace(20,10,6)
x = % output
20 18 16 14 12 10
#### Creating 3D arrays
To create a 3D array, you need to define each plane of the array separately. For example, suppose you want to create an array with dimension (2,3,3). That is, the array has 3 planes and each of them is having 2 rows and 3 columns. Therefore, we need to create three 2D arrays of size (2,3). In MATLAB, addressing the whole elements of a particular plane is done as A(:,:, plane). Here is the name of the array. The first colon (:) indicates all rows, the second colon indicates all columns and the plane indicates the plane number. Array indexing is explained in detail in later sections.
>> A(:,:,1) = [1 2 3 ;4 5 6];
>> A(:,:,2) = [7 8 9 ; 10 11 12];
>> A(:,:,3) = [-1 -2 -3 ;-4 -5 -6];
>> A % output
A(:,:,1) =
1 2 3
4 5 6
A(:,:,2) =
7 8 9
10 11 12
A(:,:,3) =
-1 -2 -3
-4 -5 -6
#### Some Built-in Arrays
There are some commonly used built-in arrays in MATLAB.
>> Z = zeros(3,4)
Z = % output
0 0 0 0
0 0 0 0
0 0 0 0
>> N = ones(4,4)
N = % output
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
>> E = eye(3)
E = % output
1 0 0
0 1 0
0 0 1
The command ‘zeros’ produces an array with all elements being zero. The size of the array is specified within the small bracket as – zeros(no_of_rows, no_of_columns).
Similar is the case for ones where all the elements are 1. The ‘eye’ command produces an identity matrix with an equal number of rows and columns specified in the input argument.
Note that within the input arguments of ‘zeros’ and ‘ones’ if you enter only one argument, it will generate a square matrix with an equal number of rows and columns specified in the argument.
We have discussed the concept of indexing in the previous section. You can go through it quickly for better understanding.
Indexing is a way of specifying the array elements. For example, consider the following vector V.
>> V = [12 9 0 2 -3 1 8 7];
If we wish to know what the 3rd element of the vector is? so that we can use it in some other array or perform some operation on it, what should we write in the command window? The answer is the following:
>> V(3)
ans =
0
So, the syntax for addressing/indexing an element is
value\_of\_the\_element = variable\_name\textbf{(}element\_position\textbf{)}
>> x = V(5)
x = % output
-3
>> y = V(1) + V(5)
y = % output
9
>> z = V(2)*V(5)
z = % output
-27
Notice the small brackets () used for indexing. Whereas we used square brackets [] for creating arrays.
#### Multiple element indexing
Suppose you need to access the 4th to 7th element from the vector V in the above example. What should be the syntax? Have a look at the following examples.
>> x = V(4:7)
x = % output
2 -3 1 8
>> y = V(3:end)
y = % output
0 2 -3 1 8 7
#### Matrix Indexing
There are two ways in which elements of a matrix can be addressed – Linear indexing & Subscript indexing. Linear indexing requires only a single parameter for indexing while subscript indexing involves multiple parameters. A (2D) matrix needs two parameters (one for the row and another for column) for indexing.
Linear indexing counts matrix elements column-wise. That is, you start counting elements of the first column. After the end of the first column, continue the counting from the top of the second column and so on until you reach the last column.
Subscript indexing is relatively simple. You take any of the elements of the array and you find its position in terms of row and column number. The (row, column) pair would be the index of the element. In the case of the 3D array, an additional parameter- ‘plane‘ will be added. So the index of an element of a 3D array would be in the form of (row, column, plane).
The following examples illustrate the indexing methods for matrices.
>> M = [3 4 9 0 ; 14 9 25 1 ; 5 0 8 -3]
M = % output
3 4 9 0
14 9 25 1
5 0 8 -3
>> M(1) % Linear Indexing
ans =
3
>> M(1,1) % Subscipt Indexing
ans =
3
>> M(5) % Linear Indexing
ans =
9
>> M(2,2) % Subscipt Indexing
ans =
9
>> x = M(2,1) - M(1,3)
x = % output
5
>> y = M(1,2) * M(3,4)
y = % output
-12
And the example of a 3D array indexing is as follows.
>> A(:,:,1) = [1 2 3 ; 4 5 6];
>> A(:,:,2) = [7 8 9 ; 10 11 12];
>> A(:,:,3) = [-1 -2 -3 ; -4 -5 -6];
>> A
A(:,:,1) = % output
1 2 3
4 5 6
A(:,:,2) =
7 8 9
10 11 12
A(:,:,3) =
-1 -2 -3
-4 -5 -6
% indexing
>> A(4)
ans =
5
>> A(2,2,1)
ans =
5
>> A(8)
ans =
10
>> A(2,1,2)
ans =
10
#### Multiple element Array Indexing
The following examples illustrate the multiple-element array indexing.
>> M = [3 4 9 0 ; 14 9 25 1 ; 5 0 8 -3]
M = % output
3 4 9 0
14 9 25 1
5 0 8 -3
>> M(1,:) % 1st row, all columns
ans =
3 4 9 0
>> M(3,:) % 3rd row, all columns
ans =
5 0 8 -3
>> M(:,1) % All rows, 1st column
ans =
3
14
5
>> M(:,3) % All rows, 3rd column
ans =
9
25
8
>> M(:,end) % All rows, last column
ans =
0
1
-3
>> M(2:3,:) % 2-3 rows, all columns
ans =
14 9 25 1
5 0 8 -3
>> M(2:3,[2 4]) % 2-3 rows, 2 & 4 columns
ans =
9 1
0 -3
You can try multiple array indexing for the 3D array also.
#### Variable Concatenation
Variable concatenation refers to the merger/joining of two or more arrays or merger/joining of one or a few elements to the existing elements of an array. For example,
>> A = [1 2 3 4];
>> A(5:7) = ones(1,3)
A = % output
1 2 3 4 1 1 1
>> A = [1 2 3 4];
>> B = ones(1,3);
>> A = [A B] % Horizontal concatenation
A = % output
1 2 3 4 1 1 1
% Matrix concatenation
>> A = [1 2 3 4 ; 5 6 7 8]
A = % output
1 2 3 4
5 6 7 8
>> A(3,:) = 2:3:11
A = % output
1 2 3 4
5 6 7 8
2 5 8 11
>> B = eye(3)
B = % output
1 0 0
0 1 0
0 0 1
>> A = [A B]
A = % output
1 2 3 4 1 0 0
5 6 7 8 0 1 0
2 5 8 11 0 0 1
#### Transpose of an array
MATLAB uses an apostrophe ( ‘ ) following the variable name for transposing an array. For example,
>> A = [1 2 3 4]
A = % output
1 2 3 4
>> B = A’
B =
1
2
3
4
>> A = [1 2 3 ; 4 5 6 ; 7 8 9 ];
A = % output
1 2 3
4 5 6
7 8 9
>> B = A’
B = % output
1 4 7
2 5 8
3 6 9
#### Deleting Elements
You can delete one or more elements from an array. All you need to do is, mention the indices of the array you wish to delete and use [] after the equality sign. For example,
>> A = [4 9 2 -1 0 3 7 1]
A = % output
4 9 2 -1 0 3 7 1
>> A(3) = [] % [] used for deleting array elements
A = % output
4 9 -1 0 3 7
>> A(4:6) = []
A = % output
4 9 -1
>> A = [4 9 2 -1 0 ; 3 7 1 -5 4 ; 0 8 9 5 1]
A = % output
4 9 2 -1 0
3 7 1 -5 4
0 8 9 5 1
>> A(:,2:4) = []
A = % output
4 0
3 4
0 1
#### Handling arrays using built-in functions
Once you have created the arrays, you can handle these using some built-in functions. This includes obtaining the size and shape of the arrays, reshaping arrays finding diagonal of a 2D array, etc.
#### Length of an array
Function – length(A)
Description – Finds the total number of elements in a vector. If the input is a matrix, it returns the largest dimension of the matrix.
>> A = [2 4 9 0 3];
>> length(A)
ans =
5
>> B = [1 2 ; 3 4 ; 5 6]
B = % output
1 2
3 4
5 6
>> length(B)
ans =
3
#### Size of an array
Function – size(A)
Description – Returns dimension of an array. If the array is a vector or a 2D matrix, the output will be a vector of length 2 containing the number of rows and columns. If the input is a 3D array, it returns a vector of length 3 containing the rows, the columns, and the planes.
>> A = [2 4 9 0 3 ; 3 1 2 -5 0]
A = % output
2 4 9 0 3
3 1 2 -5 0
>> size(A)
ans =
2 5
#### Reshaping and array
Function – reshape(A,m,n)
Description – Reshaping of array A to the dimension mxn. The process of reshaping is also carried out column-wise. That is, MATLAB takes the first column from the existing array and places it into the first column of the output array. If some of the elements are remaining from the source or the output array, it gets adjusted in the next column and so on. For example,
>> A = [2 3 4 ; 5 6 7]
A = % output
2 3 4
5 6 7
>> B =reshape(A,3,2)
B = % output
2 6
5 4
3 7
#### Diagonal of an array
Function – diag(A)
Description – If A is a vector, the output will be a diagonal matrix with diagonal elements formed using vector A and the rest of the elements of the matrix being 0. On the other hand, if A is a square matrix it returns the diagonal elements of the matrix.
>> A = [1 5 9];
>> diag(A)
ans = 1 0 0
0 5 0
0 0 9
>> B = [1 2 3 ; 4 5 6 ; 7 8 9];
B = % output
1 2 3
4 5 6
7 8 9
>> diag(B)
ans =
1 5 9
#### The inverse of an array
Function – inv(A)
Description – Evaluates inverse of a square matrix. The inverse is a very important operation used in solving mathematical problems. For example, solving linear equations. Suppose, we need to solve the following equation. We can write this equation in matrix form in two ways:
These equations can be solved as follows:
Let’s solve this using MATLAB
>> A = [2 4 2 ; -2 -3 1 ; 2 2 -3]
A = % output
2 4 2
-2 -3 1
2 2 -3
>> B = [16 ; -5 ; -3]
B = % output
16
-5
-3
>> A_inv = inv(A)
A_inv = % output
3.5 8.0 5.0
-2.0 -5.0 -3.0
1.0 2.0 1.0
>> X = A_inv*B
X = % output
1
2
3
Similarly, you can verify the other form of the solution as well.
#### Strings
Strings are the array of characters. We write strings within single quotes. For example, ‘john’, ‘abc_07’, ‘my name is Adam’, etc. Every single character is regarded as an element of the string. Therefore, all the operations and rules apply to the strings similar to a normal array.
>> a = ‘my_name’
a = % output
‘my_name’
>> a(2)
ans =
‘y’
>> a(4:end)
ans =
‘name’
>> length(a)
ans =
7
>> a(8:10) = ‘_is’
a =
‘my_name_is’
>> b = ‘_john’;
>> a = [a b]
a = % output
‘my_name_is_john’
Ok, so now we have developed a sufficient background of the MATLAB basics and implementation. In the next section, we will discuss the mathematical operations on the arrays.
Thank you.
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http://mathhelpforum.com/statistics/191900-dice-probability-print.html | 1,521,348,076,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645513.14/warc/CC-MAIN-20180318032649-20180318052649-00004.warc.gz | 186,268,400 | 2,673 | # Dice probability
• Nov 14th 2011, 11:09 AM
mathproblems
Dice probability
If you roll twofair dice, one red and one green. Whatis the probability that you get a sum of 7 given that one die is an evennumber?
Thank you.(Nerd)
• Nov 14th 2011, 11:16 AM
mr fantastic
Re: Dice probability
Quote:
Originally Posted by mathproblems
If you roll twofair dice, one red and one green. Whatis the probability that you get a sum of 7 given that one die is an evennumber?
Thank you.(Nerd)
I suggest you use a dice table (there is one here: http://www.frontiernet.net/~rhode/QingTheory.html), restrict the sample space and then calculate the required probability. | 178 | 653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-13 | latest | en | 0.846463 |
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# Thermodynamics filled in class notes_Part_60 - 4.7...
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Unformatted text preview: 4.7. COMPRESSIBILITY AND GENERALIZED CHARTS 127 Proceeding as before, we have ∂u ∂v vextendsingle vextendsingle vextendsingle vextendsingle T = T ∂P ∂T vextendsingle vextendsingle vextendsingle vextendsingle v − P, (4.284) = T parenleftbigg R v + a 2 v 2 T 3 / 2 parenrightbigg − parenleftbigg RT v − a v 2 T 1 / 2 parenrightbigg , (4.285) = 3 a 2 v 2 T 1 / 2 . (4.286) Integrating, we find u ( T,v ) = − 3 a 2 vT 1 / 2 + f ( T ) . (4.287) Here f ( T ) is a yet-to-be-specified function of temperature only. Now the specific heat is found by the temperature derivative of u : c v ( T,v ) = ∂u ∂T vextendsingle vextendsingle vextendsingle vextendsingle v = 3 a 4 vT 3 / 2 + df dT . (4.288) Obviously, for this material, c v is a function of both T and v . Let us define c vo ( T ) via df dT ≡ c vo ( T ) . (4.289) Integrating, then one gets f ( T ) = C + integraldisplay T T o c vo ( ˆ T ) d ˆ T. (4.290) Let us take C = u o +3 a/ 2 /v o /T 1 / 2 o . Thus we arrive at the following expressions for....
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# Probability question
0
251
1
what are chances for a 1/256 event to not happen after 256 times?
Mar 2, 2020
edited by Guest Mar 2, 2020
#1
+32141
+3
p = 1/256. q = 1 - 1/256 = 255/256
prob q occurs 256 times on the trot = q256
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# 3. Processor instruction set and translation of algorithm
## Exercise outline
1. Basic instructions of the processor and their description
2. Introduction and familiarize with QtRVSim architecture simulator QtRVSim
3. Test program for simple addition
4. Following tasks extending the program
5. Transformation source code in C to assembler (RISC-V instruction set).
6. Peripheral access
## What shall we do
Gain an understanding of how are the machine instructions executed/processed by a simple processor.
The simplicity, straightforwardness, and orthogonality of the instruction encoding is an essential reason why most textbooks choose as a model architecture MIPS processor. But actual license conditions are not clean and the RAM Arch64 and RISC-V architectures are more up to date. That is why the most of the leading universities switched to RISC-V architecture. Processor directly fetches and executes instructions in the binary form. PC-class personal computers do not allow us to execute RISC-V-based machine code directly on their processor (architecture X86), but there exist many simulators of this architecture. The simulator QtRVSim was developed to support the practical experience of this course.
### Part 1 - Basic instruction - description and use
Detailed description:
Instruction Instruction Syntax Operation Description
add add rd, rs1, rs2 rd ← rs1 + rs2; Add: Add together values in two registers (rs1 + rs2) and stores the result in register rd.
addi addi rd, rd1, imm12 rd ← rs1 + imm12; Add immediate: Adds a value in rs1 and a signed constant (12-bit immediate value) and stores the result in rd.
sub sub rd,rs1,rs2 rd ← rs1 - rs2 Subtract: Subtracts a value in register rs2 from value of rs1 and stores result in rd.
bne bne rs1, rs2, Label if rs1 != rs2 go to PC+2*imm12; else go to PC+4 Branch on not equal: (conditional) jump if value in rs1 is not equal to a value in rs2 (imm12 = Label - PC)
beq beq rs1, rs2, Label if rs1 == rs2 go to PC+2*imm12; else go to PC+4 Branch on equal: (conditional) jump if value in rs1 is equal to a value in rs2 (imm12 = Label - PC)
slt slt rd,rs1,rs2 rd ← (rs1 < rs2) Set on less than: set register rd to one, if the condition rs1 < rs2 is true
slli slli rd,rs1,imm5 rd ← rs1 << imm5 Shift Left Logical: Shifts value in the register to the left by imm5 bits (ekvivalent to operation multiplication by constant 2imm5 )
j j Label PC ← PC + 2*imm20 Jump: unconditional jump to the Label (imm20 = Label - PC)
lw lw rd, imm12(rs1) [rd] ← Mem[[rs1] + imm12]; Load word: Loads a word from address in memory and stores it in register rd.
sw sw rs2, imm12(rs1) Mem[[rs1] + imm12] ← [rs2]; Store word: Stores a value in register rs2 to given address in memory.
lui lui rd, imm20 [rd] ← imm20<<12 Load upper immediate: Stores given immediate value (constant) imm20 to upper part of register rd. Register is 32 bits long and imm20 is 20 bits.
li li rd, Immediate lui rd, (Immediate+0x800)[31:12];
Load Immediate: sets register rd to 32-bitovou constant. It is pseudo-instruction - it is translated into sequence of actual instructions according to the value range.
Load Address: sets register rd 32-bit value computer relative to actual PC/instruction address. It is pseudo-instruction - it is translated into sequence of actual instructions according to the value range.
The compact decription of the RISC-V instructions riscvcard.pdf
Description of the RISC-V arcitecture at Wikipedia https://en.wikipedia.org/wiki/RISC-V.
Autoritative architecture description https://riscv.org/technical/specifications
### Translation of assembler source code to machine code
There is no restriction to use any plain text editor for writing assembler source code. In the laboratory, there are installed more text editors and development environments. For example vim, Emacs, etc. We suggest to select Geany for those who have no personal preference.
Open a text editor and prepare simple assembler source file named simple-lw-sw.S. The suffix capital “.S” is crucial. This suffix is assigned by standard development tools for the source files/code in assembler language and compiler decides according to the suffix how to process the file. Other recognized suffixes are .c for C language source files. .cc or .cpp for C++, .o for object files (files which content is already translated source code into target machine native instructions but there is not defined to which address will be these code fragments located). Library files suffix .a (archive) is used for functions libraries, which are included in final executable according to their requirements/use by other files. A final executable file (program) is stored without extension on Unix class systems..
.globl _start
.option norelax
.text
_start:
loop:
lw x2, 0x400(x0)
// store the word to absolute address
sw x2, 0x404(x0)
// stop execution wait for debugger/user
// ebreak
// ensure that continuation does not
// interpret random data
beq x0, x0, loop
nop
nop
ebreak
.data
.org 0x400
src_val:
.word 0x12345678
dst_val:
.word 0
Assembly code source file consists of
• instruction mnemonics which represent an abbreviation of operation (lw - load word, sw - save word, add, sub - subtract), the operands can follow after operation mnemonics. Operands are usually numbers or another specification of registers and immediate numeric constant operands. Operands are separated by comma.
• labels (followed by a colon). It is possible to reference these labels from some instruction operands and directives to store raw words or other data into memory. The address to the memory location following the label (instruction or data) is stored in the instruction field or other memory location in the place from which is the label referenced.
• control directives (pseudo-instructions) for assembly compiler, they usually start by a dot .
• comments, source files designated by capital .S suffix are preprocessed before actual translation to machine code/object files. The rules for preprocessing are the same as for the C compiler, and the same format of comments can be used.
The following directives are used in the provided sample code
• .globl - symbols identifiers following the directive are visible/accessible even outside actual compilation unit. The symbol _start or _ _start specifies program entry point by the convention.
• .option norelax - disable link time instruction sequences optimizations (for example generate both instructions for li x2, 35 instead of single addi x2, 35)
• .set noat - disallow use of temporary assembler register which is used for the realization of some higher level instructions translated by the assembler to the sequence of machine instructions. For example la (load address), which is macro instruction translated by assembler to the sequence of lui and ori instructions.
• .set noreorder - forbid instructions reordering by assembler. Assembler is able to optimize ordering of some instructions which allows to fill delay-slots (these are not used by initial simulator setup for simplicity), reordering is not desirable for our code
• .ent jméno - the mark for start of a function
• .end jméno - the mark for end of a function
• .text - switch to fill .text section which is used to store actual program machine instructions
• .data - switch to fill .data sections which holds initialized data/variables/arrays
• .word - the directive to store into actually filed section value specified after the directive
The complete manual for GNU assembler and all its directives can be found at GNU assembleru.
The compilation is performed by cross-compiler
riscv64-unknown-elf-gcc -Wl,-Ttext,0x200 -Wl,-Tdata,0x400 -march=rv32i -mabi=ilp32 -nostdlib simple-lw-sw.S -o simple-lw-sw
The invocation requires multiple parameters, because standard programs in C language require linking of initialization sequences and library functions. But our actual goal is to describe the lowest level without this automatization the first which allows understanding how it is extended and equipped by automatic initialization sequences and constructs which allows comfortable code writing of programs at a higher level and in higher level languages. Parameters -Wl, are passed to the linker program and specify to which address is located .text section with program instructions and to where starts .data section with initialized variables. Following parameters in the order of their appearance disable automatic addition of startup and initialization sequence (-nostartfiles) and disable use of standard libraries. The file name after switch -o specifies the name of the final executable file (output). File names of one or more source files follow.
The simple-lw-sw project with appropriate Makefile can be found in /opt/apo/qtrvsim/qtrvsim_template directory.
### File buble_sort.S
Use buble_sort.S template for internal QtRVSim assembler compiler:
// Directives to make interresting windows visible
#pragma qtrvsim show registers
#pragma qtrvsim show memory
.globl _start
.globl array_size
.globl array_start
.option norelax
.text
_start:
la a0, array_start
la a1, array_size
lw a1, 0(a1) // number of elements in the array
//Final infinite loop
end_loop:
ebreak // stop the simulator
j end_loop
nop
.data
// .align 2 // not supported by QtRVSim yet
array_size:
.word 15
array_start:
.word 5, 3, 4, 1, 15, 8, 9, 2, 10, 6, 11, 1, 6, 9, 12
// Specify location to show in memory window
#pragma qtrvsim focus memory array_start
For C code todo assembly language manual translation try file start.S:
int pole[10];
int main() {
int N = 10,i;
for(i=0; i<N; i++) {
pole[i]=N-i;
}
return 0;
}
together with file start.S which executes function main from init_array.c:
.globl _start
.text
.option norelax
_start:
la x2, _end+0x4000
la x3, __global_pointer$jal main ebreak Překlad do assembleru pro MIPS a zobrazení přeložených instrukcí provedeme následujícími příkazy. riscv64-unknown-elf-gcc -march=rv32i -mabi=ilp32 -g -c init_array.c -o init_array.o riscv64-unknown-elf-gcc -march=rv32i -mabi=ilp32 -g -c start.S -o start.o riscv64-unknown-elf-gcc -Wl,-Ttext,0x200 -Wl,-Tdata,0x400 -march=rv32i -mabi=ilp32 -nostdlib init_array.o start.o -o init_array riscv64-unknown-elf-objdump --source init_array On Windows, complete MSys with make utility can be installed or plain compiler mips-elf-gcc-i686-mingw32 can be called from following batch file. PATH=%PATH%;c:\path\to\riscv64-unknown-elf-gcc\bin riscv64-unknown-elf-gcc -Wl,-Ttext,0x200 -Wl,-Tdata,0x400 -march=rv32i -mabi=ilp32 -nostdlib -g init_array.c -o init_array QtRVSim simulator is used to execute the program. Select the most simple variant of simulated processor “No pipeline no cache”. Button “Browse” is used to select executable file name (simple-lw-sw without suffix in our case) for field “Elf executable”. The diagram with the processor is opened. Use double-click on program counter register (PC) opens program listing with actual instructions. Double click on registers blocks opens a view with the list of architectural registers. Double click on data memory shows memory content. The windows layout shown on the next picture is the appropriate and intuitive starting point Select “Follow fetch” option in the “Program” listing window which highlights instruction/line actually fetched by the processor for execution. The start of listing in “Memory” windows should be set to the address 0x2000 on which data value 0x12345678 was placed. The program can be stepped through by “Machine” → “Step” menu entry or by the corresponding button on the toolbar. The value is loaded to the register first, and then it is stored to the following memory cell. Change the program to process a load-store sequence in the loop. Instruction ebreak has to be removed from the loop because its purpose is to stop program execution when reached. Test that edited value at address 0x2000 is always copied to the address 0x2004. Modify the program to add two input values on 0x2000 and 0x2004 address and stored the result at address 0x2008. Next step is to modify the program to add two vectors with a length of four words. Use assembler macroinstruction la vect_a (load address) to set registers to point to the start of the vectors. ... .data vect_a: .word 0x12345678 .word 0x12345678 .word 0x12345678 .word 0x12345678 vect_b: .word 0x12345678 ... Continue with implementation of the program to compute an average value from eight numbers. ### Automate program compilation by use of Makefile The compiler invocation is desirable to document at least and better automate. The one way is to use a script with the sequence of commands required for compilation. Such script can be written directly for shell - command line interpreter (BASH or DASH on GNU/Linux usually). But it is not practical to translate all compilation units of a larger project when only small change modifies only one or a small subset of source files. More different systems have been developed to automate exactly these tasks. Some examples are Make, Ant, qmake, Cmake, meson, etc. #### Makefile Make is the tool which allows to automate compilation of source codes, description of the compilation process is described in Makefile. Makefile template for the compilation of source files written in assembly language or C language for RISC-V simulator environment: ARCH=riscv64-unknown-elf SOURCES = change_me.S TARGET_EXE = change_me CC=$(ARCH)-gcc
CXX=$(ARCH)-g++ AS=$(ARCH)-as
LD=$(ARCH)-ld OBJCOPY=$(ARCH)-objcopy
ARCHFLAGS += -mabi=ilp32
ARCHFLAGS += -march=rv32i
ARCHFLAGS += -fno-lto
CFLAGS += -ggdb -Os -Wall
CXXFLAGS+= -ggdb -Os -Wall
AFLAGS += -ggdb
LDFLAGS += -ggdb
LDFLAGS += -nostartfiles
LDFLAGS += -nostdlib
LDFLAGS += -static
#LDFLAGS += -specs=/opt/musl/riscv64-linux-gnu/lib/musl-gcc.specs
CFLAGS += $(ARCHFLAGS) CXXFLAGS+=$(ARCHFLAGS)
AFLAGS += $(ARCHFLAGS) LDFLAGS +=$(ARCHFLAGS)
OBJECTS += $(filter %.o,$(SOURCES:%.S=%.o))
OBJECTS += $(filter %.o,$(SOURCES:%.c=%.o))
OBJECTS += $(filter %.o,$(SOURCES:%.cpp=%.o))
all : default
.PHONY : default clean dep all
%.o:%.S
$(CC) -D__ASSEMBLY__$(AFLAGS) -c $< -o$@
%.o:%.c
$(CC)$(CFLAGS) $(CPPFLAGS) -c$< -o $@ %.o:%.cpp$(CXX) $(CXXFLAGS)$(CPPFLAGS) -c $< %.s:%.c$(CC) $(CFLAGS)$(CPPFLAGS) -S $< -o$@
default : $(TARGET_EXE)$(TARGET_EXE) : $(OBJECTS)$(CC) $(LDFLAGS)$^ -o $@ dep: depend depend:$(SOURCES) $(glob *.h) echo '# autogenerated dependencies' > depend ifneq ($(filter %.S,$(SOURCES)),)$(CC) -D__ASSEMBLY__ $(AFLAGS) -w -E -M$(filter %.S,$(SOURCES)) \ >> depend endif ifneq ($(filter %.c,$(SOURCES)),)$(CC) $(CFLAGS)$(CPPFLAGS) -w -E -M $(filter %.c,$(SOURCES)) \
>> depend
endif
ifneq ($(filter %.cpp,$(SOURCES)),)
$(CXX)$(CXXFLAGS) $(CPPFLAGS) -w -E -M$(filter %.cpp,$(SOURCES)) \ >> depend endif clean: rm -f *.o *.a$(OBJECTS) (TARGET_EXE) depend #riscv64-unknown-elf --source qtrvsim_binrep -include depend tab character has to be used to indent of commands in the Makefile. Makefile does not recognize indent by spaces. Makefile consists from definitions (assignment of values to variables) and rules. The rules start by a line which defines dependency of rule target(s) on the dependencies listed after the colon. The dependencies are names of files or abstracts commands which as to be (make) available before the commands following the first rule line can create required results. File names can be complete names or their base part can be substituted by character “%” which allows specifying rule for a whole class of transformations from one compilation stage to another. Even more complete template for the compilation of assembler and C source files to RISC-V target platform with an automatic building of dependencies on header files can be found in directory /opt/apo/qtrvsim_template on the computers in the laboratory. #### Compilation A compilation is invoked by a command make (the make has to be invoked in the directory where Makefile and program source code are located). Make generates multiple output files. The file without an extension is used for execution in QtRVSIm environment. The process of compilation translated the compilation unit (one source wile together with includes header files for a simple case) into object files (.o) in relocatable form. Object files are then collected by the linker which resolves address references between compilation units and locates code to the final addresses. It is necessary to equip the actual sequences of machine instructions by the envelope which specifies where to fill references during .o files linking during the final placement on the specified addresses. Even instruction and data in the final executable form usually require some information for operating systems where they should be loaded/mapped in the memory or process address-space. The ELF (Executable and Linkable Format) is used to store these metadata in our case and generally on most of the modern systems. #### Determine variable addresses and correspondence of source code and final ELF executable Next command can be used to find addresses of the final location of variables and data entries after linking riscv64-unknown-elf-nm program The list of sections and their locations can be listed by riscv64-unknown-elf-objdump --headers program List final machine code after translation with corresponding source lines riscv64-unknown-elf-objdump --source program ## Tasks 1. learn how to use compiler and simulator 2. change program to run memory read and write in a loop (try to modify program directly in the simulator, change break to nop) 3. computation with vector - return to the source and define vectors vec_a, vec_b, vec_c - four word elements each and use lw, sw and add instructions to compute vec_c[0] = vec_a[0] + vec_b[0] 4. extend program to compute sum for each vec_c element in introduced the inner loop 5. implementation of program to compute n-th member of Fibonacci series 6. start of the work on bubble-sort algorithm • GNU MIPS-ELF corss-compiler - compiler of C language and assemblr, packages for Debian GNU/Linux x86_64, i586 and Windows, configuration and build of compiler form its source code • gcc-binutils-newlib-mips-elf_4.4.4-1_mingw32.zip - MIPS-ELF GCC 4.4.4 compiled for MinGW32 (Windows) - it requires to use switches -lm -lgcc -lc for larger programs, for simple programs in assembler without libraries, next options are required -nostdlib -nodefaultlibs -nostartfiles • MIPS processor description - including instruction set • Missouri State University - Nice simulator in Java language • It requires assembly without macro-definitions (preprocessed). gcc -E assembler.S -o preprocessed-pro-mips.s can be used for preprocessing. ### Part 2 - Transcribe a program from C to Assembler In many practical applications we have to use median filter. This median filter removes noise (obvious outliers/dead pixels) from a signal or an image. The median filter takes a neighborhood of a sample (10 samples before and 10 after), finds median value and replaces the sample value with this median. Very similar to this filter is mean filter that replaces the sample value with average value of the nearby samples. The median value is usually calculated by sorting the samples by value and picking the sample in the middle. The sorting algorithm is a cornerstone to median filter implementation. Lets assume we have 21 integers stored in array in memory. The array begins in some given address (e.g. 0x00). On integer occupies one word in the memory. The task is to sort the integers in ascending order. To do this we will implement the bubble sort algorithm. In this algorithm two adjacent values are compared and if they are in wrong order, they are swapped. And this comparisons goes repetitively through array until no swaps are done. The code for bubble sort is bellow: int pole[5]={5,3,4,1,2}; int main() { int N = 5,i,j,tmp; for(i=0; i<N; i++) for(j=0; j<N-1-i; j++) if(pole[j+1]<pole[j]) { tmp = pole[j+1]; pole[j+1] = pole[j]; pole[j] = tmp; } return 0; } The example of sorting 5 numbers is bellow: 5, 3, 4, 1, 2 –> initial state 3, 4, 1, 2, 5 –> after the first outer cycle finished 3, 1, 2, 4, 5 –> after the second outer cycle finished 1, 2, 3, 4, 5 1, 2, 3, 4, 5 1, 2, 3, 4, 5 –> after the last outer cycle finished - sorted Transcribe C code above to RISC-V assembler. Verify correctness of your implementation in RISC-V simulator. We will be using this program in the next class. So finish the program at home, if you have not finished it during the class. Here is a template, you can use: .globl array .data .align 2 array: .word 5 3 4 1 2 .text .globl start .ent start start: // TODO: Write your code here nop .end start ### How to transcribe short fragments of C code into assembler if Command if (i ==j) f = g + h; f = f – i; // s0=f, s1=g, s2=h, s3=i, s4=j bne s3, s4, L1 // If i!=j, go to label L1 add s0, s1, s2 // if block: f=g+h L1: sub s0, s0, s3 // f = f-i if-else Command if (i ==j) f = g + h; else f = f – i; // s0=f, s1=g, s2=h, s3=i, s4=j bne s3, s4, else // If i!=j, go to **else** label add s0, s1, s2 // if block: f=g+h j L2 // jump behind the **else** block else: sub s0, s0, s3 // else block: f = f-i L2: while Cycle int pow = 1; int x = 0; while(pow != 128) { pow = pow*2; x = x + 1; } // s0=pow, s1=x addi s0, 0, 1 // pow = 1 addi s1, 0, 0 // x = 0 addi t0, 0, 128 // t0 = 128 to compare (always have to compare two registers) while: beq s0, t0, done // If pow==128, end the cycle. Go to done label. slli s0, s0, 1 // pow = pow*2 addi s1, s1, 1 // x = x+1 j while done: for Cycle int sum = 0; for(int i=0; i!=10; i++) { sum = sum + i; } //Is equivalent to following while cycle: int sum = 0; int i = 0; while(i!=10){ sum = sum + i; i++; } // nut even there is do-while cycle faster // because inial sum and i are given by constants. // and execution of the fisrt cycle is guaranteed. // If they are not fixed for the first iteration // then aditional if can be used int sum = 0; int i = 0; do { sum = sum + i; i++; } while(i!=10) Read values from the data memory. // Just as an example... int a, *pa=0x80020040; int b, *pb=0x80020044; int c, *pc=0x00000124; a = *pa; b = *pb; c = *pc; // s0=pa (Base address), s1=a, s2=b, s3=c lui s0, 0x80020 // pa = 0x80020000; lw s1, 0x40(s0) // a = *pa; lw s2, 0x44(s0) // b = *pb; addi s0, 0, 0x124 // pc = 0x00000124; lw s3, 0x0(s0) // c = *pc; Increment values in an array int array[4] = { 7, 2, 3, 5 }; int main() { int i,tmp; for(i=0; i<4; i++) { tmp = array[i]; tmp += 1; pole[i] = tmp; } return 0; } Complete code for QtRVSim simulator: .globl array // label "array" is declared as global. It is visible from all files in the project. .data // directive indicating start of the data segment .align 2 // set data alignment to 4 bytes array: // label - name of the memory block .word 7, 2, 3, 5 // values in the array to increment... .text // beginning of the text segment (or code segment) .globl start start: la s0, array // store address of the "array" to the register s0 addi s1, zero, 0 // initialization instruction of for cycle: i=0, kde i=s1 addi s2, zero, 4 // set the upper bound for cycle for: beq s1, s2, done // if s1 == s2, go to label done and break the cycle lw s3, 0x0(s0) // load value from the array to s3 add s3, s3, 0x1 // increment the s3 register sw s3, 0x0(s0) // replace (store) value from s3 register addi s0, s0, 0x4 // increment offset and move to the other value in the array addi s1, s1, 0x1 // increment number of passes through the cycle (i++). j for // jump to **for** label done: nop .end start ### Peripherals mapped into memory address space QtRVSim simulator includes a few simple peripherals which are mapped into memory address space. The first is simple serial port (UART) connected to terminal window. The registers locations and bit fields is the same as for simulators SPIM and MARS. These maps serial port from address 0xffff0000. QtRVSim maps the UART peripheral to this address as well but offers alternative mapping to address 0xffffc000 which can be encoded as absolute address into LW and SW instructions with zero base register. Address Register name Bit Description 0xffffc000 SERP_RX_ST_REG Serial port receiver status register 0 Flag set to one when there is new received character in SERP_RX_DATA_REG register 1 When set to one enables interrupt from reception detailed 0xffffc004 SERP_RX_DATA_REG 7 .. 0 ASCII code of received character 0xffffc000 SERP_TX_ST_REG Status register of transmitter writing to terminal 0 When one is read, transmitter is ready to accept character 1 When set to one enables transmitter interrupt detailed 0xffffc004 SERP_TX_DATA_REG 7 .. 0 ASCII code of character to transmit The next peripherals emulates interaction with simple control elements of a real device. The registers map matches to the subset of registers of dial knobs and LËD indicators peripheral which is available for input and output on a development kits MicroZed APO which are used for your semester work. Address register name Bit Description 0xffffc104 SPILED_REG_LED_LINE 31 .. 0 The word shown in binary, decimal and hexadecimal 0xffffc110 SPILED_REG_LED_RGB1 23 .. 0 PWM duty cycle specification for RGB LED 1 components 23 .. 16 Red component R 15 .. 8 Green component G 7 .. 0 Blue component B 0xffffc114 SPILED_REG_LED_RGB2 23 .. 0 PWM duty cycle specification for RGB LED 2 components 23 .. 16 Red component R 15 .. 8 Green component G 7 .. 0 Blue component B 0xffffc124 SPILED_REG_KNOBS_8BIT 31 .. 0 Filtered values of dial knobs as 8 numbers 7 .. 0 Blue dial value B 15 .. 8 Green dial value G 23 .. 16 Red dial value R ### Analysis of Compiled Code A simple program reads position of the simulator knobs dials and converts the read values to the RGB led color and text/terminal output. Program is available from directory /opt/apo/qtrvsim_binrep on the laboratory computers. There is available archive to download as well qtrvsim_binrep.tar.gz. The C source code has been compiled by the following commands sequence riscv64-unknown-elf-gcc -D__ASSEMBLY__ -ggdb -mabi=ilp32 -march=rv32i -fno-lto -c crt0local.S -o crt0local.o riscv64-unknown-elf-gcc -ggdb -Os -Wall -mabi=ilp32 -march=rv32i -fno-lto -c qtrvsim_binrep.c -o qtrvsim_binrep.o riscv64-unknown-elf-gcc -ggdb -nostartfiles -nostdlib -static -mabi=ilp32 -march=rv32i -fno-lto crt0local.o qtrvsim_binrep.o -lgcc -o qtrvsim_binrep Alternative compilation for RISC-V when picolibc library is used. riscv64-unknown-elf-gcc -march=rv32i -mabi=ilp32 --specs=/opt/picolibc/lib/riscv64-unknown-elf/specs/picolibc.specs /opt/apo/binrep/qtrvsim_binrep/qtrvsim_binrep.c -o qtrvsim_binrep The content of the program compiled into ELF executable format is examined by objdump command riscv64-unknown-elf --source qtrvsim_binrep The same task for the MIPS architecture. mips-elf-gcc -D__ASSEMBLY__ -ggdb -fno-lto -c crt0local.S -o crt0local.o mips-elf-gcc -ggdb -Os -Wall -fno-lto -c qtmips_binrep.c -o qtmips_binrep.o mips-elf-gcc -ggdb -nostartfiles -static -fno-lto crt0local.o qtmips_binrep.o -o qtmips_binrep There is output with detailed commentaries included. qtmips_binrep: file format elf32-bigmips Disassembly of section .text: 00400018 <main>: /* * The main entry into example program */ int main(int argc, char *argv[]) { 400018: 27bdffe8 addiu29,$29,-24 allocate space on the stack for main() function stack frame 40001c: afbf0014 sw$31,20($29) save previous value of the return address register to the stack. while (1) { uint32_t rgb_knobs_value; unsigned int uint_val; rgb_knobs_value = *(volatile uint32_t*)(mem_base + SPILED_REG_KNOBS_8BIT_o); 400020: 8c04c124 lw$4,-16092($0) Read value from the address corresponding to the sum of "SPILED_REG_BASE" and "SPILED_REG_KNOBS_8BIT_o" peripheral register offset LW is instruction to load the word. Address is formed from the sum of register$0 (fixed zero) and -16092,
which is represented in hexadecimal as 0xffffc124
i.e., sum of 0xffffc100 and 0x24. The read value is
stored in register $4. 400024: 00000000 sll$0,$0,0x0 one NOP instruction to ensure that load finishes before the further value use. 400028: 00041027 nor$2,$0,$4
Compute bit complement "~" of the value in the register
$4 and store it into register$2
*(volatile uint32_t*)(mem_base + SPILED_REG_LED_LINE_o) = rgb_knobs_value;
40002c: ac04c104 sw $4,-16124($0)
Store RGB knobs values from register $4to the "LED" line register which is shown in binary decimal and hexadecimal on the QtMips target. Address 0xffffc104 *(volatile uint32_t*)(mem_base + SPILED_REG_LED_RGB1_o) = rgb_knobs_value; 400030: ac04c110 sw$4,-16112($0) Store RGB knobs values to the corresponding components controlling a color/brightness of the RGB LED 1 Address 0xffffc110 *(volatile uint32_t*)(mem_base + SPILED_REG_LED_RGB2_o) = ~rgb_knobs_value; 400034: ac02c114 sw$2,-16108($0) Store complement of RGB knobs values to the corresponding components controlling a color/brightness of the RGB LED 2 Address 0xffffc114 /* Assign value read from knobs to the basic signed and unsigned types */ uint_val = rgb_knobs_value; the read value resides in the register 4, which correspond to the first argument register a0 /* Print values */ serp_send_hex(uint_val); 400038: 0c100028 jal 4000a0 <serp_send_hex> 40003c: 00000000 sll$0,$0,0x0 call the function to send hexadecimal value to the serial port, one instruction after JAL is executed in its delay-slot, PC pointing after this instruction (0x400040) is stored to the register 31, return address register serp_tx_byte('\n'); 400040: 0c100020 jal 400080 <serp_tx_byte> 400044: 2404000a addiu$4,$0,10 call routine to send new line character to the serial port. The ASCII value corresponding to '\n' is set to argument a0 register in delay slot of JAL. JAL is decoded and in parallel instruction addiu$4,$0,10 is executed then PC pointing to the address 0x400048 after delay slot is stored to return address register and next instruction is fetch from the JAL instruction target address, start of the function serp_tx_byte 400048: 1000fff5 beqz$0,400020 <main+0x8>
40004c: 00000000 sll $0,$0,0x0
branch back to the start of the loop reading value from
the knobs
00400050 <_start>:
la $gp, _gp 400050: 3c1c0041 lui$28,0x41
400054: 279c90e0 addiu $28,$28,-28448
Load global data base pointer to the global data
base register 28 - gp.
Symbol _gp is provided by linker.
addi $a0,$zero, 0
400058: 20040000 addi $4,$0,0
Set regist a0 (the first main function argument)
to zero, argc is equal to zero.
addi $a1,$zero, 0
40005c: 20050000 addi $5,$0,0
Set regist a1 (the second main function argument)
to zero, argv is equal to NULL.
jal main
400060: 0c100006 jal 400018 <main>
nop
400064: 00000000 sll $0,$0,0x0
Call the main function. Return address is stored
in the ra ($31) register. 00400068 <quit>: quit: addi$a0, $zero, 0 400068: 20040000 addi$4,$0,0 If the main functio returns, set exit value to 0 addi$v0, $zero, 4001 /* SYS_exit */ 40006c: 20020fa1 addi$2,$0,4001 Set system call number to code representing exit() syscall 400070: 0000000c syscall Call the system. 00400074 <loop>: loop: break 400074: 0000000d break If there is not a system try to stop the execution by invoking debugging exception beq$zero, $zero, loop 400078: 1000fffe beqz$0,400074 <loop>
nop
40007c: 00000000 sll $0,$0,0x0
If even this does not stop execution, command CPU
to spin in busy loop.
void serp_tx_byte(int data)
{
00400080 <serp_tx_byte>:
400080: 8c02c008 lw $2,-16376($0)
400084: 00000000 sll $0,$0,0x0
Read serial port transmit status register,
400088: 30420001 andi $2,$2,0x1
40008c: 1040fffc beqz $2,400080 <serp_tx_byte> 400090: 00000000 sll$0,$0,0x0 Wait again till UART is ready to accept character - bit 0 is not zero. NOP in the delayslot. *(volatile uint32_t *)(base + reg) = val; 400094: ac04c00c sw$4,-16372($0) write value from register 4 (the first argument a0) to the address 0xffffc00c (SERP_TX_DATA_REG_o) serial port tx data register. } 400098: 03e00008 jr$31
40009c: 00000000 sll $0,$0,0x0
jump/return back to continue in callee program
from the return address register 32 ra
void serp_send_hex(unsigned int val)
{
004000a0 <serp_send_hex>:
4000a0: 27bdffe8 addiu $29,$29,-24
allocate space on the stack for the routine stack frame
4000a4: 00802825 or $5,$4,$0 copy value of the fisrt argument regsiter 4 (a0) to the register 5 for (i = 8; i > 0; i--) { 4000a8: 24030008 addiu$3,$0,8 set the value of the register 3 to the 8 4000ac: afbf0014 sw$31,20($29) save previous value of the return address register to the stack. char c = (val >> 28) & 0xf; 4000b0: 00051702 srl$2,$5,0x1c shift value in register 5 right by 28 bits and store result in the register 2 4000b4: 304600ff andi$6,$2,0xff abundant operation to limit value range to the character type variable and store result in the register 6 if (c < 10 ) 4000b8: 2c42000a sltiu$2,$2,10 set register 2 to one if the value is smaller than 10 c += 'A' - 10; 4000bc: 10400002 beqz$2,4000c8 <serp_send_hex+0x28>
4000c0: 24c40037 addiu $4,$6,55
if value is larger or equal (register 2 is 0/false) then add
value 55 ('A' - 10)..(0x41 - 0xa) = 0x37 = 55 to the register
6 and store result in the register 4. This operation is
executed even when the branch arm before else is executed,
but result is immediately overwritten by next instruction
c += '0';
4000c4: 24c40030 addiu $4,$6,48
add value 0x30 = 48 = '0' to the value in the register 6
and store result in the register 4 - the fisrt argument a0
serp_tx_byte(c);
4000c8: 0c100020 jal 400080 <serp_tx_byte>
4000cc: 2463ffff addiu $3,$3,-1
call subroutine to send byte to the serial port
decrement loop control variable (i) in delay-slot
for (i = 8; i > 0; i--) {
4000d0: 1460fff7 bnez $3,4000b0 <serp_send_hex+0x10> 4000d4: 00052900 sll$5,$5,0x4 the final condition of for loop converted to do {} while() loop. If not all 8 character send loop again. Shift left value in the register 5 by 4 bit positions. The compiler does not store values of local variables to the stack even does not store values in caller save registers (which requires to save previous values to the function stack frame). Compiler can use this optimization because it knows registers usage of called function serp_tx_byte(). } 4000d8: 8fbf0014 lw$31,20($29) 4000dc: 00000000 sll$0,$0,0x0 restore return address register value to that found at function start 4000e0: 03e00008 jr$31
4000e4: 27bd0018 addiu $29,$29,24
• gcc-binutils-newlib-mips-elf_4.4.4-1_mingw32.zip - GNU Compatible compiler for MIPS architecture for MS Windows with MinGW32. To compile programs for MipsIT simulator you will need to specify following parameters: -nostdlib -nodefaultlibs -nostartfiles -Wl,-Ttext,0x80020000. For more complex programs you will probably have to specify -lm -lgcc -lc parameters.
gcc -E assembler.S -o preprocessed_assembler.s | 9,733 | 35,194 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.839137 |
https://mathspace.co/textbooks/syllabuses/Syllabus-407/topics/Topic-7221/subtopics/Subtopic-96467/ | 1,638,804,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00361.warc.gz | 462,725,155 | 52,838 | # Translations
Lesson
## Translations
A translation is what occurs when we move an object or shape from one place to another without changing its size, shape or orientation. Sometimes called a SLIDE, a translation moves every point on an object or shape the same distance in the same direction.
Have a quick play with this interactive. Here you can move the balloon horizontally (left or right) or vertically (up or down). Notice how the balloon does not change shape, it just slides across the page from one place to another.
To identify and describe a translation
• firstly look for corresponding points on the object
• identify if the object has moved horizontally or vertically, and then describe that movement as left/right or up/down
• count the number of places (or units) that the object has moved
#### Example
Describe the translation from A to B.
1. Firstly look for corresponding points on the object
• I've highlighted a pair of corresponding points in yellow
2. Identify if the object has moved horizontally or vertically, and then describe that movement as left/right or up/down
• I can see that the shape has a horizontal translation and that from A to B, the object moves left.
3. Count the number of places (or units) that the object has moved
• Counting the units using my corresponding points I can see that it has moved 3 units.
So the translation from A to B is 3 units left.
This applet shows the translation of an object to its image. You can use the sliders to change the horizontal and vertical amounts.
Let's have a look at the following worked solutions.
##### Question 1
What is the translation from triangle $B$B to triangle $A$A?
1. $2$2 units left
A
$2$2 units right
B
$3$3 units left
C
$3$3 units right
D
$2$2 units left
A
$2$2 units right
B
$3$3 units left
C
$3$3 units right
D
##### Question 2
What is the translation from square $B$B to square $A$A?
1. $3$3 units up
A
$3$3 units down
B
$4$4 units up
C
$4$4 units down
D
$3$3 units up
A
$3$3 units down
B
$4$4 units up
C
$4$4 units down
D
##### Question 3
1. $2$2 units left and $4$4 units down
A
$4$4 units left and $2$2 units down
B
$2$2 units right and $4$4 units up
C
$4$4 units right and $2$2 units up
D
$2$2 units left and $4$4 units down
A
$4$4 units left and $2$2 units down
B
$2$2 units right and $4$4 units up
C
$4$4 units right and $2$2 units up
D
### Outcomes
#### GM4-8
Use the invariant properties of figures and objects under transformations (refl ection, rotation, translation, or enlargement) | 695 | 2,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-49 | longest | en | 0.897284 |
https://www.wo1bornem.be/Feb-05/many-tonne-for-meter-cube-of-crusher/8076.html | 1,586,003,395,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370521876.48/warc/CC-MAIN-20200404103932-20200404133932-00259.warc.gz | 1,222,130,969 | 6,647 | many tonne for meter cube of crusher
Language:
Currency : £ \$
# many tonne for meter cube of crusher
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1 meter square of crusher run stone to tons How many tons of crusher run in one cubic meters,Wiki Answers If one brick was one cubic meter and weighed a ton, crusher run and stone dust 2" deep 1 ton read more Convert Cubic Yards Of Crusher Run To Tons | 1,797 | 7,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-16 | latest | en | 0.817961 |
http://exercism.io/exercises/kotlin/saddle-points/readme | 1,513,184,324,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00226.warc.gz | 102,503,208 | 3,729 | Detect saddle points in a matrix.
So say you have a matrix like so:
1 2 3 4 5 0 1 2 |--------- 0 | 9 8 7 1 | 5 3 2 <--- saddle point at (1,0) 2 | 6 6 7
It has a saddle point at (1, 0).
It's called a "saddle point" because it is greater than or equal to every element in its row and less than or equal to every element in its column.
A matrix may have zero or more saddle points.
Your code should be able to provide the (possibly empty) list of all the saddle points for any given matrix.
Note that you may find other definitions of matrix saddle points online, but the tests for this exercise follow the above unambiguous definition.
## Source
J Dalbey's Programming Practice problems http://users.csc.calpoly.edu/~jdalbey/103/Projects/ProgrammingPractice.html
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise. | 237 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-51 | longest | en | 0.89211 |
https://www.powershow.com/view4/7aa9bc-NWFjN/III_Completely_Randomized_Design_CRD_powerpoint_ppt_presentation | 1,603,411,707,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00046.warc.gz | 874,427,920 | 35,179 | # III.Completely Randomized Design (CRD) - PowerPoint PPT Presentation
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## III.Completely Randomized Design (CRD)
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### III. Completely Randomized Design (CRD) III.A Design of a CRD III.B Models and estimation for a CRD III.C Hypothesis testing using the ANOVA method – PowerPoint PPT presentation
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Transcript and Presenter's Notes
Title: III.Completely Randomized Design (CRD)
1
III. Completely Randomized Design (CRD)
• III.A Design of a CRD
• III.B Models and estimation for a CRD
• III.C Hypothesis testing using the ANOVA method
• III.D Diagnostic checking
• III.E Treatment differences
2
III.A Design of a CRD
• Definition III.1 An experiment is set up using
a CRD when each treatment is applied a specified,
possibly unequal, number of times, the particular
units to receive a treatment being selected
completely at random.
• Example III.1 Rat experiment
• Experiment to investigate 3 rat diets with 6
rats
• Diet A, B, C will have 3, 2, 1 rats,
respectively.
3
Use R to obtain randomized layouts
• How to do this is described in Appendix B,
Randomized layouts and sample size computations
in , for all the designs that will be covered in
this course, and more besides.
4
R functions and output to produce randomized
layout
• gt Obtaining randomized layout for a CRD
• gt
• gt n lt- 6
• gt CRDRat.unit lt- list(Rat n)
• gt Diet lt- factor(rep(c("A","B","C"),
• gt times
c(3,2,1)))
• gt CRDRat.lay lt- fac.layout(unrandomizedCRDRat.uni
t,
• gt randomizedDiet,
seed695)
• gt CRDRat.lay
• fac.layout from dae package produces the
randomized layout.
• unrandomized gives the single unrandomized factor
indexing the units in the experiment.
• randomized specifies the factor, Diets, that is
to be randomized.
• seed is used so that the same randomized layout
for a particular experiment can be generated at a
later date. (01023)
5
Randomized layout
• Units Permutation Rat Diet
• 1 1 4 1 A
• 2 2 1 2 C
• 3 3 5 3 B
• 4 4 3 4 A
• 5 5 6 5 A
• 6 6 2 6 B
• remove Diet object in workspace to avoid using
it by mistake
• remove(Diet)
6
III.B Models and estimation for a CRD
• The analysis of CRD experiments uses
• least-squares or maximum likelihood estimation of
the parameters of a linear model
• hypothesis testing based on the ANOVA method or
maximum likelihood ratio testing.
• Use rat experiment to investigate linear models
and the estimation of its parameters.
7
a) Maximal model
• Definition III.2 The maximal expectation model
is the most complicated model for the expectation
that is to be considered in analysing an
experiment.
• We first consider the maximal model for the CRD.
8
Example III.1 Rat experiment (continued)
• Suppose, the experimenter measured the liver
weight as a percentage of total body weight at
the end of the experiment.
• The results of the experiment are as follows
• The analysis based on a linear model, that is
• Trick is what are X and q going to be?
9
Perhaps?
• Note numbering of Y's does not correspond to
Rats does not affect model but neater.
• This model can then be fitted using simple linear
regression techniques.
10
Using model to predict
• However, does this make sense?
• means that, for each unit increase in diet,
liver weight decreases by 0.120.
• sensible only if the diets differences are based
on equally spaced levels of some component
• For example, if the diets represent 2, 4 and 6 mg
of copper added to each 100g of food
• But, no good if diets unequally spaced (2, 4, and
10 mg Cu added) or diets differ qualitatively.
11
Regression on indicator variables
• In this method the explanatory variables are
called factors and the possible values they take
levels.
• Thus, we have a factor Diet with 3 levels A, B,
C.
• Definition III.3 Indicator variables are formed
from a factor
• create a variable for each level of the factor
• the values of a variable are either 1 or 0, 1
when the unit has the particular level for that
variable and 0 otherwise.
12
Indicator-variable model
• Hence
• EYi ak, varYi s2, covYi, Yj 0, (i ?
j)
• Can be written as
• Model suggests 3 different expected (or mean)
values for the diets.
13
General form of X for CRD
• For the general case of a set of t Treatments
suppose Y is ordered so that all observations
for
• 1st treatment occur in the first r1 rows,
• 2nd treatment occur in the next r2 rows,
• and so on with the last treatment occurring in
the last rt rows.
• i.e. order of systematic layout (prior to
randomization)
• Then XT given by the following partitioned matrix
(1 only ever vector, but 0 can be matrix)
14
Still a linear model
• In general, the model for the expected values is
still of the general form EY Xq
• and on assuming Y is
• can use standard least squares or maximum
likelihood estimation
15
Estimates of expectation parameters
• OLS equation is
• Can be shown, by examining the OLS equation, that
the estimates of the elements of a and y are the
means of the treatments.
• Example III.1 Rat experiment (continued)
• The estimates of a are
16
Example III.1 Rat experiment (continued)
• The estimates of the expected values, the fitted
values, are given by
17
Estimator of the expected values
• In general, where is the n-vector
consisting of the treatment means for each unit
and
• being least squares, this estimator can be
written as a linear combination of Y.
• that is, can be obtained as the product of an
matrix and the n-vector Y.
• let us write
• M for mean because MT is the matrix that replaces
each value of Y with the mean of the
corresponding treatment.
18
General form of mean operator
• Can be shown that the general form of MT is
• MT is a mean operator as it
• computes the treatment means from the vector to
which it is applied and
• replaces each element of this vector with its
treatment mean.
19
Estimator of the errors
• The estimator of the random errors in the
observed values of Y is, as before, the
difference from the expected values.
• That is,
20
Example III.1 Rat experimentAlternative
expression for fitted values
21
Residuals
• Fitted values for orthogonal experiments are
functions of means.
• Residuals are differences between observations
and fitted values
22
b) Alternative indicator-variable, expectation
models
• For the CRD, two expectation models are
considered
• First model is minimal expectation model
population mean response is same for all
observations, irrespective of diet.
• Second model is the maximal expectation model.
23
Minimal expectation model
• Definition III.4 The minimal expectation model
is the simplest model for the expectation that is
to be considered in analysing an experiment.
• The minimal expectation model is the same as the
intercept-only model given for the single sample
in chapter I, Statistical inference.
• Will be this for all analyses we consider.
24
Marginal models
• In regression case obtained marginal models by
zeroing some of the parameters in the full model.
• Here this is not the case.
• Here simply set ak m
• That is, intercept only model is the special case
where all ak s are equal.
• Clear for getting EYi m from EYi ak.
• What about yG from yT?
• If replace each element of a with m, then a
1tm.
• So yT XTa XT1tm XGm.
• Now marginality expressed in the relationship
between XT and XG as encapsulated in definition.
25
Marginality of models (in general)
• Definition III.5 Let C(X) denote the column
space of X.
• For two models, y1 ? X1q1 and y2 ? X2q2, the
first model is marginal to the second if C(X1) ?
C(X2) irrespective of the replication of the
levels in the columns of the Xs,
• That is if the columns of X1 can always be
written as linear combinations of the columns of
X2.
• We write y1 ? y2.
• Note marginality relationship is not symmetric
it is directional, like the less-than relation.
• So while y1 ? y2, y2 is not marginal to y1 unless
y1 ? y2.
26
Marginality of models for CRD
• yG is marginal to yT or yG ? yT because C(XG) ?
C(XT)
• in that an element from a row of XG is the sum of
the elements in the corresponding row of XT
• and this will occur irrespective of the
replication of the levels in the columns of XG
and XT.
• So while yG ? yT, yT is not marginal to yG as
C(XT) ? C(XG) so that yT ? yG.
• In geometrical terms, C(XT) is a
three-dimensional space and C(XG) is a line, the
equiangular line, that is a subspace of C(XT).
27
III.C Hypothesis testing using the ANOVA method
• Are there significant differences between the
treatment means?
• This is equivalent to deciding which of our two
expectation models best describes the data.
• We now perform the hypothesis test to do this for
the example.
28
Analysis of the rat example Example III.1 Rat
experiment (continued)
• Step 1 Set up hypotheses
• H0 aA aB aC m (yG ? XGm)
• H1 not all population Diet means are equal
• (yD ? XDa)
• Set a 0.05
29
Example III.1 Rat experiment (continued)
• Step 2 Calculate test statistic
• From table can see that (corrected) total
variation amongst the 6 Rats is partitioned into
2 parts
• variance of difference between diet means and
• the left-over (residual) rat variation.
• Step 3 Decide between hypotheses
• As probability of exceeding F of 3.60 with n1
2 and n2 3 is 0.1595 gt a 0.05, not much
evidence of a diet difference.
• Expectation model that appears to provide an
adequate description of the data is yG ? XGm.
30
b) Sums of squares for the analysis of variance
• From chapter I, Statistical inference, an SSq
• is the SSq of the elements of a vector and
• can write as the product of transpose of a column
vector with original column vector.
• Estimators of SSqs for the CRD ANOVA are SSqs of
following vectors (cf ch.I)
where Ds are n-vectors of deviations from Y
and Te is the n-vector of Treatment effects.
Definition III.6 An effect is a linear
combination of means with a set of effects
summing to zero.
31
• Want to show estimators of all SSqs can be
written as Y?QY.
• Is product of 1?n, n?n and n?1 vectors and
matrix, so is 1?1 or a scalar.
• Definition III.7 A quadratic form in a vector Y
is a scalar function of Y of the form Y?AY where
A is called the matrix of the quadratic form.
32
• Firstly write
• That is, each of the individual vectors on which
the sums of squares are based can be written as
an M matrix times Y.
• These M matrices are mean operators that are
symmetric and idempotent M' M and M2 M in
all cases.
33
• Then
• Given Ms are symmetric and idempotent, it is
relatively straightforward to show so are the
three Q matrices.
• It can also be shown that
34
• Consequently obtain the following expressions for
the SSqs
35
• Theorem III.1 For a completely randomized
design, the sums of squares in the analysis of
variance for Units, Treatments and Residual are
Proof follows the argument given above.
36
Residual SSq by difference
• That is, Residual SSq Units SSq - Treatments
SSq.
37
ANOVA table construction
• As in regression, Qs are orthogonal projection
matrices.
• QU orthogonally projects the data vector into the
n-1 dimensional part of the n-dimensional data
space that is orthogonal to equiangular line.
• QT orthogonally projects data vector into the t-1
dimensional part of the t-dimensional Treatment
space, that is orthogonal to equiangular line.
(Here the Treatment space is the column space of
XT.)
• Finally, the matrix orthogonally
projects the data vector into the n-t dimensional
Residual subspace.
• That is, Units space is divided into the two
orthogonal subspaces, the Treatments and Residual
subspaces.
38
Geometric interpretation
• Of course, the SSqs are just the squared lengths
of these vectors
• Hence, according to Pythagoras theorem, the
Treatments and Residual SSqs must sum to the
Units SSq.
39
Example III.1 Rat experiment (continued)
• Vectors for computing the SSqs are
• Total Rat deviations, Diet Effects and Residual
Rats deviations are projections into Rats, Diets
and Residual subspaces of dimension 5, 2 and 3,
respectively.
• Squared length of projection SSq
• Rats SSq is Y'QRY 0.34
• Diets SSq is Y'QDY 0.24
• Residual SSq is
Exercise III.3 is similar example for you to try
40
c) Expected mean squares
• Have an ANOVA in which we use F ( ratio of MSqs)
to decide between models.
• But why is this ratio appropriate?
• One way of answering this question is to look at
what the MSqs measure?
• Use expected values of the MSqs, i.e. EMSqs, to
do this.
41
Expected mean squares (contd)
• Remember expected value population mean
• Need EMSqs
• under the maximal model
• and the minimal model
• Similar to asking what is EYi?
• Know answer is EYi ak.
• i.e. in population, under model, average value of
Yi is ak.
• So for Treatments, what is EMSq?
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• i.e. an EMSq is the true mean value
• it depends on the model parameters.
42
Expected mean squares (contd)
• Need EMSqs
• under the maximal model
• and the minimal model
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• i.e. an EMSq is the true mean value and it
depends on the model parameters.
• This is analogous to saying that the expected
value of the Treatment mean vector in populations
described by the model is y.
• That is,
43
Expected mean squares (contd)
• Need EMSqs
• under the maximal model
• and the minimal model
• The EMSqs are the mean values of the MSqs in
populations described by the model for which they
are derived
• This is analogous to saying that the expected
value of the Treatment mean vector in populations
described by the model is y.
• That is,
• Put another way, what is the mean of the sampling
distribution of
• the Treatment mean vector?
• the Treatment mean square?
• (Note that both the mean vector and the mean
square are functions of Y and so are random
variables.)
• The answer is easy for the treatment mean vector
y
• But what for the Treatment mean square? Ans.
EMSq
44
EMSqs under the maximal model
• So if we had the complete populations for all
Treatments and computed the MSqs, the value of
• the Residual MSq would equal s2
• the Treatment MSq would equal s2 qT(y).
• So that the population average value of both MSqs
involves s2, the uncontrolled variation amongst
units from the same treatment.
• But what about q in Treatments EMSq.
45
The qT(y) function
• Subscript T indicates the Q matrix on which
function is based
• but no subscript on the y in qT(y),
• because we will determine expressions for it
under both the maximal (yT) and alternative
models (yG).
• That is, y in qT(y) will vary.
• Numerator is same as the SSq except that it is a
• To see what this means want expressions in terms
of individual parameters.
• Will show that under the maximal model (yT)
• and under the minimal model (yG) that
46
Example III.1 Rat experiment (continued)
• The latter is just the mean of the elements of
yT.
• Actually, the quadratic form is the SSQ of the
elements of vector
When will the SSq be zero?
47
The qT(y) function
• Now want to prove the following result
• As QT is symmetric and idempotent,
• y'QTy (QTy)'QTy
• qT(y) is the SSq of QTy, divided by (t-1).
• QTy (MT MG)y MTy MGy
• MGy replaces each element of y with the grand
mean of the elements of y
• MTy replaces each element of y with the mean of
the elements of y that received the same
treatment as the element being replaced.
48
The qT(y) function (continued)
• Under the maximal model (yT)
• Under the minimal model (yG m1n)
• MGyG MTyG yG so y'GQTyG 0 and qT(y) 0
49
Example III.1 Rat experiment (continued)
50
How qT(yT) depends on the as
• qT(yT) is a quadratic form and is basically a sum
of squares so that it must be nonnegative.
• Indeed the magnitude of depends on the size of
the differences between the population treatment
means, the aks
• if all the aks are similar they will be close to
their mean,
• whereas if they are widely scattered several will
be some distance from their mean.
51
EMsqs in terms of parameters
• Could compute population mean of MSq if knew aks
and s2.
• Treatment MSq will on average be greater than the
Residual MSq
• as it is influenced by both uncontrolled
variation and the magnitude of treatment
differences.
• The quadratic form qT(y) will only be zero when
all the as are equal, that is when the null
hypothesis is true.
• Then the EMSqs under the minimal model are
equal so that the F value will be approximately
one.
• Not surprising if think about a particular
experiment.
52
Example III.1 Rat experiment (continued)
• So what can potentially contribute to the
difference in the observed means of 3.1 and 2.7
for diets A and C?
• Obviously, the different diets
• not so obvious that differences arising from
uncontrolled variation also contribute as 2
different groups of rats involved.
• This is then reflected in EMSq in that it
involves s2 and the "variance" of the 3 effects.
53
Justification of F-test
• Thus, the F test involves asking the question "Is
the variance in the sample treatment means gt can
be expected from uncontrolled variation alone?".
• If the variance is no greater, concluded qT(y)
0
• the minimal model is the correct model since the
expected Treatment MSq under this model is just
s2.
• Otherwise, if the variance is greater, qT(y) is
nonzero
• the maximal model is required to describe the
data.
• Similar argument to examining dotplots for
Example II.2, Paper bag experiment.
54
d) Summary of the hypothesis test
• see notes
55
d) Summary of the hypothesis test
• Summarize the ANOVA-based hypothesis test for a
CRD involving t treatments and a total of n
observed units.
• Step 1 Set up hypotheses
• H0 a1 a2 ... a3 m (yG ? XGm)
• H1 not all population Treatment means are equal
(yT ? XTa)
• Set a.
56
Summary (continued)
Step 2 Calculate test statistic
• Note that EMSqs under the maximal model are
included in this table, it being recognized that
when H0 is true qT(y) 0.
• Step 3 Decide between hypotheses
• Determine probability of observed F value that
has n1 numerator d.f. t-1 and n2
denominator d.f. n-t.
• If
then the evidence suggests that the null
hypothesis be rejected and the alternative model
be used to describe the data.
57
e) Comparison with traditional one-way ANOVA
• Our ANOVA table is essentially the same as the
• the values of the F statistic from each table are
exactly the same.
• Labelling differs and the Total would normally be
placed at the bottom of the table, not at the top.
• Difference is symbolic
• Units term explicitly represents a source of
uncontrolled variation differences between
Units.
• Our table exhibits the confounding in the
experiment.
• Indenting of Treatments under Units signifies
that treatment differences are confounded or
mixed-up with unit differences.
58
f) Computation of the ANOVA in R
• Begin with data entry (see Appendix A,
Introduction to R and, Appendix C, Analysis of
designed experiments in R)
• Next an initial graphical exploration using
boxplots defer to a second example with more
data.
• ANOVA while function lm could be used, function
aov is preferred for analysing data from a
designed experiment.
• Both use a model formula of the form
• Response variable explanatory variables (and
operators)
• So far expressions on right fairly simple one
or two explanatory variables separated by a .
• Subtlety with analysis of designed experiments
• If explanatory variable is a numeric, such as a
numeric vector, then R fits just one coefficient
for it.
• For a single explanatory variable, a
straight-line relationship fitted.
• If explanatory variable is categorical, such as a
factor, a coefficient is fit for each level of
the variable indicator variables are used.
• In analyzing a CRD important that the treatment
factor is stored in a factor object, signalling R
to use indicator variables
59
f) Computation of the ANOVA in R (continued)
• There are two ways in which this analysis can be
obtained using the aov function
• without and with an Error function in the model
formula.
• Error function used in model formula to specify a
model for the Error in the experiment (a model
for uncontrolled variation).
• summary function is used and this produces ANOVA
table.
• model.tables function used to obtain tables of
means.
60
Example III.1 Rat experiment (continued)
• Following commands to perform the two analyses of
the data
• AOV without Error
• Rat.NoError.aov lt- aov(LiverWt Diet,
CRDRat.dat)
• summary(Rat.NoError.aov)
• AOV with Error
• Rat.aov lt- aov(LiverWt Diet Error(Rat),
CRDRat.dat)
• summary(Rat.aov)
• model.tables(Rat.aov, type "means")
61
Output
• gt AOV without Error
• gt
• gt Rat.NoError.aov lt- aov(LiverWt Diet,
CRDRat.dat)
• gt summary(Rat.NoError.aov)
• Df Sum Sq Mean Sq F value Pr(gtF)
• Diet 2 0.240000 0.120000 3.6 0.1595
• Residuals 3 0.100000 0.033333
• gt
• gt AOV with Error
• gt
• gt Rat.aov lt- aov(LiverWt Diet Error(Rat),
CRDRat.dat)
• gt summary(Rat.aov)
• Error Rat
• Df Sum Sq Mean Sq F value Pr(gtF)
• Diet 2 0.240000 0.120000 3.6 0.1595
• Residuals 3 0.100000 0.033333
• Analysis without the Error function parallels the
traditional analysis and that with Error is
similar to our table.
• In this course will use Error function.
62
Output (continued)
• gt model.tables(Rat.aov, type"means")
• Tables of means
• Grand mean
• 3.1
• Diet
• A B C
• 3.1 3.3 2.7
• rep 3.0 2.0 1.0
63
III.D Diagnostic checking
• Assumed the following model
• Y is distributed N(y, s2In)
• where y EY Xq and
• s2 is the variance of an observation
• Maximal model is used in diagnostic checking
• yT EY XTa
• For this model to be appropriate requires that
• the response is operating additively that the
individual deviations in the response to a
treatment are similar
• the sets of units assigned the treatments are
comparable in that the amount of uncontrolled
variation exhibited by them is the same for each
treatment
• each observation is independent of other
observations and
• that the response of the units is normally
distributed.
64
Example III.2 Caffeine effects on students
• Effect of orally ingested caffeine on a physical
task was investigated (Draper and Smith, 1981,
sec.9.1).
• Thirty healthy male college students were
selected and trained in finger tapping.
• Ten men were randomly assigned to receive one of
three doses of caffeine (0, 100 or 200 mg).
• The number of finger taps after ingesting the
caffeine was recorded for each student.
65
Entering the data
• Setting up a data frame, from scratch, for data
arranged in standard order (see App C)
• factor Students with values 130,
• factor Dose with levels 0, 100 and 200 and values
depending on whether data is entered by rows or
columns (can use rep function),
• numeric vector Taps with the 30 observed values
of the response variable.
66
Entering the data (cont'd)
gt CRDCaff.dat Students Dose Taps 1 1
0 242 2 2 100 248 3 3 200
246 4 4 0 245 5 5 100 246 6
6 200 248 7 7 0 244 8
8 100 245 9 9 200 250 10 10
0 248 11 11 100 247 12 12 200
252 13 13 0 247 14 14 100
248 15 15 200 248 16 16 0
248 17 17 100 250 18 18 200
250 19 19 0 242 20 20 100
247 21 21 200 246 22 22 0
244 23 23 100 246 24 24 200
248 25 25 0 246 26 26 100
243 27 27 200 245 28 28 0
242 29 29 100 244 30 30 200 250
• set up data.frame with factors Students and Dose
and then add response variable Taps
• CRDCaff.dat lt- data.frame(Students
factor(130),
• Dose factor(rep(c(0,100,200), times10)))
• CRDCaff.datTaps lt-
• c(242,248,246,245,246,248,244,245,250,248,247,252
,247,248,248,
• 248,250,250,242,247,246,244,246,248,246,243,245,2
42,244,250)
• CRDCaff.dat
67
Boxplots for each level of Dose
• Use function
• boxplot(split(Taps, Dose), xlab"Dose",
ylab"Number of taps")
• Average number of taps increasing as dose
increases
• Some evidence dose 0 more variable than dose 100.
68
Analysis of variance for this data
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov)
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Residuals 27 134.100 4.967
gt model.tables(Caffeine.aov, type"means") Tables
of means Grand mean 246.5 Dose Dose
0 100 200 244.8 246.4 248.3
69
The hypothesis test
• Step 1 Set up hypotheses
• H0 a0 a100 a200 m (yG XGm)
• H1 not all population Dose means are equal
• (yD XDa)
• Set a 0.05
70
The hypothesis test (continued)
• Step 2 Calculate test statistic
• The ANOVA table for the example is
Step 3 Decide between hypotheses P(F2,27 ?
6.18) p 0.006 lt a 0.05. The evidence
suggests there is a dose difference and that the
expectation model that best describes the data is
yD ? XDa.
71
Examination of the residuals, eT
• Use the Residuals-versus-fitted-values plot and
the Normal Probability plot.
• In interpreting these plots
• Note ANOVA is robust to variance heterogeneity,
if treatments equally replicated, and to moderate
departures from normality.
• Most commonly find an unusually large or small
residual.
• The cause of such extreme values requires
investigation.
• May be a recording mistake or catastrophe for a
unit that can be identified.
• But, may be valid and is the result of some
unanticipated, but important effect.
72
The Normal Probability plot
• Should show a broadly straight line trend.
__________________
73
The Residuals-versus-fitted-values plot
• Generally, the points on scatter diagram should
_____________________________ no particular
pattern
74
Problem plots
__________________________ varianc
e increases as level increases
_________________________ systematic trend in
residuals
_________________________ variance peaks
in middle
Actually, for the CRD, this plot has a vertical
scatter of points for each treatment each
should be centred on zero and of the same width.
75
R functions used in producing these plots
• resid.errors extract the residuals from an aov
object when Error function used
• fitted.errors extract the fitted values from an
aov object when Error function used
• plot to plot the fitted values against the
residuals
• qqnorm to plot the residuals against the normal
quantiles
• qqline to add a line to the plot produced by
qqnorm.
• First 2 functions are nonstandard functions from
dae package.
76
Example III.2 Caffeine effects on students
(continued)
• A violation of the assumptions would occur if all
the students were in the same room and the
presence of other students caused anxiety to just
the students that had no caffeine.
• The response of the students is not independent.
• It may be that the inhibition of this group
resulted in less variation in their response
which would be manifest in the plot.
• Another situation that would lead to an
unacceptable pattern in the plot is if the effect
becomes more variable as the level of the
response variable increases.
• For example, caffeine increases the tapping but
at higher levels the variability of increase from
student to student is greater.
• That is there is a lack of additivity in the
response.
77
R output for getting the plots
gt res lt- resid.errors(Caffeine.aov) gt fit lt-
fitted.errors(Caffeine.aov) gt data.frame(Students,
Dose,Taps,res,fit) Students Dose Taps res
fit 1 1 0 242 -2.8 244.8 2 2
100 248 1.6 246.4 3 3 200 246 -2.3
248.3 4 4 0 245 0.2 244.8 5
5 100 246 -0.4 246.4 6 6 200 248 -0.3
248.3 7 7 0 244 -0.8 244.8 8
8 100 245 -1.4 246.4 9 9 200 250 1.7
248.3 10 10 0 248 3.2 244.8 11
11 100 247 0.6 246.4 12 12 200 252
3.7 248.3 13 13 0 247 2.2 244.8 14
14 100 248 1.6 246.4 15 15 200 248
-0.3 248.3 16 16 0 248 3.2 244.8 17
17 100 250 3.6 246.4 18 18 200 250
1.7 248.3 19 19 0 242 -2.8 244.8 20
20 100 247 0.6 246.4
• Note the use of data.frame to produce a printed
list of the original data with the residuals and
fitted values.
78
Plots for the example
21 21 200 246 -2.3 248.3 22 22 0
244 -0.8 244.8 23 23 100 246 -0.4
246.4 24 24 200 248 -0.3 248.3 25
25 0 246 1.2 244.8 26 26 100 243
-3.4 246.4 27 27 200 245 -3.3 248.3 28
28 0 242 -2.8 244.8 29 29 100 244
-2.4 246.4 30 30 200 250 1.7 248.3 gt
plot(fit, res, pch 16) gt qqnorm(res, pch
16) gt qqline(res)
• The Residuals-versus-fitted-values plot appears
to be fine.
79
Normal Probability plot
• Displaying some curvature at ends.
• Indicates data heavier in tails and flatter in
the peak than expected for a normal distribution.
• Given normality not crucial and only a few
observations involved, use analysis we have
performed.
80
The hypothesis test Summary
• The ANOVA table for the example is
P(F2,27 ? 6.18) p 0.006 lt a 0.05. The
evidence suggests there is a dose difference and
that the expectation model that best describes
the data is yD ? XDa. Diagnostic checking
indicates model is OK
81
III.E Treatment differences
• So far all that our analysis has accomplished is
that we have decided whether or not there appears
to be a difference between the population
treatment means.
• Of greater interest to the researcher is how the
treatment means differ.
• Two alternatives available
• Multiple comparisons procedures
• used when the treatment factors are all
qualitative so that it is appropriate to test for
differences between treatments
• Fitting submodels
• When one (or more) of the treatment factors is
quantitative the fitting of smooth curves to the
trend in the means is likely to lead to a more
appropriate and concise description of the
effects of the factors.
• Often, for reasons explained in chapter II, a
low order polynomial will provide an adequate
description of the trend.
82
Note
• Multiple comparison procedures should not be used
when the test for treatment differences is not
significant.
• Submodels should be fitted irrespective of
whether the overall test for treatment
differences is significant.
• The difference in usage has to do with one being
concerned with mean differences and the other
with deciding between models.
83
a) Multiple comparisons procedures for comparing
all treatments
• Multiple comparisons for all treatments MCA
procedures.
• In general MCA procedures divide into those
based
• on family-wise error rates Type I error rate
specified and controlled for over all
comparisons, often at 0.05.
• those based on comparison-wise error rates Type
I error rate specified and controlled for each
comparison
• Problem with latter is probability of an
incorrect conclusion gets very high as the number
of comparisons increases.
• For comparison-wise error rate of 0.05,
family-wise error rate
• So recommend use MCA procedure based on
family-wise error rates use just Tukey's HSD
procedure.
84
Tukeys Honestly Significant Difference procedure
• determines if each pair of means is significantly
different
• is based on a family-wise error rate.
• basically for equal numbers of observations for
each mean.
• A modification that is approximate will be
provided for unequal numbers.
85
The procedure
• Each application of the procedure is based on the
hypotheses
• Ho aA aB
• H1 aA ? aB
• One calculates the statistic
qt,n,a is the studentized range with t no.
means, n Residual df and a significance level,
is the standard error of the
difference rA, rB are the number of replicates
for each of a pair of means being compared.
86
• Note that strictly speaking rA and rB should be
equal.
• When unequal rA and rB called the Tukey-Kramer
procedure.
• w(a) will depend on which means are being
compared.
• If the treatments are all equally replicated with
replication r, the formula for reduces to
87
In R
• aov has given Residual MSq and df
• model.tables produces tables of means.
• qtukey computes as follows
• q lt- qtukey(1 - a, t, n)
88
Example III.2 Caffeine effects on students
(continued)
• Have already concluded evidence suggests that
there is a dose difference. But which doses are
different?
• Output with means and q
• gt model.tables(Caffeine.aov, type "means")
• Tables of means
• Grand mean
• 246.5
• Dose
• Dose
• 0 100 200
• 244.8 246.4 248.3
• gt q lt- qtukey(0.95, 3, 27)
• gt q
• 1 3.506426
89
Decide on differences
• Any two means different by 2.47 or more are
significantly different.
• Our conclusion is that the mean for 0 and 200 are
different but that for 100 is somewhat
intermediate.
90
b) Fitting submodels
• For quantitative factors, like Dose, it is often
better to examine the relationship between the
response and the levels of the factor.
• This is commonly done using polynomials.
• Now, a polynomial of degree t-1 will fit exactly
t points.
• So a quadratic will fit exactly the three means.
• In practice, polynomials of order 2 often
sufficient.
• However, more than 3 points may be desirable so
that deviations from the fitted curve or lack of
fit can be tested
91
Polynomial models
• To investigate polynomial models up to order 2,
following models for expectation investigated
where xk is the value of the kth level of the
treatment factor, m is the intercept of the
fitted equation and g1 is the slope of the
fitted equation and g2 is the quadratic
coefficient of the fitted equation.
The X1 and X2 matrices made up of columns that
consist of the values of the levels of the factor
and their powers not the indicator variables of
before.
92
Example III.2 Caffeine effects on students
(continued)
• The X matrices for the example are
• The columns of each X matrix in the above list
are a linear combination of those of any of the X
matrices to its right in the list.
93
Marginality of models
• That is, the columns of each X matrix in the
above list are a linear combination of those of
any of the X matrices to its right in the list.
• Marginality is not a symmetric relationship in
that
• if a model is marginal to second model,
• the second model is not necessarily marginal to
the first.
• For example, EY X1g1 is marginal to EY
X2g2 but not vice-a-versa, except when t 2.
94
Equivalence of EY X2g2 and EY XDa
• C(X2) C(XD) as the 3 columns of one matrix can
be written as 3 linearly independent combinations
of the columns of the other matrix.
• So the two models are marginal to each other and
are equivalent
• However, while the fitted values are the same,
the estimates and interpretation of the
parameters are different
• those corresponding to X2 are interpreted as the
intercept, slope and curvature coefficient.
• those corresponding to XD are interpreted as the
expected (mean) value for that treatment.
• Also, in spite of being marginal, the estimators
of the same parameter differ depending on the
model that has been fitted.
• for example, for the model EY XGm,
• but for EY X1g1,
models not orthogonal
95
Hypothesis test incorporating submodels
• Test statistics computed in ANOVA table.
• Step 1 Set up hypotheses
• a) H0
• H1
• (Differences between fitted model or
•
• b) H0
• H1
• c) H0
• H1
• Set a.
96
Hypothesis test (continued)
• Step 2 Calculate test statistics
• The ANOVA table for a CRD is
Test statistics corresponds to hypothesis pairs
in Step 1.
97
Hypothesis test (continued)
• Step 3 Decide between hypotheses
• Begin with the first hypothesis pair, determine
its significance and continue down the sequence
until a significant result is obtained.
• A significant Deviations F indicates that the
description.
• Thus a model based on them would be
unsatisfactory. No point in continuing.
• If the Deviations F is not significant , then a
significant Quadratic F indicates that a 2nd
degree polynomial is required to adequately
describe the trend.
• As a linear coefficient is necessarily
incorporated in a 2nd polynomial, no point to
further testing in this case.
• If both the Deviations and Quadratic F's are not
significant, then a significant Linear F
indicates a linear relationship describes the
trend in the treatment means.
98
What if deviations is significant?
• If the Deviations are significant, then two
options are
• increase the degree of the polynomial (often not
desirable), or
• use multiple comparisons to investigate the
differences between the means.
99
Fitting polynomials in R
• Need to realize that the default contrasts for
ordered (factor) objects are polynomial
contrasts, assuming equally-spaced levels
• linear transformations of all columns of X,
except the first, such that each column
• is orthogonal to all other columns in X and
• has SSq equal to 1.
• Facilitates the computations.
• So, if a factor is quantitative
• set it up as an ordered object from the start.
• if did not make it an ordered object, then
redefine factor as an ordered.
100
Example III.2 Caffeine effects on students
(continued)
• R output for setting up ordered (factor)
• gt fit polynomials
• gt
• gt Dose.lev lt- c(0,100,200)
• gt CRDCaff.datDose lt- ordered(CRDCaff.datDose,
• levelsDose.lev)
• gt contrasts(CRDCaff.datDose) lt- contr.poly(t,
• scoresDose.lev)
• gt contrasts(CRDCaff.datDose)
• .L .Q
• 0 -7.071068e-01 0.4082483
• 100 -9.073264e-17 -0.8164966
• 200 7.071068e-01 0.4082483
101
R output for fitting polynomial submodels
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov, split
list(Dose list(L 1, Q 2)))
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Dose L 1 61.250 61.250 12.3322 0.001585
• Dose Q 1 0.150 0.150 0.0302 0.863331
• Residuals 27 134.100 4.967
this is reflected in C(X2) C(XD).
• Thus, Deviations line is redundant in this
example.
• When required, Deviations line is computed by
assigning extra powers to a single line named,
say, Dev in split function (e.g. Dev 36 see
notes).
102
Hypothesis test incorporating submodels
• Step 1 Set up hypotheses
•
• a) H0
• H1
• b) H0
• H1
• Set a 0.05.
103
Hypothesis test (continued)
• Step 2 Calculate test statistics
• gt Caffeine.aov lt- aov(Taps Dose
Error(Students), CRDCaff.dat)
• gt summary(Caffeine.aov, split
list(Dose list(L 1, Q 2)))
• Error Students
• Df Sum Sq Mean Sq F value Pr(gtF)
• Dose 2 61.400 30.700 6.1812 0.006163
• Dose L 1 61.250 61.250 12.3322 0.001585
• Dose Q 1 0.150 0.150 0.0302 0.863331
• Residuals 27 134.100 4.967
104
Hypothesis test (continued)
• Step 3 Decide between hypotheses
• The Quadratic source has a probability of 0.863
gt 0.05 and so H0 is not rejected in this case.
• The linear source has a probability of 0.002 lt
0.05 and so H0 is rejected in this case.
• It is clear quadratic term is not significant
but linear term is highly significant so the
appropriate model for the expectation is the
linear model y X1g1 where g1 m g1.
105
Fitted equation
• Coefficients can be obtained using the coef
function on the aov object, but these are not
suitable for obtaining the fitted values.
• The fitted equation is obtained by putting the
values of the levels into a numeric vector and
using the lm function to fit a polynomial of the
order indicated by the hypothesis test.
• For the example, linear equation was adequate and
so the analysis is redone with 1 for the order of
the polynomial.
106
Fitted equation for the example
• gt D lt- as.vector(Dose)
• gt D lt- as.numeric(D)
• gt Caffeine.lm lt- lm(Taps D)
• gt coef(Caffeine.lm)
• (Intercept) D
• 244.7500 0.0175
• The fitted equation is Y 244.75 0.0175 X
where X is the number of taps
• The slope of this equation is 0.0175.
• That is, taps increase 0.0175 x 100 1.75 with
each 100 mg of caffeine.
• This conclusion seems a more satisfactory summary
of the results than that the response at 200 is
significantly greater than at 0 with 100 being
intermediate.
• The commands to fit a quadratic would be
• D2 lt- DD
• Caffeine.lm lt- lm(Taps D D2)
107
Plotting means and fitted line
• Details are in the notes
• The plot produced is as follows
108
c) Comparison of treatment parametrizations
• Two alternative bases
• Lead to different parameter estimates with
different interpretations and different
partitions of the treatment SSqs.
• The total treatment SSqs and fitted values for
treatments remain the same, while the contrasts
span the treatment space.
• That is, in this case, SSqT SSqL SSqQ
109
III.G Exercises
• Ex. III.12 look at aspects of quadratic forms
• Ex. III.3 investigates the calculations with
example that can be done with a calculator
• Ex. III.4 involves producing a randomized layout
• Ex. III.5 asks for the complete analysis of a CRD
with a qualitative treatment factor
• Ex. III.6 asks for the complete analysis of a CRD
with a quantitative treatment factor | 11,259 | 40,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-45 | latest | en | 0.769398 |
http://www.jiskha.com/display.cgi?id=1316812832 | 1,495,934,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00093.warc.gz | 678,785,165 | 3,545 | # physics
posted by on .
A cyclist maintains a constant velocity of 5.9 m/s headed away from point A. At some initial time, the cyclist is 242 m from point A.
What will be his position from point A after that time?
• physics - ,
His speed is 5.9m/s
Every second he gets 5.9m farther away.
We start measuring time when he's 242m away.
So, after t seconds, he's 242 + 5.9t meters away from A. | 113 | 397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-22 | latest | en | 0.963961 |
https://mathoverflow.net/questions/359087/what-is-known-about-the-functional-square-root-of-the-riemann-zeta-function | 1,723,564,736,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00830.warc.gz | 300,728,660 | 25,707 | # What is known about the functional square root of the Riemann zeta function?
Let us consider the Riemann zeta function $$\zeta(s)$$, where $$s$$ can take on values on the domain $$\mathbb{R}_{>1}$$:
$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$
I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $$f$$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$
Questions
1. Has a closed-form solution been found for $$f$$ in equation $$(1)$$ ?
2. If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$
3. If so, I would be grateful if you have some pointers to relevant articles or other sources.
(Cross-post from MSE.)
(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $$f$$ on $$(0, \infty) \cup X$$ for some subset $$X \subset \mathbb{R} \setminus (0,\infty)$$. Then $$f$$ must map $$(1,\infty)$$ bijectively onto $$X$$, and $$X$$ iself onto $$(0,\infty)$$. Under these conditions, there is still a possibility that $$f$$ is both continuous and real-valued. I am interested in the properties of such an $$f$$.
• Are you sure this is the question you want to ask? The zeta function on the domain you consider takes values outside this domain, so it cannot have a functional square root. In fact, for z with real part greater than 1, $\zeta(z)$ may have negative real part. See e.g. arxiv.org/pdf/1112.4910.pdf. Commented May 1, 2020 at 23:37
• Well, it turns out that one needs to deal with fixpoints. I no longer have all the references on a website, but take a look at mathoverflow.net/questions/45608/… and my self-answers. I can see that there is a fixpoint in the reals between 1 and, say, 10. Commented May 2, 2020 at 3:35
• @DmitryVaintrob I should also include that I'm only considering real $s>1$, so the values $\zeta(s)$ take are positive and real too Commented May 2, 2020 at 8:17
• For real $s > 1$, $\zeta(s)$ maps $(1,\infty)$ bijectively onto itself. It is continuous and decreasing. A functional square root can therefore not be real-valued and continuous. Commented May 2, 2020 at 11:27
• @GeraldEdgar Could you please elaborate on that? I don't see why such a root can't be real-valued and continuous at once Commented May 2, 2020 at 22:00
Note that $$\zeta$$ maps $$(1,\infty)$$ bijectively onto itself, $$\zeta$$ is continuous, $$\zeta$$ is decreasing on $$(1,\infty)$$.
Suppose $$\zeta = f \circ f$$ where $$f$$ also maps $$(1,\infty)$$ onto itself. A continuous injective function $$f$$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $$f\circ f$$ is increasing, so $$f \circ f \ne \zeta$$.
Now of course we can define $$f$$ on some larger set, say $$(1,\infty) \cup X$$, where $$X$$ has the power of the continuum. Let $$f$$ map $$(1,\infty)$$ bijectively onto $$X$$ and $$X$$ bijectively onto $$(1,\infty)$$. Easy.
• Ahh, I see. Thank you! I'll modify the question according to the answer you've given, because I'm still curious about the existence of such an $f$ defined on a larger set. Commented May 3, 2020 at 16:46 | 1,030 | 3,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 38, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.881628 |
https://justaaa.com/chemistry/814439-what-is-the-wavelength-in-meters-of-photons-with | 1,709,197,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474795.48/warc/CC-MAIN-20240229071243-20240229101243-00083.warc.gz | 330,098,942 | 9,729 | Question
# What is the wavelength (in meters) of photons with the following energies? Part A.) 136 kJ/mol...
What is the wavelength (in meters) of photons with the following energies?
Part A.) 136 kJ/mol
Part C.) 4.55*10-2 kJ/mol
#### Homework Answers
Answer #1
A)
Given:
Energy of 1 mol = 1.36*10^2 KJ/mol
= 1.36*10^5 J/mol
Find energy of 1 photon first
Energy of 1 photon = energy of 1 mol/Avogadro's number
= 1.36*10^5/(6.022*10^23)
= 2.258*10^-19 J
This is energy of 1 photon
use:
E = h*c/lambda
2.258*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 8.802*10^-7 m
Answer: 8.80*10^-7 m
C)
Given:
Energy of 1 mol = 4.55*10^-2 KJ/mol
= 45.5 J/mol
Find energy of 1 photon first
Energy of 1 photon = energy of 1 mol/Avogadro's number
= 45.5/(6.022*10^23)
= 7.556*10^-23 J
This is energy of 1 photon
use:
E = h*c/lambda
7.556*10^-23J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 2.631*10^-3 m
Answer: 2.63*10^-3 m
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ADVERTISEMENT | 460 | 1,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-10 | latest | en | 0.746052 |
https://www.knowledgemax.org/how/how-does-a-dc-variable-speed-drive-work/ | 1,685,709,868,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00282.warc.gz | 902,883,588 | 12,631 | # How does a DC variable speed drive work?
### How does a DC variable speed drive work?
DC variable speed drives use a control module. This control module rectifies alternating current power and converts it to direct current for DC motor control. … The feedback signals then regulate the DC motor performance via speed, and torque.Apr 9, 2014
### Can VFD be operated on DC?
A VFD typically rectifies the 3-phase3-phaseThe conductors between a voltage source and a load are called lines, and the voltage between any two lines is called line voltage. The voltage measured between any line and neutral is called phase voltage. For example, for a 208/120 volt service, the line voltage is 208 Volts, and the phase voltage is 120 Volts.https://en.wikipedia.org › wiki › Three-phase_electric_powerThree-phase electric power – Wikipedia input to a fixed dc voltage, which is filtered and stored using large dc bus capacitors. The dc bus voltage is then inverted to yield a variable voltage, variable fre- quency output.
### Is VFD used for DC motor?
Simply No. VFD stands for variable frequency Drive. It clearly says that motor gets controlled by varying the frequency. Here in DC, there is no frequency and hence we couldn’t use VFD in DC motor.
### What is DC drive control method?
DC drive is basically a DC motor speed control system that supplies the voltage to the motor to operate at desired speed. … By using an induction motor, the DC generator was driven at a fixed speed and by varying the field of the generator, variable voltage was generated.
### Is VFD AC or DC?
– Variable frequency drives (VFDs): Variable frequency drives also control the speed of a motor, but they do so by changing the voltage and frequency and can thus only be used with AC motors.Mar 6, 2020
READ How do you read a fuel stick measuring tank?
### Does a VFD convert AC to DC?
The main two parts of a VFD are the rectifier and the inverter. The first will convert the AC voltage into a DC voltage; the second will convert this DC voltage into an AC voltage with variable magnitude and frequency.
### Why DC is used in VFD?
A VFD needs to supply its motor with a specific frequency in order to have it rotate at the desired speed. To do this, it needs a power source that is “frequency-agnostic”- one that it can mold into any frequency it desires. That would be D.C.- with proper switching, I can make any frequency I want out of a DC supply.
### Can a VFD run on battery?
The curious thing is that the VFD can work in an “emergency mode” supplied from a (really big) 48V battery. They use that to allow moving the thing quite slowly to a “controlled” position even if power fails.Oct 12, 2012
### What does a DC bus do?
The DC-BUS converts the digital input data into phase modulated signals, protected against errors generated by noise over the powerline. On the receiving side, the received signal is demodulated into the original digital data.
### What is DC bus connection?
Common dc bus Simply interlinking the dc bus between VFDs through a fused connection establishes a common bus. Linking the dc bus together between two VFDs through fused connections makes one simple form of a common bus. An oversized drive then supplies ac-to-dc rectification.Jan 11, 2016
### What is DC bus voltage in drive?
The dc bus voltage is relative to the peak voltage of the mains input. What to look for: dc bus voltage is ~1.414 x the rms line voltage (e.g., for a 480V ac drive, the dc bus should be ~678V dc. A dc voltage value that is too low can cause the drive to trip.
READ How do you pay at the pump with a debit card?
### What is the device that switches AC to DC?
A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction.
### How does VFD change voltage?
VFDs rectify a three-phasethree-phaseThe conductors between a voltage source and a load are called lines, and the voltage between any two lines is called line voltage. The voltage measured between any line and neutral is called phase voltage. For example, for a 208/120 volt service, the line voltage is 208 Volts, and the phase voltage is 120 Volts.https://en.wikipedia.org › wiki › Three-phase_electric_powerThree-phase electric power – Wikipedia input to a fixed dc voltage that is filtered and stored using dc bus capacitors. The dc bus voltage then goes through an inversion process that uses three insulated gate bipolar transistor (IGBT) pairs, one for each output phase.Jun 15, 2018
### Does a VFD put out AC or DC?
Input Voltage determines VFD Output Voltage One of the many benefits of using a variable frequency drive (VFD) to control an electric motor is the ability to adjust the voltage output to the motor for optimal efficiency. A VFD rectifies the AC input voltage to a DC voltage across the DC bus capacitors.Sep 29, 2020 | 1,112 | 4,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-23 | latest | en | 0.90728 |
http://mathopenref.com/integral.html | 1,534,540,256,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00353.warc.gz | 256,679,219 | 3,934 | # Integral
The math word "integral" is referenced in the pages listed below.
## Pages referring to 'integral'
Not all integrals have antiderivatives that can be written down using the basic functions and operations that we know. Some functions are, in fact, defined using an integral. This page shows two such functions in calculus. Interactive calculus applet.
Calculus: What happens if one of the limits of integration for a definite integral is infinity? Does the integral have a value? Or, what if the value of the integrand goes to infinity at one of the limits? This page describes how treat these cases as a use of limits. Interactive calculus applet.
The integral test provides a means to testing whether a series converges or diverges. Interactive calculus applet.
We can use integrals to find the area enclosed by a polar curve. Here, we use sectors of circles instead of rectangles to calculate the area. Interactive calculus applet.
The definite integral, the limit of a Riemann sum, can be interpreted as the area under a curve. This page explores some properties of definite integrals which can be useful in computing the value of an integral. Interactive calculus applet.
In calculus, a Riemann sum is a method for approximating the total area underneath a curve on a graph, otherwise known as an integral. It may also be used to define the integration operation. This page explores this idea with an interactive calculus applet. Interactive calculus applet.
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Become a patron of the site at patreon.com/mathopenref | 417 | 2,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-34 | longest | en | 0.877163 |
https://www.w3resource.com/python-exercises/lambda/python-lambda-exercise-29.php | 1,611,665,287,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00131.warc.gz | 1,030,960,493 | 29,601 | Python: Find the maximum and minimum values in a given heterogeneous list using lambda - w3resource
# Python: Find the maximum and minimum values in a given heterogeneous list using lambda
## Python Lambda: Exercise-29 with Solution
Write a Python program to find the maximum value in a given heterogeneous list using lambda.
Sample Solution:
Python Code :
``````def max_val(list_val):
max_val = max(list_val, key = lambda i: (isinstance(i, int), i))
return(max_val)
list_val = ['Python', 3, 2, 4, 5, 'version']
print("Original list:")
print(list_val)
print("\nMaximum values in the said list using lambda:")
print(max_val(list_val))
``````
Sample Output:
```Original list:
['Python', 3, 2, 4, 5, 'version']
Maximum values in the said list using lambda:
5
```
## Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
What is the difficulty level of this exercise?
Test your Python skills with w3resource's quiz
## Python: Tips of the Day
Python: Cache results with decorators
There is a great way to cache functions with decorators in Python. Caching will help save time and precious resources when there is an expensive function at hand.
Implementation is easy, just import lru_cache from functools library and decorate your function using @lru_cache.
```from functools import lru_cache
@lru_cache(maxsize=None)
def fibo(a):
if a <= 1:
return a
else:
return fibo(a-1) + fibo(a-2)
for i in range(20):
print(fibo(i), end="|")
print("\n\n", fibo.cache_info())
```
Output:
```0|1|1|2|3|5|8|13|21|34|55|89|144|233|377|610|987|1597|2584|4181|
CacheInfo(hits=36, misses=20, maxsize=None, currsize=20)``` | 475 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-04 | longest | en | 0.668204 |
http://driving-tests.org/beginner-drivers/how-to-calculate-gas-mileage/ | 1,369,550,953,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706635944/warc/CC-MAIN-20130516121715-00061-ip-10-60-113-184.ec2.internal.warc.gz | 82,564,953 | 8,158 | Close
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# How to Calculate Gas Mileage
With the cost of gasoline fluctuating between expensive and astronomical, many drivers are very concerned about the gas mileage their car is getting. It is quite simple to calculate your vehicle’s average gas mileage using these simple steps.
## How To Proceed
1. 1
Find out the capacity of your gas vehicle’s gas tank. This information can be found in your vehicle’s owner’s manual. Most vehicles gas tank capacity ranges from ten gallons for small compact cars to forty gallons for large trucks or SUVs.
2. 2
Fill it up. Fill your vehicle’s gas tank to full capacity. Most gas pumps have a mechanism that will stop the flow of gasoline when the tank is full. When you turn your car on, make sure that the gas tank gauge indicates that your tank is completely full.
3. 3
When you look on the gas pump to see how much money you need to pay for your gas, notice the reading of the amount of gallons you purchased. There should be an exact reading of how many gallons you bought to the hundredth of a gallon. This number should be close to the total gas tank capacity listed in your owner’s manual. Make a note of this amount. If the gas pump you used to fill up does not have this function, you can divide the total amount of your gasoline purchase by the price per gallon.
4. 4
Set your trip odometer. This is typically located on your dash board and is set by pushing a button located somewhere around the speedometer or RPM gauge. Make sure to complete this step as soon as you fill your tank with gas and before you begin to drive.
5. 5
Drive your car as you normally would.
6. 6
Drive your vehicle until the gas light comes on to indicate to you that you need to refill your tank. At this point, take note of your odometer reading for the trip you set when you filled your tank.
7. 7
Time for some math. Divide the total number of miles you drove on this tank of gas by the number of gallons of gas that you purchased when you filled your vehicle’s gas tank. The solution to this problem will be the average miles per gallon of gas that your vehicle gets.
Many drivers use particular driving strategies to improve their vehicle’s gas mileage such as slower acceleration and lower speeds. Knowing your vehicle’s average miles per gallon rating will also allow you to budget your gas expenses more accurately.
WHAT DO YOU THINK? | 534 | 2,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | longest | en | 0.946299 |
http://forums.wolfram.com/student-support/topics/19533 | 1,394,584,519,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011405327/warc/CC-MAIN-20140305092325-00003-ip-10-183-142-35.ec2.internal.warc.gz | 71,200,057 | 5,771 | Student Support Forum: 'simultaneous partial differential equation' topicStudent Support Forum > General > "simultaneous partial differential equation"
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Author Comment/Response Hiroki 11/29/08 02:47am Please help someone. Nice to meet you. I want to analayz "nonlinear reaction-diffusion equation". But, I couldn't fit the "Boundary and Initial condition". Please help and ,teach that source code. pde = {D[a[t, x, y], t] == Subscript[D, a] {D[a[t, x, y], x, x] + D[a[t, x, y], y, y]} + Subscript[\[Rho], a] a[t, x, y]^2/({1 + Subscript[k, a] a[t, x, y]^2} h[t, x, y]) - Subscript[\[Mu], a] a[t, x, y] + Subscript[\[Sigma], a], D[h[t, x, y], t] == Subscript[D, h] {D[h[t, x, y], x, x] + D[h[t, x, y], y, y]} + Subscript[\[Rho], h] a[t, x, y]^2 - Subscript[\[Mu], h] h[t, x, y] + Subscript[\[Sigma], h]}; bc = {(D[a[t, x, y], x] /. x -> -4) == 0, (D[a[t, x, y], x] /. x -> 4) == 0, (D[a[t, x, y], y] /. y -> -4) == 0, (D[a[t, x, y], y] /. y -> 4) == 0, (D[h[t, x, y], x] /. x -> -4) == 0, (D[h[t, x, y], x] /. x -> 4) == 0, (D[h[t, x, y], y] /. y -> -4) == 0, (D[h[t, x, y], y] /. y -> 4) == 0, a[0, x, y] == Exp[-(x^2 + y^2)], h[0, x, y] == Exp[-(x^2 + y^2)]}; sol = First[ NDSolve[{pde, bc}, {a[t, x, y], h[t, x, y]}, {t, 0, 2}, {x, -4, 4}, {y, -4, 4}]] Thanks URL: ,
Subject (listing for 'simultaneous partial differential equation') Author Date Posted simultaneous partial differential equation Hiroki 11/29/08 02:47am Re: simultaneous partial differential equation niels 12/03/08 10:24am Re: Re: simultaneous partial differential equat... Hiroki 12/07/08 12:42pm
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https://community.qlik.com/t5/Qlik-Sense-App-Development/Average-Standard-deviation-Confidence-interval/m-p/1315724/highlight/true | 1,561,503,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999948.3/warc/CC-MAIN-20190625213113-20190625235113-00300.warc.gz | 396,892,425 | 44,755 | # Qlik Sense App Development
Discussion board where members can learn more about Qlik Sense App Development and Usage.
Not applicable
## Average, Standard deviation, Confidence interval, ...
Hi guys !
I'm new to Qlik Sense, and i'm trying to develop an app.
One of my source data file has the same format as the one you will find attached to this message : 4 raws, the first with a category of product, the second with the type of product (can be the same type but not the same category), the third for the sales (whatever the unit is), and the fourth, the date of sale.
What I would like to do seems easy, but I couldn't find how to make it :
I need to calculate and display the average of sales per day, but making first the sub-total for each category.
I explain : for instance, the total of sales for category A and day 01/01/2010 is 50 (15+25+10)
for category B same day is 60 (30+20+10)
for category C same day 70 (20+20+30)
What I need to do is to calculate the average of sales for categories A B and C per day. For this example and the date 01/01/2010, it would be (50+60+70)/3 = 60. And so on for every days... and drawing the curve.
And the thing I was doing till now was calculating the average for day 01/01/2010 of each individual value (like (15+25+10+30+20+10+20+20+30) / 9 ) ... So not the same result at all ..
How should I do to perform my calculation ? You would help me so much guys !
Once this will be done, I'll have another problem of confidence interval :
I'll need to calculate the standard deviation between the total for categories A B and C per day (like for day 01/01/2010 : standard deviation between 50, 60 and 70, for the date 02/01/2010, standard deviation between [total category A at that day = 40] and [total category B at that date = 30] and [total category C at that date = 50] etc..), in order to display a 90% confidence interval around my average curve (of course my original data are way more complete than this example file, but the way to think is exactly the same).
So if you already have a clue for that as well ....
Thanks sooo much, waiting for your help !
Fab.
Tags (4)
9 Replies
Highlighted
MVP
## Re: Average, Standard deviation, Confidence interval, ...
Are these the numbers you expect to see?
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
Exactly what I needed ! Thanks so much.
Now I can draw my confidence interval, using the statistics normal distribution law ! like placing my upper and lower lines with the formula :
average + Z (according to Z table) * Stdev / sqrt(total number) (for the max line)
or average - Z (according to Z table) * Stdev / sqrt(total number) (for the min line)
No better solution ?
MVP
## Re: Average, Standard deviation, Confidence interval, ...
I think that is how you will have to do this... can't think of any better way to do this
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
I finally could do this, thanks again for your good help !
An easier way for a 95% interval was just to display
Avg(Aggr(Sum(Sales), Category, Date)) +/- 2 * Stdev(Aggr(Sum(Sales), Category, Date))
I just remembered this 1, 2 or 3 sigmas rule ! ahah
So the same if you want a 68% interval, just substitute factor 2 by factor 1, and for a 99% interval, substitute factor 2 by factor 3, and it's done !
Works perfectly for me !
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
Hi there !
Another question related to this previous example, that we solved last week :
The solution allowed us to draw another graph with the average and a confidence interval, that's great.
But what if now, I would like to display this average and interval in the same graph as the one representing the sales on vertical axis, and for dimensions date and category ?
The problem is that, I have two dimensions : Dates and Category, and the formulas Sunny adviced me last week are to be set as measures. However, you can't draw a graph with several dimensions AND several measures. You can just with several dimensions and one measure, or several measures and one dimension.
What do you advice me guys ?
What i want to see is the 3 curves of evolution of the sales along the time (one per category), and on the same graph, the average, and the confidence interval as we found last week. In fact, the final aim would be to be able to see when one curve goes out of this interval and is in a "risky" situation.
Thanks in advance for your precious help
MVP
## Re: Average, Standard deviation, Confidence interval, ...
Would you be able to share a sample outside of Qlik Sense to show what you are trying to get? May be create a dummy Excel sample where you can show the graph you are looking to get based on the sample data you provided or new sample data?
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
Hi Sunny !
Sorry for the delay in responding !
So, i put attached a new excel file simplified : 3 raws, one for the name, one for the date, one for the money they have.
What i want to see is on the same visualization 1) the evolution of the money for each character along the time (from day -5 till day 0 = today) and 2) the confidence interval of 95% as we did before.
The 1), I did it easily : creating two dimensions Date and Name, and one measure : Sum(Money). As you can see on "Picture_1"
The 2), thanks to what you explained previously, I do it easily on another view, creating one dimension : Date, and two measures :
Avg(Aggr(Sum(Money), Date)) + 2 * Stdev(Aggr(Sum(Money), Date))
and
Avg(Aggr(Sum(Money), Date)) - 2 * Stdev(Aggr(Sum(Money), Date))
My question is now, how could I display this confidence interval on my visualization 1), to be fast able to see which is out of the confidence interval at a D date ? (as drawn on Picture_2, with in red the lower limit and green the upper limit of the 95% interval). In fact, i have already several dimensions (2 in this example), and I need several measures (3 in this example), which is impossible ...
Thanks so much for your great help and availability !
Picture 1
Picture 2
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
Thanks for your help Sunny ! You can find all that in my message below
Not applicable
## Re: Average, Standard deviation, Confidence interval, ...
Up ! Help pleaaaaase.
Thx so much in advance to those who can help me. | 1,571 | 6,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-26 | latest | en | 0.920464 |
https://www.traditionaloven.com/tutorials/distance/convert-china-chi-unit-to-light-days-ld.html | 1,627,435,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00040.warc.gz | 1,115,455,135 | 17,453 | Convert 市尺 to ld | Chinese chǐ to light-days
# length units conversion
## Amount: 1 Chinese chǐ (市尺) of length Equals: 0.000000000000013 light-days (ld) in distance
Converting Chinese chǐ to light-days value in the length units scale.
TOGGLE : from light-days into Chinese chǐ in the other way around.
## length from Chinese chǐ to light-day conversion results
### Enter a new Chinese chǐ number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many light-days are in 1 Chinese chǐ? The answer is: 1 市尺 equals 0.000000000000013 ld
## 0.000000000000013 ld is converted to 1 of what?
The light-days unit number 0.000000000000013 ld converts to 1 市尺, one Chinese chǐ. It is the EQUAL length value of 1 Chinese chǐ but in the light-days distance unit alternative.
市尺/ld length conversion result From Symbol Equals Result Symbol 1 市尺 = 0.000000000000013 ld
## Conversion chart - Chinese chǐ to light-days
1 Chinese chǐ to light-days = 0.000000000000013 ld
2 Chinese chǐ to light-days = 0.000000000000026 ld
3 Chinese chǐ to light-days = 0.000000000000039 ld
4 Chinese chǐ to light-days = 0.000000000000051 ld
5 Chinese chǐ to light-days = 0.000000000000064 ld
6 Chinese chǐ to light-days = 0.000000000000077 ld
7 Chinese chǐ to light-days = 0.000000000000090 ld
8 Chinese chǐ to light-days = 0.00000000000010 ld
9 Chinese chǐ to light-days = 0.00000000000012 ld
10 Chinese chǐ to light-days = 0.00000000000013 ld
11 Chinese chǐ to light-days = 0.00000000000014 ld
12 Chinese chǐ to light-days = 0.00000000000015 ld
13 Chinese chǐ to light-days = 0.00000000000017 ld
14 Chinese chǐ to light-days = 0.00000000000018 ld
15 Chinese chǐ to light-days = 0.00000000000019 ld
Convert length of Chinese chǐ (市尺) and light-days (ld) units in reverse from light-days into Chinese chǐ.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: Chinese chǐ (市尺) is used for measuring length.
Second: light-day (ld) is unit of distance.
QUESTION:
15 市尺 = ? ld
15 市尺 = 0.00000000000019 ld
Abbreviation, or prefix, for Chinese chǐ is:
Abbreviation for light-day is:
ld
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Chinese chǐ and light-days ( 市尺 vs. ld ) measures exchange.
2. for conversion factors between unit pairs.
3. work with length's values and properties.
To link to this length Chinese chǐ to light-days online converter simply cut and paste the following.
The link to this tool will appear as: length from Chinese chǐ (市尺) to light-days (ld) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 870 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.808773 |
http://analyticspro.org/2016/03/14/r-tutorial-interpretation-of-r-squared-and-adjusted-r-squared-in-regression/?share=google-plus-1 | 1,591,384,066,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00418.warc.gz | 6,827,930 | 28,748 | # R Tutorial : Interpretation of R Squared and Adjusted R Squared in Regression
This tutorial talks about interpretation of the most fundamental measure reported for models which is R Squared and Adjusted R Squared. We will try to give a clear guidelines for interpreting R Squared and Adjusted R Squared
Once we have fitted our model to data using Regression , we have to find out how well our model fits the data. R gives many goodness of fit statistic out of the box when we create a model. In this tutorial we will discuss about an important statistic called R-Squared ( R² ). We will also try to bust myths that Low R Squared values are always bad and High R Squared values are always good.
By the way you should look at R Squared only once your model passes Residual analysis test as mentioned R Tutorial : Residual Analysis for Regression and R Tutorial : How to use Diagnostic Plots for Regression Models
### What is R Squared ?
R Squared is a measure which tells us how well our regression equation explains observed data values.
R Squared = ( Explained Variation in Observed Values) / (Total variation in Observed Values)
0% < = R Squared <= 100%
So R² = 67% implies that you have a regression equation which can explain 67% variation of observed values around mean.
Obviously when you add more predictor variables to regression equation which explain more variance you will get a higher R². Does it mean that when we compare 2 models on same data , the model with higher R² is always better than the model with lower R² ?
The answer is NO . Not always ! More predictor variables in a model implies more complexity which may have a side effect of Over fitting. So pure R² is not a very reliable measure. We need a measure which can tell us in absolute terms whether addition of new variable can explain variance worth of the additional Complexity.
Its for this reason that we use Adjusted R² .
### What is Adjusted R² ?
Adjusted R² is a measure derived from R² which penalizes each addition of variable for additional complexity.
N = Sample Size
p = number of predictors
Please note that p is in denominator and increased p would b=mean a decreased Adj R² if R² does not increase enough and everything else remains constant.
### Is Low R² always bad ?
NO. Desirable range of R² is highly domain dependent. Any model which attempts to predict Human behavior is seldom very precise and hence lower R² is expected. Where as for models in medicine and pharma R² values above 90% are very common.
### Is High R² always good ?
NO. As mentioned in R Tutorial : Residual Analysis for Regression and R Tutorial : How to use Diagnostic Plots for Regression Models even if you have High R² but you have some inherent Residual pattern or the residuals are Heteroscedastic or if residuals are not normally distributed then the model is not considered good enough.
As a next step you should look at interpretation of F Statistic.
Enjoy Life and Keep Learning ! | 653 | 2,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-24 | latest | en | 0.924036 |
https://socratic.org/questions/how-do-you-write-standard-form-of-equation-of-the-parabola-that-vertex-is-2-3-an | 1,719,306,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00675.warc.gz | 465,354,271 | 6,071 | # How do you write standard form of equation of the parabola that vertex is (2,3) and graph passes through the given point (0,2)?
Mar 19, 2018
The equation of parabola is $y = - \frac{1}{4} {\left(x - 2\right)}^{2} + 3$.
#### Explanation:
The vertex form of equation of parabola is
y = a(x-h)^2+k ; (h,k) being vertex , here $h = 2 , k = 3$
So the equation of parabola is y = a(x-2)^2+3 ;. The parabola
passes through point $\left(0 , 2\right) \therefore$ the point will satisfy the equation
of parabola. Putting $x = 0 \mathmr{and} y = 2$ in the equation we get,
$2 = a {\left(0 - 2\right)}^{2} + 3 \mathmr{and} 2 = 4 a + 3 \mathmr{and} 4 a = 2 - 3$or
$4 a = - 1 \mathmr{and} a = - \frac{1}{4}$. Hence ,the equation of parabola is
$y = - \frac{1}{4} {\left(x - 2\right)}^{2} + 3$.
graph{-1/4(x-2)^2+3 [-10, 10, -5, 5]} [Ans] | 339 | 837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.778704 |
https://brainor.com/math-worksheets/grade-1/addition/addition-combination-of-10/ | 1,712,975,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00862.warc.gz | 126,923,469 | 44,194 | # 15+ Addition Combinations of 10 Worksheets | Free Printables
These worksheets on addition combinations of 10 will guide you to become more proficient in calculations and increase your understanding of mathematics. Using our free printable worksheets, first-graders will practice adding numbers to strengthen their math skills.
## Making Addition Combinations of 10
In these addition combinations worksheets, students will add the desired number to make 10.
## Using Different Number of Combinations to Make 10
Use these three worksheets on addition combinations of 10 to complete the explicit addendum.
## Cut and Paste the Numbers after the Addition Combinations of 10
Cutting and pasting random numbers to make 10, will help the first graders to solve addition combinations easily.
## Counting Dots and Adding Combinations to Make 10
After counting different numbers of dots, write the correct number to make 10.
## Matching Addition Combination of 10
Here are some addends that lead to an additional combination of 10. Students need to match the exact same number which makes 10 with an arrow.
## Word Problems on Addition Combination of 10
These three worksheets are about practicing various word problems to understand and calculate the addition combinations of 10. | 253 | 1,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.891067 |
https://www.fool.co.uk/investing/2013/11/01/jw-this-model-suggests-glaxosmithkline-plc-could-deliver-a-11-annual-return/ | 1,558,646,228,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00003.warc.gz | 790,824,705 | 23,980 | # This Model Suggests GlaxoSmithKline plc Could Deliver An 11% Annual Return
One of the risks of being an income investor is that you can be seduced by attractive yields, which are sometimes a symptom of a declining business or a falling share price.
Take the UK’s largest pharmaceutical firm, GlaxoSmithKline (LSE: GSK) (NYSE: GSK.US), for example. The firm’s 4.7% prospective yield is attractive, but, 4.7% is substantially less than the long-term average total return from UK equities, which is about 8%.
Total return is made up of dividend yield and share price growth combined — but Glaxo shares are already up by 23% this year, and the firm is facing corruption allegations in China that triggered a 9% fall in Asian sales during the third quarter. Can Glaxo overcome these headwinds to deliver capital growth?
### What will Glaxo’s total return be?
Looking ahead, I need to know the expected total return from my Glaxo shares, so that I can compare them to my benchmark, a FTSE 100 tracker.
The dividend discount model is a technique that’s widely used to value dividend-paying shares. A variation of this model also allows you to calculate the expected rate of return on a dividend paying share:
Total return = (Prospective dividend ÷ current share price) + expected dividend growth rate
Rather than guess at future growth rates, I usually average dividend growth between 2009 and the current year’s forecast payout, to provide a more reliable guide to the underlying trend. Here’s how this formula looks for Glaxo:
(77.8 ÷ 1645) + 0.063 = 0.110 x 100 = 11.0%
My model suggests that Glaxo shares could deliver a 11% total return over the next few years, outperforming the long-term average total return of 8% per year I’d expect from a FTSE 100 tracker.
### Isn’t this too simple?
One limitation of this formula is that it doesn’t tell you whether a company can afford to keep paying and growing its dividend.
My preferred measure of dividend affordability is free cash flow — the cash that’s left after capital expenditure and tax costs.
Free cash flow is normally defined as operating cash flow – tax – capex.
In Glaxo’s case, the firm’s 2012 free cash flow of £1,744m was insufficient to cover its £3,814m dividend payments, but this is the exception — from 2011 back to at least 2007, Glaxo’s dividend has been amply covered by free cash flow.
### An alternative to Glaxo?
Despite the appeal of Glaxo's dividend, I am concerned about its £15bn net debt, and I believe there are better choices elsewhere for income investors, including one company that's been named by the Motley Fool's analysts as "Today's Top Income Buy".
The company concerned currently offers an inflation-linked, 5.5% dividend yield, and the Fool's analysts believe it is currently trading at almost 10% below its fair value. | 663 | 2,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-22 | latest | en | 0.919168 |
https://tutorial.math.lamar.edu/Solutions/CalcII/ImproperIntegralsCompTest/Prob2.aspx | 1,685,254,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00615.warc.gz | 664,210,496 | 17,309 | Paul's Online Notes
Home / Calculus II / Integration Techniques / Comparison Test for Improper Integrals
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### Section 7.9 : Comparison Test for Improper Integrals
2. Use the Comparison Test to determine if the following integral converges or diverges.
$\int_{3}^{\infty }{{\frac{{{z^2}}}{{{z^3} - 1}}\,dz}}$
Show All Steps Hide All Steps
Hint : Start off with a guess. Do you think this will converge or diverge?
Start Solution
The first thing that we really need to do here is to take a guess on whether we think the integral converges or diverges.
The “-1” in the denominator does not really change the size of the denominator as $$z\ gets really large and so hopefully it makes sense that we can guess that this integral should behave like, $\int_{3}^{\infty }{{\frac{{{z^2}}}{{{z^3}}}\,dz}}=\int_{3}^{\infty }{{\frac{1}{z}\,dz}}$ Then, by the fact from the previous section, we know that this integral diverges since \(p = 1 \le 1$$ .
Therefore, we can guess that the integral,
$\int_{3}^{\infty }{{\frac{{{z^2}}}{{{z^3} - 1}}\,dz}}$
will diverge.
Be careful from this point on! One of the biggest mistakes that many students make at this point is to say that because we’ve guessed the integral diverges we now know that it diverges and that’s all that we need to do and they move on to the next problem.
Another mistake that students often make here is to say that because we’ve guessed that the integral diverges they make sure that the remainder of the work in the problem supports that guess even if the work they do isn’t correct.
All we’ve done is make a guess. Now we need to prove that our guess was the correct one. This may seem like a silly thing to go on about, but keep in mind that at this level the problems you are working with tend to be pretty simple (even if they don’t always seem like it). This means that it will often (or at least often once you get comfortable with these kinds of problems) be pretty clear that the integral converges or diverges.
When these kinds of problems arise in other sections/applications it may not always be so clear if our guess is correct or not and it can take some real work to prove the guess. So, we need to be in the habit of actually doing the work to prove the guess so we are capable of doing it when it is required.
The hard part with these problems is often not making the guess but instead proving the guess! So let’s continue on with the problem.
Hint : Now that we’ve guessed the integral diverges do we want a larger or smaller function that we know diverges?
Show Step 2
Recall that we used an area analogy in the notes of this section to help us determine if we want a larger or smaller function for the comparison test.
We want to prove that the integral diverges so if we find a smaller function that we know diverges the area analogy tells us that there would be an infinite amount of area under the smaller function.
Our function, which would be larger, would then also have an infinite amount of area under it. There is no way we can have an finite amount of area covering an infinite amount of area!
Note that the opposite situation does us no good. If we find a larger function that we know diverges (and hence will have a infinite amount of area under it) our function (which is now smaller) can have either a finite amount of area or an infinite area under it.
In other words, if we find a larger function that we know diverges this will tell us nothing about our function. However, if we find a smaller function that we know diverges this will force our function to also diverge.
Therefore we need to find a smaller function that we know diverges.
Show Step 3
Okay, now that we know we need to find a smaller function that we know diverges.
So, let’s start with the function from the integral. It is a fraction and we know that we can make a fraction smaller by making the denominator larger. Also note that for $$z > 3$$ (which we can assume from the limits on the integral) we have,
${z^3} - 1 < {z^3}$
Therefore, we have,
$\frac{{{z^2}}}{{{z^3} - 1}} > \frac{{{z^2}}}{{{z^3}}} = \frac{1}{z}$
since we replaced the denominator with something that we know is larger.
Show Step 4
Finally, we know that,
$\int_{3}^{\infty }{{\frac{1}{z}\,dz}}$
diverges. Then because the function in this integral is smaller than the function in the original integral the Comparison Test tells us that,
$\int_{3}^{\infty }{{\frac{{{z^2}}}{{{z^3} - 1}}\,dz}}$
must also diverge. | 1,213 | 4,939 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | longest | en | 0.897361 |
http://www.usingenglish.com/forum/threads/120246-On-the-Corner-In-the-Corner | 1,405,156,599,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776432860.32/warc/CC-MAIN-20140707234032-00063-ip-10-180-212-248.ec2.internal.warc.gz | 574,230,732 | 13,561 | # Thread: On the Corner / In the Corner
1. Banned
Join Date
Apr 2010
Posts
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Post Thanks / Like
## On the Corner / In the Corner
Morning, I have seen “on the corner” and “in the corner”. Which is correct? Thanks
2. ## Re: On the Corner / In the Corner
Hi ArturoK2000,
Both.
“On the corner” = outside the building.
Example: We’ll meet on the corner of Torrance St.
“In the corner” = inside the building.
Example: The table is in the corner of the room.
Regards,
José Manuel Rosón Bravo
3. Senior Member
Join Date
May 2007
Posts
845
Post Thanks / Like
## Re: On the Corner / In the Corner
Would it also be a matter of AE (in the corner) and BE (on the corner)?
4. Key Member
Join Date
Mar 2008
Posts
2,129
Post Thanks / Like
## Re: On the Corner / In the Corner
Originally Posted by ArturoK2000
Morning, I have seen “on the corner” and “in the corner”. Which is correct? Thanks
I live in the big house on the corner street.
We use on here, because the type of corner ( a street corner) does not have a significant 3rd dimension (height)-- it is merely two-dimensional, so that an object would be on it. Otherwise, to indicte directions we use on/at
When I saw him, he was standing at/on the corner of that Street
If we are speaking of a box, room, or other enclosed object with a relatively significant vertical dimension, Or for indicating the position of something,
we use in:
The broom is in the corner of the room.
A vase stood in the corner of the room.
5. ## Re: On the Corner / In the Corner
Originally Posted by albertino
Would it also be a matter of AE (in the corner) and BE (on the corner)?
I don't think so, no.
#### Posting Permissions
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• | 476 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-23 | latest | en | 0.897519 |
https://learnengineering.org/working-of-steam-turbine/ | 1,531,819,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589634.23/warc/CC-MAIN-20180717090227-20180717110227-00061.warc.gz | 685,646,274 | 33,035 | Working of Steam Turbine
Working of Steam Turbine
Steam turbines are the hearts of the power plants, they are the devices which transform thermal energy in fluid to mechanical energy. In this video lecture working of steam turbine is explained in a logical manner.
A detailed webpage version of the video is given below.
Energy Absorption from fluid - Role of Rotor Blades
When high energy fluid (high pressure and high temperature) passes through series of rotor blades, it absorbs energy from fluid and starts rotating, thus it transforms thermal energy in fluid to mechanical energy.
Fig.1 Rotating blades of turbine helps in transforming thermal in fluid to mechanical energy
So series of such blade which eventually transform thermal energy are the most vital part of a steam turbine. One of such rotor set is shown in figure below.
Fig.2 A typical steam turbine rotor
If you take a close look at one of the blade, it would be clear that a blade is a collection of airfoil cross sections from bottom to top. When flow passes through such airfoils it induces a low pressure on bottom surface and high pressure on top surface of airfoil as shown in figure below.
Fig.3 Fluid flow around airfoil cross sectioned blade induces a high pressure (P) and low pressure(P) on blade surfaces
This pressure difference will induce a resultant force in upward direction, thus making the blade rotate. So some part of fluid energy will get transformed to mechanical energy of blade. Before analyzing energy transfer from fluid to blade, we will have a close look at energy associated with a fluid.
Energy Associated with a Fluid
A flowing fluid can have 3 components of energy components
• Kinetic energy - Virtue of its velocity
• Pressure Energy - Virtue of its pressure
• Internal Energy - Virtue of its temperature
Last 2 components of energy together known as enthalpy. So total energy in a fluid can be represented as sum of kinetic energy and enthalpy.
Energy Transfer to Rotors
When fluid passes through rotor blades it loses some amount of energy to the rotor blades. Due to this both kinetic and enthalpy energy of fluid come down for a typical rotor. As kinetic energy comes down velocity of flow decreases. If we directly pass this stream to next stage of rotor blades it will not transfer much energy because of low velocity of flow stream. So before passing the stream to next rotor stage we have to increase the velocity first. This is achieved with use of a set of stationary nozzle blades, also known as stator. When fluid passes through stator blades velocity of fluid increase due to its special shape thus one part of enthalpy energy will get converted into kinetic energy. Thus enthalpy of stream reduces and kinetic energy of stream increase. It is to be noted that here there is no energy addition or removal from flow, what happens here is conversion of enthaply energy into kinetic energy. Now this steam of fluid can be passed to next rotor blades and process can be repeated. Velocity and enthalpy variation of flow is shown in following figure.
Fig.4 Velocity and enthalpy variations across rotor and stator stages of a typical steam turbine
Degree of Energy Transfer
Total energy transfer to the rotor blade is sum of decrease in kinetic energy and decrease in enthalpy. Degree of contribution of each term is an important parameter in axial flow machines. This is represented by a term called of degree of reaction, which is defined as
Where both enthalpy change and kinetic energy changes are defined across the rotor blade.
0 % Reaction - Impulse Turbines
When D.O.R = 0 there will not be any enthalpy change across the rotor, such a turbine is known as impulse turbine. Blades of such a turbine would like as shown below.
Fig.5 A typical impulse turbine rotor cross section and flow pattern
Here incoming flow stream hits the blade and produces and impulse force on it. Since enthalpy across the blade does not change temperature will also remain same. There will be minor pressure drop across the rotor, but this is almost negligible. Here energy transfer to the blade is purely due to decrease in kinetic energy of fluid.
100 % Reaction Turbines
When D.O.R = 1 kinetic energy change across the rotor will be zero, energy transfer will be purely due to decrease in enthalpy. Since kinetic energy is same across the rotor absolute value of velocities remain same. This is shown in figure below.
Fig.6 A typical reaction turbine rotor cross section and flow pattern
Usually people use compromise of above two discussed cases,that is 50% D.O.R . Such turbines are known as Parson turbines, where both kinetic and enthalpy energy transfer contribute equally to power transfer to rotor. | 962 | 4,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-30 | longest | en | 0.915873 |
https://help.idecad.com/ideCAD/punching-shear-check-example-1 | 1,726,672,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00188.warc.gz | 265,148,032 | 43,682 | # Punching Shear Check Example 1
Punching Shear Design Example
In the example calculation, a punching calculation will be made for a column whose dimensions are 100x80. Other parameters in the calculation are listed as follows.
• Slab thickness: 27 cm
• Slab cover: 3 cm
• Material : C35/S420
• fctd = 1380.42 kN/m2 (Concrete design tensile strength)
The slab useful height, d , is calculated;
d = ( Slab thickness - Slab cover ) = 27 - 3 = 24 cm
Column dimensions
c1 = 100 cm c2 = 80 cm
The punching circumference (up) is calculated at a distance d/2 from the column surface and is shown in the picture below. In this case, punching is done by adding d to the column dimensions to find the perimeter's edges (b1, b2).
In this case, the punching circumference (up ) and the punching area (Az) obtained by multiplying the punching perimeter by the flooring useful height, d, is calculated as shown below.
The shear stresses plotted below are the punching stresses perpendicular to the slab plane.
The J values are the sum of the polar inertia and second moments of inertia of the surfaces forming the punching area (Az). TBDY Equation 7.28 calculates this value according to the loading direction considering the γf coefficient.
Calculation of J and γf Coefficients for Strong Axis (Major Direction)
The coefficients γf(maj) and γv(maj) are calculated as follows.
The following operations are performed to find the J (maj) value.
c (maj) is the center of gravity distance perpendicular to the moment vector, which is considered when finding the J value ( J (maj) ) on the strong axis of the section. Since the punching circumference is rectangular,
The polar inertia and second moments of inertia are calculated as follows, respectively.
In this case, the sum of the polar inertia and second moments of inertia about the section's strong (major) axis, J (maj), is found as follows.
Calculation of J and γf Coefficients for Weak Axis ( Minor Direction )
The following steps are followed to calculate the γ f(min) and γ v(min) coefficients.
The J(min) value is found as follows.
The c(min) value is the center of gravity distance perpendicular to the moment vector, which is considered when finding the J value (J (min)) on the weak axis of the section. Since the punching circumference is rectangular,
In this case, the polar and second moments of inertia are found as follows, respectively.
In this case, the sum of the polar inertia and second moments of inertia about the section's weak (minor) axis, J (min), is found as follows.
Finding Punching Stresses
The forces to be considered for punching stresses and the values obtained from the hand geometry are given below.
• Vd = 683.13 kN
• DM d(maj) = 143.53 kNm
• DM d(min) = 546.65 kNm
• Az = 1.0944 m 2
• γf(maj) = 0.579
• γv(maj) = 0.421
• γf(min) = 0.621
• γv(min) = 0.379
• J(maj) = 0.27101448 m 4
• J(min) = 0.2083328 m 4
• c(maj) = 0.62 m
• c(min) = 0.52m
• c'(maj) = 0.62 m
• c'(min) = 0.52 m
The punching stresses are found by substituting all the above values in the stress formulas and the internal forces essential to the punching design.
The absolute value of τ pd,1 , τ pd,2 is τ pd,1 = 1279.56 kN/m 2 . This obtained value is compared with the fctd value, which is the concrete design tensile strength.
τpd,1 = 1279.56 kN/m2 < fctd = 1380.42 kN/m2
Upholstery punching strength is sufficient. These values can also be compared with the report results.
Punching Shear Design Software
ideCAD Structural is an All-in-One Structural Engineering Software with Punching Shear Design and many other Structural Design Code Check tools. | 951 | 3,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.888352 |
http://mathoverflow.net/questions/9878/a-product-of-gamma-values-over-the-numbers-coprime-to-n?answertab=oldest | 1,461,918,241,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110805.57/warc/CC-MAIN-20160428161510-00081-ip-10-239-7-51.ec2.internal.warc.gz | 118,536,908 | 13,620 | # A product of gamma values over the numbers coprime to n.
Let φ(n) denote Euler's totient function and k $\perp$ n denote that k, n are integers and relatively prime. Let N = φ(n) + 1. If n is not a prime power $$\prod_{\substack{0 < k < n, \\ k \perp n }} \Gamma \left(\frac{k}{n}\right) = \sqrt{N}\prod_{ 0 < k < N}\Gamma \left(\frac{k}{N}\right) \quad (n \neq p^a) .$$ This identity was proved here as a corollary to some other identities, but the authors asked: is there some natural direct proof of this formula?´´
-
## 1 Answer
Denote $$f(n)=\prod_{k=1}^{n-1}\Gamma \left(\frac{k}{n}\right)$$ and $$F(n)=\prod_{1\le k\le n-1, k\perp n}\Gamma \left(\frac{k}{n}\right)$$ We have $f(n)=\prod_{d|n}F(n)$ and therefore by Mobius inversion $F(n)=\prod_{d|n}f(d)^{\mu(n/d)}$
By the multiplication theorem we have $f(n)=\frac{1}{\sqrt{n}}(2\pi)^{\frac{n-1}{2}}$, so if $n$ is not a prime power $$F(n)=\prod_{d|n}\left(\frac{1}{\sqrt{d}}(2\pi)^{\frac{d-1}{2}}\right)^{\mu(n/d)}=(2\pi)^{\frac{1}{2}\varphi (n)}$$ The formula $F(n)=\sqrt{\varphi(n)+1}f(\varphi(n)+1)$ follows.
- | 411 | 1,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-18 | latest | en | 0.609787 |
https://study.com/academy/answer/a-choose-the-circuit-showing-how-a-voltmeter-would-be-connected-to-measure-the-voltage-across-72-6-space-k-omega-resistor-b-assuming-the-voltmeter-to-be-ideal-what-is-its-reading.html | 1,642,653,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00177.warc.gz | 587,927,052 | 21,448 | # a. Choose the circuit showing how a voltmeter would be connected to measure the voltage across...
## Question:
a. Choose the circuit showing how a voltmeter would be connected to measure the voltage across {eq}72.6 \space k \Omega {/eq} resistor.
b. Assuming the voltmeter to be ideal, what is its reading?
## Voltmeter:
Like in most measuring instruments, the design considerations of the voltmeter are linked to the application. For example, the high internal resistance of the voltmeter, meant to minimize channeling circuit current to the meter and thus compromising the circuit parameters, also means that we connect the meter in parallel.
## Answer and Explanation: 1
Become a Study.com member to unlock this answer! Create your account
#### Part (a)
{eq}\boxed{A.} {/eq}
A voltameter is connected across the two points of which the potential difference is required.
#### Part (b)
We...
See full answer below.
Parallel Circuits: Calculating Currents with Ohm's Law
from
Chapter 9 / Lesson 6
2.6K
Understand Kirchhoff's law and Ohm's law for parallel circuits, and see how to find current in a parallel circuit using the parallel circuit current formula. | 274 | 1,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.853708 |
https://rpact.org/vignettes/planning/rpact_vs_gsdesign_examples/index.html | 1,696,033,649,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00012.warc.gz | 549,269,541 | 12,743 | # Comparing Sample Size and Power Calculation Results for a Group Sequential Trial with a Survival Endpoint: rpact vs. gsDesign
Planning
Survival
This document provides an example that illustrates how to compare sample size and power calculation results of the two different R packages rpact and gsDesign.
Author
Gernot Wassmer, Friedrich Pahlke, and Marcel Wolbers
Published
July 6, 2023
# The design
• 1:1 randomized
• Two-sided log-rank test; 80% power at the 5% significance level (or one-sided at 2.5%)
• Target HR for primary endpoint (PFS) is 0.75
• PFS in the control arm follows a piece-wise exponential distribution, with the hazard rate h(t) estimated using historical controls as follows:
• h(t) = 0.025 for t between 0 and 6 months;
• h(t) = 0.04 for t between 6 and 9 months;
• h(t) = 0.015 for t between 9 and 15 months;
• h(t) = 0.01 for t between 15 and 21 months;
• h(t) = 0.007 for t beyond 21 months.
• An annual dropout probability of 20%
• Interim analyses at 33% and 70% of total information
• Alpha-spending version of O’Brien-Fleming boundary for efficacy
• No futility interim
• 1405 subjects recruited in total
• Staggered recruitment:
• 15 pt/month during first 12 months;
• subsequently, increase of # of sites and ramp up of recruitment by +6 pt/month each month until a maximum of 45 pt/month
# Calculation with gsDesign
``````# Load the package `gsDesign`
library(gsDesign)
options(warn = -1) # avoid warnings generated by gsDesign
x <- gsSurv(
k = 3, test.type = 1, alpha = 0.025, beta = 0.2,
timing = c(0.33, 0.7), sfu = sfLDOF, # boundary
hr = 0.75,
lambdaC = c(0.025, 0.04, 0.015, 0.01, 0.007), # piecewise lambdas
S = c(6, 3, 6, 6), # piecewise survival times
eta = -log(1 - 0.2) / 12, # dropout
gamma = c(15, 21, 27, 33, 39, 45), # recruitment, pt no
R = c(12, 1, 1, 1, 1, (1405 - 300) / 45), # recruitment duration
minfup = NULL
)
print(x, digits = 5)``````
``````Time to event group sequential design with HR= 0.75
Equal randomization: ratio=1
One-sided group sequential design with
80 % power and 2.5 % Type I Error.
Analysis N Z Nominal p Spend
1 128 3.73 0.0001 0.0001
2 271 2.44 0.0074 0.0073
3 386 2.00 0.0227 0.0176
Total 0.0250
++ alpha spending:
Lan-DeMets O'Brien-Fleming approximation spending function with none = 1.
Boundary crossing probabilities and expected sample size
assume any cross stops the trial
Upper boundary (power or Type I Error)
Analysis
Theta 1 2 3 Total E{N}
0.0000 0.0001 0.0073 0.0176 0.025 385.0
0.1437 0.0175 0.4517 0.3309 0.800 329.1
T n Events HR efficacy
IA 1 26.78703 785.4162 127.3407 0.516
IA 2 38.62360 1318.0620 270.1171 0.743
Final 50.80093 1405.0000 385.8810 0.816
Accrual rates:
Stratum 1
0-12 15
12-13 21
13-14 27
14-15 33
15-16 39
16-40.56 45
Control event rates (H1):
Stratum 1
0-6 0.025
6-9 0.040
9-15 0.015
15-21 0.010
21-Inf 0.007
Censoring rates:
Stratum 1
0-6 0.0186
6-9 0.0186
9-15 0.0186
15-21 0.0186
21-Inf 0.0186``````
# Calculation with rpact
## Design
``````# Load the package `rpact`
library(rpact)
packageVersion("rpact")``````
``````design <- getDesignGroupSequential(
sided = 1, alpha = 0.025, beta = 0.2,
informationRates = c(0.33, 0.7, 1),
typeOfDesign = "asOF"
)
kable(summary(design))``````
Sequential analysis with a maximum of 3 looks (group sequential design)
O’Brien & Fleming type alpha spending design, one-sided overall significance level 2.5%, power 80%, undefined endpoint, inflation factor 1.015, ASN H1 0.8656, ASN H01 0.9826, ASN H0 1.0127.
Stage 1 2 3
Information rate 33% 70% 100%
Efficacy boundary (z-value scale) 3.731 2.440 2.000
Stage levels (one-sided) <0.0001 0.0074 0.0227
Cumulative alpha spent <0.0001 0.0074 0.0250
Overall power 0.0175 0.4691 0.8000
Design parameters and output of group sequential design
User defined parameters
• Type of design: O’Brien & Fleming type alpha spending
• Information rates: 0.330, 0.700, 1.000
Derived from user defined parameters
• Maximum number of stages: 3
Default parameters
• Stages: 1, 2, 3
• Significance level: 0.0250
• Type II error rate: 0.2000
• Two-sided power: FALSE
• Test: one-sided
• Tolerance: 0.00000001
• Type of beta spending: none
Output
• Cumulative alpha spending: 0.00009549, 0.00738449, 0.02499999
• Critical values: 3.731, 2.440, 2.000
• Stage levels (one-sided): 0.00009549, 0.00735097, 0.02274488
## Sample size / timing of interim analyses
``````piecewiseSurvivalTime <- list(
"0 - <6" = 0.025,
"6 - <9" = 0.04,
"9 - <15" = 0.015,
"15 - <21" = 0.01,
">= 21" = 0.007
)
accrualTime <- list(
"0 - <12" = 15,
"12 - <13" = 21,
"13 - <14" = 27,
"14 - <15" = 33,
"15 - <16" = 39,
">= 16" = 45
)
y <- getPowerSurvival(
design = design, typeOfComputation = "Schoenfeld",
thetaH0 = 1, directionUpper = FALSE,
dropoutRate1 = 0.2, dropoutRate2 = 0.2, dropoutTime = 12,
allocationRatioPlanned = 1,
accrualTime = accrualTime,
piecewiseSurvivalTime = piecewiseSurvivalTime,
hazardRatio = 0.75,
maxNumberOfEvents = x\$n.I[3],
maxNumberOfSubjects = 1405
)
kable(summary(y))``````
Power calculation for a survival endpoint
Sequential analysis with a maximum of 3 looks (group sequential design), overall significance level 2.5% (one-sided). The results were calculated for a two-sample logrank test, H0: hazard ratio = 1, power directed towards smaller values, H1: hazard ratio = 0.75, piecewise survival distribution, piecewise survival time = c(0, 6, 9, 15, 21), control lambda(2) = c(0.025, 0.04, 0.015, 0.01, 0.007), maximum number of subjects = 1405, maximum number of events = 386, accrual time = c(12, 13, 14, 15, 16, 40.556), accrual intensity = c(15, 21, 27, 33, 39, 45), dropout rate(1) = 0.2, dropout rate(2) = 0.2, dropout time = 12.
Stage 1 2 3
Information rate 33% 70% 100%
Efficacy boundary (z-value scale) 3.731 2.440 2.000
Overall power 0.0175 0.4702 0.8009
Expected number of subjects 1354.8
Number of subjects 785.4 1318.1 1405.0
Expected number of events 328.9
Cumulative number of events 127.3 270.1 385.9
Analysis time 26.8 38.6 50.8
Expected study duration 44.9
Cumulative alpha spent <0.0001 0.0074 0.0250
One-sided local significance level <0.0001 0.0074 0.0227
Efficacy boundary (t) 0.516 0.743 0.816
Exit probability for efficacy (under H0) <0.0001 0.0073
Exit probability for efficacy (under H1) 0.0175 0.4526
Legend:
• (t): treatment effect scale
Design plan parameters and output for survival data
Design parameters
• Information rates: 0.330, 0.700, 1.000
• Critical values: 3.731, 2.440, 2.000
• Futility bounds (non-binding): -Inf, -Inf
• Cumulative alpha spending: 0.00009549, 0.00738449, 0.02499999
• Local one-sided significance levels: 0.00009549, 0.00735097, 0.02274488
• Significance level: 0.0250
• Test: one-sided
User defined parameters
• Direction upper: FALSE
• lambda(2): 0.025, 0.040, 0.015, 0.010, 0.007
• Hazard ratio: 0.750
• Maximum number of subjects: 1405
• Maximum number of events: 385.9
• Accrual time: 12.00, 13.00, 14.00, 15.00, 16.00, 40.56
• Accrual intensity: 15.0, 21.0, 27.0, 33.0, 39.0, 45.0
• Piecewise survival times: 0.00, 6.00, 9.00, 15.00, 21.00
• Drop-out rate (1): 0.200
• Drop-out rate (2): 0.200
Default parameters
• Theta H0: 1
• Type of computation: Schoenfeld
• Planned allocation ratio: 1
• kappa: 1
• Drop-out time: 12.00
Sample size and output
• lambda(1): 0.01875, 0.03000, 0.01125, 0.00750, 0.00525
• Total accrual time: 40.56
• Expected number of events: 328.9
• Overall reject: 0.8009
• Reject per stage [1]: 0.01754
• Reject per stage [2]: 0.45262
• Reject per stage [3]: 0.33070
• Early stop: 0.4702
• Analysis times [1]: 26.79
• Analysis times [2]: 38.62
• Analysis times [3]: 50.80
• Expected study duration: 44.87
• Maximal study duration: 50.80
• Number of events per stage [1]: 127.3
• Number of events per stage [2]: 270.1
• Number of events per stage [3]: 385.9
• Number of subjects [1]: 785.4
• Number of subjects [2]: 1318.1
• Number of subjects [3]: 1405
• Expected number of subjects: 1354.8
• Critical values (treatment effect scale) [1]: 0.516
• Critical values (treatment effect scale) [2]: 0.743
• Critical values (treatment effect scale) [3]: 0.816
Legend
• (i): values of treatment arm i
• [k]: values at stage k
# Comparison: Analysis time of rpact vs. gsDesign
Absolute differences:
``````timeDiff <- as.data.frame(sprintf("%.5f", (x\$T - y\$analysisTime)))
rownames(timeDiff) <- c("Stage 1", "Stage 2", "Stage 3")
colnames(timeDiff) <- "Difference analysis time"
kable(timeDiff)``````
Difference analysis time
Stage 1 -0.00000
Stage 2 0.00004
Stage 3 -0.00011
# Remark
Obviously, there is a difference in the calculation of the necessary number of events which are, in rpact, calculated as
``````(qnorm(0.975) + qnorm(0.8))^2 / log(0.75)^2 * 4 *
getDesignCharacteristics(getDesignGroupSequential(
sided = 1, alpha = 0.025,
kMax = 3, typeOfDesign = "asOF", informationRates = c(0.33, 0.7, 1)
))\$inflationFactor``````
``[1] 385.0479``
which is slightly different to the maximum number of events in gsDesign which is
``x\$n.I[3]``
``[1] 385.881``
Therefore, running
``````getSampleSizeSurvival(
design = design, typeOfComputation = "Schoenfeld",
thetaH0 = 1,
dropoutRate1 = 0.2, dropoutRate2 = 0.2, dropoutTime = 12,
allocationRatioPlanned = 1,
accrualTime = accrualTime,
piecewiseSurvivalTime = piecewiseSurvivalTime,
hazardRatio = 0.75,
maxNumberOfSubjects = 1405
)\$analysisTime``````
`````` [,1]
[1,] 26.76183
[2,] 38.57834
[3,] 50.63114``````
is not exactly equal to getPowerSurvival from above. This, however, has definitely no consequences in practice but explains the slight differences in rpact and gsDesign.
System: rpact 3.4.0, R version 4.2.2 (2022-10-31 ucrt), platform: x86_64-w64-mingw32
To cite R in publications use:
R Core Team (2022). R: A language and environment for statistical computing. R Foundation for Statistical Computing, Vienna, Austria. URL https://www.R-project.org/.
To cite package ‘rpact’ in publications use:
Wassmer G, Pahlke F (2023). rpact: Confirmatory Adaptive Clinical Trial Design and Analysis. https://www.rpact.org, https://www.rpact.com, https://github.com/rpact-com/rpact, https://rpact-com.github.io/rpact/. | 3,681 | 10,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | latest | en | 0.84839 |
https://www.bartleby.com/questions-and-answers/an-organization-purchased-a-fouryear-dollar10000-bond-paying-a-7percent-annual-coupon-and-interest-r/870a2be4-18f0-416f-bd03-cc9e55856b3c | 1,580,084,030,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00064.warc.gz | 776,421,021 | 24,407 | # An organization purchased a four-year \$10,000 bond paying a 7% annual coupon and interest rates (for present value purposes are currently at 4%. This works out to a present value of the payment streams of \$673.08 + \$647.19 + \$622.30 + \$9,146.40 = \$11,088.97, which is the market price of the bond. What is the duration of this bond? 75 years65 years75 years0 years
Question
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• An organization purchased a four-year \$10,000 bond paying a 7% annual coupon and interest rates (for present value purposes are currently at 4%. This works out to a present value of the payment streams of \$673.08 + \$647.19 + \$622.30 + \$9,146.40 = \$11,088.97, which is the market price of the bond. What is the duration of this bond?
1. 75 years
2. 65 years
3. 75 years
4. 0 years
check_circle
Step 1
Duration of a bond measures the interest rate sensitivity of the bond. However, Macaulay duration measures the number of years it takes to repay the investors the price of the bond. It takes the cash flows in the form of coupon and redemption value at the end of the maturity period.
Step 2
Duration of the bond is calculated as below:
Here, r = yield to maturity = 4%
Step 3
Excel wroki...
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Tagged in | 379 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-05 | latest | en | 0.933346 |
https://stats.stackexchange.com/questions/396493/why-does-central-limit-theorem-break-down-in-my-simulation/396496 | 1,601,463,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00414.warc.gz | 639,670,735 | 40,376 | # Why does Central Limit Theorem break down in my simulation?
Let say I have following numbers:
4,3,5,6,5,3,4,2,5,4,3,6,5
I sample some of them, say, 5 of them, and calculate the sum of 5 samples. Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.
But when they are following numbers, I just replaced 4 with some big number:
4,3,5,6,5,3,10000000,2,5,4,3,6,5
Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?
• It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below. – oemb1905 Mar 10 '19 at 0:50
• @JimSD the CLT is an asymptotic result (i.e. about the distribution of standardized sample means or sums in the limit as sample size goes to infinity). $n=5$ is not $n\to\infty$. The thing you're looking at (the approach toward normality in finite samples) is not strictly a result of the CLT, but a related result. – Glen_b Mar 10 '19 at 11:30
• @oemb1905 n=30 is not sufficient for the sort of skewness OP is suggesting. Depending on how rare that contamination with a value like $10^7$ is it might take n=60 or n=100 or even more before the normal looks like a reasonable approximation. If the contamination is about 7% (as in the question) n=120 is still somewhat skew – Glen_b Mar 10 '19 at 11:42
• – Sextus Empiricus Mar 10 '19 at 21:32
• Think that values in intervals like (1,100,000 , 1,900,000) will never be reached. But if you make means of a decent amount those sums, it will work! – David Mar 11 '19 at 7:42
Let's recall, precisely, what the central limit theorem says.
If $$X_1, X_2, \cdots, X_k$$ are independent and identically distributed random variables with (shared) mean $$\mu$$ and standard deviation $$\sigma$$, then $$\frac{X_1 + X_2 + \cdots + X_k}{k\frac{\sigma}{\sqrt{k}}}$$ converges in distribution to a standard normal distribution $$N(0, 1)$$ (*).
This is often used in the "informal" form:
If $$X_1, X_2, \cdots, X_k$$ are independent and identically distributed random variables with (shared) mean $$\mu$$ and standard deviation $$\sigma$$, then $$X_1 + X_2 + \cdots + X_k$$ converges "in distribution" to a standard normal distribution $$N(k \mu, \sqrt{k} \sigma)$$.
There's no good way to make that form of the CLT mathematically precise, since the "limit" distribution change, but it's useful in practices.
When we have a static list of numbers like
4,3,5,6,5,3,10000000,2,5,4,3,6,5
and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.
• Identically distributed is no problem: each number in the list is equally likely to be chosen.
• Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.
So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.
So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.
Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.
(*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.
• Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway. – JimSD Mar 9 '19 at 7:09
• don't know if applicable here, but theorem of CLT convergence regulated by skewness en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem – seanv507 Mar 9 '19 at 15:56
• I'm a bit confused by @MatthewDrury's definition of the CLT. I think that $\frac{\sum X_k}{k}$ converges to a constant by the LLN, not a normal distribution. – JTH Mar 9 '19 at 17:41
• @seanv507 absolute third moment, rather than skewness; the two are related but note that for a symmetric distribution with finite third moment that the Berry-Esseen bound on $|F_n(x)-\Phi(x)|$ is not 0 because $\rho/\sigma^3$ is not skewness – Glen_b Mar 10 '19 at 11:28
• @Glen_b Yah, I was being a bit informal (which I perhaps should not have been), but I can fix that up this afternoon since it's led to a bit of confusion. – Matthew Drury Mar 10 '19 at 16:46
In general, the size of each sample should be more than $$5$$ for the CLT approximation to be good. A rule of thumb is a sample of size $$30$$ or more. But, with the population of your first example, $$5$$ is OK.
pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 5
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
In your second example, because of the shape of the population distribution (for one thing, it's too much skewed; read the comments by guy and Glen_b bellow), even samples of size $$30$$ won't give you a good approximation for the distribution of the sample mean using the CLT.
pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 30
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
But, with this second population, samples of, say, size $$100$$ are fine.
pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 100
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")
• It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem. – guy Mar 9 '19 at 4:02
• Perfect. Added. Tks. – Zen Mar 9 '19 at 4:06
• Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed. – JimSD Mar 9 '19 at 7:05
• @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you. – JimSD Mar 9 '19 at 7:25
• @guy the Berry Esseen theorem is about third absolute moment about the mean $\rho=E[|X-\mu|^3]$ not just the third moment about the mean $\mu_3=E[(X-\mu)^3]$. This makes it responsive to not just skewness but also heavy tails. – Glen_b Mar 10 '19 at 11:47
I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.
Let's write the random variable you're sampling as $$\mu+\sigma Z$$, where $$\mu$$ is the mean and $$\sigma$$ the standard deviation so $$Z$$ has mean $$0$$ and variance $$1$$. The cumulant-generating function of $$Z$$ is $$-\frac{1}{2}t^2-\frac{i\gamma_1}{6}t^3+o(t^3)$$. Here $$\gamma_1$$ denotes the skew of $$Z$$; we could write it in terms of the skew $$\kappa_3$$ of the original variable $$\mu+\sigma Z$$, viz. $$\gamma_1=\sigma^{-3}\kappa_3$$.
If we divide the sum of $$n$$ samples of $$Z$$'s distribution by $$\sqrt{n}$$, the result has cgf $$n\left(-\frac{1}{2}\left(\frac{t}{\sqrt{n}}\right)^2-\frac{i\gamma_1}{6}\left(\frac{t}{\sqrt{n}}\right)^3\right)+o(t^3)=-\frac{1}{2}t^2-\frac{i\gamma_1}{6\sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $$t$$ for the graph to look right, we need sufficiently large $$n$$. This calculation motivates $$n\propto\gamma_1^2$$. The two samples you considered have very different values of $$\gamma_1$$.
Short answer is, you don't have a big enough sample to make central limit theorem apply.
• That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small. – whuber Mar 9 '19 at 15:45
• @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern... – Henry Mar 9 '19 at 16:49
• @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size – Henry Mar 9 '19 at 16:51
• @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further. – whuber Mar 9 '19 at 18:33 | 3,010 | 10,715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-40 | latest | en | 0.903577 |
https://thekidsworksheet.com/solving-equations-with-variables-on-both-sides-worksheet-kuta/ | 1,657,011,549,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00416.warc.gz | 621,569,247 | 20,973 | # Solving Equations With Variables On Both Sides Worksheet Kuta
Solving Equations With Variables On Both Sides Worksheet Kuta. 1 name_____ solving equations with variables on both sides of the equal sign solve each equation. *click on open button to open and print to worksheet.
Use m 7 and n 8 2 8 x y. These worksheets for grade 10 linear equations class assignments and practice. 1 name_____ solving equations with variables on both sides of the equal sign solve each equation.
Source: lbartman.com
23) 35 + 7x = −3(−5 − 4x) {4} 24) 8(5 − 6v) = −7 − v {1} 25) −7 − 5a. Byjus online calculator tool makes calculations faster and easier.
### Pdf Solving Equations With Variables On Both Sides Of The.
Worksheet by kuta software llc. 2 2 gmxamdle i fw diht shd 4ihn rfmitnwirtce 4 zadlzgqe xbprval b2 0.s worksheet by kuta software llc 11) −3. All scripture is about jesus
### Solving Equations With Variables On Both Sides Worksheet From Briefencounters.ca.
On the other side of the balance, place 14 jelly beans and three empty. 1 name_____ solving equations with variables on both sides of the equal sign solve each equation. Solve each inequality and graph its solution.
### This Worksheet Will Help Students Practice Solving And Graphing Multistep Inequalities.
Equations w/ variable on both sides solve each equation. Sep 28, 2017 — worksheet by kuta software llc. (2) solve −7x − 3x + 2 = −8x − 8.
### Problems With The 17Th Amendment;
1) 8 5 p = p 8. Equations with variables on both sides worksheet. Kuta equations with variables on both sides 3.
### 1 Date_____ Period____ ©M Z2M0A1L5F Qkyuwtyai Jszopfdtmwqatrteo Llxlqcw.s P ]Amlnli \Rjiigphmtgsr Wrteosne\Rpvqejdj.
Examples are shown to help guide students through the process. E 5afl 4l l wrhiqg0hat nsc 9r oeqsse urlvme 8d3. Solving equations with variables on both sides worksheets help students to learn how to solve equations mentally by using the multiplication table and also learn how to identify a solution to an equation with given numbers as well as by using inverse operations. | 580 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | latest | en | 0.791817 |
http://nob.cs.ucdavis.edu/classes/ecs010-2014-02/outlines/2014-04-17.html | 1,708,583,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00205.warc.gz | 28,802,089 | 1,992 | # Outline for April 17, 2014
Reading: text, § 5, 7
Due: April 17, 2014
1. Why you don’t count with floating point numbers [roundoff.py]
2. Simultaneous assignment [swap.py]
1. Simple assignment: variable = expression
2. Simultaneous assignment: variableA, variableB = expressionA, expressionB
3. Decision structures
1. If statement [if0.py]
2. Executes once, based on condition
3. Syntax
4. Conditions
1. Resolves to boolean value
2. Literal booleans: True (1), False (0)
3. Relational operators
1. Use two arithmetic expressions connected with relational operatorsto create a boolean
2. Relational operators: >, >=, <, <=, ==, \lstinline/!=/
3. Precedence: resolved after arithmetic operators
4. Connectives: and, or, not
5. 6 > 2 + 3; "UCD" == "Sac State"
5. Two-way decisions [if1.py]
1. if-else statements
2. One condition, two possible code blocks
3. else very powerful when the positive condition is easy to describe but not the negative
6. Multi-way decisions [if2.py]
1. Can execute code based on several conditions
2. elif (else if)
3. else only reached if all previous conditions false
4. Nested if statements
7. Indefinite loops: execute until a general condition is false (while)
1. while [while.py]
2. Contrast with for
3. break causes program to fall out of loop (works with for too) [loop1.py]
4. continue causes program to start loop over immediately (works with for too) [loop1.py]
8. Definite loops: execute a specific (definite) number of times (for)
1. General form: for i in iterator
2. Iterator is either list or something that generates a list
3. Very common form: for i in range(1, 10)
9. range() in detail [for.py]
1. range(10) gives 0 1 2 3 4 5 6 7 8 9
2. range(3, 10) gives 3 4 5 6 7 8 9
3. range(2, 10, 3) gives 2 5 8
4. range(10, 2, -3) gives 10 7 4
10. Program: counting to 10 [toten.py]
11. Program: sum the first 10 squares [sumsq.py]
You can also obtain a PDF version of this. Version of April 16, 2014 at 10:59PM | 600 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.793003 |
https://www.emathzone.com/tutorials/calculus/average-and-instantaneous-rate-of-change.html | 1,566,692,529,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00242.warc.gz | 784,954,527 | 12,812 | # Average and Instantaneous Rate of Change
A variable which can assign any value independently is called the independent variable, and the variable which depends on some independent variable is called the dependent variable.
For Example:
If $x = 0,1,2,3, \ldots$ etc, then
We see that as $x$ behaves independently, we call it the independent variable. But the behavior of $y$ or $f(x)$ depends on the variable $x$, so we call it the dependent variable.
Increment:
Literally the word increment means increase, but in mathematics this word covers both increase as well as decrease because the increment may be positive or negative. Simply put, the word increment in mathematics means “the difference between two values of variables.”
The final value minus the initial value is called an increment in the variable. The increment in $x$ is denoted by the symbols $\delta x$ or $\Delta x$ (read as “delta $x$”).
If $y = f(x)$, and $x$ changes from an initial value ${x_0}$ to the final value ${x_1}$, then $y$ changes from an initial value ${y_0} = f({x_0})$ to the final value ${y_1} = f({x_1})$.
Thus, the increment in ‘$x$
$\Delta x = {x_1} - {x_0}$
produces a corresponding increment in ‘$y$
$\Delta y = {y_1} - {y_0} = f({x_1}) - f({x_0})$
Average Rate of Change:
If $y = f(x)$ is a real valued continuous function in the interval $({x_0},{x_1})$, then the average rate of change of ‘$y$’ with respect to ‘$x$’ over this interval is
$\frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}}$
But $\Delta x = {x_1} - {x_0}$
$\Rightarrow {x_1} = {x_0} + \Delta x$
$\therefore \frac{{f({x_0} + \Delta x) - f({x_0})}}{{\Delta x}}$
Instantaneous Rate of Change:
If $y = f(x)$ is a real valued continuous function in the interval $({x_0},{x_1})$, then the average rate of change of ‘$y$’ with respect to ‘$x$’ over this interval is
$\mathop {\lim }\limits_{{x_1} \to {x_0}} \frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}}$
But $\Delta x = {x_1} - {x_0}$
$\Rightarrow {x_1} = {x_0} + \Delta x$
This shows that $\Delta x \to 0$, as ${x_1} \to {x_0}$
Average or Instantaneous Rate of Change of Distance
OR
Average or Instantaneous Velocity:
Suppose a particle (or an object) is moving in a straight line and its positions (from some fixed point) after times ${t_0}$ and ${t_1}$ are given by $S({t_0})$ and $S({t_1})$, then the average rate of change or the average velocity is
Also, the instantaneous rate of change of distance or instantaneous velocity is | 760 | 2,447 | {"found_math": true, "script_math_tex": 41, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 46, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2019-35 | longest | en | 0.786781 |
https://se.tradingview.com/script/mheEtfmN-A-Useful-MA-Weighting-Function-For-Controlling-Lag-Smoothness/ | 1,660,066,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00650.warc.gz | 473,812,173 | 163,495 | # A Useful MA Weighting Function For Controlling Lag & Smoothness
So far the most widely used moving average with an adjustable weighting function is the Arnaud Legoux moving average ( ALMA ), who uses a Gaussian function as weighting function. Adjustable weighting functions are useful since they allow us to control characteristics of the moving average such as lag and smoothness.
The following moving average has a simple adjustable weighting function that allows the user to have control over the lag and smoothness of the moving average, we will see that it can also be used to get both an SMA and WMA .
A high-resolution gradient is also used to color the moving average, makes it fun to watch, the plot transition between 200 colors, would be tedious to make but everything was made possible using a custom R script, I only needed to copy and paste the R console output in the Pine editor.
Settings
• length : Period of the moving average
• -Lag : Setting decreasing the lag of the moving average
• +Lag : Setting increasing the lag of the moving average
Estimating Existing Moving Averages
The weighting function of this moving average is derived from the calculation of the beta distribution, advantages of such distribution is that unlike a lot of PDF , the beta distribution is defined within a specific range of values (0,1). Parameters alpha and beta controls the shape of the distribution, with alpha introducing negative skewness and beta introducing positive skewness, while higher values of alpha and beta increase kurtosis .
Here -Lag is directly associated to beta while +Lag is associated with alpha. When alpha = beta = 1 the distribution is uniform, and as such can be used to compute a simple moving average .
Moving average with -Lag = +Lag = 1, its impulse response is shown below.
It is also possible to get a WMA by increasing -Lag, thus having -Lag = 2 and +Lag = 1.
Using values of -Lag and +Lag equal to each other allows us to get a symmetrical impulse response, increasing these two values controls the heaviness of the tails of the impulse response.
Here -Lag = +Lag = 3, note that when the impulse response of a moving average is symmetrical its lag is equal to (length-1)/2.
As for the gradient, the color is determined by the value of an RSI using the moving average as input.
I don't promise anything but I will try to respond to your comments
Patreon: https://www.patreon.com/alexgrover
Become a Patreon and get access to exclusive technical indicators!
You can also check out some of the indicators I made for luxalgo : https://www.tradingview.com/u/LuxAlgo/#published-scripts
Skript med en öppen källkod
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Vill du använda det här skriptet i ett diagram? | 780 | 3,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.909216 |
https://rdrr.io/cran/powdist/man/ReversalPowerLaplace.html | 1,695,359,894,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00750.warc.gz | 516,249,290 | 7,103 | # ReversalPowerLaplace: The Power Reversal Laplace Distribution In powdist: Power and Reversal Power Distributions
## Description
Density, distribution function, quantile function and random generation for the power reversal Laplace distribution with parameters mu, sigma and lambda.
## Usage
```1 2 3 4 5 6 7 8 9``` ```drplaplace(x, lambda = 1, mu = 0, sigma = 1, log = FALSE) prplaplace(q, lambda = 1, mu = 0, sigma = 1, lower.tail = TRUE, log.p = FALSE) qrplaplace(p, lambda = 1, mu = 0, sigma = 1, lower.tail = TRUE, log.p = FALSE) rrplaplace(n, lambda = 1, mu = 0, sigma = 1) ```
## Arguments
`x, q` vector of quantiles. `lambda` shape parameter. `mu, sigma` location and scale parameters. `log, log.p` logical; if TRUE, probabilities p are given as log(p). `lower.tail` logical; if TRUE (default), probabilities are P[X ≤ x ], otherwise, P[X > x]. `p` vector of probabilities. `n` number of observations.
## Details
The reversal power Laplace distribution has density
a,
where -∞<μ<∞ is the location paramether, σ^2>0 the scale parameter and λ>0 the shape parameter.
## Examples
```1 2 3 4``` ```drplaplace(1, 1, 3, 4) prplaplace(1, 1, 3, 4) qrplaplace(0.2, 1, 3, 4) rrplaplace(5, 2, 3, 4) ```
powdist documentation built on May 1, 2019, 10:11 p.m. | 426 | 1,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-40 | latest | en | 0.62657 |
http://at.metamath.org/ileuni/df-sub.html | 1,601,330,064,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00389.warc.gz | 10,805,060 | 3,329 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > df-sub Structured version GIF version
Definition df-sub 6961
Description: Define subtraction. Theorem subval 6980 shows its value (and describes how this definition works), theorem subaddi 7074 relates it to addition, and theorems subcli 7063 and resubcli 7050 prove its closure laws. (Contributed by NM, 26-Nov-1994.)
Assertion
Ref Expression
df-sub − = (x ℂ, y ℂ ↦ (z ℂ (y + z) = x))
Distinct variable group: x,y,z
Detailed syntax breakdown of Definition df-sub
StepHypRef Expression
1 cmin 6959 . 2 class
2 vx . . 3 setvar x
3 vy . . 3 setvar y
4 cc 6689 . . 3 class
53cv 1241 . . . . . 6 class y
6 vz . . . . . . 7 setvar z
76cv 1241 . . . . . 6 class z
8 caddc 6694 . . . . . 6 class +
95, 7, 8co 5455 . . . . 5 class (y + z)
102cv 1241 . . . . 5 class x
119, 10wceq 1242 . . . 4 wff (y + z) = x
1211, 6, 4crio 5410 . . 3 class (z ℂ (y + z) = x)
132, 3, 4, 4, 12cmpt2 5457 . 2 class (x ℂ, y ℂ ↦ (z ℂ (y + z) = x))
141, 13wceq 1242 1 wff − = (x ℂ, y ℂ ↦ (z ℂ (y + z) = x))
Colors of variables: wff set class This definition is referenced by: subval 6980 subf 6990
Copyright terms: Public domain W3C validator | 499 | 1,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-40 | longest | en | 0.351658 |
https://money.stackexchange.com/questions/16870/how-to-determine-the-interest-rate | 1,718,960,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00168.warc.gz | 364,239,085 | 36,833 | # How to determine the interest rate?
I'm really not good at math. So please forgive me for this noob question that I have.. Just trying to learn basic percentage calculation with money lending..
Here's an example of how the APR works. You see an advertisement offering a 30-year fixed-rate mortgage at 7 percent with one point. You see another advertisement offering a 30-year fixed-rate mortgage at 7 percent with no points. Easy choice, right? Actually, it isn't. Fortunately, the APR considers all of the fine print.
Say you need to borrow \$100,000. With either lender, that means that your monthly payment is \$665.30. If the point is 1 percent of \$100,000 (\$1,000), the application fee is \$25, the processing fee is \$250, and the other closing fees total \$750, then the total of those fees (\$2,025) is deducted from the actual loan amount of \$100,000 (\$100,000 - \$2,025 = \$97,975). This means that \$97,975 is the new loan amount used to figure the true cost of the loan. To find the APR, you determine the interest rate that would equate to a monthly payment of \$665.30 for a loan of \$97,975. In this case, it's really 7.2 percent.
from howstuffsworks
How to calculate the interest rate that would equate to a monthly payment of \$665.30 for a loan of \$97,975 in 360 months?
What is the formula for this?
• The advertised APR assumes that you keep the loan for the full 30 years; never have a late charge; never have a pre-payment penalty; and never re-finance. If you pay off the loan ahead of schedule, your effective average interest rate will be higher, because the closing costs will be spread out over less time. The advertised APR also does not account for your money that is tied up in the escrow account, not earning any interest. Commented Dec 8, 2015 at 20:33
Let's start by saying that of all the things to solve in a mortgage equation, the rate is the toughest. It's the least friendly to solving with pencil and paper. While I never tire of expressing my love for my Texas Instruments BA-35 financial calculator, it's no longer in production, and most folk won't have access, but Excel is right there.
The financial function pops up for you with RATE as a choice for solving.
I filled in the cells to show the numbers. The -665.30 is a typical convention for money flows as it's a payment from you not a credit for interest you earn. Note, some mortgage calculators leave this entry as positive. It takes a second to see how one's calculation or spreadsheet does this.
Last, you can see the software wants a "guess." This is because the software has a loop, guessing and getting closer to the solution. I entered .005 to guess 6% per year. The correct solution is .006 or 7.2% per year.
Class dismissed.
• Wait! My Excel doesn't have those funny blue scroll bars? (Seriously, thank you for a great answer) Commented Sep 16, 2012 at 16:08
• Funny. How's this? (Disclaimer - PC users may not see the blue scroll bars. I am on a Mac.) Thank-you for the kind words. I was afraid I got carried away with the level of detail, but maybe not, I hope OP returns to see how to run the numbers. Commented Sep 16, 2012 at 17:55
• I still don't understand.. I don't use excel so I don't know how to do this... Commented Oct 5, 2012 at 16:24
• My understanding is that unlike calculating the payment, given the other details, calculating the rate is not straightforward. A rate is guessed, and moved up or down until the number is found. You guess the rate, calculate payment, and lower or raise the rate until the payment is correct to the penny. Commented Oct 5, 2012 at 20:11
• Also, some scenarios can have more than one rate. For example, if the borrower receives large amounts of cash at different times (especially before and after making payments), there can be different rates that solve the equation. The initial guess encourages the program to find an answer close to the guess. Commented Dec 8, 2015 at 20:38 | 953 | 3,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-26 | latest | en | 0.956411 |
http://www.solutioninn.com/a-random-sample-of-n-12-fouryearold-red-pine | 1,506,016,556,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687834.17/warc/CC-MAIN-20170921172227-20170921192227-00277.warc.gz | 553,598,751 | 7,757 | # Question: A random sample of n 12 four year old red pine
A random sample of n = 12 four-year-old red pine trees was selected, and the diameter (in inches) of each tree’s main stem was measured. The resulting observations are as follows:
a. Compute a point estimate of s, the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate?
b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate?
c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use?
d. Suppose that the diameter distribution is normal. Then the 90th percentile of the diameter distribution is m = 1.28s (so 90% of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of m in this case; then use it along with your estimate of s from Part (a).)
Sales0
Views254 | 258 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-39 | latest | en | 0.922758 |
http://mathhelpforum.com/number-theory/128918-sequence.html | 1,527,171,611,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00335.warc.gz | 180,682,424 | 9,780 | 1. ## Sequence...
Find anfor this sequence:
5, 6, 7, 5, 6, 7, 5, 6, 7, ...
2. Originally Posted by bearej50
Find anfor this sequence:
5, 6, 7, 5, 6, 7, 5, 6, 7, ...
you can use a piece-wise definition: $\displaystyle a_n = \left \{ \begin{matrix} 5 & \text{ if } n \equiv 1 \bmod 3 \\ 6 & \text{ if } n \equiv 2 \bmod 3 \\ 7 & \text{ if } n \equiv 0 \bmod 3 \end{matrix} \right.$
are you familiar with the notation $\displaystyle n \equiv m \bmod 3$? It means $\displaystyle n = m + 3k$ for some integer $\displaystyle k$
3. I am familiar with this notation but I was looking for a more generalized formula. For example, for the sequence: 5, 7, 5, 7, .... an = 6 - (-1)^(n+1) | 262 | 681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-22 | latest | en | 0.708045 |
https://sportsbizusa.com/20-2nd-grade-phonics-worksheets-free/ | 1,620,290,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00211.warc.gz | 570,991,059 | 11,569 | HomeTemplate ➟ 20 20 2nd Grade Phonics Worksheets Free
20 2nd Grade Phonics Worksheets Free
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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Prepositions of place free printable prepositional phrase worksheet, free printable preposition worksheets for middle school, free printable prepositions of time worksheets, free printable preposition worksheets for grade 1, esl free printable worksheets prepositions, via: hu.pinterest.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind […]
20 Homophones Worksheet High School
Homophones English ESL Worksheets for distance learning homophones lesson plan high school, homophones quiz high school, homonyms and homophones worksheets for high school, homophones exercises for high school, free printable homophone worksheets high school, via: en.islcollective.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason […] | 1,385 | 6,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-21 | latest | en | 0.96302 |
https://static.hlt.bme.hu/semantics/external/pages/m%C3%A1sodrend%C5%B1_aritmetika_($Z_2$)/en.wikipedia.org/wiki/Uniformization_(set_theory).html | 1,669,877,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00059.warc.gz | 567,456,342 | 11,973 | # Uniformization (set theory)
In set theory, the axiom of uniformization, a weak form of the axiom of choice, states that if ${\displaystyle R}$ is a subset of ${\displaystyle X\times Y}$, where ${\displaystyle X}$ and ${\displaystyle Y}$ are Polish spaces, then there is a subset ${\displaystyle f}$ of ${\displaystyle R}$ that is a partial function from ${\displaystyle X}$ to ${\displaystyle Y}$, and whose domain (in the sense of the set of all ${\displaystyle x}$ such that ${\displaystyle f(x)}$ exists) equals
${\displaystyle \{x\in X|\exists y\in Y(x,y)\in R\}\,}$
Such a function is called a uniformizing function for ${\displaystyle R}$, or a uniformization of ${\displaystyle R}$.
Uniformization of relation R (light blue) by function f (red).
To see the relationship with the axiom of choice, observe that ${\displaystyle R}$ can be thought of as associating, to each element of ${\displaystyle X}$, a subset of ${\displaystyle Y}$. A uniformization of ${\displaystyle R}$ then picks exactly one element from each such subset, whenever the subset is nonempty. Thus, allowing arbitrary sets X and Y (rather than just Polish spaces) would make the axiom of uniformization equivalent to AC.
A pointclass ${\displaystyle {\boldsymbol {\Gamma }}}$ is said to have the uniformization property if every relation ${\displaystyle R}$ in ${\displaystyle {\boldsymbol {\Gamma }}}$ can be uniformized by a partial function in ${\displaystyle {\boldsymbol {\Gamma }}}$. The uniformization property is implied by the scale property, at least for adequate pointclasses of a certain form.
It follows from ZFC alone that ${\displaystyle {\boldsymbol {\Pi }}_{1}^{1}}$ and ${\displaystyle {\boldsymbol {\Sigma }}_{2}^{1}}$ have the uniformization property. It follows from the existence of sufficient large cardinals that
• ${\displaystyle {\boldsymbol {\Pi }}_{2n+1}^{1}}$ and ${\displaystyle {\boldsymbol {\Sigma }}_{2n+2}^{1}}$ have the uniformization property for every natural number ${\displaystyle n}$.
• Therefore, the collection of projective sets has the uniformization property.
• Every relation in L(R) can be uniformized, but not necessarily by a function in L(R). In fact, L(R) does not have the uniformization property (equivalently, L(R) does not satisfy the axiom of uniformization).
• (Note: it's trivial that every relation in L(R) can be uniformized in V, assuming V satisfies AC. The point is that every such relation can be uniformized in some transitive inner model of V in which AD holds.)
## References
• Moschovakis, Yiannis N. (1980). Descriptive Set Theory. North Holland. ISBN 0-444-70199-0. | 664 | 2,624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-49 | latest | en | 0.855012 |
http://math.stackexchange.com/questions/64192/the-fundamental-group-of-k-3-3-relationship-between-its-generators-and-em | 1,469,324,043,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00314-ip-10-185-27-174.ec2.internal.warc.gz | 162,236,098 | 18,038 | # The fundamental group of $K_{3,3}$ — relationship between its generators and embedding into manifolds
So I've been reading this wonderful PDF textbook on algebraic topology:
http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf
In particular, I'm very interested in the chapter on graphs. This corollary seems to give the construction of the fundamental group of a graph in fairly simple terms (here $\chi = V - E$):
Corollary. If X is a connected graph, then $\pi_1(X)$ is a free group with one generator for each edge not in a given maximal tree. If X is finite, then $\pi_1(X)$ is free on 1 − $\chi(X)$ generators; in particular, $\chi(X) \le$ 1, with equality if and only if X is a tree.
So that gives two ways to get the fundamental group: a formula giving the number of generators $1 - \chi(X)$ (and it's a free group), and an algorithm for removing edges of the maximal tree and attributing one generator to each edge remaining thereafter.
When I use this corollary on $K_{3,3}$, I get 4 generators. But, I was only expecting to get two generators, since I know that $K_{3,3}$ (the so-called "utility graph") can be embedded on the torus.
What am I missing? What exactly is the relationship between the number of generators in the free group $\pi_1(X)$, and the surfaces on which it can be embedded? I'm pretty sure there's something I'm not understanding here, and I'm very keen to learn what it could be! Thanks in advance.
-
I don't think there is any relationship.
For example, a tree embeds into any surface so $0$ generators can embed in anything.
on the other hand, consider the graph obtained as follows. Start with an n-gon. At each vertex, have an edge emanate out. From each of these "spokes", stick a small triangle. (For n=8, think of an octopus holding a triangle in each arm.)
Then that graph embeds in every surface, but has $n+1$ generators for it's fundamental group.
Hence, given a natural number $n$, every surface has graphs embedded whose fundamental group is free on $n$ generators.
-
Jason didn't spell it out, so I will. The map on fundamental groups induced by the embedding of a utility graph into a torus is not injective on fundamental groups.
- | 546 | 2,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.927125 |
https://pointerwriters.com/how-do-you-solve-ln-x-2-16-3/ | 1,611,664,469,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00609.warc.gz | 516,726,930 | 11,945 | # How Do You Solve Ln X 2 16
I found: ##x=e^8=2,980.95##
We can use the property of the logs that allows you to take out the exponent of the argument and place it as multiplier in front of the log to get:
##2ln(x)=16##
rearrange:
##ln(x)=16/2=8##
use the definition of log to get:
##x=e^8=2,980.95##
4.6/5
Price (USD)
\$ | 112 | 329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-04 | latest | en | 0.637229 |
http://math.stackexchange.com/questions/57985/where-are-the-zeros-of-the-complex-sine-and-cosine | 1,469,379,412,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00161-ip-10-185-27-174.ec2.internal.warc.gz | 152,462,480 | 18,527 | # Where are the zeros of the complex sine and cosine?
Do sin(z) and cos(z) have any zeroes where the imaginary part of z is non-zero? How could I prove that (or show that it's reasonable)?
-
$\cosh\,x \geq 1$ if $x$ is real... – J. M. Aug 17 '11 at 6:49
Use de Moivre's to write sin and cos as complex exponentials. Multiply through to get a quadratic in terms of $e^{ix}$. Solve over all branches. – anon Aug 17 '11 at 6:51
@JM: How is cosh, x≥1 related to the complex sin and cos? – Sara Aug 17 '11 at 6:51
$\sin(x+iy)=\sin\,x\cosh\,y+i\cos\,x\sinh\,y$. You know where the zeroes of real-valued $\sin$, $\cos$, and $\sinh$ are, so... – J. M. Aug 17 '11 at 6:54
@anon: Thanks! I solved the quadratic. – Sara Aug 17 '11 at 6:56
We can use $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$ Set this equal to $0$. A little manipulation yields $(e^{iz})^2=1$.
If the imaginary part of $z$ is non-zero, the norm of $e^{iz}$ is greater than $1$, contradicting the fact that $(e^{iz})^2=1$. A mild variant of the same argument works for $\cos z$.
-
The norm is actually $\exp(-\text{Im}(z))$ which can be less than $1$, but it is equal to $1$ only if $\text{Im}(z)=0$. – Minethlos Nov 18 '14 at 11:03
There is none. It follows, for example, from the Weierstrass products $$\cos z=\prod_{n=1}^\infty \left(1-\frac{4z^2}{\pi^2(2n-1)^2}\right),$$ $$\sin z=z\prod_{n=1}^\infty \left(1-\frac{z^2}{(\pi n)^2}\right),$$ which are valid for all $z\in \mathbb C$.
-
Looks a bit like using a sledge-hammer to crack a nut... – Did Aug 17 '11 at 9:51
@anon - btw often in the literature I've seen that referred to as "diverging" to zero. – John M Aug 17 '11 at 12:34
Yes, and these infinite products converge for all complex $z$: when we say that we mean, in particular, that they do not diverge to zero. So they have value zero only if one of the factors vanishes. – GEdgar Aug 17 '11 at 12:44
@GEdgar, John M: For some strange reason I've only heard of convergence refer to the existence of a limit, but I see the definition you refer to right on Wikipedia. Yes, I know the Wallis product converges, so it was meaningless to point out an irrelevant technicality. – anon Aug 17 '11 at 18:11
@anon - actually (& I've only recently learned this myself) the reason for this is that most of the theorems regarding convergence of infinite products have to rule out the "converge to 0" case as exceptional. It's easy enough to see why if you take the log. – John M Aug 18 '11 at 2:20 | 805 | 2,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-30 | latest | en | 0.846797 |
http://stackoverflow.com/questions/3876354/algorithm-for-determining-if-2-graphs-are-isomorphic/3876438 | 1,406,581,211,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510261958.8/warc/CC-MAIN-20140728011741-00162-ip-10-146-231-18.ec2.internal.warc.gz | 282,657,764 | 16,334 | # Algorithm for determining if 2 graphs are isomorphic
Disclaimer: I'm a total newbie at graph theory and I'm not sure if this belongs on SO, Math SE, etc.
Given 2 adjacency matrices A and B, how can I determine if A and B are isomorphic.
For example, A and B which are not isomorphic and C and D which are isomorphic.
``````A = [ 0 1 0 0 1 1 B = [ 0 1 1 0 0 0
1 0 1 0 0 1 1 0 1 1 0 0
0 1 0 1 0 0 1 1 0 1 1 0
0 0 1 0 1 0 0 1 1 0 0 1
1 0 0 1 0 1 0 0 1 0 0 1
1 1 0 0 1 0 ] 0 0 0 1 1 0 ]
C = [ 0 1 0 1 0 1 D = [ 0 1 0 1 1 0
1 0 1 0 0 1 1 0 1 0 1 0
0 1 0 1 1 0 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 0 1
0 0 1 1 0 1 1 1 0 0 0 1
1 1 0 0 1 0 ] 0 0 1 1 1 0 ]
(sorry for this ugly notation, I'm not quite sure how to draw matrices on SO)
``````
Here's how I've started my algorithm (pardon my lack of mathematical rigor) please help me complete/correct!
1. If size (number of edges, in this case amount of 1s) of A != size of B => graphs are not isomorphic
2. For each vertex of A, count its degree and look for a matching vertex in B which has the same degree and was not matched earlier. If there is no match => graphs are not isomorphic.
3. Now that we cannot quickly prove that A and B are not isomorphic, what's the next step? Would it be correct try every permutation of lines in A until A matches B? Really not sure about this one...
-
I'm sure it's terrible, but you could always brute force it: keep the nodes in A in order, then go through every permutation of the labeling of nodes in B until they match or there are no more. Of course, there's almost certainly a better way... like this... – JoshD Oct 6 '10 at 20:07
en.wikipedia.org/wiki/… seems that no-one knows any polynomial time algorithm. So it's ok to just brute-force. try every permutation of nodes of same degree etc. – Oleg Grenrus Oct 6 '10 at 20:08
## 2 Answers
That's a quite difficult problem to solve. There is a Wikipedia page about it:
According to that page there are a number of special cases that have been solved with efficient polynomial time solutions, but the complexity of the optimal solution is still unknown.
-
Thanks for the reference. Strangely enough, I have an intuition that graph isomorphism should be an easy problem to solve since it seems quite easy for my brain to visually determine if 2 graphs are isomorph. Maybe I haven't tried on a big enough graph... – Olivier Lalonde Oct 6 '10 at 20:45
Haha, I have the exact opposite problem. Can't see if two graphs are isomorphic even if they are very small. – Gleno Oct 6 '10 at 20:48
@olivier Lalonde: How long does your brain take to check for isomorphism in dense graphs with 50, 100 or more nodes? – MAK Oct 7 '10 at 10:53
My project - Griso - at sf.net: http://sourceforge.net/projects/griso/ with this description:
Griso is a graph isomorphism testing utility written in C++ and based on my own algo.
See Griso's sample input/output on this page: http://funkybee.narod.ru/graphs.htm
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# Estimating Planar Areas Using Analogue Methods
This math problem determines the areas of simple and complex planar figures using measurement of mass and proportional constructs. Materials are inexpensive or easily found (poster board, scissors, ruler, sharp pencil, right angle), but also... (View More)
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Learners will calculate the diameter of the Moon using proportions. This activity is in Unit 1 of the Exploring the Moon teachers guide, which is designed for use especially, but not exclusively, with the Lunar Sample Disk program.
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Chat Online | 2,689 | 12,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-39 | latest | en | 0.863508 |
https://www.calculatoratoz.com/en/perimeter-of-icosahedron-given-insphere-radius-calculator/Calc-39149 | 1,718,852,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00700.warc.gz | 621,078,027 | 56,140 | ## Perimeter of Icosahedron given Insphere Radius Solution
STEP 0: Pre-Calculation Summary
Formula Used
Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5)))
P = (360*ri)/(sqrt(3)*(3+sqrt(5)))
This formula uses 1 Functions, 2 Variables
Functions Used
sqrt - A square root function is a function that takes a non-negative number as an input and returns the square root of the given input number., sqrt(Number)
Variables Used
Perimeter of Icosahedron - (Measured in Meter) - Perimeter of Icosahedron is the sum of the total distance around all the edges of the Icosahedron.
Insphere Radius of Icosahedron - (Measured in Meter) - Insphere Radius of Icosahedron is the radius of the sphere that is contained by the Icosahedron in such a way that all the faces just touching the sphere.
STEP 1: Convert Input(s) to Base Unit
Insphere Radius of Icosahedron: 7 Meter --> 7 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
P = (360*ri)/(sqrt(3)*(3+sqrt(5))) --> (360*7)/(sqrt(3)*(3+sqrt(5)))
Evaluating ... ...
P = 277.865506064835
STEP 3: Convert Result to Output's Unit
277.865506064835 Meter --> No Conversion Required
277.865506064835 277.8655 Meter <-- Perimeter of Icosahedron
(Calculation completed in 00.020 seconds)
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## < 11 Perimeter of Icosahedron Calculators
Perimeter of Icosahedron given Surface to Volume Ratio
Perimeter of Icosahedron = (360*sqrt(3))/((3+sqrt(5))*Surface to Volume Ratio of Icosahedron)
Perimeter of Icosahedron given Circumsphere Radius
Perimeter of Icosahedron = (120*Circumsphere Radius of Icosahedron)/(sqrt(10+(2*sqrt(5))))
Perimeter of Icosahedron given Lateral Surface Area
Perimeter of Icosahedron = 30*sqrt((2*Lateral Surface Area of Icosahedron)/(9*sqrt(3)))
Perimeter of Icosahedron given Insphere Radius
Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5)))
Perimeter of Icosahedron given Space Diagonal
Perimeter of Icosahedron = (60*Space Diagonal of Icosahedron)/(sqrt(10+(2*sqrt(5))))
Perimeter of Icosahedron given Total Surface Area
Perimeter of Icosahedron = 30*sqrt(Total Surface Area of Icosahedron/(5*sqrt(3)))
Perimeter of Icosahedron given Face Area
Perimeter of Icosahedron = 30*sqrt((4*Face Area of Icosahedron)/sqrt(3))
Perimeter of Icosahedron given Volume
Face Perimeter of Icosahedron = 30*((12*Volume of Icosahedron)/(5*(3+sqrt(5))))^(1/3)
Perimeter of Icosahedron given Midsphere Radius
Perimeter of Icosahedron = (120*Midsphere Radius of Icosahedron)/(1+sqrt(5))
Perimeter of Icosahedron given Face Perimeter
Perimeter of Icosahedron = 10*Face Perimeter of Icosahedron
Perimeter of Icosahedron
Perimeter of Icosahedron = 30*Edge Length of Icosahedron
## Perimeter of Icosahedron given Insphere Radius Formula
Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5)))
P = (360*ri)/(sqrt(3)*(3+sqrt(5)))
## What is an Icosahedron?
An Icosahedron is a symmetric and closed three dimensional shape with 20 identical equilateral triangular faces. It is a Platonic solid, which has 20 faces, 12 vertices and 30 edges. At each vertex, five equilateral triangular faces meet and at each edge, two equilateral triangular faces meet.
## What are Platonic Solids?
In three-dimensional space, a Platonic solid is a regular, convex polyhedron. It is constructed by congruent (identical in shape and size), regular (all angles equal and all sides equal), polygonal faces with the same number of faces meeting at each vertex. Five solids who meet this criteria are Tetrahedron {3,3} , Cube {4,3} , Octahedron {3,4} , Dodecahedron {5,3} , Icosahedron {3,5} ; where in {p, q}, p represents the number of edges in a face and q represents the number of edges meeting at a vertex; {p, q} is the Schläfli symbol.
## How to Calculate Perimeter of Icosahedron given Insphere Radius?
Perimeter of Icosahedron given Insphere Radius calculator uses Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5))) to calculate the Perimeter of Icosahedron, The Perimeter of Icosahedron given Insphere Radius formula is defined as the sum of the total distance around all the edges of the Icosahedron and is calculated using the insphere radius of the Icosahedron. Perimeter of Icosahedron is denoted by P symbol.
How to calculate Perimeter of Icosahedron given Insphere Radius using this online calculator? To use this online calculator for Perimeter of Icosahedron given Insphere Radius, enter Insphere Radius of Icosahedron (ri) and hit the calculate button. Here is how the Perimeter of Icosahedron given Insphere Radius calculation can be explained with given input values -> 277.8655 = (360*7)/(sqrt(3)*(3+sqrt(5))).
### FAQ
What is Perimeter of Icosahedron given Insphere Radius?
The Perimeter of Icosahedron given Insphere Radius formula is defined as the sum of the total distance around all the edges of the Icosahedron and is calculated using the insphere radius of the Icosahedron and is represented as P = (360*ri)/(sqrt(3)*(3+sqrt(5))) or Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5))). Insphere Radius of Icosahedron is the radius of the sphere that is contained by the Icosahedron in such a way that all the faces just touching the sphere.
How to calculate Perimeter of Icosahedron given Insphere Radius?
The Perimeter of Icosahedron given Insphere Radius formula is defined as the sum of the total distance around all the edges of the Icosahedron and is calculated using the insphere radius of the Icosahedron is calculated using Perimeter of Icosahedron = (360*Insphere Radius of Icosahedron)/(sqrt(3)*(3+sqrt(5))). To calculate Perimeter of Icosahedron given Insphere Radius, you need Insphere Radius of Icosahedron (ri). With our tool, you need to enter the respective value for Insphere Radius of Icosahedron and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Perimeter of Icosahedron?
In this formula, Perimeter of Icosahedron uses Insphere Radius of Icosahedron. We can use 9 other way(s) to calculate the same, which is/are as follows -
• Perimeter of Icosahedron = 30*Edge Length of Icosahedron
• Perimeter of Icosahedron = (120*Circumsphere Radius of Icosahedron)/(sqrt(10+(2*sqrt(5))))
• Perimeter of Icosahedron = 30*sqrt((4*Face Area of Icosahedron)/sqrt(3))
• Perimeter of Icosahedron = 30*sqrt((2*Lateral Surface Area of Icosahedron)/(9*sqrt(3)))
• Perimeter of Icosahedron = (120*Midsphere Radius of Icosahedron)/(1+sqrt(5))
• Perimeter of Icosahedron = 10*Face Perimeter of Icosahedron
• Perimeter of Icosahedron = (60*Space Diagonal of Icosahedron)/(sqrt(10+(2*sqrt(5))))
• Perimeter of Icosahedron = (360*sqrt(3))/((3+sqrt(5))*Surface to Volume Ratio of Icosahedron)
• Perimeter of Icosahedron = 30*sqrt(Total Surface Area of Icosahedron/(5*sqrt(3)))
Let Others Know | 2,192 | 7,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.799977 |
http://www.sitepoint.com/functional-programming-and-php/ | 1,462,104,183,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860115836.8/warc/CC-MAIN-20160428161515-00034-ip-10-239-7-51.ec2.internal.warc.gz | 797,945,880 | 32,694 | Many programmers like to talk about functional programming, but if you ask them if they’ve ever done it, most of their replies will be “No”. The reason is quite simple: we are taught to think in an imperative manner when we first start learning to program, in terms of flow charts and steps to be followed in the program. So in this article I’ll explain some important concepts to functional programming and how to write functional code in PHP.
## Important Concepts of Functional Programming
Starting with the jargon, Wikipedia defines functional programming as “a programming paradigm that treats computation as the evaluation of mathematical functions and avoids state and mutable data.” In functional programming, functions are treated as first-class citizens, whereas in imperative programming we are mostly concerned with the data and the steps to alter it to reach the desired result.
When we say functions are first-class, it means we can use functions just as we use values in imperative programming. They can be passed as an argument to a function, defined inside another function, and can even be returned as a result. In other words, “functions are values”.
We’ll come back to this again later, but there are many other important concepts of functional programming. To mention a few:
### Immutability
Immutability is the behavior that a value of a variable cannot be changed once it is defined. Different languages have different ways of achieving this; in PHP, for instance, the only way to make a variable immutable is to define it as a constant.
### Recursion
Recursion is also very prominent in functional programming. In imperative programming, we can use looping constructs like `for` and `foreach` when we need to manipulate collections or arrays, walking through each element and keeping a temporary variable to hold the current value. But because of immutability, this approach is not possible in functional programming. Recurssion is the answer because such book keeping is done implicitly with the call stack.
Suppose we want to write a function to find the sum of all elements in an array (forget that `array_sum()` exists for the time being). In a functional style, we would write:
``````<?php
function sum(\$array) {
if (empty(\$array))
return 0;
else
return \$array[0] + sum(array_slice(\$array, 1));
}
\$total = sum(array(1, 2, 3)); // 6``````
An empty list will return 0, which is our base condition. For an array with more than one value, it will return the results of adding the first element with the recursive sum of all other elements.
### Pure Functions and Referential Transparency
A function is said to be free from side effects if it does not change the value of an object outside itself, such as a global or static variable, and does not have any I/O effects like writing into file, database, and so on. Such functions are otherwise called pure functions.
The output of a pure function will always be the same for a given set of arguments, which leads to another property called referential transparency. When a function is referentially transparent, we can replace that function with its value without affecting the behavior of the program. All mathematical functions are pure functions, whereas date functions, `rand()`, etc. are impure.
### Higher Order Functions
The concepts above can be achieved in almost any programming language, but first class functions and higher order functions are the two most distinguishing features of functional programming. I explained that first class functions means that functions can be treated as values. Higher order functions are functions that can take functions as arguments and can return a function as their result. Two important features were added relatively recently which enabled us to write higher order functions in PHP: lambdas and closures.
### Lambda Functions
A lambda function (also known as anonymous function) is nothing but a function that has no name. When we define an anonymous function, a reference to the function is returned which is stored in a variable for later use. We use this variable to call the function whenever it’s needed.
This concept has been adopted in many different languages. In fact, you’re probably using lambda functions in your day-to-day JavaScript programming, passing them as callback functions for different user interactions and Ajax calls.
``````\$("#myButton").click(function () {
// do something
});``````
This piece of code is so simple and easy to understand, which might make us forget its functional aspect.
PHP introduced this awesome feature in version 5.3, which lets us write PHP code in a similar fashion:
``````<?php
\$square = function (\$arg) {
return \$arg * \$arg;
};
\$value = \$square(2); // 4``````
When talking about functions, and anonymous functions in particular, its important to understand how variable scope is handled. JavaScript, for example, allows you to access a variable from an outer scope inside the lambda, whereas PHP does not. Inside the lambda is its own scope, just as with regular PHP functions.
### Closures
Sometimes you’ll want to reference a variable from the parent scope inside your function. Closures are similar to lambda functions with the minor difference that you can access variables from an outer scope. We can use “reach out” and bind an outer variable using PHP’s `use` keyword, also introduced in PHP 5.3.
``````<?php
\$rate = .12;
\$findInterest = function (\$value) use (\$rate) {
return \$value * \$rate;
};
\$interest = \$findInterest(100);``````
In this case we don’t pass the interest rate each time we call the function. Instead, we’ve defined it outside and made it available inside the function with the `use` keyword.
### Partial Functions and Currying
A partial function, simply put, is a function created from an existing function by partially applying its arguments. You only need to pass the remaining arguments when you call the created function.
We can create partial functions in PHP with the use of closures. Here’s an example to find the volume of a box given its length, width and height. All arguments are optional; if you don’t supply all of the arguments, the function will return another function to accept the necessary values that remain.
``````<?php
\$volume = function (\$length = 0, \$width = 0, \$height = 0) use (&\$volume) {
\$args = func_get_args();
\$numArgs = func_num_args();
if (\$numArgs == 3) {
return \$length * \$width * \$height;
}
else if (\$numArgs < 3) {
return function() use(&\$volume, \$args) {
\$newArgs = array_merge(\$args, func_get_args());
return call_user_func_array(\$volume, \$newArgs);
};
}
else {
}
};``````
All of the parameters are optional. First a check is made to determine whether the caller passed all arguments. In that case, we can directly return the volume by multiplying length, width and height. If the number of arguments is less than the parameters, a new function is returned to find the volume “pre-seeded” with the given arguments.
Now suppose that most of the time we’re finding the volume of box with a fixed length, say 10. This can be easily done by passing 10 as the first argument, or we could create the partial function by fixing 10 as the first argument and then only ask for the remaining values.
``````<?php
\$standardVolume = \$volume(10);
\$vol = \$standardVolume(5, 5); // 250``````
Currying is a special case of partial functions where you convert a function that takes multiple arguments into to multiple functions that will each take a single argument. For example, something like f(x,y,z) to f(x)(y)(z) (although PHP syntax doesn’t allow nesting of function calls like this). Timothy Boronczyk has written an excellent article on currying with a practical example if you’re interested in seeing more.
There are many practical uses of functional programming features in PHP. For instance, lambda functions are widely used when working with callbacks. Using the Slim framework, for example, you can define a route like this:
``````<?php
\$app = new SlimSlim();
\$app->get("/home", function () {
});``````
Slim invokes the callback function when the request URL matches this route. Vance Lucas wrote about some other interesting use cases of Lambda functions a while back.
Safe programming is encouraged by avoiding state and mutable data. In functional programming you should write functions that do exactly one thing each and do not cause any side effects. The paradigm’s emphasis on modularity and conciseness of functions can make its easier to reason about your program in terms of different, small sub programs as well.
Functional programming can also help you write code that focuses what you want to achieve instead of explicitly managing the incidentals in the process (compare recursion with having to manage loop counter variables).
But keep in mind however that some of the advantages traditionally associated with functional programming are not applicable to PHP as it’s not designed to be a functional programming language. For example, side-effect-free functions lend well to parallel processing, but PHP scripts are not run in this manner.
It’s not always easy to calculate the cost of recursion and lazy functions, either, and there can be significant performance issues due to the internal overhead. Sometimes it makes more sense to write programs in terms of mutability for better efficiency.
Perhaps the biggest disadvantage of functional programming is its steep learning curve for those trained in an imperative manner. But over all, functional programming is interesting, and learning it will give you the tools to think about old problems in a new light, helping you to grow as a programmer. It’s not a one-size fits all solution, but it can be applied as appropriate for cleaner, more eloquent PHP code.
## Summary
Functional programming is more than just a programming paradigm; it’s a way of thinking and reasoning about your program. If you can think functionally, you can do functional programming in almost any language.
In this article we discussed the basics of functional programming, leveraging features of PHP to write provide their examples. Although, the example given in this article may not be practically useful to you, you’ll find many situations where the functional style can significantly improve the quality of code you are writing. Try to find such cases, think functionally, and have fun!
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Tags: Expert
• Gonçalo
Thank you, I’ve learned a lot.
• Shameer C
Hi Gonçalo,
Thanks for the comment, glad that article was useful to you. :)
Awesome Article Guys , You Have Forgotten About Silex Micro Framework :D
• Shameer C
Thanks Faizal!
We haven’t forgotten Silex; of course that’s a great framework. But I use Slim most of the time :)
• Harlin
Good job, Shameer, on making this very simple.
• http://www.vancelucas.com Vance Lucas
Shameer – Thanks for the link and mention to my PHP 5.3 closures blog post. Slim and other micro-frameworks use closures as callbacks (which you mentioned), and that’s pretty much as far as they go with them. I wrote a specifically more “functional” micro-framework called Bullet that takes the concept of functional programming style in routes much further: http://bulletphp.com/ – you might want to take a look as it is relevant to this article, and is also very easy to understand for newcomers to this concept.
• Shameer C
Hey Lucas,
That’s awesome!. I will definitely take a look at this framework soon.
• Elizabeth M Smith
Just one thing to add so people aren’t confused – a “lambda function” in PHP has a specific meaning (a function created via create_function() ) to the language and is not the same as the new anonymous functions in 5.3
So referring to an anonymous function as a lambda in PHP will have people scratching their heads.
• Shameer C
Hi Elizabeth,
Thanks for pointing that out. I didn’t know this difference in PHP, but I think in practice lambda functions served the same purpose when anonymous functions was not available. Is there any difference?
• http://habilbozali.com Habil BOZALİ
Thank you for important points!
• Zero
I thought I would have a go with lambda functions and this is what I found:
// example 1, failure
\$f = function(\$arg) {
return \$arg * 1.5;
};
function somefunc() {
\$v = \$f(25); // I get error: function name must be a string
}
// example 2 – success
function somefunc() {
\$f = function(\$arg) {
return \$arg * 1.5;
};
\$v = \$f(25); // this works
}
It seems the lambda function has to be within the scope of the calling function. Any comments?
• Shameer C
Hey Zero,
In the first case you got error as \$f is not available iniside somefunc(). To get this working, you can re-write the function as
``` \$somefunc = function() use(\$f) { \$v = \$f(25); }; ```
• http://www.whiteroomsolutions.com Rob S
Really good article, clear and well explained.
Thanks very much Shameer!
• http://dirtyhandsphp.com Shiv
Awesome Article. Very good stuff to learn. Keep it up!!!
• Dmitry
Hi!
Partial Functions and Currying was difficult to understand :) Functional programming needs to write code beautifully or don’t write completely.
• http://heymanengineering.com Sam
Thank you Shameer for a great article.
There’s a very interesting course on coursera.org called ‘Programming Languages’ given by Dan Grossman at the University of Washington. He starts off with ML for exactly the same reasons you mention: functional programming can help you become a better programmer in any language.
• Shameer C
Thanks Sam for the info, I will take a look at the course details. :)
• santanu
good stuffs…
• http://www.zachis.it/blog Zach Smith
this was actually informative – not like most of what i see online. sharing :)
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» 5 years ago, # | +8 good contest!
» 5 years ago, # | +12 nice!f2 is a interesting problem,and i learned a lot from it.
» 5 years ago, # | 0 What's the proof of optimality for D for given coffee distribution-strategy ? Is it just intuitive or we can prove it ?
• » » 5 years ago, # ^ | ← Rev. 2 → 0 Yes we can prove it.The approach is greedy. Consider the caffeine values 6 5 5 3 1 1 in order and consider the distribution for 2 days and in each day you are taking 3 cups.6 + 5 + (5 — 1) + (3 — 1) + max(0, 1 — 2) + max(0, 1 — 2) = 17If you consider this case 6 5 1 1 3 5 in order6 + 5 + (1 — 1) + (1 — 1) + (3 — 2) + (5 — 2) = 15This is because in the first case max(0, -1) + max(0, -1) hides subtracting 2 from the final output.Sorry for my bad example.
• » » 5 years ago, # ^ | ← Rev. 2 → +25 If you can bear some algebra then you can better appreciate the proof. ;) So again let's sort it in decreasing order a0 ≥ a1 ≥ ... ≥ an - 1, and we consider the sum . Now there exists an m such that this sum is equal to : is strictly decreasing as i increases. In other words we can 'assume' that a0 through am - 1 are strictly greater than , and the rest at most . To show that this is optimally, let b0, ... , bn - 1 be a permutation of these ai's, and since we spread it over k days, at most k of the coffee cups are the first of the days (hence 0 penalty), at most k of them has penalty 1, and so on. Hence the sum is now at most . Choosing only those i's that are relevant, we see that this sum is also equal to . Now if p is the number of those i's satisfying , we can split the sum into two categories: the bi's contain at most m integers from a0, ... , am - 1 and the rest are not more than ; so . On the other hand, the sum (considering only those i's that are relevant) at at least so , as desired.
• » » » 5 years ago, # ^ | 0 Amazing! That's what I was looking for. Thanks.
» 5 years ago, # | ← Rev. 2 → 0 .
» 5 years ago, # | +24 E has another simple solution.Traverse all pairs in increasing order (starting from (1, 2), (1, 3), ...) and print each pair followed by its flipped pair (not repeating any). 1 2 2 1 1 3 3 1 1 4 4 1 2 3 3 2 ...
• » » 5 years ago, # ^ | 0 Yeah I also used the same strategy.
• » » 5 years ago, # ^ | 0 i used this too, nice!
• » » 5 years ago, # ^ | ← Rev. 3 → 0 There is another solution where in we could start with say pair (1,2) and keep incrementing both the elements of the pair until the second element of the pair reaches k.After that, we flip the pair and keep decrementing both x and y until y becomes 1. Repeat this process and we get the correct answer.
• » » 5 years ago, # ^ | 0 Yeah ! It is the simplest approach.
» 5 years ago, # | +6 Can someone please explain me the tutorial of F1. It would be of great help.
• » » 5 years ago, # ^ | ← Rev. 2 → +4 I solved it in following steps: Suppose root of the tree is 1. Then by single dfs we find how many red and blue vertices are present in subtree of each vertex. Then we see each parent to child edge, if child's subtree contains 0 blue vertices and all red vertices(or 0 red vertices and all blue vertices) then this edge is good.
» 5 years ago, # | +10 I think this contest have much tricky questions then previous div3 contest.
» 5 years ago, # | -31 the contest was terrible and now seeing from this editorial you really didn't care about us. it's a joke, especially problem D. i'm at a loss of words trying to explain myself why you keep mixing math with programming. what connection do you see between the two? genuinely
» 5 years ago, # | 0 In Problem B for n=1 ans should be 0, bcoz if there is only one candy, tanya will give that's one to his father, since now there is no candy she could not eat any day.(There is no candy). Shouldn't it be 0 ? (My soln got hacked just due to only this case which i made myself ans = 0 if n==1) :(
• » » 5 years ago, # ^ | 0 No, because if dad eats the 1-th candy, sum of weights of candies Tanya eats in even days = sum of weights of candies Tanya eats in odd days = 0.
» 5 years ago, # | 0 i've learned a lot from this contest, specially that i must read the entire problem set.
» 5 years ago, # | 0 Can someone suggest some other approach for F2. Thanks.
• » » 5 years ago, # ^ | +1 Hi, my solution is still similar to the editorial but maybe it can help. I first run a DFS to do a minimal merge of the components of the same color: I colorize the entire path between each vertex of color, and check for collisions. You only have to keep a count of number of components per color to know if the child has to give its color to his parent. Then instead of counting the ways of cutting, i count the number of valid colorings, where a color for a node is either "in subtree" or "not in subT". Then for the second DFS I just compute the possibilities for these two "colors". The formulas are similar to editorial. Solution in O(n log mod) runs in 0.5s, because I took modular inverse to obtain the product of "all but one". Hope it helps, my code here: 50352250
• » » » 5 years ago, # ^ | 0 Could You explain how counting works? I haven't really understood it from the tutorial. Although I understand all the rest.
• » » » » 5 years ago, # ^ | +1 Either you already have made # of colors in subtree cuts in this subtree or you have made this number minus one. In my case corresponds to "is this node color in the subree?". You can think of DP0 as "am i done here? = has color of parent" so naturally you can keep a color and get from color of child to color of parent but not the converse. Then like classical tree DP you sum possibilities of disjoint events (like choosing one children color) and multiply the ones for independent events (like choosing a color for each subtree). This is not an easy DP pb because you have to keep track of both. Try to get the intuition behind the formulae before looking at tricks for computing it faster.
• » » » » » 5 years ago, # ^ | 0 Okey, I think I am starting to understand it now. One more question. How do we know in our dp that we cut exactly k-1 times? No more no less?I will try solving some easier problems with tags (trees, combinatorics) and then come back to this one. Thanks
• » » » » » » 5 years ago, # ^ | +1 1: you checked there were no collisions during the minimal merging 2: at each cut your separating the "top" color from the "bottom" one, the components stay connected. When you already cut it, a color disappears from the possibilities (DP1->DP0) and becomes a color not in subtree. DP0 means we are done w/ all colors of subtree, DP1 there is one left at the root of subtree. 3: Since you want the root node to have a color from its subtree, you return DP1. Each color except this one has been cut -> K-1.
» 5 years ago, # | 0 can anybody take a look to my post hereand help me to optimize my dp solution for the D1 to solve D2 thanks in advance
• » » 5 years ago, # ^ | 0 Its painful to read code of other people sometimes. Let me know if you haven't understood from editorial.
• » » » 5 years ago, # ^ | 0 I can’t figure out the counting part. Logic behind dp[v][1] and dp[v][0]
» 5 years ago, # | ← Rev. 2 → 0 1118D1 — Coffee and Coursework (Easy version) Let the current number of days be k. The best way to distribute first cups of coffee for each day is to take k maximums in the array How do you know that it's the best way to distribute? I don't get that
• » » 5 years ago, # ^ | ← Rev. 2 → 0 Checking on samples and doing some casework and observations, It is better to use maximum values initially because they may get depleted later on.
» 5 years ago, # | 0 Need help with the problem F2. What I've done is- For every color, tried to build $k$ same-colored subtrees. If not possible (collision occurs), the answer is trivially $0$. For simplicity, let's call uncolored vertices as black vertices. When an answer exists, there will be many (possibly $0$) subtrees induced by black vertices, whose leaves will have many colored components(sub-trees) adjacent to them. Let's call the set of these colored component/sub-trees $S_T$ and the black components $B_T$. Instead of cutting $k-1$ edges, I tried to color the black vertices in a way such that: a. only the trees in $S_T$ get largerb. no new subtrees form.c. all the vertices get colored.Clearly, doing step 2 for all the black sub-trees in $B_T$ will lead us to the result. The answer will be the product of $count(coloring_.a_.black_.component_.in_.B_T)$. So, all I need is to find a way to count the colorings of $1$ black component. It should be an easy DP but somehow I'm failing to grasp it. The tutorial did not help (sorry). Any help is welcome. Thanks.
» 4 years ago, # | 0 The way Mike has explained F2 is remarkable. Thanks Mike and Vovuh.
» 4 months ago, # | 0 For problem F2 I have an approach. However I am stuck at some part Please read this blog for my approach. Note: For good coders you will find one more interesting problem to solve. | 2,750 | 9,795 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.783938 |
http://physics.stackexchange.com/questions/59803/galaxies-moving-away-at-the-speed-of-light | 1,469,420,866,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00135-ip-10-185-27-174.ec2.internal.warc.gz | 201,859,851 | 18,873 | Galaxies moving away at the speed of light
As an arts student, I really find those cosmological questions hard to understand and hence come here to seek your kind help.
The Hubble constant $H_0$ is estimated to be about 65 km/s/Mpc, where 1 Mpc (megaparsec) is around 3.26 million light-years. At what distance would galaxies be moving away at exactly the speed of light? (I found that there is something called Hubble Radius, but is this the same as Hubble Radius?) If there were galaxies farther than the Hubble radius, how would they appear to us?
-
Possible duplicate: physics.stackexchange.com/q/12049/2451 – Qmechanic Apr 2 '13 at 13:36
That's a very, very good question! Actually, the point you are addressing is the reason why physicists coined the term "observable universe". Those galaxies moving away from us with a speed bigger than the speed of light will never be visible (in the light they emit right now) to us and are outside the so-called "Hubble sphere". The distance you are looking for is $\frac{c}{H_0}$.
So as sad as it may sound, every second more and more galaxies and astronomical objects are leaving the observable universe, never to be seen again.
-
The Hubble sphere is not a horizon - in fact there are galaxies receding faster than $c$ now that we will be able to see eventually. – Chris White Jul 26 '13 at 19:15
The Hubble sphere is the locus of points where the Hubble flow is equal to c. The Hubble sphere is not an event horizon.
All observed objects with a redshift greater than about 1.46 are outside the Hubble sphere. In other words, these objects are receding superluminally. Furthermore, these objects were outside of the Hubble sphere, and were receding superluminally at the time the light was emitted.
The diameter of the observable universe is greater than the diameter of the event horizon. We will never observe events that are currently happening outside of the event horizon, but we can still see events from galaxies that are currently outside of the event horizon via the light they emitted long ago.
- | 475 | 2,064 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-30 | latest | en | 0.961645 |
https://learn.adafruit.com/all-about-leds/forward-voltage-and-kvl | 1,721,444,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00873.warc.gz | 320,737,864 | 17,410 | For every LED, in order to use it properly, we need to know the Forward Voltage. What is this forward voltage? Lets explain it in a photo:
In our three-piece circuit, we have the battery (which generates voltage) and the resistor+LED (which uses up the voltage). I will now tell you a very key 'law' of electronics:
In any 'loop' of a circuit, the voltages must balance: the amount generated = the amount used
This "Voltage Loop" law was discovered by a fellow named Kirchhoff (thus it is called Kirchhoff's Voltage Law = KVL). And we can see the loop above, where one part is made of the +9V battery. The other half must use up the +9v (making it -9V so that both halves of the loop equal out).
So what does this have to do with the Forward Voltage of an LED? Well, the Forward Voltage is the 'negative voltage', used by the LED when it's on. Kinda like a 'negative battery'! So lets modify our diagram slightly.
Whenever the LED is on, the voltage it uses it up is somewhere between 1.85V and 2.5V. We'll say 2.2V for average - that's a good assumption for most red, yellow, orange and light-green LEDs. If we subtract that from 9V we get about 6.8V left. This is the voltage that must be 'absorbed' by the resistor.
## Quick Quiz!
Let's say we have the same circuit above, except this time its a 5V battery and an LED with a forward voltage of 2.5V, how much voltage must be 'absorbed' by the resistor?
Voltages Generate = Voltages used, so 5V = 2.5V + ResistorVoltage. The voltage across the resistor is 2.5V.
Let's say we have the same circuit above, except this time its a 5V battery and an LED with a forward voltage of 3.4V, how much voltage must be 'absorbed' by the resistor?
Voltages Generate = Voltages used, so 5V = 3.4V + ResistorVoltage. The voltage across the resistor is 1.6V.
## Ohm's Law
What is interesting about the law we just learned (KVL) is that in no place do we use the resistance of the resistor. It never shows up in the equation. Yet from our previous experiements we know for a fact that changing the resistance affects how bright the LED is. There must be something else going on, lets keep working on understanding the details….
Next we're going to throw in another important law. This one is called Ohm's Law- and it describes how resistors work.
Voltage across a resistor (volts) = Current through the resistor (amperes)* The Resistance of the resistor (ohms)
There's a more common shorthand notation which you'll see very often:
V = I * R
Or the two other ways of writing to solve for current or resistance:
I = V / R
R = V / I
The V is for voltage, the R is for resistance and the I, confusingly, is for current. Yeah, that I is a little annoying isn't it, since theres not even a single I in the word current? Unfortunately, there are 100 years working against us here, so just bear with us on that one.
Quick Quiz!
If I have a 3 ohm resistor (R) with a current of 0.5 Amperes (I) going through it. What is the voltage (V) across the resistor?
We'll use the V = I * R form of Ohm's Law. V = 0.5 A * 3 ohm = 1.5 Volts.
Now I have a 1000 ohm resistor (R), and a voltage across it of 6.8V (V), what is the current (I) going through the resistor?
We will use the I = V / R form of Ohm's Law. Current = 6.8 V / 1000 = 6.8 milliAmps.
Ohm's law is very important and its worth drilling a bit to become familiar with it. We suggest coming up with other mix+match numbers of resistances, currents and voltages and using them to solve for the unknown value. If you're working with a friend, quiz each other and check your answers! There are also 'calculators' online you can check yourself against.
## Solving for the current
We'll now combine both KVL and Ohm's Law with our diagram. Our LED is connected to a 1000 ohm resistor (you should verify this by checking the resistor color stripes!), and the voltage across that resistormust be 6.8V (the law of KVL) so the current through that resistor must be 6.8V / 1000ohm = 6.8 mA (Ohm's law).
Our diagram is getting a little dense, but we're pretty much done. The resistor current is 6.8mA and that current is also going through the LED, so the LED current is 6.8mA. "Big whoop," you may be saying. "What do I care about the LED current?" The reason you should care is that:
The amount of current (I) going through an LED is directly proportional to how bright it appears.
Aha! Finally, the last piece of the puzzle. If we increase the current, the LED will be brighter. Likewise, if you decrease the current, the LED will be dimmer. By picking the correct resistor, you have full control over how the LED appears.
Whenever using an LED, make sure to always have a resistor! The resistor limits the current, which will keep the LED from burning out!
Most of the time, you'll want to have a really bright LED so you'll be calculating the smallest resistor you can get away with and not damage the LED. But note that the more current used by the LED, the quicker you'll drain the battery. So there are good reasons for wanting to control the brightness if say you have a small battery and you want the lights to last a long time.
Since as we have seen, too much current will make the LED go poof, what is the best amount of current we should use? For some very big 'power LEDs', the current can be as high as 1 or 2 Amperes, but for pretty much every 3mm, 5mm or 10mm LED, the amount of current you're expected to use is 20mA. You can see this in the datasheet we talked about earlier. See the right-most column? IF is the Forward Current (I) and they use 20mA.
For 99% of LEDs you will encounter, the optimal current is 20 milliAmperes (0.02 A) but don't be too scared to push it up to 30mA if you need a litle more brightness.
This guide was first published on Feb 11, 2013. It was last updated on Feb 11, 2013.
This page (Forward Voltage and KVL) was last updated on Feb 11, 2013.
Text editor powered by tinymce. | 1,538 | 5,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.921361 |
https://www.mrexcel.com/board/threads/sum-of-numbers.223016/ | 1,695,791,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00633.warc.gz | 979,849,543 | 20,691 | # Sum of Numbers
#### Peter100
##### Well-known Member
Hi
Is there a function to calculate the total of the figures between 1 and a given number ?
ie.
the sum of 1 - 24 = 300
the sum of 1 - 36 = 666
In reality I want to enter 24 in cell A1 and in B1 display the result 300
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Good evening Peter100
I don't believe there is, but there's a cheeky little formula that statisticians use to work it out :
((n*(n+1))/2
Where n is the number you want to sum up to:
the sum of 1 - 24 = 300
24*25=600
600/2=300
HTH
DominicB
EDIT : The formula you would want is :
=(A1*(A1+1))/2
=SUMPRODUCT(ROW(INDIRECT("1:"&A1)))
=(A1*(A1+1))/2
is far superior.
little story for illustration
about 100 years ago, when Albert Einstein was 10-12 years old or less ...
Teacher (who wants to have some free-time to prepare the next lesson): Guys & gals, take a little sheet of paper, write down all numbers from 1 to 100 and calculate the sum
Little Albert, crossing his arms and looking to the sky for 5 seconds, takes his pencil and writes down the result. Ready!
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
...
50 + 51 = 101
hmm... 50 * 101 = 5050
=SUMPRODUCT(ROW(INDIRECT("1:"&A1)))
I get #REF
this works for me
=SUMPRODUCT(ROW(1:24))
using INDIRECT in SUMPRODUCT, is it possible ?
do I miss something ?
kind regards,
Erik
Cheers All yep they both work
now I can finish my calculator
=SUMPRODUCT(ROW(INDIRECT("1:"&A1)))
I get #REF
It works for me. Of course, it's inferior to (N*(N+1))/2. Also bounded by 65536...
this works for me
=SUMPRODUCT(ROW(1:24))
Not robust: Insert a row before the formula cell to see...
using INDIRECT in SUMPRODUCT, is it possible ?
do I miss something ?
...
Sure, it's possible.
Cheers All yep they both work
now I can finish my calculator
Don't opt for the SumProduct version though...
found the solution
it was related to an error on my sheet
as a side note, this story is actually about Karl Gauss, and I believe it was Grade 2. besides, we all know that Einstein stunk at grade school math
cheers. ben.
little story for illustration
about 100 years ago, when Albert Einstein was 10-12 years old or less ...
Teacher (who wants to have some free-time to prepare the next lesson): Guys & gals, take a little sheet of paper, write down all numbers from 1 to 100 and calculate the sum
Little Albert, crossing his arms and looking to the sky for 5 seconds, takes his pencil and writes down the result. Ready!
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
...
50 + 51 = 101
hmm... 50 * 101 = 5050
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## Homework Statement
#56.) Someone makes necklaces and sells them for 10 dollars each. His average sales were 20 per day. When he raises the price to 11 dollars per day, the average sales drops 2.
a.)Find the demand function, assuming it is linear.
b.)If the material to make each necklace costs 6 dollars, what should the selling price be to maximize profit?
#14.)A rectangular container with an open top is to have a volume of 10m^3. The length of it's base is twice the width. Materials for the base cost 10$per square meter. Material for the sides cost 6$ per meter. Find the cost of materials for the cheapest such container.
## The Attempt at a Solution
For #56 I have solved part a.) I found the demand function to be
$$y=-2x$$
For part b. I am a bit confused.
It seems to me like I should just let x = (x-6), and then maximize that function? Is that the correct idea?
#14.)For this one I made a box and labeled it's dimensions:
$$h(w)(2w)$$
Then I got thinking about it, and I don't know how to set this one up at all.
I worked out the following:
$$LWH = 10$$
$$L = 2W$$
$$2W^{2}H = 10$$
Area for the sides is:
$$2(2WH)+2(HW)$$
Area for the base is:
$$2W^{2}$$
I just don't know how to relate all of these together? I think it's a similar style to the other problem I posted here, which is why I grouped them together. What is the next step to relate these functions to their appropriate costs so that I can maximize? For the container one, I thought I would just have to find the minimum surface area of the box that still had the volume of 10, but the two different costs ruined that idea for me.
gb7nash
Homework Helper
## Homework Statement
#56.) Someone makes necklaces and sells them for 10 dollars each. His average sales were 20 per day. When he raises the price to 11 dollars per day, the average sales drops 2.
a.)Find the demand function, assuming it is linear.
b.)If the material to make each necklace costs 6 dollars, what should the selling price be to maximize profit?
#14.)A rectangular container with an open top is to have a volume of 10m^3. The length of it's base is twice the width. Materials for the base cost 10$per square meter. Material for the sides cost 6$ per meter. Find the cost of materials for the cheapest such container.
## The Attempt at a Solution
For #56 I have solved part a.) I found the demand function to be
$$y=-2x$$
Part a) is wrong. Do a sanity check on this. If you sell for 0 dollars, by your equation, you sell 0 necklaces.
I also can't tell if your x represents demand or quantity. Usually, for profit problems, you'll want the independent variable to be quantity and the dependent variable to be price. Just so there's no ambiguity as to what you're dealing with, you might want to replace your equation with p's and q's.
Now looking at part a), try to find the equation of the line that fits through the two points: (q1,p1) = (20,10) and (q2,p2) = (18,11)
I see what you are saying, I made a mistake there.
If two points on the demand function are (10,20) and (11,18)
then
$$\frac{\Delta Q}{\Delta P} =\frac{18-20}{11-10} = -2$$
$$Q = -2P + k$$
$$18 = -2(11) + k$$
$$18 = -22 + k$$
$$18 + 22 = k$$
$$k = 40$$
So the demand function for part a. of that problem is given by:
$$Q = -2P + 40$$
Now, it costs him 6 dollars per necklace, what should his selling price be to maximize profits? I'm still not sure how to relate his cost of 6 per item made to the above formula.
If Q is how many he is selling, then wouldn't Q(P-6) give his profits?
and so I could write:
$$F(P)=(P-6)(-2P+40)$$
Which would be a polynomial like:
$$F(P)=-2P^{2} + 52P - 240$$
which is opening downwards, so there will be a max!
$$\frac{d}{dp}-2P^{2} + 52P - 240 = -4P+52$$
$$F'(P)= -4P+52 = 0$$
$$-4P = -52$$
$$P = \frac{52}{4}$$
$$P = 13$$
That seems reasonable for that question to me? Or am I mistaken again?
I guess it's time to tackle the other problem (thanks for the help on the first one):
#14.)A rectangular container with an open top is to have a volume of 10m^3. The length of it's base is twice the width. Materials for the base cost 10$per square meter. Material for the sides cost 6$ per meter. Find the cost of materials for the cheapest such container.
I'm trying to figure out how to set this one up, when I come to something usable I will post here. | 1,222 | 4,350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-49 | latest | en | 0.941152 |
https://brian-rose.github.io/ClimateLaboratoryBook/courseware/assignment-cesm-climate-change.html | 1,718,713,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00245.warc.gz | 128,590,017 | 8,425 | # Assignment: Climate change in the CESM simulations#
This notebook is part of The Climate Laboratory by Brian E. J. Rose, University at Albany.
## Part 1#
Following the examples in the lecture notes, open the four CESM simulations (fully coupled and slab ocean versions).
Calculate timeseries of global mean ASR and OLR and store each of these as a new variable. Recall that ASR is called FSNT in the CESM output, and OLR is called FLNT.
Plot a timeseries of (ASR - OLR), the net downward energy flux at the top of the model, along with a 12 month rolling mean, analogous to the plot of global mean surface air temperature in the lecture notes.
Note that the rolling mean is important here because, just like with surface air temperature, there is a large seasonal cycle which makes it harder to see evidence of the climate change signal we wish to focus on.
## Part 2#
Calculate and show the time-average ASR and time-average OLR over the final 10 or 20 years of each simulation. Following the lecture notes, use the 20-year slice for the fully coupled simulations, and the 10-year slice for the slab ocean simulations.
## Part 3#
Based on your plots and numerical results from Parts 1 and 2, answer these questions:
1. Are the two control simulations (fully coupled and slab ocean) near energy balance?
2. In the fully coupled CO2 ramp simulation, does the energy imbalance (ASR-OLR) increase or decrease with time? What is the imbalance at the end of the 80 year simulation?
3. Answer the same questions for the slab ocean abrupt 2xCO2 simulation.
4. Explain in words why the timeseries of ASR-OLR look very different in the fully coupled simulation (1%/year CO2 ramp) versus the slab ocean simulation (abrupt 2xCO2). Think about both the different radiative forcings and the different ocean heat capacities.
## Part 4#
Does the global average ASR increase or decrease because of CO2-driven warming in the CESM?
Would you describe this as a positive or negative feedback?
## Part 5#
In the previous question you looked at the global average change in ASR. Now I want you to look at how different parts of the world contribute to this change.
Make a map of the change in ASR due to the CO2 forcing. Use the average over the last 20 years of the coupled CO2 ramp simulation, comparing against the average over the last 20 years of the control simulation.
## Part 6#
Repeat part 5, but this time instead of the change in ASR, look at the just change in the clear-sky component of ASR. You can find this in the output field called FSNTC.
The FSNTC field shows shortwave absorption in the absence of clouds, so the change in FSNTC shows how absorption and reflection of shortwave are affected by processes other than clouds.
## Part 7#
Discussion:
• Do your two maps (change in ASR, change in clear-sky ASR) look the same?
• Offer some ideas about why the clear-sky map looks the way it does.
• Comment on anything interesting, unusual or surprising you found in the maps.
## Credits#
This notebook is part of The Climate Laboratory, an open-source textbook developed and maintained by Brian E. J. Rose, University at Albany.
It is licensed for free and open consumption under the Creative Commons Attribution 4.0 International (CC BY 4.0) license.
Development of these notes and the climlab software is partially supported by the National Science Foundation under award AGS-1455071 to Brian Rose. Any opinions, findings, conclusions or recommendations expressed here are mine and do not necessarily reflect the views of the National Science Foundation. | 798 | 3,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.898367 |
https://qmlt.readthedocs.io/en/latest/_modules/qmlt/numerical/regularizers.html | 1,571,255,530,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00140.warc.gz | 678,944,975 | 5,168 | # Source code for qmlt.numerical.regularizers
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# Unless required by applicable law or agreed to in writing, software
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
"""
Regularizers
========================================================
**Module name:** :mod:qmlt.numerical.regularizers
.. currentmodule:: qmlt.numerical.regularizers
A collection of regularizers to facilitate experiments with the numerical circuit learner.
Summary
-------
.. autosummary::
l2
l1
Code details
------------
"""
import numpy as np
[docs]def l2(circuit_params):
r"""L2 regulariser :math:0.5 \sum_{i=1}^N w_i^2 for a vector :math:w = (w_1,...,w_N) of circuit parameters.
Args:
circuit_params (ndarray): 1-d array containing the values of the circuit parameters to regularize.
Returns:
float: Scalar l2 loss.
"""
circuit_params = np.array(circuit_params)
if circuit_params.ndim > 1:
raise ValueError("Regulariser expects a 1-dimensional array, got {} dimensions".format(circuit_params.ndim))
return 0.5*np.dot(circuit_params, circuit_params)
[docs]def l1(circuit_params):
r"""L1 regulariser :math:\sum_{i=1}^N |w_i| for a vector :math:w = (w_1,...,w_N) of circuit parameters.
Args:
circuit_params (ndarray): 1-d array containing the values of the circuit parameters to regularize.
Returns:
float: Scalar l1 loss.
"""
circuit_params = np.array(circuit_params)
if circuit_params.ndim > 1:
raise ValueError("Regulariser expects a 1-dimensional array, got {} dimensions".format(circuit_params.ndim))
return np.sum(np.absolute(circuit_params)) | 448 | 1,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.525422 |
https://math.stackexchange.com/questions/2596040/finding-a-particular-solution-to-a-second-order-differential-equation | 1,563,728,248,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00332.warc.gz | 462,923,147 | 36,894 | # Finding a particular solution to a Second Order Differential Equation
I am attempting to do the following problem (8.6.26 in Mary Boas, mathematical methods in physical sciences 3rd Edition). It provides the following Auxiliary equation:
$$(D^2+1)y=8x \sin(x)$$ I know that a general solution can be written:$$y=y_c+y_p,$$ where $y_p$ is a particular solution, and $y_c$ is a characteristic equation. I have obtained the characteristic equation to be: $$y_c=c_1e^{-ix}+c_2e^{ix},$$ where $c_1$ and $c_2$ are arbitrary constants.
We are considering: $$y''+y=8x \sin(x)$$ to find the particular solution we can consider the imaginary part of: $$Y''+Y=8xe^{ix}$$ My problem is coming when looking at the particular solution later. I can see that the roots of the auxiliary equation are $\pm i,$ so it makes sense to assume a particular solution of form: $$Cxe^{ix}.$$ From this: $$y_p'=Cixe^{ix}+Ce^{ix},$$ $$y_p''=-Cxe^{ix}+2Ce^{ix}$$ plugging these back in i find $$C=-4xi$$ and $$y_p=-4x^2ie^{ix}$$ 'expanding this', and considering only the imaginary pary, I get: $$y_p=-4x^2 \cos(x)$$ However this is clearly wrong, as when I re-substitute it back in, i get: $$-8\cos(x)+8x\sin(x)=8x \sin(x),$$ I am assuming I am missing a term somewhere, but I have repeated the question at least five times now and cannot spot it. I am posting it here to see if someone can see where I have gone wrong. Thanks in advance
• You should have $$y_p''=-Cxe^{ix}+{2\color{red}{i}Ce^{ix}}$$ – John Doe Jan 7 '18 at 20:27
The mistake happened when you calculated $y_p''$. It should be $$y_p''=-Cxe^{ix}+2iCe^{ix}$$
Also, the current $y_p$ provides a solution to $(D^2+1)y=8\sin(x)$, so you are looking for a particular solution of one order higher for $(D^2+1)y=8x\sin(x)$: $$y_p=Cxe^{ix}+Dx^2e^{ix}$$
Solving this correctly should give us that $$Im(y_p)=-2x^2\cos(x)+2x\sin{x}$$
As I mentioned in a comment, you computed $y_p''$ wrong. You should have $y_p''=-Cxe^{ix}+2iCe^{ix}$, which would give $$2iC e^{ix}=8xe^{ix}$$ Once you reached this point, you declared that $C=-4xi$. You cannot do this, since you assumed $C$ is a constant, and so it cannot depend on $x$. So clearly, we have not got a particular solution of this form. If we add a term though, we can get a solution. Take $$y_p=Cxe^{ix}+Dx^2e^{ix}$$ Then $$y_p''+y_p=2iCe^{ix}+2De^{ix}+4iDxe^{ix}=8xe^{ix}$$This gives that $D=-2i$, and $C=2$. So $$y_p=2xe^{ix}-2ix^2e^{ix}$$ The imaginary part of this is $$2x\sin x-2x^2\cos x$$(in agreement with the other answer) | 829 | 2,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-30 | latest | en | 0.86302 |
https://board.mfwbooks.com/viewtopic.php?f=3&t=1956&p=101262 | 1,576,201,833,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540547536.49/warc/CC-MAIN-20191212232450-20191213020450-00368.warc.gz | 303,920,389 | 23,702 | ## Math - Help Getting Started
Learning God's Story
Joy1139
Posts: 25
Joined: Wed Apr 26, 2006 10:03 am
### Math - Help Getting Started
JL wrote: I've read the math pages in the TM, but I'm a very visual person and sometimes just reading it doesn't click with me. Does anyone have any pointers, or could you just briefly walk me through what a couple of math lessons are like? I hope this makes sense, it's late and that's probably part of my problem!
I will try to help with the first grade math question. I am also very visual and had to read the math section many times. I probably still didn't do everything exactly right. Still, I was very pleased with the math in first grade and I am also glad that I did not use another math curriculum either in place of it or alongside it. In the beginning, we tried several different things and this ended up being what we stuck with.
For the number of the day (your TM will tell you the number of the day), my son would write it on his 100 chart. This was the first page in his math binder. Next, he would go to the page where they write out the equations and he would write the days number there. Then, he would count out the number with his beans and beans sticks. He would show me that he understood what the number meant... 12 is one bean stick and two single beans, etc . After that, he would complete the equation page. He would draw a picture of his bean sticks and beans and then complete the place value and equation section. Then, he would get his penny for the day and count the money out to me. I'm sorry I don't have this right in front of me so I hope I'm not confusing you.
After that, we would do some type of math activity. We did a variety of things. Some things that we did came from the TM and some did not. Just look through your TM and put a little pencil mark by the things you want to do. You do not have to do it all. I ended up using several commercial games that I purchased ready-made. One was called Sum Swamp. It covers addition and subtraction. The other was called Little Spender. It's about money. We used the Complete Book of Math but it was really just a supplement. I would just pick and choose pages for him to complete. I tried to pick pages that had him doing some kind of activity. He loved rolling his dice and adding the numbers together.
I also did a good bit of oral math with him. During math time and at other times throughout the day, I would ask him questions about math. Like, "If I had seven cookies and ate three of them, how many would I have left?" He really enjoyed doing this.
So, that's all we really did, I think. He is doing very, very well with math in 2nd grade and will complete it quickly. I found that the math in first grade was a perfect foundation. 1st is very focused on place value and once they understand that, everything just clicks together very well.
JL
Posts: 1
Joined: Wed Aug 02, 2006 8:26 pm
Thank you, Joy1139, for kind of 'walking me through' your math lesson. I had read the math section a couple of times, but it just wasn't clicking at all! After reading your post, and reading the section again, I can kind of 'see' how it will work.
Thanks so much!
JL
Lucy
Posts: 442
Joined: Mon Nov 08, 2004 9:37 am
### Complete Book of Math
momma2boys wrote:I have ds (8) and ds (6). I am excited to get started!! One question - I ordered the Complete Book of Math for my 6 year old - should I go through it front to back, one section at a time? Or should I skip around? Thanks!!!
Anna
Hi Anna,
In the first grade manual the topics are all cycled through twice. The manual suggest that you use a page or 2 from each topic as you move through the book. She spends more time on such topics as addition and subtraction. You will see that some topics have more pages than others so I would let this be a guide as to how many weeks you should spend on a topic. Once the topic looks like it is moving into a new level of the concept move on to the next topic saving it for the 2nd time you cycle back around.
This is all laid out in a schedule in the first grade manual. Remember too that this book is used to compliment the first grade math program and is not considered by MFW to be a complete program.The math intro. even states that the pages are supplementary and that the hands-on activities in the manual are more important. Be sure to add lots of everyday, real-life math activities for your first grader and to use the hands-on activities suggested in the Complete Book of math to teach the topics.
I hope that helps you make your plans.
Lucy
wife to Lee and mom to Twila 18 (girl) and Noel 16(boy). Happy MFW user since 2002.
Cyndi (AZ)
Posts: 543
Joined: Mon Jun 18, 2007 4:22 pm
### Help with math materials
groovymama wrote:Where is the calender? I have looked and looked.
Also, what day do we start using the yellow/blue circles?
Any information will be so helpful.
Actually, we didn't get the neat pre-printed fill-in-the-number calendars with MFW1 like we had with MFWK -- and we were BUMMED about it! We bought a big wall calendar and used it instead, but Crystal's idea of using one of those freebie calendars or one with cute pictures works great, too. My dd got one as a present that had a coloring page each month and she put it up by her school desk herself! The charts with days of the week and months of the year are good tools and easy to make.
As far as I can tell about the Math Boxes, it's your pick when to do what. I never had a firm schedule with it. (And probably didn't do a good enough job.) The TM says, "Spend some time most school days on addition and subtraction." You get to pick which hands-on activity you'd like to use. I just flipped through the Grid and didn't find mention of the Math Boxes.
2018/19: US1877
used MFW from K through WHL
cbollin
### Re: Help with math materials
[quote="groovymama""] Also, what day do we start using the yellow/blue circles?
Any information will be so helpful.[/quote]
The yellow/blue circles are one suggested hands on activity for teaching Addition and Subtraction. The detailed information about how to use them will be in the manual around pages 16-17 called Addition and Sub,traction Activities, math boxes. These are the pages just before the start of Day 1's detailed instructions.
Those hands on activities involving the yellow/blue circles will be scheduled in the Math Grid weekly schedule starting on Day 33 and then again later in the year. So, when the topic in the grid is about Addition and Subtraction is when you can do those activities in addition (a hem) to other times that you would like to do them.
-crystal
RachelT
Posts: 352
Joined: Thu Aug 03, 2006 2:45 pm
### Re: Help with math materials
Hi Daisy! If you can't find your calendars, we have found it fun to design and print calendars each month on the website below:
www.starfall.com
Rachel
Rachel, wife to Doug ~ 1995, mom to J (17) and B (15)
MFW K (twice), 1st (twice), Adv., ECC, & CtG 2006-2010,
Classical Conversations 2010-2016,
ECC/AHL 2016-17, eclectic 2017-18, WHL & US1 2018-19
http://rachelsreflections-rachelt.blogspot.com/
Lainie
Posts: 65
Joined: Sat Aug 11, 2007 2:33 am
Location: Tualatin, OR
Contact:
### Math - Help Getting Started with the math
lschalk wrote:I have no idea when to incorporate the Complete Book of Math! The only math activities I [see] in my TM are calendar, number of the day, and 100 chart.
Linda
Posted Fri Aug 15, 2008 2:47 pm by Lainie
Hi Linda,
First of all {{{big hug}}} and don't forget to breathe.
I purchased MFW 1st last year too so hopefully we are looking at the same TM.
In the first two days of the yellow section (YS), you are right that it simply refers to Calendar, Number of the Day and 100 Chart.
Right above days 3 through 7 there is a box titled Sequences. Underneath is a Hands-On activities section and Workbook section. For days 3-7 your child should complete pages 4-12 of the CBoM as well as pages 13-20. How much your child does each day is based on you and your child. The exact amount for each day isn't scheduled just the amount for the week since each child is different and capable of different lengths of seat work.
Does that make sense? After the first week, above the Day__ grid you will see a box above it with assigned CBoM pages, library book suggestions (from the list starting on page 222 of the TM) as well games to play to reinforce math (see page 220 of TM)
Don't worry!
Lainie (Oregon)
"Sanctify them in truth; Thy word is truth" John 17:17
Have completed 1st, entire 5 year cycle, and all high school! Whoo hoo!
Have graduated one with MFW, 1 dd- junior, and 1ds- freshmen
http://mishmashmaggie.com/
Poohbee
Posts: 394
Joined: Sat Feb 24, 2007 10:38 pm
Location: North Dakota
### Re: Math - Help Getting Started with the math
Posted Fri Aug 15, 2008 3:39 pm by Poohbee
Lainie explained the yellow grid section very well, I thought. I just wanted to add a couple of things.
You will cycle through all of the topics covered in 1st grade math twice. So, the first time through, you spend a week on Sequences, a week on Sorting and Classifying, 2 weeks on Numbers and Counting, a week on Comparing, etc. Along with Calendar, Number of the Day, and 100 Chart, you choose something for each day. One day, you might read a children's literature selection that relates to the topic. Another day, you might do a sheet from the CBoM. Another day, you might play a math game or do a hands-on activity for fun reinforcement. (The CBoM and the TM have great ideas for hands-on activities and games). One day a week, if you ordered the deluxe pkg., you use pattern blocks and the Pattern Animals book. (The pattern blocks and animals were my dd's favorite thing to do each week)!
And, since you will cover each topic twice, you don't have to feel that you need to use ALL of the children's lit. book suggestions or ALL of the CBoM sheets the first time around. Know that you will do some of them when you go through the topic a second time, in more depth or at a higher level.
The wonderful thing about the yellow grid pages is that it gives you a choice. It allows you to incorporate hands-on activities, children's literature, and worksheets, and you get to choose when you want to do what. I would say, don't do hands-on and a worksheet and a game all on the same day. It would probably be too much for a first grader. Mix it up a bit and do one activity one day and a different activity the next. I found this to be especially important during the first time through the topics. We tended to do more hands-on and literature the first time through and only a few worksheets. The second time through, we did more worksheets. That's just how it worked for us.
I'm sure the office will get back to you and be very helpful! Don't give up! You can do it!
Jen
happily married to Vince (19 yrs)
blessed by MFW since 2006
have used every year K-1850MOD
cbollin
kugoi wrote:Last week was box day, yay!! I got 1st for ds and neither of us can wait to get started. Anyways, I was reading through the teacher's manual and I'm a little confused about the math. This is what I'm getting.
Everyday we do the calendar, 100 chart, # of the day activities on pages 9 and 10. Then we do the appropriate hands on activities, library books, workbook pages at my discretion, keeping in mind we'll go through the cycle again. Is that right?
My confusion comes in here, is there any time that I do the activities and they aren't listed? Like it won't say every week to do a graph, but under graphs it says to do one a week. What about the addition and subtraction activities. Do I wait to do any of those until they're scheduled, starting on Day 33? Or do I start them right away? And then once we've explored the properties, I need to add in drill right?
The K and 1st grade manuals are very differently organized. I think as you get started in the first grade, it will click.
I would wait to start Addition/Subtraction on day 33. Some children will need into 2nd grade to learn those math facts. It’s ok. Drill is started on Day 33 as part of the activities. After the add/subt focus unit is completed, I would find ways to work for 2-5 minutes on some drills on as many other days as it seems right to do so. So, it would become part of my daily routine even after the 4 week focus unit is done. I’m not sure what MFW’s intent was.
{In the Kindy program you will start some basic addition and subtraction. so it's not like they will never see addition/subtraction before day 33 in first grade.)
Are there times you do the activities and they aren’t listed?
I’m assuming you mean the section called Other math Games and Activities (p.13-15 in my manual). I would wait to introduce them until they are listed in the schedule. Then, when you come to the Exploration Day (when math is lightly scheduled) you can add in for review and fun some of your child’s favorite activities done to that point. So if they’ve enjoyed one of those activities, let them play again. Exploration Day is an easy day for that because the math is lighter. (every 5th day)
On the times where the activities are listed, you will have some freedom to select and mix/match during the week. For example, on the weeks where Addition and Subtraction is the focus of the math unit, you do not have to play all 4 “games” and all of the Add/Subt activities on each day. That would be overwhelming and take too much time. Mixing and matching will allow for your child to enjoy a wide variety of stuff for those 4 weeks of Addition/Subtraction focus.
This seems to be the part you understand.
Math is set up for 5 days per day.
On one of those days, which is called Exploration Day, you have a break in the routine and for math you do the Pattern Animals block book. (you can add in some favorite activities that you have already done.)
On the other days the routine is:
*Calendar,
*Number of the Day (and it will list in the math schedule what the number of the day is. This is a little confusing because you don’t do number of the day on Exploration Day. So you’ll hit Number 100 as the number of the day on school day 124. but don’t worry. It works out.
*100 Chart
*Hands On activities (well, that starts on school day as scheduled in the grid
*Workbook Pages (also starting on school day optional
(and the real life math, which is done during life, not necessarily school time)
The hands on activities and workbook pages are scheduled in the Math Grid. All of that you understood correctly. They don't have to do all of those workbook pages the first time through, or the 2nd time either.
-crystal
Last edited by cbollin on Fri Feb 13, 2009 7:46 pm, edited 3 times in total.
cbollin
kugoi wrote:Like it won't say every week to do a graph, but under graphs it says to do one a week.
You mentioned the Weekly Graph/Chart says to "try to make a new one each week"
Several things:
*some charts will be done in the context of other math topics. One example is with the Addition/Subtraction activities to show different ways to add up to a number. So, even though Graphing is not the topic those weeks, you will have scheduled opportunities to make charts in some other math topics.
*Depending on how you read that suggestion in the section called Weekly Graph/Craft, it may be possible that Marie didn’t intend it to be a “do this once a week for the whole year.” I completely understand if you or anyone reads it that way because I did too. But if that were the case, Marie would have scheduled it that way. That’s just her style. (or no one has ever said "hey did you intend to leave that out?!)
So, that makes me look at that sentence in the context of the whole year of plans, and I wonder out loud (and could be wrong):
**make graphs/chart of many things including some that cover a week’s worth of information such as “how many days this week did it rain?” Some people might enjoy expanding it to bigger stuff such as a month's worth of information.
**If you want to do some kind of weekly graph each week of school, you have the freedom to do that. I personally plan to wait until it is scheduled and then consider doing more graphs once in a while when there is interest.
**and my interpretation is that you’d wait until the topic is introduced in the weekly math guide before doing the graphs, then as interest and time permit, try to do a new graph each week after that. (but remember, you’ll do some graphs in some other activities.) So maybe you don't have to do a new one each and every week of the whole school year.
-crystal
kugoi
Posts: 22
Joined: Sun Dec 24, 2006 12:28 pm
Thanks Crystal!
I understand what you're saying, probably because it's pretty much how I was thinking of things, but wasn't sure if that was the way it was supposed to be. I just needed someone to tell me, lol.
The good thing is that we're already pretty far along. DS can already count by 2s, 5s and 10s. He started doing it by himself several months ago, so we've been practicing since them. He's also pretty good at addition and subtraction already because we've been adding it into daily life. Even my 4yo is doing addition up to 5. So I'm not really worried about it, was just a little confused.
Mom of 7, ages 14, 10, 8, 6, 4, 1 and new baby.
Have used K, 1st, Adv, ECC, and CtoG
hollybygolly
Posts: 38
Joined: Thu Aug 24, 2006 7:42 pm
### math activities in First?
my3boys wrote:I've been planning out a schedule for the first few weeks of First. I am finding the number of suggested daily math activities overwhelming. I am assuming that I'm supposed to choose a few to do for a while and then switch it up with a few different ones? - I'm just not even sure where to start. Can someone help me sort through it?
Hi there~
I don't have my manual downstairs right now...I know the math LOOKED overwhelming to me too at first. What I had to do was read and reread those math pages and the daily grid section about 3 times thoroughly before it "clicked." From what I can remember, each day you do the following:
1. Calendar
2. 100 sheet (writing the number of the day/counting) and gluing a bean onto a stick to eventually teach tens and ones
3. That little daily math sheet where they write the number and an equation
4. Daily game or activity (using the ideas in the math section of the manual) that go along with the topic in the daily grid
5. For extra practice, you can add any pages from the Complete Book of Math that correspond to the topic being taught.
It seems like a lot, but it actually flowed very nicely! What I had to really drill into my head was that the workbook was IN ADDITION to the math, extra practice, optional, not required. It was so much easier for me to focus on the worksheets but the games and hands-on stuff is where the real learning happened. I'm sure some other mamas will be along to help you. : ) Holly
Have a blessed day loving our Savior-Holly
Mommy to: Annie and Lynne (11), Maely (8), Gracie (6) and one precious one waiting in Heaven
Currently using~MFW 1st grade (again!); Rome to Reformation
cbollin
### Re: math activities in First?
Hi Alison,
I'm wondering.. do you mean the games and things from about pages 13-17 ish?
I'm feeling the organization pain myself this time around in 1st grade math. In my case, it clicked 4 years ago. hmm... I got older and more stressed recently. <grin>
anyway. Lots of those p. 9-17 activities will actually be scheduled in the yellow pages. One such quick example: take a quick glance at yellow math schedule for days 13-22. You'll see that is a good time to play specific games from pages 13-15. When addition/subtraction is the focus such as days 98 through whatever it is, you'll get a note in the yellow grid to do stuff from pages. 16-17. And it's like that all through it. So, that will help.
my help and hint to overcome that flipping around feeling from how it is printed?
1. use the cardstock summary sheet from the student supplement material to help organize the daily routines. (that's the same stuff that is on p. 9-12 in summary format.) I will put it in a page protector in a few minutes... and it stays on the bulletin board next to my dd's desk. Those are done 4 days each week. on exploration day, you don't.
3. then turn to the instructions pages in the manual when the grid says to
Praying for both of our brains to just let it flow!
amen.
-crystal
my3boys
Posts: 149
Joined: Thu Oct 25, 2007 12:50 pm
### Re: math activities in First?
Thanks Crystal! It is the stuff on pages 9-17 that I was talking about. All those ideas are great and I want to get them in, but I know that between teaching two kids First, managing CtG for my 10yo and a bunch of therapy appointments for my 7yo - if it's not scheduled I won't do it. I'm trying to schedule out everything ahead of time so that I actually use the materials I bought this year. I haven't even looked at the student sheets so I'll have to check out that page you were talking about. I feel better knowing I'm not the only one too
Alison
Mom to 3 busy boys ages 11, 8, and 6
finished K, First, ECC, and CtG - currently using RtR
inHisgrip
Posts: 16
Joined: Thu Mar 20, 2008 3:33 pm
### Re: math activities in First?
Thank you for asking this Alison! We just purchased 1st this weekend and I've been pouring over the TM. Anyone who is worried about there not being enough math in 1st is crazy (or hasn't looked at the TM). I'm excited to be able to teach so much math in such a hands on way.
And... thanks Crystal for the reminder to take it as it comes. I always need to hear that.
Now, to finish K
- Amy
Married since 98 to my best friend
ds 2003
dd 2005
angel baby October 2008
ds 2009
ds 2014
romans8x28
Posts: 23
Joined: Fri Jun 01, 2012 10:35 am
### Mom struggling with 1st grade math
kimber79 wrote:Hi everyone! I'm preparing for MFW-1st and have been reading through the math portion and making a list of what would be required of my son daily in addition to what would be specific to each day such as the hands-on activities and workbook pages. According to the daily math activities listed in the TM, it seems like those alone will take a while to complete in order to be sure that my son understands what we're doing (he has only done the math provided in MFW-K along with supplemental games and math worksheets.) It looks like the daily math along with the other activities could take almost an hour to get through just based on my calculations. Does this sound about right?
Also, is there more of a teaching method for some concepts like counting by 2's, 5's and 10's or do we just do that each day? I'm feeling like it's a lot to grasp just in the daily activities that I don't want to rush through it and not have my son get it just for the sake of going through it all daily. I've looked in the Saxon 1 TM that was given to me as a comparison and it also includes some of the same daily activities like calendar, counting, etc. but also spreads out some of the concepts over a time period rather than having everything happen daily. Am I missing something about MFW math? Or am I just not understanding and/or over-analyzing how this is supposed to happen? The Complete Book of Math is more of just worksheets to practice concepts along with hands-on activities but isn't actually teaching concepts just practicing, or is that how it will be learned, through practicing? Thanks in advance for your help in understanding this!
What I did was break all the activities down to usually one but sometimes two a day, depending on how many were listed and if it was a one or two week time frame for the overall concept. Then, I broke down the number of pages in the CBoM to do as even number each day as possible. So far, it hasn't been taking very long. The most CBoM we've done is a front and back page so far and the pages have been really easy. Of course, we're only on day 18 so my plan might not ending up working when there are days with more pages and the concepts are getting harder. The Teacher's Manual emphasizes that the most important part of the math is the activities and if you aren't getting the CBoM done that is okay. Also, you do each concept twice--the second time at a harder level, so some of the pages from the first round can be saved for the second round, though there are new pages listed for the second round, too.
Just for example the next one week long concept is Comparing Sizes on Days 23-27. The two activities listed are Number War & Guess My Number from the TM and pages 65-70 in the CBoM. Day 23--Number War & CBoM 65; Day 24--Guess My Number & CBoM 66-67; Day 25--Exploration Day-No Math; Day 26--Number War & CBoM 68-69; Day 27--Guess My Number & CBoM 70.
Are you talking about counting by 2's, 5's, and 10's as an activity for after you've reached Day 100? I took all of those "after 100 chart is done" activities and assigned one of them for each day. (I made count by 2's a separate day from count by 5's, etc.) Use your 100 chart or find/make a blank one and fill in all the "by 2's" with one color and they can practice using the chart until they are able to say them orally without help.
Hopefully someone who has made it through the whole year will be able to chime in with their ideas, too!
Rhonda is teaching:
Girl - 9, Boy - 7, Girl - 2, Baby in September!
2011-2012 - K-1st Ed.
2012-2013 - First-1st Ed.
2013-2014 - Adventures-2nd Ed., K-2nd Ed.
2014-2015 - ECC-2nd Ed., ECC & First-2nd Ed.
kimber79
Posts: 18
Joined: Thu May 19, 2011 2:01 pm
### Re: Mom struggling with 1st grade math
Thanks for your help! I didn't realize that counting by 2's, 5's and 10's was to be done once the 100 chart had been completely filled out - that makes more sense now. I'll go back and look at the pages in the CBoM to see how you've lined things out here and see if I can figure out a plan too. Thanks!
Mom2theteam
Posts: 184
Joined: Thu Nov 11, 2010 12:33 pm
### Re: Mom struggling with 1st grade math
We are on day 29 of 1st. I'm not having an issue getting the math in as far as the amount of time it takes. The daily activities go really fast after just a couple of weeks and I can definitely tell he is learning from them even though we go through them very quickly. In fact, he usually does his calendar while I'm making breakfast. From the counter, I ask him, "Okay, what is today? What was yesterday? What is tomorrow?" Then, he writes his number in the 100 chart and adds his bean. Often, he even does his number of the day sheet while I'm making breakfast. I will go over and fill in the boxes and then he will go from there. I'll ask him how many bean sticks and how many single beans and he will draw them. I stop from breakfast for a minute to help him with the expanded number. (But, he is really not going to need help with that much longer and he barely does now.) I help him with the equations from the counter too. He does his own equation using beans as manipulative. He counts out the number of total beans he needs to come up with and then separates into two piles and counts the piles and writes the equation. (I often have him count the beans by 2's ) He really has learned quite a bit just by the daily activities already. (We also have a weather chart that he fills in.) I have mine count by 2's, 5's or 10's to the day we are on on the 100 chart. It only takes a few seconds. Doing it this way helps with the overall amount of time it takes. At first, I sat down with him to do it all. Now, I don't have to.
We are more into workbook pages than hands on activities. We do the workbook pages and then add the hands on as we have time, but you could definitely do the opposite. For the "games" I put a time limit on them. Like number war, I put a timer on for 10 minutes or so and we play for that long and stop. That keeps us to the time limit. I make sure to focus on stuff he hasn't mastered. If the planned theme is something he already has down very well, I review it for a day or two and work on something he doesn't know as well for the rest of the week. I think last week was comparing or something like that. We worked on place value for most of the week because he hadn't mastered that. We probably spend a total of 20-30 minutes on math each day in addition to the 10 minutes or so on the daily activities.
Hope that helps!
Heather
Wife to an amazing man
Mom to 6, ages 10, 7, 7, 5, 5, 3
Zack, 10 CtG
Samantha & Blake, twins, 7, CtG
Matthew & Joshua, twins, 5, MFW K
Nicholas, 3 derailing and tagging along
MelissaB
Posts: 368
Joined: Sun May 09, 2010 10:01 pm
### Re: Mom struggling with 1st grade math
Hi, Kimber -
To save time each day, we put together a 1st grade Math binder with tabs for: calendar; 100 chart; store coupons; etc. There was a clear pocket in the front of the binder with several wood shape pieces (for calendar pattern work) and the crayons/markers needed. We could flip quickly through the binder, which saved a lot of time.
It may take a few days to get into the "groove" of the MFW math, but we thought it was an excellent(!) program. By the end of the year, our dd was counting money, adding, subtracting, completely comprehending place value, counting by tens, etc..
Enjoy!
Melissa B.
Melissa B. (Arkansas)
Girls ages 16 & 13
Completed K, 1st, and Investigate {ECC; CTG; RTR; Expl.-1850; and 1850-Mod. Times}
"That they may teach the young women to be sober, to love their husbands, to love their children,.." Titus 2:4
MelissaB
Posts: 368
Joined: Sun May 09, 2010 10:01 pm
### Struggling with 1st grade math
mtmom9107 wrote:
Mon Oct 19, 2015 4:00 pm
So saying my 7 year old dd is struggling is often a gross understatement. We are using Learning God's Story as written and just started week 10. She is doing acceptably well with all other areas (reading, Bible, etc)
We have been doing all the math activities every single day they are assigned. She really isn't getting it and it's so frustrating! For example: when we do the 100 chart she can tell me what is 1 number more than whatever number i'm pointing at, and she does okay with it. But then we go to do addition and try to add 1 to something and I word it exactly the same and she is completely clueless! I break out that 100 chart again and we go over it again with the math problem and it takes 4-5 times over the same problem before she really just repeats the answer back to me that i've been telling her without actually understanding. I have also tried using beans, counting bears and the unifix cubes as visuals. She struggles with simply identifying numbers, especially the teens. 13 is very often 30 and vise versa for example. I don't think she is dyslexic because she does pretty good with her reading lessons, sure there's the normal 'b' vs 'd', but she usually catches it herself and reads fine.
Has anyone else had similar struggles that has found success with something else? Or should I just continue to push through, try to not exasperated and hope that eventually something will click.
I look forward to hearing from other Moms on this, but my first instinct is to say: Slow down. Keep moving forward.
Are you doing the workbook? I might put it aside for a few months - - just a few months! You'll be surprised how much children that age develop mentally in a short time. Do the activities she's strongest in during most of the class time, and introduce one thing she struggles about halfway through.
She WILL get it!
Melissa B. (Arkansas)
Girls ages 16 & 13
Completed K, 1st, and Investigate {ECC; CTG; RTR; Expl.-1850; and 1850-Mod. Times}
"That they may teach the young women to be sober, to love their husbands, to love their children,.." Titus 2:4
Julie in MN
Posts: 2925
Joined: Mon Jun 28, 2004 3:44 pm
Location: Minnesota
### Re: Struggling with 1st grade math
I agree with Melissa to allow more time. Like she said, you can either set it aside or keep introducing new lessons daily. Math teaching materials cannot anticipate each child's "lightbulb moment" exactly, and it's great in homeschool that we're free to slow down and speed up, to speed forward and step back. There will be many math bumps over the years, even into college, and kids will jump ahead here and need to repeat something there.
Here is some homeschool wisdom I tried to remember when my kids didn't get something:
1. Ruth Beechick used to say that the best program was "the third one" -- because by that time, the child was ready for whatever you had been trying to teach.
2. The Bluedorns (of Trivium Pursuit, a Christian classical ed book) didn't introduce formal math until age 10, and their kids were very advanced.
I remember one day on the boards we were discussing kids who weren't grasping math methods. At the time, I was a grandma of a kindergartner and observed that kids repeat over and over until they know it so well that they want to go faster. Here it is, in case that didn't make sense: http://board.mfwbooks.com/viewtopic.php ... 353#p97353
Anyways, my point is that as a grandma, I saw no need to rush; as a mom, I might have answered differently LOL.
As her mom and teacher, you both care and feel responsible -- <hugs> for that.
Julie
Julie, married 29 yrs, finding our way without Shane
(http://www.CaringBridge.org/visit/ShaneHansell)
Reid (21) college student; used MFW 3rd-12th grades (2004-2014)
Alexandra (29) mother; hs from 10th grade (2002)
Travis (32) engineer; never hs
kw4blessings
Posts: 166
Joined: Sat Feb 11, 2012 12:56 pm
### Re: Struggling with 1st grade math
Hi Laura,
I could've written your post. My oldest has struggled with math from the beginning. I totally agree with the other ladies, take it SLOW. I don't recommend switching curriculums. We've tried 5 or so and...the curriculum isn't the problem. MFW's 1st grade math, IMO, is the best way for children to learn math, especially children who struggle in this area. I've learned to slow things way down and be patient. I'll admit that I wasn't as patient as I would've liked to be in past years. I was a decent math student in school and could not wrap my mind around that fact that my sweet girl just didn't "get" it.
Some encouragement?.... She'll get it. Eventually. My dd is technically a 4th grader, but we're trucking through a 2nd grade level math book. (She doesn't know that bc we use a non-graded curriculum ) It used to freak me out that she was so "behind", but now we just move along at her pace. She still hates math for the most part, but there's always that occasional day (this morning, in fact), when she says, "You know, Mommy, math isn't SO bad. It's sometimes kind of cool to figure out the answer." Now ask me again tomorrow and it will probably be the opposite response!
Kelly, blessed mama to
sweet girl 10, busy boys 8, 6, 3
2016-17 CTG, K, and All Aboard!
mtmom9107
Posts: 6
Joined: Tue Feb 26, 2013 10:38 am
### Re: Struggling with 1st grade math
Thanks everyone! It's still going a little rough, but if anything your advise and encouragement to chill out and keep plugging along is making it not quite so rough!
Laura
Wife to my darling Jeramy for 8 years
Mommy to:
Josie~age 7 ~ Currently using: Learning God's Story.
Eva~age 4 ~ Currently using: Sonlight P4/5, Rod and Staff workbooks.
Judah~22 months
Baby girl coming early November, 2015
### Who is online
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https://www.sanfoundry.com/data-structure-questions-answers-priority-queue/ | 1,713,873,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00174.warc.gz | 871,191,757 | 21,326 | Data Structure Questions and Answers – Priority Queue
This set of Data Structure Multiple Choice Questions & Answers (MCQs) focuses on “Priority Queue”.
1. With what data structure can a priority queue be implemented?
a) Array
b) List
c) Heap
d) Tree
Explanation: Priority queue can be implemented using an array, a list, a binary search tree or a heap, although the most efficient one being the heap.
2. Which of the following is not an application of priority queue?
a) Huffman codes
b) Interrupt handling in operating system
c) Undo operation in text editors
d) Bayesian spam filter
Explanation: Undo operation is achieved using a stack.
3. Select the appropriate code that inserts elements into the list based on the given key value.
(head and trail are dummy nodes to mark the end and beginning of the list, they do not contain any priority or element)
a)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
temp.setNext(trail);
}
else
{
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(temp);
temp.setNext(dup);
}
count++;
}
}
b)
Note: Join free Sanfoundry classes at Telegram or Youtube
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
temp.setNext(trail);
}
else
{
Node cur = dup;
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(temp);
temp.setNext(dup);
}
count++;
}
}
c)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
temp.setNext(trail);
}
else
{
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(dup);
temp.setNext(cur);
}
count++;
}
}
d)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
temp.setNext(trail);
}
else
{
while((key>dup.getKey()) && (dup!=trail))
{
dup = cur
cur = cur.getNext();
}
cur.setNext(dup);
temp.setNext(cur);
}
count++;
}
}
Explanation: Have two temporary pointers ‘dup’ and ‘cur’ with ‘cur’ trailing behind ‘dup’. Traverse through the list until the given key is greater than some element with a lesser key, insert the new node ‘temp’ in that position.
4. What is the time complexity to insert a node based on key in a priority queue?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n2)
Explanation: In the worst case, you might have to traverse the entire list.
5. What is the functionality of the following piece of code?
public Object delete_key()
{
if(count == 0)
{
System.out.println("Q is empty");
System.exit(0);
}
else
{
Node dup = cur.getNext();
Object e = cur.getEle();
count--;
return e;
}
}
a) Delete the second element in the list
b) Return but not delete the second element in the list
c) Delete the first element in the list
d) Return but not delete the first element in the list
Explanation: A pointer is made to point at the first element in the list and one more to point to the second element, pointer manipulations are done such that the first element is no longer being pointed by any other pointer, its value is returned.
6. What is not a disadvantage of priority scheduling in operating systems?
a) A low priority process might have to wait indefinitely for the CPU
b) If the system crashes, the low priority systems may be lost permanently
c) Interrupt handling
d) Indefinite blocking
Explanation: The lower priority process should wait until the CPU completes the processing higher priority process. Interrupt handling is an advantage as interrupts should be given more priority than tasks at hand so that interrupt can be serviced to produce desired results.
7. Which of the following is not an advantage of a priority queue?
a) Easy to implement
b) Processes with different priority can be efficiently handled
c) Applications with differing requirements
d) Easy to delete elements in any case
Explanation: In worst case, the entire queue has to be searched for the element having the highest priority. This will take more time than usual. So deletion of elements is not an advantage.
8. What is the time complexity to insert a node based on position in a priority queue?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n2)
Explanation: In the worst case, you might have to traverse the entire list.
Sanfoundry Global Education & Learning Series – Data Structure.
To practice all areas of Data Structure, here is complete set of 1000+ Multiple Choice Questions and Answers.
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# How many sq ft in a table 8 ft by 2 ft?
Updated: 9/25/2023
Wiki User
9y ago
8 x 2 = 16 sq ft
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Q: How many sq ft in a table 8 ft by 2 ft?
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Related questions
### How many square inches are in a floor tile that measures 2 square feet?
2 sq ft = 2*144 = 288 sq inches.2 sq ft = 2*144 = 288 sq inches.2 sq ft = 2*144 = 288 sq inches.2 sq ft = 2*144 = 288 sq inches.
### How many sq ft is 2' x 8''?
Convert both units to feet and multiply: 12 in = 1 ft → 2 ft × 8 in = 2 ft × 8 ÷ 12 ft = 4/3 sq ft = 1⅓ sq ft ≈ 1.33 sq ft
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### How many yards to cover a 2.5 x 8ft banquet table?
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2
### How many sq ft are in a sq mt?
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### How many 30 sq meters equal how many sq feet?
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### How many sq ft in a circle with a diameter of 3 ft?
Area = pi*r2 where r = d/2 = 3/2 or 1.5 ft So area = 7.07 sq ft
### How many sq m of 26000 sq ft?
26 000 (sq feet) = 2 415.47904 sq meters | 593 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.911711 |
https://electronics.stackexchange.com/questions/212794/confusion-with-current-in-voltage-out-amplifier | 1,718,494,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00697.warc.gz | 205,340,896 | 46,061 | # Confusion with current in voltage-out amplifier
Why do we call some amplifiers this way? I mean lets say one call an amplifier as: "Current in voltage-out amplifier". I dont understand this because if current is in then inevitably voltage is also in.
What makes us naming the type of input here(current or voltage)?
For example is common-emitter BJT NPN amplifier voltage in voltage out? But some current is also in and some current is also out. Why choose one?
What is an example of current in voltage out amplifier? And if current is in then some voltage also must be in why we call it current in?
EDIT: Please if possible give examples with single BJT transistor amplifiers. Because that creates my current confusion.
• Why do we call some amplifiers this name? or Why does this BJT setup entail this type of amplifier? If the former, then in principal it shouldn't matter whether you use a BJT or an Op-Amp to explain this concept Commented Jan 22, 2016 at 12:24
• in bjt ce a little current is in which controls Ic but voltage is also in between base and emitter. so which one is in? Commented Jan 22, 2016 at 12:27
• The question if the BJT is a current or voltage controlled device was INTENSIVELY and CONTROVERSELY discussed elesewhere!
– LvW
Commented Jan 22, 2016 at 12:36
• "transimpedance amplifier" is a common term for them.
– user16324
Commented Jan 22, 2016 at 13:02
## 4 Answers
What is an example of current in voltage out amplifier? And if current is in then some voltage also must be in why we call it current in?
A transimpedance amplifier (TIA) is "current in" and "voltage out" circuit. The input impedance is theoretically zero ohms so if you tried to put a milli-volt at the input the current taken (theoretically) would be infinite.
They are quite commonly used for amplifying photodiodes: -
Reason: the output from a photodiode (PD) is current for a given light-power-density on the device. It produces a current out and, if you connected the PD output to a resistor, it would still try to push the same current through that resistor for the same light input. The side effect of this is that there will be a voltage produced across that resistor to keep ohms law happy.
So why not just feed the PD output to a resistor and use a voltage amplifier? You can do this but what you will find is that if the light is modulated (i.e. a typical data transmission from say a fiber optic), the presense of the resistor (and the self-capacitance of the PD) will form a low pass filter and make the data look very non-ideal.
So a TIA is used because its input impedance is zero ohms i.e. it is a current-in device with zero volts produced. With zero volts produced at the input, the self-capacitance of the PD is defeated and you can get significantly higher bandwidths thru the amplifier.
For example is common-emitter BJT NPN amplifier voltage in voltage out? But some current is also in and some current is also out. Why choose one?
Strictly speaking (with respect to the physics) a BJT is a voltage-input device and any current taken into the base is all part and parcel of the Shockley diode equation. Regarding the output, it is generally accepted that it is a current-out device because of this: -
For a given base voltage (that gives rise to a given base current), the collector current is largely constant (flat) for a vast change is colelctor-emitter voltage. This is normally where a linear amplifier uses a BJTs characteristic i.e. it's a current out device BUT that current out converts to a voltage out for a given collector resistance.
The Schockley equation for a diode tells you what current flows in the diode but a variant of this is the ebers-moll equation. This tells you the collector current for a given Vbe: -
The important thing to note is that the BJT is a voltage-in device and a current-out device. Look at the top equation - what drives collector current is Vbe and Vbc BUT Vbc is negative (collector and base reverse biased) so the dominant part of the equation is governed by Vbe and not Ib - base current is a by-product that is convenient to use because base current (as a side-effect) and collector current conveniently appear to be proportional to each other. As far as I know (and I'm no physics expert), this happy accident is just that and has "convenienced" the use of hFE as a measure of Iout/Iin.
• is there a transistor equivalent of this? Let me expound on my confusion here. Consider a BJT common emitter amplifier. What if I know both input current and voltage? Will it be voltage in or current in??? Commented Jan 22, 2016 at 11:59
• A BJT (getting down to the nitty gritty in the physics) is ultimately a voltage input device but current also flows because of the base-emitter region is a forward biased diode. An op-amp is a bunch of BJTs with negative feedback (Rf) so, in a way there kind of is a BJT equivalent. Commented Jan 22, 2016 at 12:01
• do you mean this terminology makes sense with opamps rather than transistor amplifiers? Commented Jan 22, 2016 at 12:04
• SPICE is a very versatile tool to demonstrate that a designed circuit works as desired, but it should not be used for designiung the circuit. This would be nothing else than a trial-and-error game with questionable results.
– LvW
Commented Jan 22, 2016 at 13:37
• @Andy aka-ohh no, it was my reply to user`s16307 question "do you use them when designing transistor circuits"? You can trust me, I am aware of your technical competence. Sorry, I forgot to write down to whom it was concerning (linguistic OK?).
– LvW
Commented Jan 22, 2016 at 14:07
Current and voltage are related by the impedance of whatever they are applied to. You can only chose two of voltage, current, and impedance. Put another way, there are only two degrees of freedom between these three parameters.
Suppose a amplifier had 0 input impedance, and the parameter it measured to amplify was the current. The voltage would always be 0.
In fact, it is possible to do this:
Assuming ideal parts, consider what happens when a current is dumped onto IN. That makes the - input higher than the + input, so the opamp output goes down. Eventually it goes down enough so that the current thru R1 matches the current being dumped onto IN. In fact, the current onto IN always goes thru R1, producing a voltage across it. Due to how the feedback around the opamp works, the - input is always held at the same voltage as the + input.
This circuit is a amplifier with current input and voltage output. Such a thing has the special name of transimpedance amplifier. Note that its gain is not dimensionless. The gain is the output voltage divided by the input current. Voltage divided by current has units of resistance. This makes sense since one way to look at a resistor is as a voltage to current converter.
In the above circuit, this transimpedance amplifier has a gain of -R1. If R1 is 3.3 kΩ, for example, and you put 1 mA in, then the output will be (1 mA)(-3.3 kΩ) = -3.3 V.
• This is a great example of a TIA, but the o.p. doesn't understand the difference between "current-mode" and "voltage-mode" operation. Commented Jan 22, 2016 at 12:10
• im not asking about opamps. please see my edit Commented Jan 22, 2016 at 12:16
• @user: You asked about current-in voltage-out amplifiers. I used a opamp circuit to illustrate the concept and show that these things are actually possible to useful approximations. Commented Jan 22, 2016 at 12:20
• yes great answer indeed but if you read my question im writing about CE single transistor amplifier before the edit. thanks Commented Jan 22, 2016 at 12:22
• user16307: OK, I see your point. But a general answer is not possible, because we have to define: What is the "stage" we are referring to? The BJT is regarded as a current source - however, the collector resistor produces a corresponding voltage output. Now, it is up to you to decide if you allocate this resistor to the stage or not. In some other cases the output current is used directly as a base current for another tzransistor (Example: Darlington transistor). In this case, I think it is appropriate to say: I out for the 1st transistor and I in for the second.
– LvW
Commented Jan 22, 2016 at 14:15
Why do we call some amplifiers this way? I mean lets say one call an amplifier as: "Current in voltage-out amplifier".
It's because the voltage output is, ideally, controllably correlated to the current input.
One case is when you're measuring the magnitude of a current coming in using voltage meter. You feed current-to-voltage amplifier a current then it converts it to a voltage. Offering a somewhat linear transformation.
One application for example. When you're quantifying glucose in a blood sample on a test strip, you combine an enzyme, a electroactive mediator, and a blood sample. By applying some sort of a stimulus (a voltage) with maybe an DAC. The redox reaction emits a current in a certain direction with a certain magnitude.
So then one way for an embedded software engineering to analyze it, is by using a transimpedance amplifier (current-to-voltage) amplifier. Then read it in with an ADC.
Note this is just one example, and here's microchip's friendly App Notes for glucose design
• sorry but i do not agree this way of explanation. consider a BJT common emitter amplifier. What if I know both input current and voltage? Will it be voltage in or current in??? Commented Jan 22, 2016 at 11:57
• You have to ask yourself what sort of linearity system this bjt common emitter amplifier offers. If the voltage in and voltage out are linearly correlated, then yes it's a voltage in and voltage out amplifier Commented Jan 22, 2016 at 12:01
• "It's because the voltage output is, ideally, linearly correlated to the current input." If input current and voltage are also linear what will we do? Commented Jan 22, 2016 at 12:02
• Then you my friend have discovered how to amplify power Commented Jan 22, 2016 at 12:03
• you dont get my point. Im saying if input current and input voltage are also linear it means both are linear with the output voltage. What do you say about CE transistor amplifier? is it current in or voltage in? Commented Jan 22, 2016 at 12:05
The inputs and outputs are named with respect to what is the signal-carrying property.
So an ideal voltage-in voltage-out amplifier makes the promise my output voltage will be gain times the input voltage, I will supply any current there necessary to keep it so, and this is independent of whatever current enters my input.
Then, an ideal current-in current-out amplifier will instead expect that the current entering its input is the signal, so it promises my output current will be gain times the input current, I will drive the output voltage to whatever value necessary to keep it so, and this is independent of the voltage in my inputs.
Note that the other property, i.e. current for voltage input and so on, is always unspecified, it is whatever necessary to satisfy to specified property.
The mixed combinations should obvious from these examples.
A major technical difference caused by what does the amplifier promise to do, is that the ideal input impedance of a voltage-in amplifier is infinite so no current is needed to drive it, whereas the ideal input impedance of a current-in amplifier is zero, such that an arbitrarily small voltage can be used to provide the input current. Conversely, the ideal output impedance of a voltage out amplifier is zero, so that it provides any amount of current without a change in voltage, whereas the ideal output impedance of a current-out amplifier is infinite, such that it only needs infinitely small adjustments of voltage to supply the output current. | 2,727 | 11,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920896 |
https://math.stackexchange.com/questions/444988/looking-for-a-clear-definition-of-the-geometric-product | 1,718,449,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861586.40/warc/CC-MAIN-20240615093342-20240615123342-00804.warc.gz | 359,313,540 | 58,077 | # Looking for a clear definition of the geometric product
In brief: I'm looking for a clearly-worded definition1 of the geometric product of two arbitrary multivectors in $$\mathbb{G}^n$$.
I'm having a hard time getting my bearings in the world of "geometric algebra", even though I'm using as my guide an introductory undergraduate-level2 book (Linear and geometric algebra by Macdonald).
Among the general problems that I'm running into is that most definitions and theorems that I find (either in this book, or online) seem to apply to some multivectors (e.g. to $$k$$-vectors, or to blades), not all. Sometimes it is not clear to me whether a definition or result refers to all multivectors in $$\mathbb{G}^n$$ or only to a distinguished subset (e.g. blades), since these definitions/theorems are expressed in terms the word "vector". This leads to the pervading doubt as to whether this word "vector" is being meant as synonymous with "multivector"—i.e. an object in the so-called "vector space $$\mathbb{G}^n\;$$")—, or with "$$1$$-vector", or with "$$k$$-vector", or something else entirely.
(Hence the specification "clearly-worded" in my question above. A more accurate specification would have been "unambiguously-worded", but it would have been puzzling on first encounter.)
Case in point is the definition of the geometric product in $$\mathbb{G}^n$$. Macdonald gives a very partial definition of this product for "vectors" (and only in $$\mathbb{G}^3$$)3, but far as I can tell Macdonald never defines this product in general, even though he uses it freely throughout much of the book! I find this astonishing, to put it mildly. But, please correct me if I'm wrong.
1In his answer below Alan Macdonald writes "I do not think it possible to give a quick definition of the general geometric product." In light of this remark, I want to stress that succinctness is not among the requirements in my specifications what I'm looking for.
2The original version of this post incorrectly described this book as being written for "high-school students", but the author pointed out this error in his answer below. I apologize for the (now-amended) inaccuracy.
3 On p. 82, Macdonald gives a definition for the geometric product of two $$1$$-vectors in $$\mathbb{G}^3$$, and later explicitly states: "We have defined the geometric product of two vectors, but not for example, the geometric product of a vector and a bivector. This will be taken up in the next chapter, where we will learn to take the geometric product of any two multivectors." As far as I can tell, however, the "next chapter", which is called simply $$\mathbb{G}^n$$, never fulfills this promise. Or at least, it never gives a definition for the geometric product of any two multivectors in $$\mathbb{G}^n$$.
• As an aside, let me add that in my (admittedly very brief) acquaintance with "geometric algebra" I have gotten the impression not only that it's a field is plagued with an excess of unclear wording (as described in my post), but also that it is relegated to some sort of "alternative math" ghetto. I.e. the field is kept alive by a small but feisty fringe of diehards. I wonder if these two "first impressions", to the extent that they are correct, are related somehow. Does the sheer complexity of the geometric algebra formalism necessarily keep it outside of the "mainstream" of math education?
– kjo
Commented Jul 16, 2013 at 13:08
• I just went to fetch my copy of L&GA and it has a section (6.1, page 93) on the geometric product in $\mathbb G^n$ for arbitrary multivectors. Is there something particular with this section that you find unclear? Commented Jul 16, 2013 at 15:31
• @kjo For sure, the approach geometric algebra is not "mainstream," (yet?) but it's definitely valid. Another thing that bodes well for it is that people are apparently applying it in computers. One thing to keep in mind is the extreme conservatism of modern education. They basically always stick to "the old way" because it is less controversial to do so. They just don't want to have to go through the ordeal of defending a new method. (continued...) Commented Jul 18, 2013 at 12:27
• @kjo (continued...) I've been evaluating the literature on geometric algebra for about three years now. I don't think it suffers terminology problems (much) more than any conventional field. I did find a lot of versions written by nonmathematicians hard to follow, but that is an understandable difference in exposure to certain prose. I don't know if it's quite ready for highschoolers yet, but certainly it is a very interesting approach. At least it complements the mainstream approach very well. Commented Jul 18, 2013 at 12:30
• @kjo Earlier I said "valid" but the word I was searching for was "viable." Commented Jul 18, 2013 at 12:35
There is a definition of the geometric product that applies to general multivectors in any Clifford algebra. It follows directly from the definition of the Clifford algebra. To define a Clifford algebra you need a vector space $V$ and a symmetric bilinear form $B(u,v)$ defined for any $u,v\in V$. The Clifford algebra is the quotient of the tensor algebra of $V$ with respect to the two-sided ideal generated by all elements of the form $u\otimes v+v\otimes u -2B(u,v)$ where $u,v\in V$. The geometric product is the product in the quotient algebra. It is standard and you can find the definition of that in any textbook on abstract algebra. Basically, the geometric product is the product in the tensor algebra of $V$ modulo the ideal.
AN EXAMPLE:
To illustrate, consider $\mathbb R^2$ and the bilinear form defined by $B(e_1,e_1)=1$, $B(e_2,e_2)=1$, $B(e_1,e_2)=0$, where $e_1=(1,0)$ and $e_2=(0,1)$. The two-sided ideal generated by $u\otimes v+v\otimes u -2B(u,v)$ is infinite dimensional as the tensor algebra itself. It contain the following elements among others:
$e_1\otimes e_1-1,\quad e_2\otimes e_2-1,\quad \text{and}\quad e_1\otimes e_2+e_2\otimes e_1$.
This can be used to compute the following products:
$e_1e_1 = e_1\otimes e_1=e_1\otimes e_1 -(e_1\otimes e_1-1)=1$,
$e_2e_2 = e_2\otimes e_2=e_2\otimes e_2 -(e_2\otimes e_2-1)=1$,
$e_1e_2=e_1\otimes e_2= \tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)+\tfrac{1}{2}(e_1\otimes e_2+ e_2\otimes e_1)=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
In short, $e_1^2=1$, $e_2^2=1$, and $e_1e_2=-e_2e_1$.
Even though the tensor algebra is infinite-dimensional, the quotient algebra is finite-dimensional. See what happens if you try to get to grade 3. For instance, consider this product
$e_1(e_1\wedge e_2)$ where $e_1\wedge e_2=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
It is again a straightforward application of the tensor product modulo the ideal:
$e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_1\otimes e_1\otimes e_2 - e_1 \otimes e_2\otimes e_1)$
but $e_1\otimes e_1\otimes e_2=(e_1\otimes e_1-1)\otimes e_2 +e_2=e_2$ and
$e_1\otimes e_2\otimes e_1=(e_1\otimes e_2+e_2\otimes e_1)\otimes e_1 -e_2\otimes e_1\otimes e_1=$
$=e_2\otimes e_1\otimes e_1=e_2+e_2\otimes(e_1\otimes e_1-1)=e_2$.
So, we get $e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_2+e_2)=e_2$, that is we are back to grade 1. Since the quotient algebra is finite-dimensional, every element can be expressed in term of the basis, which consists of $1, e_1, e_2, e_1e_2$ in the example we are considering. So, every multivector $A$ can be expressed as follows:
$A=s+xe_1+ye_2+pe_1e_2$.
If you have two such multivectors you can compute the product simply by using associativity, distributivity, and the properties we have derived above: $e_1^2=1, e_2^2=1, e_1e_2=-e_2e_1$.
You can easily repeat this exercise for other dimensions and for different bilinear forms.
To come back to the definition of the geometric product, here is how you can understand its significance. In geometry, you are dealing with certain geometric structures. For instance, you might want to find a line passing through two points, or you might want to find a point at the intersection of two lines. These kinds of problems can be dealt with efficiently by applying the exterior structure. You also might want to find, say, a line which passes through a given point and is perpendicular to another line. This kind of problem is related to the orthogonal structure. The tensor product is too general. By using the quotient algebra you are effectively eliminating any part of the tensor product which is not related to exterior or orthogonal structure. What is left has a clear geometric significance. In a way, the geometric product does a lot of work for you behind the curtains, so that you can concentrate on the relevant geometric structures. The expression $uv=u\cdot v+u\wedge v$ is not really the definition of the product. It is just a property that the geometric product of two vectors has.
Alan Macdonald does not use the definition of the geometric product I described above because he does not presume his readers are familiar with the tensor algebra, ideals, or quotients. Instead, he wants to concentrate on applications, geometric properties of the algebra, and on computation. If you are not satisfied with his approach, perhaps you need to read another book. This one
Clifford Algebras and Lie Theory, by Meinrenken
is recent and it uses the same definition I used. There are other equivalent ways to define Clifford algebras. If you are interested, check out these books as well
Quadratic Mappings and Clifford Algebras, by Helmstetter and Micali,
Clifford Algebras: An introduction, by Garling.
Perhaps, after trying to read these books you will appreciate Alan's book more.
Clifford algebra is a well-established part of standard mathematics. It is used in differential geometry and Clifford Analysis, not to mention various applications in physics. No one questions its validity. People who refer to it as Geometric algebra simply want to help promote it in engineering, applied mathematics, and physics. The focus is on applications rather than mathematical rigour. As Alan has pointed out, you don't need to know how the algebra is defined in general in order to use it. You can always compute the product in the basis. It gets tedious to do it by hand as the dimension of the underlying vector space increases, but it can be implemented on a computer quite easily.
1. The book is not intended for high school students. According to the preface it is intended for "the introductory linear algebra course", a sophomore college course. The preface also recommends a calculus course for "mathematical maturity".
2. Definition 5.9 defines the geometric product of two vectors. The first paragraph of Section 6.1 gives reasons for not giving a definition of the geometric product of arbitrary multivectors. (It also cites a paper which gives a definition.) Instead, Theorem 6.1 gives the fundamental rules for manipulating the geometric product of multivectors.
3. The answer from Muphrid starts by assuming that for vectors $a, b, c,\, a(bc) = (ab)c$. Of course this is true. But it cannot be used to define the general geometric product. For
$(ab)c$ = (scalar + bivector)(vector)
and (bivector)(vector) has not been defined. The answer from ahala starts by assuming that the geometric product is linear without giving a reason. Of course this is true. But an unjustified assumption cannot be used to define the general geometric product.
I do not think it possible to give a quick definition of the general geometric product.
• "I do not think it possible to give a quick definition of the general geometric product." I take your word for it, but I never asked for a "quick" definition. That said, this would be the only instance (in my admittedly narrow experience) in which such foundational specifications are not amenable to a succinct statement. (This is just to say that your statement is somewhat rattling to my provincial expectations...) Thank you for correcting my statement regarding the book's intended audience. I have reworded my post accordingly. I apologize for the error.
– kjo
Commented Jul 19, 2013 at 20:50
• And thank you also for your critique of the other answers.
– kjo
Commented Jul 19, 2013 at 20:53
• Welcome, Alan :) hope to see you around! Commented Jul 19, 2013 at 22:45
• Ignore my previous comment. You are begging the question: Of course, if you can take the product of any number of vectors, then you can take the product of, say, two. But you have not shown how to this. In my paper I first define the product AB of two multivectors. Then I can define (AB)C as the product of the multivectors AB and C. The paper is complicated. If I knew an easier way, I would use it. I know from thirty years teaching experience that ideas such as these are hard at first. I applaud you for trying to thinks these ideas through; most would simply ignore them and move on. Commented Jul 23, 2013 at 0:10
• Let's take something more familiar: the real number system. They are defined in terms of Dedekind cuts or Cauchy sequences. The definitions are complicated. But you can use real numbers without understanding (or even knowing about) the definitions. What you need to know is their properties: commutativity, associativity, etc. It is the same with multivectors and the geometric product: you can use them if you know their properties from my Theorem 6.1, even though you don't know their definitions. Commented Jul 27, 2013 at 16:49
Here is an attempt at an informal, hand-waving definition. Let's start by defining the relevant terms:
A multivector is an element of a geometric algebra. It can be represented by a linear combination of blades.
Blades are the combined geometric products between scalars and vectors of an orthonormal basis in the associated vector space. The geometric product is defined such that the range of blade products is finite in a given (finite) geometric algebra. Products with $$>n$$ vectors can always be reduced to a product with $$\leq n$$ vectors, where $$n$$ is the dimension of the associated vector space. A blade's grade is the number of basis vectors that are multiplied together (after it is reduced). A blade of grade $$k$$ is called a $$k$$-blade. Permutations of the same product end up being scaled versions of each other, so it is possible to define a set of blades as a basis for the vector space of multivectors by choosing one of each permutation set. The standard basis of $$\mathbb{R}^3$$ $$\{e_1,\, e_2,\, e_3\}$$ leads to this standard basis of blades for $$\mathbb{G}^3$$: $$\{1,\, e_1,\, e_2,\, e_3,\, e_1e_2,\, e_1e_3,\, e_2e_3,\, e_1e_2e_3\}$$.
A $$k$$-vector is a linear combinations of $$k$$-blades. Sometimes, $$2$$-vectors are called bivectors, $$3$$-vectors are called trivectors, etc. Sometimes $$1$$-vectors are simply called vectors, since they have a one-to-one correspondence with the vectors in the associated vector space.
Here are some examples of each term, using the standard basis of $$\mathbb{R}^3$$ as the defining vectors. The second examples are not in reduced form.
• $$0$$-blade : $$\:2$$, $$2(2)$$
• $$1$$-blade : $$\:e_1$$, $$\,3e_3e_1e_1$$
• $$2$$-blade : $$\:4e_1e_3$$, $$\,e_1e_3e_3e_2$$
• $$0$$-vector : $$\:3$$, $$\,3 + 2$$
• $$1$$-vector : $$\:4e_1 + 5e_2$$, $$\,6e_1 + 5e_3 - 2e_3$$
• $$2$$-vector : $$\:-13e_2e_3 + 3e_1e_3$$, $$\,12e_2e_3 + 3e_1e_3 - 3e_2e_3$$
• $$3$$-vector : $$\:-2e_1e_2e_3$$, $$\,3e_1e_2e_3 + 5e_1e_2e_3$$
• Multivector : $$\:3 + 4e_1 + 4e_3 - 13e_2e_3 + 3e_1e_3 - 2e_1e_2e_3$$, $$\,8 + 2e_1 + 4e_1e_2 - 2e_1$$
The geometric product is distributive over addition. So, to perform a geometric product between multivectors we can just represent them in the same basis of blades and distribute the terms: \begin{align} u &= u_1 + u_2e_1 + u_3e_2 + ... + u_5e_1e_2 + ... \\ v &= v_1 + v_2e_1 + v_3e_2 + ... + v_5e_1e_2 + ... \end{align} \begin{align} uv &= u(v_1 + v_2e_1 + v_3e_2 + ... + v_5e_1e_2 + ...)\\ &= u(v_1) + u(v_2e_1) + u(v_3e_2) + ... + u(v_5e_1e_2) + ...\\ &= (u_1 + u_2e_1 + u_3e_2 + ... + u_5e_1e_2 + ...)(v_1) \\ &\quad + (u_1 + u_2e_1 + u_3e_2 + ... + u_5e_1e_2 + ...)(v_2e_1) \\ &\quad + (u_1 + u_2e_1 + u_3e_2 + ... + u_5e_1e_2 + ...)(v_3e_2) \\ &\quad + ... \\ &\quad + (u_1 + u_2e_1 + u_3e_2 + ... + u_5e_1e_2 + ...)(v_5e_1e_2)\\ &\quad + ... \\ &= (u_1)(v_1) + (u_2e_1)(v_1) + (u_3e_2)(v_1) + ... + (u_5e_1e_2)(v_1) + ...\\ &\quad+ (u_1)(v_2e_1) + (u_2e_1)(v_2e_1) + (u_3e_2)(v_2e_1) + ... + (u_5e_1e_2)(v_2e_1) + ...\\ &\quad + (u_1)(v_3e_2) + (u_2e_1)(v_3e_2) + (u_3e_2)(v_3e_2) + ... + (u_5e_1e_2)(v_3e_2) + ...\\ &\quad + ... \\ &\quad + (u_1)(v_5e_1e_2) + (u_2e_1)(v_5e_1e_2) + (u_3e_2)(v_5e_1e_2) + ... + (u_5e_1e_2)(v_5e_1e_2) + ...\\ &\quad + ... \\ \end{align} We are left with a sum of products of blades. The geometric product is associative, so we don't actually need the parenthesis: \begin{align} &= u_1v_1 + u_2e_1v_1 + u_3e_2v_1 + ... + u_5e_1e_2v_1 + ...\\ &\quad + u_1v_2e_1 + u_2e_1v_2e_1 + u_3e_2v_2e_1 + ... + u_5e_1e_2v_2e_1 + ...\\ &\quad + u_1v_3e_2 + u_2e_1v_3e_2 + u_3e_2v_3e_2 + ... + u_5e_1e_2v_3e_2 + ...\\ &\quad + ...\\ &\quad + u_1v_5e_1e_2 + u_2e_1v_5e_1e_2 + u_3e_2v_5e_1e_2 + ... + u_5e_1e_2v_5e_1e_2 + ...\\ &\quad + ...\\ \end{align} These are just blades that haven't been reduced. To reduce them, we apply the identities that define the geometric product:
• scalars commute with basis vectors: $$ae_i = e_ia$$
• the product of a basis vector with itself is 1: $$e_ie_i = 1$$
• basis vectors anticommute with other basis vectors: $$e_ie_j = -e_je_i$$
Continuing our computation. First we bring the scalars together: \begin{align} &= u_1v_1 + u_2v_1e_1 + u_3v_1e_2 + ... + u_5v_1e_1e_2 + ...\\ &\quad + u_1v_2e_1 + u_2v_2e_1e_1 + u_3v_2e_2e_1 + ... + u_5v_2e_1e_2e_1 + ...\\ &\quad + u_1v_3e_2 + u_2v_3e_1e_2 + u_3v_3e_2e_2 + ... + u_5v_3e_1e_2e_2 + ...\\ &\quad + ...\\ &\quad + u_1v_5e_1e_2 + u_2v_5e_1e_1e_2 + u_3v_5e_2e_1e_2 + ... + u_5v_5e_1e_2e_1e_2 + ...\\ &\quad + ...\\ \end{align} Then we reduce the vector products into our basis blades. We do this by swapping pairs of vectors (and negate the blade every time we do) to match the order of our basis blades. When we have a product between the same vector, we can reduce it to 1 and drop it from the blade: \begin{align} &= u_1v_1 + u_2v_1e_1 + u_3v_1e_2 + ... + u_5v_1e_1e_2 + ...\\ &\quad + u_1v_2e_1 + u_2v_2 - u_3v_2e_1e_2 + ... - u_5v_2e_2 + ...\\ &\quad + u_1v_3e_2 + u_2v_3e_1e_2 + u_3v_3 + ... + u_5v_3e_1 + ...\\ &\quad + ...\\ &\quad + u_1v_5e_1e_2 + u_2v_5e_2 - u_3v_5e_1 + ... - u_5v_5 + ...\\ &\quad + ... \end{align} Finally, after we combine like terms, we are left with a new linear combination of blades that represents our resulting multivector: \begin{align} &= (u_1v_1 + u_2v_2 + u_3v_3 + ... + u_5v_5 + ...)\\ &\quad + (u_2v_1 + u_1v_2 + u_5v_3 + ... - u_3v_5 + ...)e_1\\ &\quad + (u_3v_1 + u_1v_3 - u_5v_2 + ... + u_2v_5 + ...)e_2\\ &\quad + ...\\ &\quad + (u_5v_1 - u_3v_2 + u_2v_3 + ... + u_1v_5 + ...)e_1e_2\\ &\quad + ... \end{align} In this way, we only really need to define the product between a set of basis vectors to define the entire geometric product. As Elias said, once you have these identities along with the distributive and associative propreties, the rest of it just sort of falls out. Other definitions in geometric algebra can get away with only defining operations on blades or $$k$$-vectors because they likewise distribute over addition and so trivially apply to multivectors via the same sort logic we just followed: distribute the operation then apply the definition.
However, some simplified results such as the fact that the geometric product between $$1$$-vectors equals the sum of the inner and outer product $$$$uv = u\cdot v + u\wedge v$$$$ do not extend up to multivectors by the same logic. Being unable to see the difference was one of the biggest challenges I had in learning about geometric algebra. I think not finding a reference with full definitions of each term and how they relate to each other contributed to my confusion. It's hard to know when a term is used for convenience or a restrictive reason until you can see the whole picture!
• Thanks in particular for pointing out that $uv=u⋅v+u∧v$ does not hold in general for multivectors; that got me un-stuck, since I had misread Hestene's Space-Time Algebra book (formula 3.3) and thought that was the definition of geometric product for multivectors-- it's not, and mistakenly assuming it is leads to undesirable results. Your definition works much better :-) Commented Oct 15, 2021 at 19:44
• I would suggest a minor rephrasing, in order to make your presentation seem less circular. That is, instead of saying "The geometric product is distributive over addition" and "the geometric product is associative", etc., in the middle of defining geometric product, I would say, instead, that what we are working with here are formal linear sums of the basis elements, modulo the identities that we would like to be true about the geometric product we are defining: that is, distributivity, associativity, etc. It became crystal clear to me as soon as I started thinking of it that way. Commented Oct 15, 2021 at 19:55
• Is there a meaningful geometric interpretation of the geometric product of two blades? Commented Nov 23, 2023 at 14:35
This uses the usual properties. For vectors $a, b, c$, $aa$ is a scalar, $a(bc) = (ab)c$, and $a(b+c) = ab + ac$. For a scalar $\alpha$, $(\alpha a)b = \alpha(ab)$.
• A $k$-blade is a geometric product of $k$ anticommuting vectors.
Typically this is written in terms of wedge products, which is why this can be confusing, but you can always take a wedge product of several vectors and orthogonalize those vectors to turn those wedges into geometric products.
I use the word $k$-blade here, rather than $k$-vector. Usually, the two would be considered equivalent. But here I'm just sticking to "vector" meaning "1-blade".
• Because blades are formed from some number of anticommuting vectors, the geometric product of blades is well-defined in terms of the product of vectors and associativity.
So if I have two blades $K = abc$ and $L = defgh$ where all $a,b,c$ are orthogonal and $d,e,f,g,h$ are orthogonal to each other, then $KL = (abc)(defgh)$ is a meaningful geometric product.
• A multivector is a linear combination of blades, which aren't necessarily all of the same grade. Using linearity, the geometric product of multivectors merely involves geometric products of component blades.
Let $M = ijkl$ and $N= mn$. See that $(K+L)(M+N)$ is given by
\begin{align*}(K+L)(M+N) &= KM + LM + KN + LN \\ &= (abc)(ijkl) + (defgh)(ijkl) + (abc)(mn)+ (defgh)(mn)\end{align*}
Because of linearity, we only have to consider the definition of $A_rB_s$. And Because of the associativity, $A_rB_s =aA_{r-1}B_s = a(A_{r-1}B_s)$ for some $a$ and $A_{r-1}$, we only to consider the definition of $aA_r$.
below is not a brief word definition per se, but my current understanding of the construction of geometric products.
1). starting from the axiom $u^2$ is a real scalar, we have $u\cdot v$ defined as real.
2). From $u\cdot v$, we can define orthogonality such that if $u\cdot v = 0$, $uv = -vu=u\wedge v$.
3). from 2), we can build orthogonal basis $\{e_i\}$. The geometry products of $\{e_i\}$ are well defined according to 2) above. And then $\{e_i\}$ expands to canonical basis $\{1,e_i, e_ie_j,...\}$ with geometry products defined.
4).The general definition of $aA_r$ can be obtained from the linearity and from the geometry products of canonical basis.
I get the impression from the other answers that people have two different ideas of what constitutes a "definition" of the geometric product.
The Clifford algebra of $$V$$ is usually defined as the tensor algebra of $$V$$ modulo $$v^2=\pm\left\lVert v\right\rVert^2$$, or words to that effect. This completely defines the product. Given arbitrary expressions $$A, B, C$$ from the algebra, $$AB=C$$ iff you can transform $$AB$$ into $$C$$, or vice versa, using only manipulations permitted by the axioms. You can show that any expression can be reduced to a canonical form of $$2^n$$ terms (some perhaps zero), and that expressions are equal iff they have identical canonical forms.
Alternately, you can define "Clifford numbers" as elements of $$F^{2^n}$$ with an inclusion $$V\hookrightarrow F^{2^n}$$, define multiplication as a binary operation on this space, and prove that it satisfies the algebraic rules.
This is completely analogous to definitions of other number systems; e.g. you can define complex numbers as $$\mathbb R[i]$$ modulo $$i^2=-1$$, or as $$\mathbb R^2$$ with $$(a,b)*(c,d) = (ac-bd,ad+bc)$$. The latter is useful for software implementation and for proving that complex numbers actually exist, but the former is easier to remember and is closer to how people think about them in practice.
Macdonald's book takes the view that the product is an operation on Clifford numbers, and indeed it never seems to define it (instead citing a paper of his that is freely available online). Practically speaking it doesn't matter because the rules in Theorem 6.1 are enough to work out any product. But I'm not convinced it was a good idea pedagogically.
He wrote in his answer, "I do not think it possible to give a quick definition of the general geometric product." I disagree; I think it can be defined in a sentence. It's not so easy to prove that the definition is consistent and nontrivial, but it's still a definition.
I have exactly this problem, and have resorted to Python code to do this (for orthogonal basis vectors). The geometric product algorithm should be no more than a page or so of Python code for handling an arbitrary number of orthogonal basis vectors.
This code should be easily defined for reasonably large dimensions (ie: 64) because it takes advantage of bit ordering. There are $2^n$ entries for a good reason. The reason is that the bits set indicate the use of a basis vector.
def gamul(d1,m1, d2,m2, r):
maxd = max(d1,d2)
if maxd < 0: return
r[0b000] += m1[0b000] * m2[0b000] #0b000 is scalar part
if maxd < 1: return
r[0b001] += m1[0b001] * m2[0b000] #0b001 is e1
r[0b001] += m2[0b001] * m1[0b000] #0b010 is e2
if maxd < 2: return
r[0b000] += m1[0b001] * m2[0b001]
r[0b000] += m1[0b010] * m2[0b010]
r[0b010] += m1[0b010] * m2[0b000]
r[0b010] += m2[0b010] * m1[0b000]
r[0b011] += m1[0b001] * m2[0b010] #0b011 is e1^e2
r[0b011] -= m1[0b010] * m2[0b001]
And this is an example of m4 as vector (4,6) times m5 as vector (-2,-1) written into m6:
m4[0b001] = 4 # 4*e1
m4[0b010] = 6 # 6*e2
print m4
m5[0b001] = -2
m5[0b010] = -1
print m5
gamul(2,m4, 2,m5, m6)
print m6 #array has values at entry 0b00 and 0b11, scalar+bivector
This is a brute force definition that is handling dimension 0,1,2. I could calculate 3,4,5 by brute force as well and call it a day. But I highly suspect that there is a pattern that calculates it all, does it very efficiently, and works fine even for unbounded dimensions (where d1,d2 only get passed in to limit how far the code will look for non-zero items).
The other beautiful part about this is that if I ask questions like, "is this multivector zero?" just means that all components are zero. "is this scalar" means that everything except entry 0 is zero. "is this a vector?" means that every entry not having exactly 1 bit set for its index is zero, etc. In general, the blade of an entry is the number of bits set on its index. For instance, 0b011 is a bivector, 0b111 is a trivector, 0b100 is vector e3, etc. This is why 0b000 is the scalar entry. There is probably no other reasonable ordering for this.
So, noting that every entry so far is multiplying a single entry from each m and either adding it or subtracting it is interesting. I am going to guess that the entire thing works like this. Furthermore, with this definition, the basis vectors are sorted by definition:
e1 ^ e2 #just add your value
e2 ^ e1 #sort with adjacent swaps, and number of swaps determines sign
e3 ^ e2 ^ e1
What makes the standard definition seem so vague is that typically, only vector times multivector is arbitrarily defined, and there is a lot of stringing together distributing over the components before getting down to something concrete.
A bivector is JUST a multivector in which it is asserted that all other elements are set to zero. The tricky thing about GA is that the output type depends on the input type and their values, etc. It's necessary to have a simple calculator for doing concrete calculations, without all the overhead of things like SymPy, let alone not knowing how to re-implement it in another environment (JavaScript, Go, etc).
Less than a page of Python that defines Geometric Product similar to Foundations of Geometric Algebra Computing, except using binary indexing and the xor operation to dramatically simplify the explanation. The xor (per vector, with each vector in a bit) is why the product of two vectors splits into a scalar and bivector part. 01 xor 10 is 11, 10 xor 10 is 00, 01 xor 01 is 0, etc. Dot and wedge product are not explicitly defined anywhere, yet they fall out of the definition as a side-effect (computationally, obvious without symbolic manipulation):
https://github.com/rfielding/gaMul
Mr Macdonald, thank you very much for the link correction. Your construction is very simple and it is in all similar to that proposed by Mr. John W.Bales in 2011 https://arxiv.org/abs/1107.1375v1
Now, regarding the question about a simple way for 'defining' the product of two multivectors consider the following:
1) any multivector 'a' of dimension n can be put in the form of an ordered pair (a0,a1) of multivectors of dimension n-1, such that a=a0+gamma(n)*a1. With the order of the bases gamma as proposed by Mr. Bales
2) any n-dimensional multivector 'a' can be transformed into a column vector by the recursión formula psi(a,n)=[psi(a0,n-1);psi(a1,n-1)]. The inverse transformation is also possible allways.
3) to any n-dimensional multivector 'a' one can associate the matrices:
omega1(a,n)=[[omega1(a0,n-1),gamma(n)^2*omega2(a1,n-1)];[omega2(a1,n-1),omega1(a0,n-1)]] omega2(a,n)=[[omega2(a0,n-1),-gamma(n)^2*omega1(a1,n-1)];[omega1(a1,n-1),-omega2(a0,n-1)]]
Then, the geometrical product of two general multivectors 'a' and 'b' can be calculated simply by the formula: ab=omega1(a,n)*psi(b,n) and transforming into an ordered pair by 2.
I hope you agree with these assertions. An interesting thing about this method is that all Cayley-Dickson numbers (as generalized by A.A.Albert in 1942 for any signature) can also be multiplied by a similar construction but with three matrices. If you are interested in seen this construction, please search my full name in the academia edu page. With respect.
• @kjo Maybe you would like the algorithm proposed here to calculate the product of any two multivectors in any algebra (I mean: with any number of dimensions and any signature of the imaginary bases, including complex ei^2=-1, splits ei^2=1 or duals ei^2=0, and mixing of these) Commented Jun 4, 2020 at 17:13
I recommend consulting the writings of Dr. David Hestenes and especially "Clifford Algebra to Geometric Calculus".
Geometric Calculus R&D http://geocalc.clas.asu.edu
"The most comprehensive treatment of the mathematical theory is given in the book Clifford Algebra to Geometric Calculus [1984]. Except for the last chapter added in 1979, the manuscript was ready for publication by 1976 but did not appear in print until 1984, owing to an unfortunate series of publishing mishaps. Papers with extensions and improvements in the calculus are collected in Universal Geometric Calculus[Editor: there was a hyperlink to the Universal Geometric Calculus webpage of the website here]. With general arguments and many examples, the case is made there and elsewhere that geometric algebra is a more efficient general computational tool than matrix algebra. It is therefore a candidate to supplant (or subsume) matrix algebra in computer software and software designs for scientific computation." http://geocalc.clas.asu.edu/html/Evolution.html
An informal definition may be made in terms of the basis vectors $e_1, e_2, \dots e_n$: It is the product you get by requiring that it fulfils a couple of usual rules for products, notably bilinearity, distributivity and associativity (but not commutativity), as well as the following rule: $$e_i e_j = \begin{cases} 1 & i = j \\ -e_j e_i & i \neq j. \end{cases}$$ This definition may not be entirely rigourous, but it is very simple, and in practice these are the rules you would use for calculating the geometric product of arbitrary multivectors, written as sums of basis blades.
A basic goal of the geometric product is that given the product and one of the terms used to create it the other term can be recovered. The exclusive or of logic and the symmetric difference of set theory both have this property. The geometric product is analogous but differs because it is over a vector space.
To begin consider only basis blades. These represent subspaces. A scalar represents the subspace that is the origin. A vector represents a one-dimensional subspace, an infinite line. A bivector represents a two-dimensional subspace, an infinite plane. A trivector represents a three-dimensional subspace. And so forth up to as many dimensions as you please.
First an imprecise version of the geometric product.
• Let $$s_1$$ and $$s_2$$ each be a subspace corresponding to a basis element.
• Let S be the smallest space that contains both subspaces.
• Let s be the intersection of the two subspaces.
• The geometric product of the two subspaces is the largest subspace of S that excludes s. We might write $$s_1*s_2$$=S-s.
Next in the more exact jargon of vector algebra
• Let $$s_1$$ and $$s_2$$ each be a subspace corresponding to a basis element.
• Let S be the span of $$s_1$$ and $$s_2$$. S = span($$s_1$$,$$s_2$$)
• Let s be the largest subspace that is a subspace of both $$s_1$$ and $$s_2$$.
• $$s_1*s_2$$ is the dual of s in S.
An example is $$e_{12}$$*$$e_{23}$$ = $$e_{13}$$.
• S is $$e_{123}$$
• s is $$e_2$$
• $$s_1*s_2$$ is the dual of $$e_2$$ in $$e_{123}$$, which is $$e_{13}$$.
Subspaces have no sign so this approach won't give you the signs of the terms. To get the correct sign it is necessary to do the algebra. For example $$e_{2}*e_{12}$$=$$-e_{1}$$.
Multivectors are represented as sums of weighted basis elements. Now that we know the geometric product for basis elements we just multiply everything out to get the geometric product of arbitrary multivectors. The scalar in the product is the scalar product. The sum of all the other terms is the commutator product.
Recall at the very beginning 'twas written "A basic goal of the geometric product is that given the product and one of the terms used to create it the other term can be recovered." This goal is not always achieved. In order to do this said term must have an inverse. Many multivectors don't have inverses. An example is $$e_{12}$$+$$e_{34}$$. It may appear to be a bivector but it is not. It isn't a blade. It doesn't represent a subspace. One may still algebraically produce a geometric product from any two multivectors but it doesn't always have an apparent geometric meaning. This is why there is an emphasis on blades: they can represent subspaces and are clearly geometric objects.
It often isn't obvious whether a multivector is a blade. If the product of the multivector and its dual is the space then the multivector is a blade. | 10,399 | 35,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 101, "wp-katex-eq": 0, "align": 6, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.946727 |
https://capitalsportsbets.com/ZuluCodeBets4/what-does-the--and-mean-in-sports-betting-money.html | 1,576,253,675,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00186.warc.gz | 306,142,635 | 10,966 | For beginning sports gamblers, moneylines (sometimes called money lines or American odds) can be confusing. Unlike point spreads, which are concerned with who wins and by how much, a moneyline is solely dependent upon who wins. Moneylines are used most commonly in low-scoring games like baseball or hockey, but they may also be used in boxing and other sports.
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Thanks to the modern language and a constantly updated Webster's dictionary, there is a word for pretty much anything you can think of. Furthermore, every study, no matter the obscurity has its own word to describe the resident expert. You know, words like paleontologist, botanist, astrologer, just to name a few. In the sports betting world, there are a few words that can be used to describe serious bettors who spend most of their time breaking down bets and looking for advantages that give them the best chance at winning their wager. The word I want to focus on right now is "handicapper". Read More >>
Many reasons contribute to why point spread betting is one of the most favored by NFL players, certainly one reason it is so attractive to the recreational player is that essentially you have a 50-50 proposition on every team no matter who they are playing meaning you have the opportunity to bet your favorite team no matter who their opponents and in theory have a 50% chance of winning your bet at reasonable odds. Take the 2008 Detroit Lions who went through the regular season without winning a game, now betting them on the NFL money line odds would have been a disaster whereas their point spread record for that season was 7 wins and 9 losses. Similarly the 2007 Dolphins went through the NFL regular season with only one win however their point spread record stood at 5 wins, 7 losses and 4 pushes against the closing line.
There are several very important terms and phrases you need to know if you have your sights set on becoming a sports bettor who has even the slightest idea of what you are about to bet on. Doc's Sports "How To" section has you covered for everything you need to know, but this may be one of the most important phrases to know when betting on sports that use a point-spread. Sure, money line, juice, totals, covering the spread and bad beat are all important words and phrase to know, but "laying the points" is something the general public loves to do. Read More >>
Betting “against the spread” (ATS) just means you’re betting on the point spread in a particular matchup as opposed to the moneyline, or some other type of wager. Bettors often use a team’s ATS record to gauge its performance against the spread. For example, the New England Patriots were 11-5 ATS in the 2017 regular season, meaning they covered the posted point spread 11 times, and failed to cover five times.
When it's not NFL season, BetOnline keeps on chugging along with point spreads for numerous other sports including men's + women's basketball (pro + college), along with run lines for baseball (full game + 5-inning), and they even have goal lines for several hockey leagues worldwide. BetOnline excels when it comes to betting on any sport, visit them today and give them a chance to prove it...it will not cost a cent!
Firstly you really need to understand the basics of what sports betting is all about, and what's involved with placing wagers. These basics are relatively straightforward, so thankfully it doesn't take long to get up to speed. It's definitely advisable to familiarize yourself with them though. Our beginner's guide to sports betting is the perfect resource for this. Here's a selection of some of the topics it covers.
A teaser is a bet that alters the spread in the gambler's favor by a predetermined margin – in American football the teaser margin is often six points. For example, if the line is 3.5 points and bettors want to place a teaser bet on the underdog, they take 9.5 points instead; a teaser bet on the favorite would mean that the gambler takes 2.5 points instead of having to give the 3.5. In return for the additional points, the payout if the gambler wins is less than even money, or the gambler must wager on more than one event and both events must win. In this way it is very similar to a parlay. At some establishments, the "reverse teaser" also exists, which alters the spread against the gambler, who gets paid at more than evens if the bet wins.
In the United States, it was previously illegal under the Professional and Amateur Sports Protection Act of 1992 for states to authorize legal sports betting, hence making it effectively illegal. The states of Delaware, Montana, Nevada, and Oregon—which had pre-existing sports lotteries and sports betting frameworks, were grandfathered and exempted from the effects of the Act.[6]
When betting off the board a calculation will be made according to the odds on each event in the parlay. You will win the same amount as if you bet each event separately and parlayed all winnings as you went. An exception is if every event you pick is at -110 odds, in which case predetermined nice round odds will be used. Except for a three-leg parlay, these preset odds are not as generous as if the calculation method were used. For this reason, it is a good idea to have at least one event in the parlay that isn't -110, which will force the calculation method to be used.
“Limitations. Subject to the foregoing, and subject to all of the terms and conditions of this Compact, the Tribe shall establish, at its discretion, by tribal law, such limitations as it deems appropriate on the amount and type of Class III Gaming conducted, the location of Class III Gaming on Indian Lands, the hours and days of operation, and betting and pot limits, applicable to such gaming.”
###### Straight bet - Amid all the fancy and lucrative-looking bets that are available, never lose sight of the value in a standard straight bet. You probably should learn and practice this bet often before learning any others, and it should be noted that people who bet for a living or a large portion of their income place straight bets almost exclusively.
A. It is unlawful for any person to, directly or indirectly, knowingly accept for a fee, property, salary or reward anything of value from another to be transmitted or delivered for gambling or pari-mutuel wagering on the results of a race, sporting event, contest or other game of skill or chance or any other unknown or contingent future event or occurrence whatsoever.
In today's world, online shopping is all the rave. Sites like eBay and Amazon have taken the world by storm and people do whatever they can to avoid setting foot in a retail store. Unfortunately, there will be times where you receive your order and it's completely wrong or it looks nothing like you saw in the picture. Thankfully, you have the ability to get your money back and try again. To put that situation into sports betting context, let's talk about baseball. Baseball offers a unique bet called "listed pitchers" where your bet only goes live if the listed pitchers start the game. If not, just like Amazon, you will get a refund. Read More >>
The National Football League is fully against any sort of legalization of sports betting, strongly protesting it as to not bring corruption into the game. On the other hand, the CEO of the International Cricket Council believe sports betting, in particular in India, should be legalized to curb illegal bookies where match fixing has occurred from nontransparent bookmakers. Many of the illegal proceeds also allegedly go to fund terror, drugs and other illegal activities.[citation needed]
```What this spread means is that, for the purposes of wagering, the Chargers will have 3.5 points deducted from their final score. For a bet on them to be successful, they would therefore have to win by four points or more. The Colts, on the other hand, will have 3.5 points added to their final score. A bet on them would be successful if they won the game, or lost by 3 points or less.
```
The wager is that the two teams will combine to score more or less than 43 points. If the total score is 43, then the wager is a push and you get your money back. It doesn’t matter which team scores the points. It could be 24-20 for either teams and “Over” wins the wager. Likewise, the final score could be Jets 45-0 (Yeah!) and the “Over” wins. On the flip side, the score can be 21-20 and the “Under” wins. Likewise, the Jets could win 42-0 (Yeah!) and the “Under” also wins.
For years the sports handicappers at Maddux Sports have been making money for us and our clients betting the strongest sports picks on the Internet. Maddux Sports feels that in this industry the best sports handicappers can charge less for their picks than the big name scamdicappers. Maddux Sports makes its money from repeat sports bettors that know the value of their bankroll. Featured on:
For those who are only really interested in wagering for a bit of fun, you'll be ready to go as soon as you've finished reading this page. For those of you with ambitions of making long term profits, we also point you towards a wide range of additional information and advice that will help you to achieve such goals. We're not promising that we'll make you an instant expert, but we'll certainly give you the chance to become one.
If your sports betting experience consists mostly of office pools during March Madness or a casual wager between you and a friend while you watch the Super Bowl, the transition to serious sports betting means learning how to read betting lines. The biggest difference between making the kind of casual bets mentioned above and placing wagers with online sportsbooks or at brick-and-mortar bookshops is the use of sports betting lines. Casual wagers usually involve each person in the bet picking one team to win, then wagering an equal amount, say \$20 or \$30. Professional bookmakers, online sports betting exchanges, and sports betting facilities in casinos have a more complex system for offering wagers on sporting events, in part to ensure profit on the part of the book, and in part to present a standardized representation of odds.
The basic principle of point spreads and totals is that you have roughly a 50% chance of winning, so technically a "fair" return on these types of wagers should be equal to the amount staked. However, they are usually priced up by bookmakers at odds of -110 (1.91 in decimal odds format, 10/11 in fractional odds format), which means for every \$110 staked you stand to win \$100. You don't have to stake as much as \$110 of course, but the point is that a successful wager will only return 90% of the amount staked (plus the initial stake of course).
You may often notice that the spread is sometimes set at an even number such as 3, 6 , 10, etc. In this case if the favored team won by the exact amount set for the spread the bet would be pushed, and all bets would be returned. For example, if the Patriots were 3 point favorites and they won by a FG (3 points) than this would results in a push, meaning no matter which side you bet on you would get your money returned to you.
```All individuals are banned from advertising or promoting any football betting activity in which FA regulations prohibit them from engaging. This, however, only applies to individuals in their personal capacities. For example, if a club is sponsored by a betting company and said company places its logo on the club's kit, the team's players are not in violation of the betting rules.
```
For example: If it is universally believed that Alaska State is better than Hawaii Tech, a standard win/loss wager wouldn't work, since most likely people would bet more money on Alaska State. Therefore, the Sports Book could put the point spread on Alaska State at 6.5 points. This would mean if you bet on Alaska State, not only would they have to win the game, they'd have to win by at least 7 points to make you a winner, otherwise a ticket on Hawaii Tech, even though the team might have loss, would be a winner.
Easily the most popular type of betting for NFL football is “spread” betting or more commonly known as betting against the spread. Bettors who are new to NFL betting or betting in general may be a little confused with NFL spread betting, but it is pretty easy to understand once it is explained to you. We will explain what betting against the spread means below.
Sportsbooks also offer wagering on the total or Over/Under – the cumulative number of points scored by both teams in a game. You can bet on if the final score will go either Over or Under that set total. For example, in a game pitting Team A vs. Team B, the total is 48. A wager on Over wins if the total points scored exceeds 48, and a wager on Under wins if the total points scored falls below 48. If the total lands on 48, it’s a push and wagers are refunded.
A favorite (e.g. Patriots -280) on the money line works just like our bet price example above. In our new example, the Patriots are listed at -280, meaning you would need to risk \$280 for a return of \$100 on them. It follows that a winning bet on the Pats pays \$100 (plus your initial investment of \$280 back). This added risk is why betting the spread is usually more popular, especially on favorites.
Point spreads focus on a margin of victory between the two teams and again, what you’re looking for is the positive and negative signs. If there is a minus sign next to a team’s spread, that suggest that they are favored and have to win by or cover that amount. If there is a positive sign, that tells you that they are the underdog and they are getting points. For example: let’s say that the New England Patriots are playing the Buffalo Bills and the Patriots are -5.5 and the Bills are therefore +5.5. If you bet the Patriots, they have to win by six points or more to cover. If you bet the Bills, they can lose by five points or less, or they can win the game outright and you would still win your bet.
The moneyline works differently. With this type of wager whichever team wins outright pays off. There is no spread. How does the line work? The favorite is listed with a minus sign and a number. That number is the amount of cash that must be bet in order to win \$100. The underdog is posted with a plus sign in front of a number. The number is how much a sports bettor wins on a \$100 wager.
Totalizators. In totalizators (sometimes called flexible-rate bets) the odds are changing in real-time according to the share of total exchange each of the possible outcomes have received taking into account the return rate of the bookmaker offering the bet. For example: If the bookmakers return percentage is 90%, 90% of the amount placed on the winning result will be given back to bettors and 10% goes to the bookmaker. Naturally the more money bet on a certain result, the smaller the odds on that outcome become. This is similar to parimutuel wagering in horse racing and dog racing.
Mississippi became the fourth state in the United States to launch sports betting operations on August 1, 2018 when Gold Strike Casino Resort in Tunica Resorts and Beau Rivage in Biloxi started taking wagers.[35] On August 30, 2018, West Virginia became the fifth state to launch sports betting, with Hollywood Casino at Charles Town Races the first casino to offer sports betting.[36] New Mexico became the sixth state to offer sports betting on October 16, 2018 with the launch of sports betting at the Santa Ana Star Casino in Bernalillo.[37]
The first thing you’ll notice with moneyline odds is that there is either a positive or negative sign in front of the number. What that sign denotes is how much you’ll win betting on each side. If there’s a positive sign next to the odds, that indicates the amount of money you would win if you bet \$100. If the odds on a tennis player said +150, that means that for a \$100 bet, you would win \$150. Now if there is a minus sign in front of the odds, that is the number that you would have to bet in order to win \$100. For example, if a football team was -250, that means you’d have to bet \$250 to win \$100.
```The biggest advantage of the moneyline for the NBA is that your team doesn't have to overcome the point spread for you to win your game. If your handicapping leads you to believe that one team is likely to win but you can be less certain that they will win by as much as the point spread then the moneyline may be attractive. You are sacrificing some potential return because the moneyline won't pay as much for the favorite as the point spread will, but it's obviously better to make a small profit than it is to lose a bet. This is particularly attractive in basketball because the favorites can often face large point spreads and teams can win comfortably and effectively without covering the spread.
```
Sports betting would be easy — or maybe just easier — if all that was required was to correctly pick the winning team. Gambling institutions, sportsbooks and bookies fall back on point spreads to make the process a little more difficult and to create the ultimate wagering challenge. You'll need a solid understanding of the point spread system if you hope to have a profitable season.
When you see a moneyline value associated with the point spread, it is the percentage amount you must pay in order to book the bet. Also known as the juice or vig, if you see -11.5 (-115), it means you have to bet \$115 to win \$100 — a 15 percent commission for the sportsbook. The underdog may see a value such as +11.5 (+105), which means you’ll have to bet \$100 to win \$105 if your team successfully covers the spread. | 4,852 | 22,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-51 | latest | en | 0.956863 |
https://equadratures.org/_documentation/tutorial_3.html | 1,624,574,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00408.warc.gz | 208,650,521 | 6,914 | Foundations III: Solving linear systems for model fitting¶
In this final foundation tutorial, let’s apply the tools we introduced in the previous two tutorials. Consider the Rosenbrock’s function in 2D
$f(x_1, x_2) = (1 - x_1)^2 + 100 (x_2 - x_1^2)^2$
with $$x_1, x_2$$ uniformly distributed in $$[-1, 1]$$. Suppose that we want to fit a polynomial model for this function, e.g. to estimate its output moments. In EQ, models are composed using the Poly object, which encapsulates the function
$g(\mathbf{x}) = \sum_{i=1}^P c_i \psi_i(\mathbf{x}) \qquad (1)$
where $$\psi_i(\mathbf{x})$$ are the orthogonal polynomials defined on a certain basis, as we mentioned in the last tutorial. Fitting the polynomial amounts to solving for the coefficients $$c_i$$. In practice, this is equivalent to solving a linear system
$\mathbf{Ac} = \mathbf{b} \qquad (2)$
where $$\mathbf{A}(i, j) = \psi_j(\mathbf{x}_i)$$ (similar to the Vandermonde type matrix in the previous tutorial), and $$b(i) = f(\mathbf{x_i})$$ contain function evaluations. In most practical problems, function evaluations are expensive, and we wish to obtain the most accurate model with the fewest amount of function evaluations. How many function evaluations do we need to solve the linear system? Let’s explore this question by examining two example solver methods in EQ. Before we dive in, let’s define the function and set up the preliminaries:
import equadratures as eq
import numpy as np
def rosenbrock(x):
return (1.0 - x[0])**2 + 100.0 * (x[1] - x[0]**2)**2
x0 = eq.Parameter(distribution='uniform', order=4, lower=-1.0, upper=1.0)
x1 = eq.Parameter(distribution='uniform', order=4, lower=-1.0, upper=1.0)
my_param_list = [x0, x1]
First, coefficients can be solved in EQ using numerical integration. Why integration? If we multiply the equation (1) above with $$\psi_j(\mathbf{x})$$ and integrate, we get
$\int g(\mathbf{x}) \psi_j(\mathbf{x}) dx = \int \sum_{i=1}^P c_i \psi_i(\mathbf{x}) \psi_j(\mathbf{x}) dx = c_j$
Owing to the orthogonality of the basis functions we chose, every term vanishes except for the j-th term. Thus, coefficients can be solved effectively by integration, provided that the integral is evaluated accurately. This is handled via various quadrature methods in EQ. For example, if we choose to use tensor grid, Gauss-Legendre quadrature is a proven heuristic that solves the system to a high degree of accuracy. In code, this is handled with the following.
my_basis = eq.Basis('tensor-grid')
my_poly = eq.Poly(parameters=my_param_list, basis=my_basis, method='numerical-integration')
my_poly.set_model(rosenbrock)
The first line declares the suitable basis for us to use, which is fed to the Poly object in the second line. We specify that we want to solve the coefficient via numerical integration. The third line initiates and executes the solver within the Poly object to set the coefficients in Poly.
How good is the fit? We can evaluate the goodness of fit on a select number of points to test how well our model does. The following code evaluates $$R^2$$ scores (between 0 and 1) for the goodness of fit on the training points (points used to fit the polynomial model) and some testing points which we set as a uniform grid on the input domain. The latter is a good indication of how well the model performs in the domain of interest and whether the model has overfitted—of course, at the cost of extra model evaluations.
test_pts = np.mgrid[0:1.1:0.1, 0:1.1:0.1]
test_pts = np.vstack([test_pts[0].flatten(), test_pts[1].flatten()]).T
test_evals = eq.evaluate_model(test_pts, rosenbrock)
train_r2, test_r2 = my_poly.get_polyscore(X_test=test_pts, y_test=test_evals)
print(train_r2, test_r2)
We should find that both the train and test scores are very close to 1. This is expected, since the function we are fitting is a polynomial!
Note that this method relies entirely on numerical quadrature methods. An implication of this is that we are restricted to certain types of basis, namely the tensor grid and sparse grid. In addition, one function evaluation is required per basis term we have. In this example, we needed $$5^2 = 25$$ evaluations, but in high dimensions, the number of terms of these grids can be very prohibitive. Suppose that we cannot afford 25 evaluations. Can we solve the coefficients with fewer?
Total-order least squares¶
The least squares method focuses on the linear system (2) and finds coefficients that minimises the 2-norm error, i.e. the total squared error between prediction ($$\mathbf{Ac}$$) and true function evaluations ($$\mathbf{b}$$). An immediate advantage here is that we are not restricted to any type of basis. Instead of using the tensor grid, let’s use the total order basis, which only has 15 basis terms.
my_basis = eq.Basis('total-order')
X_train = np.random.uniform(-1.0, 1.0, size=(20, 2))
y_train = eq.evaluate_model(X_train, rosenbrock)
my_poly = eq.Poly(parameters=my_param_list, basis=my_basis, method='least-squares',
sampling_args={'mesh':'user-defined', 'sample-points':X_train,
'sample-outputs':y_train})
my_poly.set_model()
Note the other salient difference here: we are free to set the points where we evaluate the function. A simple—but often effective—approach is to simply select points randomly according to the input distribution, i.e. the uniform distribution. The selected points and corresponding function evaluations are fed to the Poly object instead of the function itself.
How many function evaluations do we need? Here, we used 20. For least squares generally, the number of evaluations needs to be at least larger than the number of basis terms. The more function evaluations used, the more stable the solution is going to be against small perturbations. Again, we can see how this model performs by using the get_polyscore method.
test_pts = np.mgrid[0:1.1:0.1, 0:1.1:0.1]
test_pts = np.vstack([test_pts[0].flatten(), test_pts[1].flatten()]).T
test_evals = eq.evaluate_model(test_pts, rosenbrock)
train_r2, test_r2 = my_poly.get_polyscore(X_test=test_pts, y_test=test_evals)
print(train_r2, test_r2)
which should give us values close to 1 again.
Conclusions¶
This tutorial showcased two solution methods in EQ, but this merely scratches the surface of the many coefficient solving strategies in EQ. To mention a few more examples:
• Compressed sensing allows us to surpass the restriction that we need more function evaluations than basis terms, but we need to assume some special structure in the solution.
• Elastic net promotes sparsity in the solution, which can improve the model’s generalisation capabilities.
• Subsampling strategies (such as QR column pivoting) allow us to maximise the utility of a limited number of function evaluations.
• Dimension reduction methods exploit special low-dimensional structures in functions to drastically reduce the number of function evaluations required to fit a model.
In the other tutorials, these methods are explored in further details. | 1,731 | 6,997 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-25 | latest | en | 0.732517 |
http://dict.youdao.com/w/eng/tolerance_zone/ | 1,591,515,459,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00054.warc.gz | 33,270,460 | 9,245 | go top
## tolerance zone
• [机] 公差带；公差范围
### 网络释义专业释义
[机] 公差带
... 5 有关公差和公差带的术语及定义 殴观碾昝喙纪浴薯怦仙厦披厨卤邹框狍艋诘疑漏汀传妨亠锭蠊肱弁绽绦青届犹蕉镣脾镤耸嗌溥薏远掏黏凋汀耦悖餮 30 公差带(tolerance zone)是由代表上极限偏差和下极 限偏差或上极限尺寸和下极限尺寸的两条直线所限定的 一个区域。
公差区
... 公差单位 tolerance unit 公差区 tolerance zone 公差区位置 tolerance zone position ...
差范围
fit tolerance zone 配合公差带
更多收起网络短语
• 公差带 - 引用次数：11
When there are several tolerance on one feature, the constrain relationship between each tolerance zone was established.
当同一要素存在多个形位公差要求时，建立各公差带之间的约束关系。
参考来源 - 基于数学定义的圆柱要素公差数学建模与分析技术的研究
• 公差带 - 引用次数：4
In this meantime, also introduces the relative tooth and trace error & tolerance zone, designed tooth and trace, the curvilinear figure of the designing tooth & trace.
同时，文中还介绍了与此相关的齿形、齿向误差及公差带，设计齿形、齿向和设计齿形、齿向曲线图。
参考来源 - 渐开线圆柱齿轮鼓形齿齿形和齿向误差的测量(一) A Measurement Error of the Tooth Profile and Tooth Trace for the Involute Cylindrical Gear's Crowned Teeth(Ⅰ)
·2,447,543篇论文数据，部分数据来源于NoteExpress
### 柯林斯英汉双解大词典
#### tolerance zone
• 1.
N an designated area where prostitutes can work without being arrested 红灯区
• [机]公差带；公差范围
### 双语例句权威例句
• Technical drawings. Tolerancing of orientation and location. Projected tolerance zone.
技术图纸定向定位公差投影公差
• The concept of tolerance zone comes from the documents of service management and consumption behavior.
容忍概念来自服务管理消费行为两方面文献
• Information service can be broken down into different classification while different service classification has different management of tolerance zone.
信息服务分为不同类型不同服务类型容忍管理不同
\$firstVoiceSent
- 来自原声例句 | 639 | 1,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-24 | latest | en | 0.398387 |
https://openpress.usask.ca/introtoappliedstatsforpsych/chapter/16-3-paired-sample-sign-test/ | 1,725,906,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00223.warc.gz | 403,604,172 | 24,396 | 16. Non-parametric Tests
# 16.3 Paired Sample Sign Test
Here we have two measurements from each subject, typically before and after. If the difference between measurements is , assign a , if , assign a , if 0 assign a 0. (Be sure to keep the direction of subtraction consistent with the hypothesis.) We again have 2 cases, for small () and large () samples, as with the median sign test. The critical and test statistics are the same as the median sign test. We’ll work through an example with a small sample.
Example 16.5 : We have the following data on number of ear infections on swimmers before and after taking a medication that is hypothesized to prevent infections :
Swimmer Infections before, Infections after, Difference () A 3 2 + B 0 1 – C 5 4 + D 4 0 + E 2 1 + F 4 3 + G 3 1 + H 5 3 + I 2 2 0 J 1 3 –
In the last column, we have assigned when , when and when . We are interested in reduced infections so is “good” for this situation. Test if the reduction in infections is significant.
1. Hypothesis.
: MD difference 0
: MD difference 0
2. Critical statistic.
Use the Sign Test Critical Values Table with = (no. of ) + (no. of ) = 7 + 2 = 9 and with a one-tailed test to find
3. Test statistic.
4. Decide.
so do not reject .
5. Interpretation.
There is not enough evidence to say that there is a reduction in the number of infections. | 360 | 1,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.912837 |
https://ece.vt.edu/undergrad/courses/3004.html | 1,723,554,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00606.warc.gz | 174,519,776 | 16,262 | # ECE 3004 - AC Circuit Analysis (3C)
Course Description
Application of the basic laws and techniques of circuit analysis to ac circuits. Complex numbers and algebra with an emphasis on phasor representation of circuits. Calculation of the frequency response of circuits with R, L, and C components, independent sources, controlled sources, and operational amplifiers. Analysis of AC steady-state circuits and determination of average power. Magnetically coupled circuits. Laplace and Fourier transforms. Representation of circuits by two-port models.
Why take this course?
Circuit analysis and design using discrete R, L, and C components is the most fundamental skill for electrical engineers. This course stresses the AC analysis of basic circuit elements and teaches modeling, design, and analysis skills for AC circuits. It also covers such topics as: magnetically coupled circuits and their application to transformers; frequency response and active and passive filters; the application of Laplace transforms and Fourier series and transforms to analyzing circuits; and two-port circuits.
Learning Objectives
• Manipulate complex numbers and develop the equations for AC circuits using phasors
• Apply Kirchhoff’s laws and Ohm’s law to analyze RLC circuits in the frequency domain
• Calculate power and energy in single and multiple frequency AC circuits
• Determine the transfer characteristics of common AC filter circuits
• Describe and calculate the response for circuits with mutual coupling and with transformers
• Apply Laplace and Fourier transforms in the analysis of AC circuits
• Model basic circuit structures with 2-port equivalent circuits
• Model and analyze basic circuits using MatLab and PSpice | 322 | 1,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.857767 |
https://www.jiskha.com/display.cgi?id=1223250168 | 1,503,054,594,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104634.14/warc/CC-MAIN-20170818102246-20170818122246-00047.warc.gz | 915,838,174 | 4,075 | # math
posted by .
The greater of two numbers is 1 more than twice the smaller. Three times the greater exceeds 5 times the smaller by 10. Find the numbers.
The second of three numbers is 1 less than the first. The third number is 5 less than the second. If the first number is twice as large as the third, find the three numbers.
These are two separate problems and we are supposed to write an equation for each and solve it for the numbers, but only using one variable!! HELP!
• math -
L-1=2S
4L-10=5s
I will critique your work on the second.
• math -
but we are only supposed to use 1 variable??
• math -
The second of three numbers is 1 less than the first. The third number is 5 less than the second. If the first number is twice as large as the third, find the three numbers.
• math -
What is 2 plus 2?
## Similar Questions
1. ### Math
Can someone please help me write this statement out in formula form. There should be two formulas. Question.... Three times the larger of two numbers is equal to four times the smaller. The sum of the numbers is 21. Find the numbers. …
2. ### algebra
the larger of two numbers is 1 more than twice the smaller.the sum of the numbers is 20 less than three times the larger.find the numbers
3. ### Reiny! Reiny! Reiny! Answer Check and HW Help!
Translate to a system of equations and solve: 1) The sum of two numbers is 36. One number is 2 more than the other. Find the numbers. Answer: (17,19) 2) Find two numbers whose sum is 66 and whose difference is 18. Answer: (37,29) These …
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5. ### MATh ( translate words into number)
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How would you set this problem up and work it: The difference between two numbers is 16. Three times the larger number is seven times the smaller. What are the numbers?
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the greater of two numbers is 1 more than twice the smaller. three times the greater exceeds 5 ties the smaller by 10. find the numbers and write the equation. please help me
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Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3 of the smaller and 1/5 of the greater numbers are together 21.The numbers are:
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Five times the larger of two numbers plus four times the smaller is 271. Three times the larger less twice the smaller is 57. find the numbers
10. ### algebra
Three times the difference of two numbers is 3 more than the larger and three times their sum is less than 8 Times the smaller by 2 .find the numbers
More Similar Questions | 713 | 2,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-34 | latest | en | 0.927404 |
https://www.embibe.com/exams/cbse-ncert-solutions-for-class-10-maths-chapter-15/ | 1,716,972,825,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00256.warc.gz | 642,354,441 | 68,036 | • Written By Abhishek_verma
# NCERT Solutions for Class 10 Maths Chapter 15 Probability
NCERT Solutions for Class 10 Maths Chapter 15 Probability: The NCERT Solutions for Class 10 Maths Chapter 15 deals with Probability. It provides insight into important questions students must follow to prepare for the exams. Academic experts design all the NCERT Solutions as per the latest syllabus. Therefore, students can practice these solutions thoroughly to score better marks in the exam.
Probability is quite a fun and important chapter to study. Students will learn about elementary and complementary events, probability of different events, probability of a sure event and other related concepts. Since Probability is not a theoretical topic, students can practice over 1000+ questions at Embibe. They must practice the in-text questions provided by Embibe to enhance their preparation for exams. Read the article to know more about NCERT Solutions for Class 10 Maths Chapter 15 Probability.
## NCERT Solutions for Class 10 Maths Chapter 15: Important Topics
Probability is one of the most fun chapters to study in Class 10. This is the first time students will be preparing for this chapter. Therefore, there are chances that the students prepare for the exams. It is advisable that the students focus on every chapter thoroughly. The NCERT Solutions for Class 10 Maths Chapter 15 will make it easier for the students. Furthermore, students need to understand the preparation strategy accordingly.
The Class 10 Maths Chapter 15 has only one unit with over 1000+ questions that will make it easier for the students to prepare for the exam. These questions are curated from 40 books. Therefore, students can start their practice. Students may refer to the Embibe Explainers and Embibe 3D videos for exam preparations. Embibe also provides NCERT Exemplars. Therefore, the students can focus on all the subtopics and start their preparation.
### NCERT Solutions for Class 10 Maths Chapter 15: Points to Remember
• In the theoretical approach to probability, we try to predict what will happen without performing the experiment.
• An elementary event is said to be favourable to compound event A if it satisfies the definition of the compound event.
• The sum of the probabilities of all the outcomes (elementary events) of an experiment is 1.
• An outcome of a random experiment is called an elementary event.
• An event A associated with a random experiment is said to occur if any one of the elementary events associated with event A is an outcome.
Students can get more important points on Class 10 Maths Chapter 15 from Embibe.
## NCERT Solutions for Class 10 Maths: All Chapters
The detailed NCERT Solutions for Class 10 Maths for all chapters are given below:
## FAQs on NCERT Solutions for Class 10 Maths Chapter 15
Q.1: Which topics are covered in Class 10 Maths Chapter 15?
Ans. The Class 10 Maths Chapter 15 introduces students to Probability and its related concepts.
Q.2: How can students score good marks in Class 10 Maths Chapter 15 in exams?
Ans. Students have to practice the topic efficiently and consistently to ensure that they score well in the exams. For Class 10 Maths Chapter 15, students can solve practice questions for Probability on Embibe. Solving these questions will help students to secure good marks.
Q.3: What are the key benefits of NCERT Solutions for Class 10 Maths Chapter 15 by Embibe?
Ans. NCERT Solutions for Class 10 Maths Chapter 15 by Embibe offers solutions to all the exercise questions in the NCERT class 10 Maths Chapter 15. All the questions provided in this article are solved by experts of Embibe. The problems are solved in a step-by-step manner to enable better understanding amongst students.
Q.4: How many questions are there in Class 10 Maths Chapter 15?
Ans. Embibe provides over 1000+ questions for Class 10 Maths Chapter 15.
Q.5: Can I take mock tests for Class 10 Maths?
Ans. Yes, students can mock tests for Class 10 for. These mock tests are available on Embibe. | 856 | 4,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.926869 |
http://www.i-programmer.info/bookreviews/218-data-science/9401-bayesian-methods-for-hackers.html | 1,529,424,576,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863100.8/warc/CC-MAIN-20180619154023-20180619174023-00073.warc.gz | 429,588,442 | 12,999 | Bayesian Methods for Hackers
### New Book Reviews!
Bayesian Methods for Hackers
Author: Cameron Davidson-Pilon
Publisher: Addison-Wesley
Pages: 256
ISBN: 978-0133902839
Print: 0133902838
Kindle: B016060UHA
Audience: Statisticians wanting to know about probabilistic programming
Rating: 3
Reviewer: Mike James
Probabilistic programming with Bayesian inference could be the next ground breaking technology, so a book on the topic is welcome.
The book's subtitle is fairly accurate "Probabilistic Programming and Bayesian Inference". The problematic part of the title is in the use of the term "Hackers". What exactly is a "hacker" when used in connection with advanced statistics and programming? From the contents of the book I have to deduce that it means a reader wanting to use advanced ideas without the need to dig deep into the theory - and especially into the math. In fact there is a lot of math in this book but it doesn't take a leading role in anything. Instead it is introduced where it cannot be avoided. This is a shame as personally I find clear mathematics to be the best way of explaining and understanding any complicated or difficult idea. The author clearly could have taken a different approach as math seems to leak out more and more as the book goes on.
The most important thing to say about this book is that it is on an important topic. Bayesian statistics is an alternative approach to the problem of inference in uncertain conditions. There are huge philosophical difficulties with the Bayesian approach. Many Bayesians are happy to work with probability as if it was a measure of belief.
Adherents of the most quoted alternative approach, the frequentists, only use probability in its more simple and direct physical interpretation. A probability is the long run frequency of an event. A frequentist would never apply a probability to an event that can only occur once. To them the question "what is the probability of life on other planets" is not about probability, but to a Bayesian it is.
There are lots of other problem with the Bayesian approach. For example, you start with how much you believe something and convert this into an estimate of your new belief after incorporating some data that either makes your belief stronger or weaker. The problem is how much is your initial belief? This is the problem of selecting a "prior" and it is both practically and philosophically difficult.
All of this deep thinking and worrying about the philosophical difficulty of the Bayesian approach is described in just a few pages. This isn't adequate, but if you are a "hacker" it is just possible that all you want to get on with using the techniques rather than worrying about the deeper idea.
What the first chapter does is to give you a small experience of using Bayesian reasoning along with PyMC - a Python library for probabilistic programming. Providing just a flavor of a topic is typical of the book, even if it is sometimes hard to detect. If you put the philosophical problems aside, the real problem with Bayesian stats is working things out. Probabilistic programming, which is based on a fairly recent approach called Monte Carlo Markov Chain or MCMC, makes general Bayesian modelling much, much easier. In fact, you could say that it has, and is, bringing about a revolution in the usefulness of the Bayesian approach. The example used is interesting, especially if you are a frequentist, because of the ease with which a complex question is quickly formulated as a Bayesian model. However, if you are hoping for a clear explanation of how PyMC works, or should be used, then you are going to be disappointed. You are given the flavor, but this is not a clear systematic explanation.
Chapter 2 promises more on PyMC but it is still confused. Part of the problem is terminology. When the term "variable" is used it could mean statistical variable, Python variable or PyMC "variable" which is in fact an object/class. I'm sure it is possible to explain the logic that is inherent in PyMC, but this chapter, and overall this book, doesn't do it. At best you are going to be getting a rough idea about PyMC from using it.
By Chapter 2 you also encounter another problem with the book's approach. Most of the descriptions try to avoid deep mathematical explanations by making general observations about how things work and yet there are lots of equations in the text which you are expected to understand. I think that if you can understand the equations you could also benefit from having them used in explanations of how it all works. Math doesn't have to be obscure.
Chapter 3 moves on the inner workings of MCMC - only it doesn't. The description of MCMC is recognizable as such only if you already have some idea what MCMC is all about. Trying to fathom in what sense things "converge" to a solution, or even what generating samples is all for, is difficult. Then to move on and discuss ways of improving convergence or diagnosing convergence is a bit of a waste of time. Of course, if you do have an idea of what MCMC is all about then it's, surprisingly useful information.
From here the book becomes increasingly idiosyncratic and personal. Chapter 4 introduces the law of large numbers as a guiding principle. Then on to matters that connect the frequentist approach to Bayesian practice and how small samples can mislead. Chapter 5 introduces the idea of a loss function, but in a half-hearted way with most of the math stripped out - but not enough of the math to make it easy to read.
Chapter 6 deals with the problem of what prior to use and the distinction between objective and subjective priors is introduced. The emphasis seems to be on how to pick a good prior but the discussion doesn't go far enough for you to solve the problem - that's because there isn't a solution that everyone agrees on. The end of the chapter suggests that the prior you actually choose doesn't matter as long as the sample size is large. The idea that Bayesian models are often used in an iterative context, and in this case you can think of the prior as a sort of transient behavior that dies away, isn't really discussed.
The final chapter focuses on the very specific application of A/B testing and the Bayesian t test in particular.
You might think that I hated this book. Not so. There was a lot I enjoyed reading and it made me think. In particular, the examples did succeed in making me understand how powerful PyMC is and probabilistic programming in general. However, I start out being reasonably well versed in statistical theory - I'm a frequentist who is happy to use Bayes methods. I am not so clear on MCMC and PyMC is new to me. As a result I got a lot from reading this book, but it was very clear that if I knew nothing about statistics then I would be struggling to understand the important ideas.
For a potential practitioner with a good background in stats, this book gives a taste of what it is like to go down the probabilistic programming road. The sad thing is that if it had embraced the math more whole-heartedly the book would be much more useful. Only buy a copy of this book if you already know a lot of statistics and want to move into Bayesian probabilistic programming and still be prepared to do a lot of additional reading around the topics mentioned.
To keep up with our coverage of books for programmers, follow @bookwatchiprog on Twitter or subscribe to I Programmer's Books RSS feed for each day's new addition to Book Watch and for new reviews.
SQL and Relational Theory (3rd Edition) Author: C J DatePublisher: O'ReillyPages: 582 ISBN: 978-1491941171Print:1491941170Kindle: B017S0YPOG Audience:SQL DevelopersRating: 4Reviewer:Kay Ewbank This updated edition of a classic on relational theory has added NoSQL to the mix. + Full Review Modern Web Development Author: Dino EspositoPublisher: Microsoft PressPages: 448 ISBN: 978-1509300013Print: 1509300015Kindle: B01C33LL6IAudience: ASP .NET developersRating: 5Reviewer: Mike James Modern web development from a Microsoft point of view is a mystery to many. Can this book [ ... ] + Full Review More Reviews
Last Updated ( Tuesday, 02 February 2016 )
RSS feed of book reviews only RSS feed of all content Copyright © 2018 i-programmer.info. All Rights Reserved. Joomla! is Free Software released under the GNU/GPL License. | 1,799 | 8,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-26 | latest | en | 0.949937 |
http://americanhistory.si.edu/collections/search/object/nmah_694664 | 1,475,236,481,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662166.99/warc/CC-MAIN-20160924173742-00011-ip-10-143-35-109.ec2.internal.warc.gz | 8,786,577 | 12,740 | # Painting - Geometric Mean (Archytas)
Description
This painting demonstrates a construction for finding the geometric mean of two line segments credited to the Greek mathematician Archytas (flourished 400–350 BC), an admirer of Pythagoras. Place the line segments end to end, and draw a circle with this length as radius. Erect a perpendicular at the point where the line segments meet (d in the figure), and consider this to be the altitude of a right triangle inscribed in the semicircle. By similar triangles, the length of the perpendicular of a triangle inscribed in a semicircle is the geometric mean of the two lengths into which it divides the diameter of the circle. Hence the length of d is the mean of the segments e and f.
This painting an orange-red background, and shows a triangle inscribed in an orange semicircle. The perpendicular from the right angle of the triangle divides the triangle into triangles similar to it, painted in black and white.
The painting, and the attribution of the theorem to Archytas, are based on a passage from Evans G. Valens, The Number of Things: Pythagoras, Geometry and Humming Strings (1964), p. 118. The figure on this page of this book from Crockett Johnson's library is annotated.
This oil painting on masonite is #65 in the series. It is inscribed on the back: GEOMETRIC MEAN (ARCHYTAS) (/) Crockett Johnson 1968. It has a wooden frame.
Location
Currently not on view
Object Name
painting
date made
1968
referenced
Archytas
painter
Johnson, Crockett
Physical Description
masonite (substrate material)
wood (frame material)
Measurements
overall: 38.4 cm x 62 cm x 2.5 cm; 15 1/8 in x 24 7/16 in x in
ID Number
1979.1093.40
accession number
1979.1093
catalog number
1979.1093.40
subject
Art
Science & Mathematics
Crockett Johnson
See more items in
Medicine and Science: Mathematics
Crockett Johnson
Data Source
National Museum of American History, Kenneth E. Behring Center
Credit Line
Ruth Krauss in memory of Crockett Johnson
Additional Media
## Visitor Comments
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https://www.papercamp.com/essay/65586/Supply-And-Demand-Simulation-Eco-365 | 1,560,955,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00177.warc.gz | 857,555,439 | 5,606 | # Supply and Demand Simulation Eco 365
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Supply and Demand Simulation
ECO 365
Supply and Demand Simulation
Goodlife Management is a company which rents two-bedroom apartments in the town of Atlantis on a month-to-month basis. (University of Phoenix, 2012) The simulation illustrates the change in supply and demand of the rentals, the implementation of price ceilings, changes in population, and changes in consumer’s preferences. This paper will discuss the key facts listed above as well analyze the results of the simulation.
Change in Supply and Demand he first scenario of the simulation dealt with how to lower the vacancy rate while maximizing revenue. (University of Phoenix, 2012) It was suggested that a vacancy rate of 15% would be ideal however I determined that at a rental rate of \$950 dollars a month I could lower my vacancy rate to 5% thus maximizing my amount of revenue at \$1.81 million. This scenario implements the law of demand. “Quantity demanded rises as price falls, keeping other things constant”. (University of Phoenix, 2012)
The second scenario proposes leasing 2,500 two-bedroom apartments at the most profitable rate. (University of Phoenix, 2012) The current rental rate is \$ 1, 100; this rate proved not to be the most profitable. Raising the rental rate to \$ 1, 550 thus allowed me to increase the quantity of our supply of rooms to 2,500 as proposed. This utilizes the law of supply; which states that quantity supplied rises keeping all other things constant. (Colander, 2010, p.239) Also indicating an upward slope of the supply curve.
Equilibrium Price
Scenario three presented making a decision of determining the equilibrium monthly rental rate for a two-bedroom apartment on a temporary lease. (University of Phoenix, 2012) Lowering the current rental rate to \$ 1,050signified a surplus of 0 thus making quantity supplied and quantity demanded reach the point of equilibrium of 2,000. (University of Phoenix, 2012)Equilibrium price is identified as that price/point where quantity... | 483 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-26 | latest | en | 0.887194 |
https://www.jiskha.com/questions/1422914/a-solution-of-acetic-acid-1-0-mol-l-1-ch3cooh-contains-sodium-acetate-nach3coo-the | 1,638,143,149,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00554.warc.gz | 908,697,334 | 5,459 | # Chemistry
A solution of acetic acid, 1.0 mol L–1 CH3COOH, contains sodium acetate, NaCH3COO. The percent ionization of the above solution is 0.25%. The Ka for acetic acid is 1.8 x 10–5. Which of the following is the concentration of NaCH3COO in acetic acid?
So i use Ka = [H+][CH3COO] / [CH3COOH]
after this im guessing
but am i to go 1 - 0.0025 = 0.9975
and then, sub for ka =
1.8 x 10–5. = (0.0025x)^2 / 0.9975x
1. 👍
2. 👎
3. 👁
1. Your error is in thinking the Ac is the same as HAc. They are equal in a solution of acetic acid but the problem tells you that NaAc has been added. Now the solution has H^+ one number and the Ac is whatever you've added. So make that
Ka = (0.0025)(Ac^-)/(0.9975)
and solve for Ac^-. That will be the concn of the sodium acetate in that solution.
1. 👍
2. 👎
so after solving for [CH3OO-] = 7.182x10^-3
but the answer provided is, any ideas?
4.7 x 10–3 mol L–1
1. 👍
2. 👎
## Similar Questions
1. ### College Chemistry
A buffer solution contains 0.120M acetic acid and 0.150M sodium acetate. a. How many moles of acetic acid and sodium acetate are present in 50.0 ml of solution? b. if we add 5.55 mL of 0.092M NaOH to the solution in part (a) how
2. ### chem
Which of the following could be added to a solution of sodium acetate to produce a buffer? ? A) acetic acid only B) acetic acid or hydrochloric acid C) potassium acetate only D) sodium chloride or potassium acetate E) hydrochloric
3. ### chemistry-
How do I construct an ICE table for this question. CH3COOH + NaOH --> CH3COONa + H2O Fill in the equilibrium line of the table. [Think about the initial pH and think about the pH at the end point of the titration. Initial pH 2.72
4. ### Chemistry
An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.01 mol of acetic acid. How many moles of sodium acetate will you need to add to the solution? The pKa of acetic acid is 4.74.
1. ### Chemistry
Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of
2. ### Buffer Solution
You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 93 mL of 1.00 M NaOH solution?
3. ### chemistry
A buffer is prepared using acetic acid, CH3COOH, (a weak acid) and sodium acetate, CH3COONa (which provides acetate ions, the conjugate base), according to the following proportions: Volume of CH3COOH(aq): 131.0 mL Concentration
4. ### chemistry
A buffer is prepared using acetic acid, CH3COOH, (a weak acid, pKa = 4.75) and sodium acetate, CH3COONa (which provides acetate ions, the conjugate base), according to the following proportions: Volume of CH3COOH(aq): 100.0 mL
1. ### school
What is the pH of 0.1 M formic acid solution? Ka=1.7„e10-4? What is the pH value of buffer prepared by adding 60 ml of 0.1 M CH3COOH to 40 ml of a solution of 0.1 M CH3COONa? What is the pH value of an acetate buffer (pK=4.76)
2. ### chemistry
The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is __________. The Ka of acetic acid is 1.76 × 10-5.
3. ### Gen CHEM 2
I need to produce a buffer solution that has pH 5.28. I have already 10 mmol of acetic acid. How much solid sodium acetate will i need to add to this solution? pKa of acetic acid is 4.74. I plugged in all my info to the
4. ### Biochemistry
Show Calculations used in mkaing 250mL of 0.5M sodium acetate ph 4.7. You are given: 1M acetic acid (weak acid) 1M sodium hydroxide (strong base) Sodium acetate buffer: ph 4.7 acetic acid pKa: 4.7 So far this is what I've gotten: | 1,221 | 3,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-49 | latest | en | 0.87084 |
https://www.cleariitmedical.com/2020/10/angular-displacement-velocity-and.html | 1,643,140,939,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00461.warc.gz | 726,274,163 | 119,597 | ## Angular Displacement, Velocity and Acceleration-Advanced
Compared to other sections, Physics is considered to be the most scoring section. If prepared thoroughly, Physics can help students to secure a meritorious position in the exam. These questions are very important in achieving your success in Exams after 12th.
Q1. A particle B is moving in a circle of radius a with a uniform speed u. C is the centre of the circle and AB is diameter. The angular velocity of B about A and C are in the ratio
• 1 : 1
• 1 : 2
• 2 : 1
• 4 : 1
1 : 2
Q2. A strap is passing over a wheel of radius 30 cm. During the time the wheel moving with initial constant velocity of 2 rev/sec. comes to rest the strap covers a distance of 25 m. The deceleration of the wheel in rad/s2 is
• 0.94
• 1.2
• 2.0
• 2.5
0.94
Q3. A particle starts rotating from rest. Its angular displacement is expressed by the following equation where is in radian and t is in seconds. The angular acceleration of the particle is
• 0.5 rad/sec2 at the end of 10 sec
• 0.3 rad/sec2 at the end of 2 sec
• 0.05 rad/sec2 at the end of 1 sec
Q4. The planes of two rigid discs are perpendicular to each other. They are rotating about their axes. If their angular velocities are 3 rad/sec and 4 rad/sec respectively, then the resultant angular velocity of the system would be
Q5. A sphere is rotating about a diameter
• The particles on the surface of the sphere do not have any linear acceleration
• The particles on the diameter mentioned above do not have any linear acceleration
• Different particles on the surface have different angular speeds
• All the particles on the surface have same linear speed
The particles on the diameter mentioned above do not have any linear acceleration
Q6. When a ceiling fan is switched on, it makes 10 rotations in the first 3 seconds. How many rotations will it make in the next 3 seconds (Assume uniform angular acceleration)
• 10
• 20
• 30
• 40
30
Q7. When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many more rotations will it make before coming to rest (Assume uniform angular retardation)
• 36
• 24
• 18
• 12
12
## Want to know more
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Angular Displacement, Velocity and Acceleration
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true
7783647550433378923
UTF-8 | 937 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-05 | longest | en | 0.86315 |
https://gis.stackexchange.com/questions/324381/statistical-downscale-of-raster-with-sum-function | 1,563,869,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00248.warc.gz | 403,462,060 | 34,573 | # Statistical downscale of raster with sum function
Usually raster dowscaling methods exploit interpolation techniques. However, I am not sure how to proceed (in R, QGIS, GEE, or Python) if a raster needs to be downscaled (=i.e. its resolution increased, e.g. from 1 km to 100 m) such that the sum of the underlying pixels yields the value of the original lower resolution pixel. Of course I would like to to this in a way that not all pixels underlying the coarser resolution pixel have the same value, but based on the distribution of pixels on another raster layer (i.e. with a statistical regression approach).
the easiest approach is to increase the resolution of your raster (i think downscaling is an awkward term myself) and then to perform the regression or redistribution or further analysis on your new raster with the secondary data.
When restricting the higher resolution cells to be the sum of the original cell, in R it's very straightforward;
``````require(raster)
# a dummy raster of 10 x 10 cells
r <- raster(xmn=0, xmx=100, ymn=0, ymx=100, ncol=10, nrow=10)
r[] <- sample(1:4, 100, replace=T)
# disaggregation factor, in this case we're going to increase the res by 10
fact <- 10
# disaggregate and increase resolution
# by not setting an interpolation method, we force the new cells to hold the value of the 'parent' cell
r.inc <- disaggregate(r, fact)
# so you need to divide the higher res cells by the disaggregation factor squared
r.inc <- r.inc/fact^2
# check
> cellStats(r,sum) == cellStats(r.inc,sum)
[1] TRUE
``````
then go on to perform your regression, glm etc using a stack | 395 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-30 | latest | en | 0.856925 |
http://mymathforum.com/economics/329945-finding-effective-annual-rate-financial-management-class.html | 1,566,529,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00322.warc.gz | 135,234,135 | 9,488 | My Math Forum Finding Effective Annual Rate (Financial Management Class)
Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance
April 3rd, 2016, 01:23 AM #1 Newbie Joined: Apr 2016 From: Korea Posts: 1 Thanks: 1 Finding Effective Annual Rate (Financial Management Class) This is a Perpetuity question and I'm having a hard time finding the EAR(Effective Annual Rate) It states the following: Interest rate: 6 % compounded semi-annually $100 payed every three months per year So how would you calculate the EAR in this situation? The amount of cash flows per year is greater than the number of compounding per year (PP>CP) Please help........ Thanks from manus April 3rd, 2016, 05:30 PM #2 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Quote: Originally Posted by samkb8 Interest rate: 6 % compounded semi-annually$100 payed every three months per year So how would you calculate the EAR in this situation? The amount of cash flows per year is greater than the number of compounding per year (PP>CP)
The EAR calculation has nothing to do with the frequency of the payment:
r = .06 (annual rate/100)
n = 2 (compounding periods)
EAR = (1 + r/n)^n - 1
= (1 + .06/2)^2 - 1
= 1.0609 - 1
= .0609 which means 6.09%
If the payment frequency does not match the compounding frequency,
then the rate used in order to pay interest on payment dates needs to
be calculated, and must result such that it equals the EAR:
EAR = .0609
p = 4 (payment frequency)
k = required rate
(1 + k)^p = 1 + EAR
1 + k = (1 + EAR)^(1/p)
k = (1 + EAR)^(1/p) - 1
k = 1.0609^(1/4) - 1
k = 1.014889... - 1
k = .014889... which means ~1.4889%
In other words, crediting the interest every 3 months using
rate of ~1.4889 results in effective annual rate of 6.09 %
Tags annual, class, effective, financial, finding, management, rate
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Contact - Home - Forums - Cryptocurrency Forum - Top | 654 | 2,304 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-35 | latest | en | 0.902731 |
https://devsolus.com/2023/08/14/iterating-pandas-df-finding-sum-after-loop/ | 1,696,225,542,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00253.warc.gz | 220,588,312 | 26,933 | # Iterating Pandas DF, Finding Sum After Loop
I’m trying to find the sum of the variable "candle" in the code below, but not sure how to write the code. I know someone is going to suggest using NumPy, but I don’t even know where to begin with that.
``````for index, row in df.iterrows():
openPrice = row['open']
closePrice = row['close']
if closePrice - openPrice >= 0:
candle = 0
elif closePrice - openPrice <= 0:
candle = 1
``````
I’m looping through a dataframe that contains stock prices (open and close). I want to find the difference between the two and find the sum of the times the output is =1. Let’s say the rows are: 1, 0, 0, 1. The sum should be 2.
### >Solution :
Consider following dataframe:
`````` open close
0 20 10
1 20 30
2 20 30
3 20 5
``````
Then you can do:
``````df["col3"] = (df["close"] - df["open"] < 0).astype(int)
final_sum = df["col3"].sum()
print(df)
print(f"{final_sum=}")
``````
This prints:
`````` open close col3
0 20 10 1
1 20 30 0
2 20 30 0
3 20 5 1
final_sum=2
`````` | 362 | 1,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.840644 |
https://endless-sphere.com/forums/viewtopic.php?f=3&t=100189 | 1,566,608,099,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00118.warc.gz | 442,036,165 | 12,303 | 36v vs. 48v Range Question.
General Discussion about electric bicycles.
Electric Earth 1 W
Posts: 59
Joined: Nov 11 2018 8:02pm
36v vs. 48v Range Question.
I apologize in advance because I'm sure this has been answered, but I'm having trouble searching in such a way as to find the answer. Assuming you drove the same course under the same conditions, same motor efficiency, etc, would a 48v 1000w motor use more battery power than a 36v 500w motor simply because it's a bigger, more power hungry motor? Or is it more like, since it's a set course, it would take the same amount of energy to complete under the same conditions, regardless of motor? Or would the 48v be more efficient since it could do things like climb small hills encountered in real world riding more easily? If you're trying to optimize your spending to get the most mileage, is there a benefit to a 36v 500w vs. 48v 1000w system?
motomech 1 GW
Posts: 3287
Joined: Sep 11 2010 12:21am
Location: Punta Cana Baja Mexico
Re: 36v vs. 48v Range Question.
Your problem is you are trying to evaluate an electric motor system as if it is an internal combustion engine system and they are quite different.
You certainly could spend time studying Ohm's Law and things like Induction and Capacitance for your own edification, but for the sake of brevity here, I will simply point out the "real World" considerations that are important when building an Ebike system.
First off, the motor rating is just that. It's a guide to the amount of Power (Watts= Volts X Amps) that can can applied to it (load), over time, before it starts to produce enough heat to damage it. This is very involved and includes Voltage, Amps, ambient temps, load and RPM and perhaps the best way to start to wrap your head around of this system works is to spend time at the simulator @ EbikesCA, plugging in variables and studying the results.
A more practical consideration of motor rating along the "how much power" line is that experience has shown that the manufacturer's rating can be exceeded 1 1/2 to 2 times, if not continuous, for a substantial amount of time.
Motomech
'03 Rocky Mountain Edge 2WD 260 Q100H frt and Ezee V1 rear 2 Elifebike 20A & 25A 9-FET controllers 12S/10Ah Multistar Lipo rear 4Ah Turnigy frt Luna Cyclops Extra lite Alex 24DM rims, Crazy Bobs run ghetto tubeless. 25 mph. Mean Well HLG-320H-54A
https://endless-sphere.com/forums/viewt ... =3&t=83430
'07 GT Idive 4 4.0, Q100C 201 14S LiPoly elifebike 9-FET 20A controller. 23 MPH.
https://endless-sphere.com/forums/viewt ... 4#p1378484
wturber 1 MW
Posts: 1896
Joined: Aug 23 2017 8:52pm
Location: Fountain Hills,AZ
Contact:
Re: 36v vs. 48v Range Question.
The bike that has a motor and/or drive train that is best matched to the course and speed the course is ridden would be the most efficient. The voltage run doesn't tell you much if anything about general efficiency.
You optimize efficiency by operating the motor at efficient motor RPMs, by staying below 20 mph and/or using a recumbent and/or fairings so that wind resistance doesn't consume lots of power.
You can run simulations here to see how things work with different motors, controllers, bikes, conditions, etc.
https://www.ebikes.ca/tools/simulator.html?
"Commuter - DC Booster"
Iron Horse 3.0 hardtail - 48V / 1000W / 470rpm generic Chinese DD Hub motor (ebay)
8 x 36v 4.3ah 10s 2P battery packs - 1500W 30A DC Boost Converter delivers 54v and about 1000 watts peak
53T/42T Sakae Road cranks - 30mph+ on flats
viewtopic.php?f=2&t=90369
motomech 1 GW
Posts: 3287
Joined: Sep 11 2010 12:21am
Location: Punta Cana Baja Mexico
Re: 36v vs. 48v Range Question.
The question of system efficiency is way over-rated for your "run of mill" ebike and the differences are so small that an extra oz. or two of battery capacity can make it up on a 15 or 20 pound battery pack. So little as to be inconsequential. Suffice to say, for what it is worth in terms of system efficiency, it's better to run a lower speed range motor of higher voltage, than run a higher speed range motor on lower Voltage.
Matching the motor speed range to the desired over the road speed by applying the correct Voltage is the real reasoning behind selecting a system Voltage.
The majority of first ebike builders are probably looking to cruise in the low to high 20's mph range and, by chance have selected a mid-speed range motor which is going into a 26" whl. This dictates a 48 Volt pack which, w/ a geared motor would produce a low to mid 20's ebike or w/ a DD motor, would be in the mid to high 20's mph.
Since to cost and complication of a 36V vs a 48V system is basicly the same, there is little reason to select a 36 V system unless the bike is to be ridden on multi-use trails or boardwalks only.
Since range is a matter of Watts, the only real way a 36V pack can offer any real range advantage is, one will be moving slower and conserving in that regard.
There are cases, say if one is forced to use a low-speed range motor or a sm. rear whl., where the extra cost of a 52 V pack would be worthwhile to bring up the speed.
Going above 52V's not only increases the cost, but complexity as well, generally increasing at an exponential rate as one moves up.
So study as you care to for your knowledge, but follow the Speed/motor speed/Volts/ whl. size rule, w/ an eye to what expected hills are like, in your build.
P.S. Wturber sneaked in between my posts, but basicly, I'm elaborating on his statements.
Motomech
'03 Rocky Mountain Edge 2WD 260 Q100H frt and Ezee V1 rear 2 Elifebike 20A & 25A 9-FET controllers 12S/10Ah Multistar Lipo rear 4Ah Turnigy frt Luna Cyclops Extra lite Alex 24DM rims, Crazy Bobs run ghetto tubeless. 25 mph. Mean Well HLG-320H-54A
https://endless-sphere.com/forums/viewt ... =3&t=83430
'07 GT Idive 4 4.0, Q100C 201 14S LiPoly elifebike 9-FET 20A controller. 23 MPH.
https://endless-sphere.com/forums/viewt ... 4#p1378484
Electric Earth 1 W
Posts: 59
Joined: Nov 11 2018 8:02pm
Re: 36v vs. 48v Range Question.
Thanks, guys, for the replies.
Your elaboration was what really got at the things I was wondering. I'm glad you did so. It all makes more sense now, and now I know more what I need to research in order to decide on upgrading my system.
It's crazy how much there is to learn with these things. "Oh, you're interested in an ebike? Don't suppose you have a degree in electronics? If not, you will soon!"
"You want a new battery, huh? Do you know about all of the different aspects of the different ratings? After you learn all that, you could build your own battery. You just need to get a spot welder, and learn about circuit boards and BMS, etc, etc." Five hours later, still barely scratched the surface... After that you can start on motors. Oh, and good luck on the controllers...
I've loved bikes my whole life, and the regular mechanical ones are so simple to deal with. Now that I've decided to start into ebikes, I can search and read endlessly on this forum learning and trying to answer my questions. Half the time it just leads to a deeper question... It's fun, but I feel like I just found my last hobby for a lifetime. This one is gonna take me until I'm dead to learn about.
wturber 1 MW
Posts: 1896
Joined: Aug 23 2017 8:52pm
Location: Fountain Hills,AZ
Contact:
Re: 36v vs. 48v Range Question.
The trick is in getting a few fundamentals down and then keeping it simple. There is a ton of detail that you can learn about motors and how they work, but what you need to know from the standpoint of buying or building a bike isn't really all that much. The same goes for batteries and controllers. Search and read through these forums and you'll learn more than you need to know.
"Commuter - DC Booster"
Iron Horse 3.0 hardtail - 48V / 1000W / 470rpm generic Chinese DD Hub motor (ebay)
8 x 36v 4.3ah 10s 2P battery packs - 1500W 30A DC Boost Converter delivers 54v and about 1000 watts peak
53T/42T Sakae Road cranks - 30mph+ on flats
viewtopic.php?f=2&t=90369
dogman dan 100 GW
Posts: 34929
Joined: May 17 2008 12:53pm
Location: Las Cruces New Mexico USA
Re: 36v vs. 48v Range Question.
If you ride the same, same speed, same pedaling, same weather, etc, you can ride the 48v motor quite efficiently. Same energy used.
But often, you will ride faster, take off from stops faster, and use more. Your choice to ride different, or not. I often ride a 2000w 48v bike, which uses a very heavy motor, at 200-300w. But having that 2000w there when a hill gets steep results in better efficiency up that hill, vs overheating a tiny motor. This bike is mostly ridden for exercise in fact, but it can also tow a heavy trailer if I want that. It does not make me less efficient when I ride it slow on flat ground at all.
Depending on the total load on the system, a big motor can be much more efficient than a tiny overloaded one. Overkill your motor if you weigh a lot, have really tough hills, or tow trailers.
donn 1 kW
Posts: 364
Joined: Aug 13 2018 10:30am
Location: Seattle
Re: 36v vs. 48v Range Question.
Electric Earth wrote:
May 14 2019 8:18pm
It's fun, but I feel like I just found my last hobby for a lifetime. This one is gonna take me until I'm dead to learn about.
Only to the extent you're so inclined. The other way to approach is to just have some confidence that as all these different configurations are going out there and working for people, they are all OK for general purposes. Geared hub, direct drive, crank drive, which is best? The industry can't decide, they use them all. 36V, 48V? Etc. After using one for a while, you'll have a lot better basis for knowing what suits your particular needs, and all the information will start to fit into place in that perspective.
I think maybe recognizing that, one important part of the initial setup is a good display that gives you data you need to understand how things are working out - power usage at the moment in watts, and average in watt-hours per mile; peak amperes draw, stuff like that. Then you know what your experience means.
Dave Sloan 100 W
Posts: 128
Joined: Jun 22 2009 9:05am
Re: 36v vs. 48v Range Question.
I know that when I bought my hub motor that they were all rated at 36v 500w = at least that what was stamped on the outside. Most of the people that bought these ran 56v or better batteries. These hubs all came with stainless steel spokes so that you could probably do 3500w which you may need if climbing a tree. Efficiency wise - my hub is much more efficient at any voltage now that the bearings have slightly loosened up after 10000 miles. I can run this hub now at 25 miles an hour continuously at 90 degrees with very little heat dissipation. If you live in flat country and do not haul a lot then 36v is fine but with hills and load and better control then go the extra mile and get 48v (56v lifepo4) and enjoy 10 plus years of riding. Also go for a good bike and battery carrier and seat = look at the whole equation.
Best of five Ebikes is Etrek 820 with
front 9c - 48v 20ah LiFePO4 on rear
Schwinn spring seat - Rockshox seat post
25a GM controller. | 2,976 | 11,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | longest | en | 0.959078 |
http://mathhelpforum.com/calculus/130810-separation-variables-problem.html | 1,524,339,294,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945317.36/warc/CC-MAIN-20180421184116-20180421204116-00163.warc.gz | 199,894,306 | 10,779 | 1. Separation of Variables Problem
Hi. I am trying to solve dy/dx= -6y - 8.
I get:
-1/6Ln (-6y-8)=x +c.
Dividing by -1/6, you get Ln (-6y-8)= -6x-6c.
I next take the e of both sides to get -6y-8= e^(-6x-6c).
How do I proceed from here, as the correct answer is y= -4/3+ce^-6x?
2. Originally Posted by jaijay32
Hi. I am trying to solve dy/dx= -6y - 8.
I get:
-1/6Ln (-6y-8)=x +c.
Dividing by -1/6, you get Ln (-6y-8)= -6x-6c.
I next take the e of both sides to get -6y-8= e^(-6x-6c).
How do I proceed from here, as the correct answer is y= -4/3+ce^-6x?
Try dividing both sides by $\displaystyle 3y + 4$.
So you get
$\displaystyle \left(\frac{1}{3y + 4}\right)\frac{dy}{dx} = -2$
$\displaystyle \int{\left(\frac{1}{3y + 4}\right)\frac{dy}{dx}\,dx} = \int{-2\,dx}$
$\displaystyle \int{\frac{1}{3y + 4}\,dy} = -2x + C_1$
$\displaystyle \frac{1}{3}\ln{|3y + 4|} + C_2 = -2x + C_1$
$\displaystyle \frac{1}{3}\ln{|3y + 4|} = -2x + C$, where $\displaystyle C = C_1 - C_2$
$\displaystyle \ln{|3y + 4|} = -6x + 3C$
$\displaystyle |3y + 4| = e^{-6x + 3C}$
$\displaystyle |3y + 4| = e^{3C}e^{-6x}$
$\displaystyle 3y + 4 = \pm e^{3C}e^{-6x}$
$\displaystyle 3y + 4 = Ae^{-6x}$ where $\displaystyle A = \pm e^{3C}$
$\displaystyle 3y = Ae^{-6x} - 4$
$\displaystyle y = \frac{A}{3}e^{-6x} - \frac{4}{3}$.
Now by letting $\displaystyle \frac{A}{3}$ equal some other constant, the problem is solved.
3. How are you seperating this equation? Did your instructor say you HAVE to use seperation of variables? The differential is linear:
$\displaystyle \frac{dy}{dx}+6y=8$
$\displaystyle \mu = e^{\int 6dx}\Rightarrow \mu = e^{6x}$
$\displaystyle (y*e^{6x})'=\int 8e^{6x}dx$
$\displaystyle y*e^{6x}=\frac{4}{3} e^{6x}+C$
$\displaystyle y=\frac{4}{3}+ Ce^{-6x}$
The thing about these differentials, is that more often than not there are several ways of solving it. Some are way more easier to solve than not.
4. Actually, I would consider "separating variables" to be easier than finding an integrating factor!
jaijay32, from -6y-8= e^(-6x-6c), you can separate the exponential as -6y- 8= e^(-6x)e^(-6c). Since c is an "unknown constant", e^(-6c) is also some unknown constant so just call it "C": -6y- 8= Ce^(-6x).
(In fact, that is better: the integral of 1/x dx is ln|x|, not ln(x). What you should have got was ln|-6y-8|= ln|6y+8|= -6x- 6c so |6y+ 8|= e^(-6x-6c)= e^(-6x)e^(-6c). The point is that, since an exponential is always positive, "-6y- 8= e^(-6x)e^(-6c)" implies that -6y- 8 is positive. But what you really have is |6y+8|= e^(-6x)e^(-6c) so that 6y+8 can be either positive or negative. Writing "-6y- 8= Ce^(-6x)" or "6y- 8= Ce^(-6x)" includes C being either positive or negative.)
Now, with -6y- 8= Ce^(-6x), -6y= 8+ Ce^(-6x) and then y= -4/3+ (C/-6)e^(-6x)= -4/3+ C'e^(-6x) with C'= C/-6.
If you had written it 6y+ 8= Ce^(-6x), you would have 6y= -8+ Ce^(-6x) and then y= -4/3+ (C/6)e^(-6x) or y= -4/3+ C"e^(-6x), exactly the same answer with C"= C/6 now. | 1,216 | 2,980 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-17 | latest | en | 0.733977 |
https://neo4j.com/docs/graph-algorithms/current/labs-algorithms/harmonic-centrality/index.html | 1,579,706,008,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607118.51/warc/CC-MAIN-20200122131612-20200122160612-00077.warc.gz | 579,534,512 | 6,493 | ### 9.2.6. The Harmonic Centrality algorithm
This section describes the Harmonic Centrality algorithm in the Neo4j Labs Graph Algorithms library.
Harmonic centrality (also known as valued centrality) is a variant of closeness centrality, that was invented to solve the problem the original formula had when dealing with unconnected graphs. As with many of the centrality algorithms, it originates from the field of social network analysis.
The Harmonic Centrality algorithm was developed by the Neo4j Labs team and is not officially supported.
This section includes:
#### 9.2.6.1. History and explanation
Harmonic centrality was proposed by Marchiori and Latora in Harmony in the Small World while trying to come up with a sensible notion of "average shortest path".
They suggested a different way of calculating the average distance to that used in the Closeness Centrality algorithm. Rather than summing the distances of a node to all other nodes, the harmonic centrality algorithm sums the inverse of those distances. This enables it deal with infinite values.
The raw harmonic centrality for a node is calculated using the following formula:
`raw harmonic centrality(node) = sum(1 / distance from node to every other node excluding itself)`
As with closeness centrality, we can also calculate a normalized harmonic centrality with the following formula:
```normalized harmonic centrality(node) = sum(1 / distance from node to every other node excluding itself) / (number of nodes - 1)```
In this formula, ∞ values are handled cleanly.
#### 9.2.6.2. Use-cases - when to use the Harmonic Centrality algorithm
Harmonic centrality was proposed as an alternative to closeness centrality, and therefore has similar use cases.
For example, we might use it if we’re trying to identify where in the city to place a new public service so that it’s easily accessible for residents. If we’re trying to spread a message on social media we could use the algorithm to find the key influencers that can help us achieve our goal.
#### 9.2.6.3. Harmonic Centrality algorithm sample
The following will create a sample graph:
``````MERGE (a:Node{id:"A"})
MERGE (b:Node{id:"B"})
MERGE (c:Node{id:"C"})
MERGE (d:Node{id:"D"})
MERGE (e:Node{id:"E"})
The following will run the algorithm and stream results:
``````CALL algo.closeness.harmonic.stream('Node', 'LINK') YIELD nodeId, centrality
RETURN nodeId,centrality
ORDER BY centrality DESC
LIMIT 20;``````
The following will run the algorithm and write back results:
``````CALL algo.closeness.harmonic('Node', 'LINK', {writeProperty:'centrality'})
Calculation:
`k = N-1 = 4`
``` A B C D E
---|-----------------------------
A | 0 1 2 - - // distance between each pair of nodes
B | 1 0 1 - - // or infinite if no path exists
C | 2 1 0 - -
D | - - - 0 1
E | - - - 1 0
---|------------------------------
A | 0 1 1/2 0 0 // inverse
B | 1 0 1 0 0
C |1/2 1 0 0 0
D | 0 0 0 0 1
E | 0 0 0 1 0
---|------------------------------
sum |1.5 2 1.5 1 1
---|------------------------------
*k |0.37 0.5 0.37 0.25 0.25```
Instead of calculating the farness, we sum the inverse of each cell and multiply by `1/(n-1)`.
#### 9.2.6.4. Huge graph projection
The default label and relationship-type projection has a limitation of 2 billion nodes and 2 billion relationships. Therefore, if our projected graph contains more than 2 billion nodes or relationships, we will need to use huge graph projection.
Set `graph:'huge'` in the config:
``````CALL algo.closeness.harmonic('Node', 'LINK', {graph:'huge'})
#### 9.2.6.5. Cypher projection
If node label and relationship type are not selective enough to describe your subgraph to run the algorithm on, you can use Cypher statements to load or project subsets of your graph. This can also be used to run algorithms on a virtual graph. You can learn more in the Section 2.2, “Cypher projection” section of the manual.
Set `graph:'cypher'` in the config:
``````CALL algo.closeness.harmonic(
'MATCH (p:Node) RETURN id(p) as id',
'MATCH (p1:Node)-[:LINK]-(p2:Node) RETURN id(p1) as source, id(p2) as target',
{graph:'cypher', writeProperty: 'centrality'}
);``````
#### 9.2.6.6. Syntax
The following will run the algorithm and write back results:
``````CALL algo.closeness.harmonic(label:String, relationship:String,
{write:true, writeProperty:'centrality', graph:'huge', concurrency:4})
Table 9.34. Parameters
Name Type Default Optional Description
label
string
null
yes
The label to load from the graph. If null, load all nodes.
relationship
string
null
yes
The relationship type to load from the graph. If null, load all relationships.
write
boolean
true
yes
Specifies if the result should be written back as a node property.
concurrency
int
available CPUs
yes
The number of concurrent threads used for running the algorithm. Also provides the default value for 'readConcurrency' and 'writeConcurrency'.
int
value of 'concurrency'
yes
writeConcurrency
int
value of 'concurrency'
yes
The number of concurrent threads used for writing the result.
writeProperty
string
'centrality'
yes
The property name written back to.
graph
string
'huge'
yes
Use 'huge' when describing the subset of the graph with label and relationship-type parameter. Use 'cypher' for describing the subset with cypher node statement and relationship statement.
Table 9.35. Results
Name Type Description
nodes
int
The number of nodes considered.
int
evalMillis
int
Milliseconds for running the algorithm.
writeMillis
int
Milliseconds for writing result data back.
The following will run the algorithm and stream results:
``````CALL algo.closeness.harmonic.stream(label:String, relationship:String, {concurrency:4})
YIELD nodeId, centrality``````
Table 9.36. Parameters
Name Type Default Optional Description
label
string
null
yes
The label to load from the graph. If null, load all nodes.
relationship
string
null
yes
The relationship type to load from the graph. If null, load all relationships.
concurrency
int
available CPUs
yes
The number of concurrent threads used for running the algorithm. Also provides the default value for 'readConcurrency'.
int
value of 'concurrency'
yes
Table 9.37. Results
Name Type Description
node
long
Node ID.
centrality
float
Closeness centrality weight.
#### 9.2.6.7. Graph type support
The Harmonic Centrality algorithm supports the following graph types:
• ✓ undirected, unweighted
• ❏ undirected, weighted | 1,727 | 6,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-05 | latest | en | 0.916353 |
https://www.karlin.mff.cuni.cz/~pyrih/e/e2001v1/c/ect/node93.html | 1,716,699,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00715.warc.gz | 720,090,502 | 6,056 | Next: Dyadic Selenoid Up: New examples Previous: Whyburn's Curve
## Pseudo-arc
A. Lelek, ZBIORY (Sets), PZWS Warszawa 1966, pages 90-95.
The first hereditarily indecomposable continuum has been constructed by B. Knaster in 1922. This continuum K, discovered by Knaster, is just named the pseudo-arc or Knaster's continuum. The pseudo-arc K is obtained as the common part of a decreasing sequence of continua in the plane, every of which is the union of finitely many discs, forming a chain of sets.
We will consider chains
of discs Di in the plane, having the property that any two consecutive disc Di and Di+1 have interior points in common, i.e., that their intersection -- nonempty by assumption -- is not contained in the boundary of any of them (Figure A). Discs Di are called links of the chain . The union of all links of the chain is a continuum. Observe that if we delete from the chain a link which is not an end link, i.e., a link Di such that 1 < i < k, then the union of the remaining links is a not connected set, having two components.
Let two chains of discs and be given. We say that the chain refines the chain if each link D's of the chain is contained in the interior of at least one link Di of the chain , i.e., if for every there exists a link such that and that the disc D's has no common points with the boundary of the disc Di (Figure B).
We say that the chain is crooked in the chain if it refines and if, for every pair of links Di and Dj of the chain such that
i + 2 < j, (1)
and for every pair of links D's and D'v of the chain , intersecting the links Di and Dj respectively, i.e.,
there are, in the chain , links D't and D'u, lying between the link D's and D'v in the same order, i.e.,
such that
(2)
In other words, a part of the chain which is contained between the links D's and D'v has to form a fold in the chain between the links Di and Dj (Figure C). The chain , to go from the link Di to the link Dj has first to come to the link Dj-1, next go back to the link Di+1, and only after this it can reach the link Dj. Such a fold must exist for every two links Di and Dj having no adjoining (i.e. neighbor) link in common.
Now, let there be given in the plane two points a and b, and an infinite sequence
of chains of discs, satisfying, for every , the following conditions:
1o
the point a belongs to the first, and the point b to the last link of the chain ;
2o
the diameter of every disc in the chain is less than ;
3o
the chain is crooked in the chain .
Thus, for example, the chain is crooked in the chain , and the chain is crooked in the chain (on Figure D only a part of the chain is presented). The foldings are more and more condensed and they overlapped themselves.
Denote by Kn the union of all links of the chain Dn. From condition 3o it follows in particular that the each next chain refines the previous one, and therefore the continua Kn form a decreasing sequence
The Pseudo-arc K is defined as the common part of all continua Kn, i.e.,
By virtue of condition 1o points a and b belong to the continuum K, thus it is nondegenerate. We shall prove that the continuum K is hereditarily indecomposable.
Let be an arbitrary continuum. To prove that the continuum K is indecomposable it is enough to show that every continuum distinct from K' is nowhere dense in K'. Consider an arbitrary point and number .
Since , hence there is a point which therefore is at a distance at least from each point , where is a fixed number, independent from y. In fact, if not, then some points of the set would be arbitrarily close to the point q, and so the point q would belong to the closure of the set Y, which is impossible, because the set Y is closed, and it does not contain the point q. Let us take a natural number n so large that the inequalities are satisfied
Denote by Di and Dj (with ) the links of the chain whose union contains the points p and q. Since , hence , (here means the metric in the plane) and thus it follows that between the links Di and Dj there are at least two links of the chain . Indeed, in the opposite case the links Di and Dj would have a neighbor common link, and choosing suitable points p' and q' in the intersections of this link with the links Di and Dj we would have
by condition 2o. Thus we can assume that the indices i and j satisfy the inequality (1). Further, we can assume also that the point p belongs to one, and the point q to the other of the links Di and Dj.
Denote by D's and D'v the links of the chain which contain points p and q, respectively. Additionally we assume that . In the opposite case, i.e. if , the further part of the proof runs in the same way (with changing of the roles of the links Di and Dj as well as Di+1 and Dj-1).
So,
whence we infer, by virtue of condition 3o and the definition of folding of chains, that there are two links D't and D'u in the chain which lie between the links D's and D'v in the same order and which satisfy the inclusions (2). The union of the links of the chain distinct from the link D'u is a not connected set (see above) containing the point p in one component, and the point q in the other. Since the continuum K' joins the points p and q and is contained in the union Kn+1 of all links of the chain , hence K' must pass thru the link D'u. Thus there is a point
(Figure E). We shall prove that the point x does not belong to the continuum Y. Indeed, if it would be , then by the same reason as previously for the continuum K' the continuum Y would have to pass thru the link D't. Then there would exist a point
which would be in the link Dj-1 according to (2). But the neighbor links Dj and Dj-1 have their diameters less than by condition 2o, and therefore we would have
which is impossible because .
So, the point x belongs to the set and to the link Di+1 simultaneously. The neighbor links Di and Di+1 have diameters less than , whence it follows that
Since was an arbitrary positive number, hence, in this way we have proved in that there are points of the set which lie arbitrarily closely to the point p. The point p was an arbitrary point of the continuum Y. Thus we have proved that the continuum Y is nowhere dense in K', and the proof of hereditary indecomposability of the continuum K is finished.
An example of a chain crooked in a chain with 7 links is on Figure F.
Figure ( A ) a chain with links, construction of the Pseudo-arc
Figure ( B ) refining a chain, construction of the Pseudo-arc
Figure ( C ) crookedness used in the construction of the Pseudo-arc
Figure ( D ) crookedness used in the construction of the Pseudo-arc
Figure ( E ) crookedness used in the construction of the Pseudo-arc
Figure ( F ) an example of a chain crooked in a chain with 7 liks, construction of the Pseudo-arc
Source files: a.cdr . a.eps . a.gif . a.txt . b.cdr . b.eps . b.gif . b.txt . c.cdr . c.eps . c.gif . c.txt . d.cdr . d.eps . d.gif . d.txt . e.cdr . e.eps . e.gif . e.txt . f.cdr . f.eps . f.gif . f.txt . latex.tex . title.txt .
Here you can read Notes or write to Notes.
Next: Dyadic Selenoid Up: New examples Previous: Whyburn's Curve
Janusz J. Charatonik, Pawel Krupski and Pavel Pyrih
2001-02-21 | 1,792 | 7,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.951795 |
https://kb.paessler.com/en/topic/9563-what-are-percentiles-and-what-differences-do-they-make-in-prtg-reports | 1,723,642,956,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00326.warc.gz | 260,907,182 | 6,448 | ### What is this?
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# What are percentiles and what differences do they make in PRTG reports?
#### 0
In PRTG reports, statistics can be shown as percentiles. But I do not know what percentiles are, how PRTG calculates them, and what kind of effect they have on the data I get.
Created on Oct 1, 2010 2:23:17 PM by
Last change on May 6, 2019 11:54:54 AM by
4 Replies
# Percentiles in reports
In reports in PRTG, you get statistically condensed information about the data that various sensors collect. In the sensor report settings, you can choose to get percentile calculation, as opposed to raw data or normal percentages for example.
Percentiles can be computed according to various mathematical methods.
## How percentiles are calculated in PRTG
1. Sort the measurements: The values of a data sample have to be ordered in a row from the lowest to the highest value. You get the individual values and the number of rows N.
2. Compute the row numbers of the percentile value RN using the percentile value P and the number of rows N:
`RN = 1 + ((N-1) * P)`
There are two types of RNs: FRN = floor (RN) and CRN = ceiling (RN).
3. Determine the result:
`If (CRN = FRN = RN) \\then (value of expression from row at RN) \\else (value of expression for row at FRN) + (RN - FRN) * (CRN row - FRN row value)`
Note: You can define if you want to get the then or the else result. If you choose Continuous in the report percentile settings, the result is always interpolated like in the else part. If you choose Discrete, the next smaller discrete value, which is the FRN, is taken as a result, according to the then part.
## Example
A 95th-percentile says that in 95% of all cases, the data is below a certain value, while in only 5% of all cases, the data is above a certain value.
• Step 1: Let us assume that we have the following number of rows and measured values:
Row Number Value 1 2 3 4 5 6 7 8 1 3 7 21 25 26 66 72
• Step 2: `RN = 1 + ((8-1) * 0.95) = 7,65` with FRN = 7 and CRN = 8
• Step 3: Since CRN = FRN = RN is not true, we calculate the else-part:
`Percentile value = 66 + ((7,65-7)* (72-66)) = 69.9`
So 69.9 is our continuous 95th-percentile. If we want to have a discrete 95th-percentile value, it will be the FRN, which is 66.
As you can see, in a 95th-percentile calculation of our measurements, the value number 8 (72) would be discarded because it belongs to the 5% of peaks. In our case this is true for the 95th continuous percentile as well as for the 95th discrete percentile.
The main advantage of using percentiles is that unusually high values (like whiskers in boxplots) are not included into the averaging calculations. This means that statistics include more relevant data. In the example of the 95th-percentile, 5% of the highest measured values are discarded for the statistical report.
If you prefer not to exclude extremes, because you do not want to disguise the effect they have on your monitored IT infrastructure, do not choose percentiles in the report settings.
Note: For more details on this percentile calculation, please see archive.org.
Created on Mar 3, 2015 4:24:30 PM by
Last change on Jun 30, 2022 12:40:57 PM by
#### 0
Should say Column Number instead of Row Number .... ?
Created on May 20, 2016 5:38:08 AM
#### 0
And Percentile value = 66 + ((7,95-7)* (72-66)) = 69.9 should be Percentile value = 66 + ((7.65-7)* (72-66)) = 69.9
Created on May 20, 2016 5:40:57 AM
#### 0
Hi Raymond,
Thank you, I fixed the typo. "Row Number" does not mean "table row" but refers to an ordered row of values, so it is correct here. The table in the article just shows values in a row in this format, it is not a "data table".
Regards,
Created on May 20, 2016 11:34:41 AM by
Disclaimer: The information in the Paessler Knowledge Base comes without warranty of any kind. Use at your own risk. Before applying any instructions please exercise proper system administrator housekeeping. You must make sure that a proper backup of all your data is available. | 1,158 | 4,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.901726 |
https://blog.ericzheng.org/normcdf | 1,660,932,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00669.warc.gz | 158,094,405 | 7,528 | # console.blog()
a "technical" "blog" by Eric Zheng
Let's Play---Implementing a normcdf
Summary: pretty normal, if you ask me...
### introduction
In stat class, one of our favorite calculator functions is the good old normcdf, which finds the probability that, when we sample a random variable, we’ll get something within a certain range. As implemented on the trusty TI-nspire, it can seem like black magic: you put some numbers in, and out pops your perfect AP Stat score! I wanted to write a post demystifying this process a bit. First, we’ll clarify what we mean by the normcdf function. Then, we’ll take a look at the mathematical building blocks for the function. Finally, we’ll implement it in Python.
This post should be accessible to anyone who is currently taking AP Stat, has a cursory familiarity with Python, and has an AP Calc BC-level understanding of series. Even without the Python, the math part is (hopefully) interesting.
### prelude: what’s normal, anyway?
We generally have a vague notion that a “normal” distribution is a probability distribution that’s bell-shaped. For example, take the heights of students in a school: when we say that this variable (height) is normally distributed, we mean that the frequencies sort of look like this:
The above distribution is what we call discrete, since there are finitely many data points. Now imagine that we have an infinitely large population: the distribution would get smoother and smoother until we have a continuous curve like the one below:
Here, the y-axis is the relative frequency of a given x-value’s occurrence. Let’s get slightly more technical with how we describe this distribution. When we talk about a normal distribution, there are really two parameters that we need to take care of: $\mu$ (the mean, or the height we would expect from an average person) and $\sigma$ (the standard deviation, or roughly how far we would expect any given person to fall away from our mean). You can also think of $\sigma$ as how spread out the height values are. To make this clearer, here’s the same continuous distribution with the mean height and a height one standard deviation above the mean labeled:
For this simulation, I arbitrarily set $\mu = 69$ and $\sigma = 4$.The key takeaway is that with a continuous distribution, we’re not really interested in the relative frequency of a given height (let’s just say the mean height $\mu$ as an example). Since there are infinitely many possible heights, the probability of any given height being exactly 69 is zero! What is meaningful is the total probability within a given interval—in the plot above, let’s say the probability between the mean $\mu$ and one standard deviation above the mean, or $\mu + \sigma$. Since the curve is normalized (i.e. the total blue area is equal to 1), the probability of any randomly selected height $x$ falling within $\mu \le x \le \mu + \sigma$ is simply the cumulative area between the red and orange lines. To give a concrete example, for the above distribution, the probability of randomly selecting a height $x$ such that $69 \le x \le 73$ is the cumulative distribution function1 under this normal curve. In other words, the normcdf from 69 to 73.
(note: the code I used to generate these graphs can be found here)
### mathy stuff
Now, if you’ve taken AP Stat, I probably haven’t told you anything new yet. I just wanted to provide a quick refresher on the normal distribution; here’s where the new stuff comes in. As you probably guessed when I mentioned the area under the curve, the normcdf function is really just the integral of the probability function (shown as the black curve in the graph above and referred to as a normpdf on the calculator). So, if we know the equation for the black curve, we should be able to integrate it and get an expression for normcdf, right?
Well, reality is unfortunately slightly more complicated than that. You may have heard the normal distribution referred to as the “Gaussian” distribution2. That’s because the pdf function for generating it (technically called the “kernel function”) is the Gaussian function, or:
where $\mu$ and $\sigma$ are our friends mean and standard deviation from before. Really, we just want the area under the curve from $a$ to $b$:
That doesn’t look too troublesome to integrate, but if you try, you’re guaranteed to fail: there isn’t any elementary antiderivative. This means that there is no combination of “basic” functions like sine, log, etc. whose derivative is our buddy $p(x)$. Whelp. We’re going to have to use numerical methods to approximate it.3
### implementing erf(x)
Let’s use the time-honored method of getting unstuck and take a step back to look at a “completely different” function, which we’ll call the error function, or erf. (Quite an unfortunate name, really…) This function is defined as:
Well, this is slightly better, but we’re still no closer to actually getting the antiderivative. Gee, if only the integrand were something easy, like a polynomial. We can do polynomials all day long…
Huh. What if we used the Taylor series of $e^x$ to re-express the integrand as a polynomial? Then we could easily get an infinite series that converges to the value we want, erf(x)! Recall from AP Calc BC that the series expansion for $e^x$ is:
So that pesky little integrand is really just:
That doesn’t look so intimidating any more. Let’s just integrate it to get:
Or, with the ever-so-elegant summation notation:
Now, this is really an approximation of the true value of erf(x); we’d have to sum an infinite number of terms to get the real value. That’s not very useful, but the good thing is that we can get arbitrarily close to the true value of erf(x) by adding just a finite number of terms. Because this is an alternating series, the maximum possible error of the approximation up to $n$ terms is simply bounded by the absolute value of the nth term, a famous result from AP Calc BC.
To recap this section, I can’t really find a “closed-form” expression for erf(x), but I can get arbitrarily close to it by adding up the terms of the series shown above. Once the terms I’m adding get below my desired precision $\epsilon$, I’m guaranteed to have the value of erf(x) to within $\epsilon$.
(Yes, we could have jumped directly to the Taylor series for normcdf, but it isn’t as insightful, and the erf function is important in its own right.)
### putting the math together
Okay, so we have a great approximation for erf(x), a function very close to the Gaussian integral that we were trying to approximate in the first place. How do we proceed from here?
Well, let’s revisit another friend from AP Statistics, the z-score. The z-score tells us how many standard deviations a given x-value is from the mean and is defined by:
Note that this looks suspiciously like the exponent in the Gaussian function: it’s really just a scaling factor for the normal distribution. All normal curves are equal, but there’s one that’s more equal than the rest: the standard normal curve, which is just a normal distribution with $\mu = 0$ and $\sigma = 1$. We really like this one because the numbers are pretty easy to work with (if you’re in Mrs. Brown’s AP Stat class, the standard normal distribution is what’s on the infamous green sheet).
It turns out that any normal distribution is really just the standard normal distribution, scaled up by a factor of $z$. Taking advantage of the $z$ substitution and that $\sigma^2 = 1$ for the standard normal, let’s simplify our normcdf(a,b) function to look like:
Now, we’ll just use some basic integral rules and the substitution $u = z/\sqrt{2}$ to get:
Aha! We can now see that:
### putting the code together
Honestly, I could probably stop here, but this is supposed to be a coding blog, so I might as well write some code. (Also, I have a lovely Markdown code highlighting thing here, and it’d be a shame not to use it.) Let’s implement our original erf(x) function in Python:
# computes error function to desired accuracy epsilon
def erf(x, epsilon=0.000001):
# since erf(x) is symmetric, we only focus on x > 0
y = abs(x)
# adjust epsilon to account for normalization at the end
k = 1.1283791670955125739 # this is 2/sqrt(pi)
epsilon /= k
# compute the Taylor series about zero
# use the alternating error term to know when to stop
term = y
n = 0.0
sign = +1.0
factorial = 1.0
power = y
result = term
while term > epsilon:
n += 1
factorial *= n
sign = -sign
power *= y * y
term = power / (factorial * (2 * n + 1))
result += sign * term
# normalize and return result
if x < 0:
return -k * result
return k * result
Let’s test it:
>>> erf(0.13)
0.14586711478
# according to Wolfram Alpha, the result should be 0.145867
And now for the finishing touches:
# computes normal cumulative distribution function from lower bounds a,
# upper bounds b, mean avg, standard deviation stdev, and desired accuracy
# epsilon
def normcdf(a, b, avg=0.0, stdev=1.0, epsilon=0.000001):
# get z scores
z_a = (a - avg) / stdev
z_b = (b - avg) / stdev
k = 0.7071067811865475244008 # this is 1/sqrt(2)
# because we multiply by 0.5, we shouldn't need to scale epsilon
return 0.5 * (erf(k * z_b, epsilon) - erf(k * z_a, epsilon))
With fingers crossed, let’s try it:
>>> normcdf(-0.3, 0.1, 0.8, 0.5)
0.0668532280857
# according to the calculator, the result should be 0.06685331227
# perfectly within our 0.000001 precision
Yay, it works! (If you don’t want to copy/paste, you can grab a copy of the code from the comfort of your own home here.)
### conclusion
Math, and especially statistics, can seem like a black box at times. But with the power of Python, you can peer into the black box!
If you’re interested in this sort of stuff, here’s some food for thought: the Taylor series we used to approximate erf(x) was centered around 0, which means it really works best for small values of x. It would still give us the right answer for really large x, but it would be terribly inefficient. Can you come up with a better approximation for larger numbers?
### footnotes
1Sometimes, normcdf is called the normal cumulative density function, but for the purposes of this post, the distinction between a probability density function and probability mass function isn’t important. I’ll just call it the distribution function.
2What is it with this Gauss guy? Wikipedia has an entire page devoted to things named after him!
3Interestingly, even though the indefinite integral of $p(x)$ has no elementary form, the definite integral from negative infinity to infinity can be found using polar trickery. It turns out to be the reciprocal of that coefficient in front, which really just exists to normalize the function. | 2,514 | 10,730 | {"found_math": true, "script_math_tex": 29, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-33 | latest | en | 0.934214 |
https://www.geeksforgeeks.org/maximize-big-when-both-big-and-small-can-be-exchanged/ | 1,653,554,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00086.warc.gz | 898,512,532 | 26,899 | # Maximize big when both big and small can be exchanged
• Difficulty Level : Easy
• Last Updated : 11 May, 2021
Given N Big Candies and M Small Candies. One Big Candy can be bought by paying X small candies. Alternatively, one big candy can be sold for Y small candies. The task is to find the maximum number of big candies that can be bought.
Examples:
Input: N = 3, M = 10, X = 4, Y = 2
Output:
8 small candies are exchanged for 2 big candies.
Input: N = 3, M = 10, X = 1, Y = 2
Output: 16
Sell all the initial big candies to get 6 small candies.
Now 16 small candies can be exchanged for 16 big candies.
In first example, Big candies cannot be sold for profit. So, only the remaining small candies can be exchanged for big candies.
In second example, Big candies can be sold for profit.
Approach: If initial big candies can be sold for profit i.e. X < Y then sell the big candies and update the count of small and big candies. Then, sell all of the updated small candies in order to buy big candies.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach``#include ``using` `namespace` `std;` ` ``// Function to return the maximum big`` ``// candies that can be bought`` ``int` `max_candies(``int` `bigCandies,`` ``int` `smallCandies,``int` `X, ``int` `Y)`` ``{`` ``// If initial big candies`` ``// can be sold for profit`` ``if` `(X < Y)`` ``{`` ``smallCandies += Y * bigCandies;`` ``bigCandies = 0;`` ``}` ` ``// Update big candies that can be bought`` ``bigCandies += (smallCandies / X);` ` ``return` `bigCandies;`` ``}` ` ``// Driver code`` ``int` `main()`` ``{`` ``int` `N = 3, M = 10;`` ``int` `X = 4, Y = 2;`` ``cout << (max_candies(N, M, X, Y));`` ``return` `0;`` ``}`
## Java
`// Java implementation of the approach``class` `GFG {` ` ``// Function to return the maximum big candies`` ``// that can be bought`` ``static` `int` `max_candies(``int` `bigCandies, ``int` `smallCandies,`` ``int` `X, ``int` `Y)`` ``{`` ``// If initial big candies can be sold for profit`` ``if` `(X < Y) {` ` ``smallCandies += Y * bigCandies;`` ``bigCandies = ``0``;`` ``}` ` ``// Update big candies that can be bought`` ``bigCandies += (smallCandies / X);` ` ``return` `bigCandies;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `N = ``3``, M = ``10``;`` ``int` `X = ``4``, Y = ``2``;` ` ``System.out.println(max_candies(N, M, X, Y));`` ``}``}`
## Python3
`# Python3 implementation of the approach` `# Function to return the maximum big candies``# that can be bought``def` `max_candies(bigCandies, smallCandies, X, Y):`` ` ` ``# If initial big candies can`` ``# be sold for profit`` ``if``(X < Y):`` ` ` ``smallCandies ``+``=` `Y ``*` `bigCandies`` ``bigCandies ``=` `0`` ` ` ``# Update big candies that can be bought`` ``bigCandies ``+``=` `(smallCandies ``/``/` `X)` ` ``return` `bigCandies` `# Driver code``N ``=` `3``M ``=` `10``X ``=` `4``Y ``=` `2``print``(max_candies(N, M, X, Y))` `# This code is contributed by Code_Mech`
## C#
`// C# implementation of the approach``using` `System;` `class` `GFG``{`` ` ` ``// Function to return the maximum`` ``// big candies that can be bought`` ``static` `int` `max_candies(``int` `bigCandies,`` ``int` `smallCandies,`` ``int` `X, ``int` `Y)`` ``{`` ``// If initial big candies`` ``// can be sold for profit`` ``if` `(X < Y)`` ``{`` ``smallCandies += Y * bigCandies;`` ``bigCandies = 0;`` ``}` ` ``// Update big candies that can be bought`` ``bigCandies += (smallCandies / X);` ` ``return` `bigCandies;`` ``}` ` ``// Driver code`` ``static` `public` `void` `Main ()`` ``{`` ``int` `N = 3, M = 10;`` ``int` `X = 4, Y = 2;`` ``Console.WriteLine(max_candies(N, M, X, Y));`` ``}``}` `// This Code is contributed by ajit...`
## PHP
``
## Javascript
``
Output:
`5`
My Personal Notes arrow_drop_up | 1,462 | 4,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-21 | latest | en | 0.712249 |
https://theknolwedgehub.com/what-information-is-indexed-by-the-graph-coinbase/ | 1,638,230,940,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00011.warc.gz | 631,971,354 | 10,352 | Categories
# what information is indexed by the graph coinbase
The graph coinbase is an algorithm that uses information about the graph of Bitcoin transactions to calculate the graph entropy of Bitcoin transactions. It provides a measure of the “weight of an edge” in the graph.
This is great because most of us never knew the graph entropy of Bitcoin…
The graph entropy, as its name implies, measures the amount of information that is necessary to describe the state of a node in the graph. For example, if you were to look at a tree, you could see that there was a path through the branches of the tree that had the same path to the root. But you’d need more information to describe the state of the node at the end of the path.
The graph entropy is used in the Bitcoin network to calculate the cost of a transaction. The graph entropy is a way to measure the cost of a single transaction, rather than the cost of the entire network. But that doesn’t mean the graph entropy is useless. It can help us predict the likely outcome of a transaction by showing us the number of possible paths through the network that can be used to describe the transaction.
That’s the thing about cryptocurrencies, the more you know the more you can use. The more accurate we can predict the likely outcome of a transaction, the more likely we are to use that prediction in a transaction. So if you wanted to send a large amount of Bitcoin, you’d want to send it through the Bitcoin network. If you wanted to send more than you could afford to send through the Bitcoin network, you’d put it on a Bitcoin mining pool.
Graphcoin, a Bitcoin startup, is doing just that. It’s been building a network of Bitcoin mining pools, each one storing a portion of Bitcoin. This allows each pool to determine its own hash rate and thus its own bitcoin holdings. So if you wanted to send Bitcoins to someone, youd send them through the graphcoin network. But if you wanted to send a lot more Bitcoins than you could afford to send through the graphcoin network, youd put them into a Bitcoin mining pool.
Graphcoin is already worth a whopping \$10 million dollars. Because of this, I expect they will continue to do what many of you already have been doing, which is building a Bitcoin mining pool and keep adding those coins to the network.
I’m not sure where you’ll get the 1 million bitcoin you have in your Bitcoin mining pool if you try to send 1 million Bitcoin through the graphcoin network. We don’t yet know how many coins are in a Bitcoin mining pool, but it’s also likely that that number is probably much less than 1 million. So for now, we can wait and hope you’re using your mining pool to send more Bitcoins to your Bitcoin address.
As you can see from the graph, we do have more than 1 million Bitcoins in our mining pool, but that number is hard to know for sure. The graph shows us the current total of Bitcoins in the Bitcoin network, but does not give us the total number of Bitcoins in a Bitcoin mining pool.
It is interesting to note that the total number of Bitcoins in the Bitcoin network is about 14 billion. This represents the total number of Bitcoins that exist today, minus those that have been halved or are currently being mined. So in theory, we could theoretically be mining about 14 billion Bitcoins using our mining pool right now. But of course, the number is going to be much lower than that. | 711 | 3,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | latest | en | 0.955015 |
http://www.lookingforananswer.net/the-number-176-is-divisible-by-what-number.html | 1,477,062,141,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718284.75/warc/CC-MAIN-20161020183838-00364-ip-10-171-6-4.ec2.internal.warc.gz | 568,004,455 | 7,484 | # The number 176 is divisible by what number?
A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor ... 176. If the thousands ... in that all such numbers ... - Read more
... and "Number 2" redirect here. For other ... the maximum number of consecutive odd numbers that are prime ... 176; 177; 178; 179; 180; 181; 182; 183; 184; 185; 186 ... - Read more
## The number 176 is divisible by what number? resources
### The Math Dude : How to Tell If a Number is Divisible by 7 ...
Now that we know how to check for divisibility by all the numbers from ... The quick and dirty tip for checking if a number is divisible by 9 is to add up the digits ...
### The Momen Blog: What is the smallest number divisible by ...
What is the smallest number divisible by each of the numbers 1 ... Any number which is divisible 20 is also divisible by ... Any number divisible by 11 ...
### Each digit in the n-digit number N is 1. What is the ...
Without invoking Fermat's little: Note that (at least in recreational mathematics), a number where each digit is one is called a Repunit, and a number where each ...
### Divisibility Rules: How to test if a number is divisible ...
... are divisible by 4, the number 112 ... Examples of numbers that are divisible by 5 ... Rule A number is divisible by 9 if the sum of the digits are evenly ...
### Divisibility by 2, Odd number - Math10.com
If only one of the addends is not divisible by 2, the number ... number and 597 is not divisible because its last digit is odd number. Numbers divisible by 2 ...
### Numbers Divisible by 3 - Code85
Numbers Divisible by 3 Any number is divisible by 3 if the sum of the digits is divisible by 3. ... and d are divisible by 3, allowing a 3 to be factored out, ...
### The Math Dude : How to Tell If a Number is Divisible by 4 ...
... last time and learn three more tips that you can use to test if numbers are divisible by 4 ... A number is divisible by 3 only if its digits add up to a ...
### Divisibility by: 2 3 If the sum of the digits is divisible ...
4 If the last two digits form a number divisible by 4, the number is also. 5 If the last digit is a 5 or a 0, the number is divisible by 5.
### divisibility - Number of numbers divisible by 5 and 6 ...
For example let's take only tree-digit numbers, how many of them are divisible by both 5 and 6? ... How can I count the number of numbers divisible by both 5 and 6?
### Is 176 divisible by 3? - Research Maniacs
Is 176 divisible by 3? In other words, if you divide 176 by 3, will you get a whole number with no remainder? Of course, you could use a calculator to find out if 176 ...
### Numbers Divisible by 6 - AAA Math
Numbers are evenly divisible by 6 if they are evenly divisible by ... which is evenly divisible by 3 but 3627 is an odd number so the number 3627 is not evenly ...
### Numbers Divisible by 7 - AAA Math
Numbers Divisible by 7. To determine if a number is divisible by 7, take the last digit off the number, ...
### Numbers Divisible by 4 - AAA Math
Numbers Divisible by 4. ... which is evenly divisible by 4 so the number 3628 is evenly divisible by 4. Why is this always true? ...
### How to Find the Smallest Number That a Number Is Divisible By
How to Find the Smallest Number That a Number Is Divisible By. All numbers are divisible by ... of the prime numbers, the first divisible number is your ...
### Numbers evenly Divisible by 3 - AAA Math
Numbers are divisible by 3 if the sum of all the individual digits is evenly divisible by ... which is evenly divisible by 3 so the number 3627 is evenly divisible by 3.
Related Questions
Recent Questions | 916 | 3,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2016-44 | latest | en | 0.907742 |
https://howtodoinjava.com/python/numpy/python-numpy-tutorial/ | 1,695,429,248,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00462.warc.gz | 345,832,552 | 38,210 | # Python Numpy 101: A Beginners Guide
NumPy is a open-source Python library that stands for Numerical Python. NumPy is widely used for working with large, multi-dimensional arrays (ndarray), matrices of numerical data and it provides extremely fast and efficient numerical operations on arrays. NumPy is commonly employed in data science and machine learning to perform mathematical operations on large datasets.
By the end of this article, we will have a good understanding of the basics of NumPy. We will learn about the installation, NumPy data types, and why NumPy is used. Finally, we will learn about NumPy arrays in detail, data visualization using NumPy, and some of its limitations.
So, without further ado, let’s get started!
## 1. Why use NumPy?
NumPy arrays are more efficient and faster. Numpy has a built-in data structure called an array similar to the usual Python list, but it can store and operate on data much more efficiently.
Here are some reasons to use NumPy:
• NumPy operations perform faster than equivalent operations on Python lists. This is because NumPy arrays are stored in memory in a way optimized for faster access.
• NumPy provides various functions to perform common operations on arrays without loops. This makes the code easier to read and understand.
• NumPy code is clearer because you can perform operations on arrays at once using the built-in functions, which makes the code more readable and understandable.
## 2. Installing NumPy
To use NumPy, we first need to install it on the system. There are several methods, but let’s take a look at two simple methods.
### 2.1. Installing NumPy with ‘pip‘
The ‘pip‘ is a Python’s package installation manager that makes it easy to install Python libraries or frameworks. If we have Python version 3.4 or higher, then Pip comes by default. Otherwise, we will need to install pip before installing NumPy.
Now, first, launch the command prompt and type the following command:
pip install numpy
Hit Enter and we will see that NumPy will start installing. Now we can use NumPy in Python programs.
After the installation is finished, we have to import NumPy into a Python program. We can either use import numpy or import numpy as np.
import numpy
//or
import numpy as np
### 2.2. Installing NumPy with Anaconda
Another way to install NumPy is to install Anaconda. Anaconda is a Python distribution that provides us with access to different tools.
When we install Anaconda, it installs all the major libraries automatically. To install Anaconda, first download it using from its download page.
Now, launch it, but remember to check the following boxes:
Just click on the Install button and wait for the installation to complete. Once Anaconda is installed, we can use NumPy in the Windows command prompt, VS Code editor, or PowerShell prompt (one of the tools available in the Anaconda Navigator).
If we are going to use NumPy, it is a best practice to use it in a Jupyter Notebook. Jupyter Notebook is a web-based, interactive computing notebook.
To use Jupyter, open Anaconda Navigator in our system and open the Jupyter Notebook. We can see the Jupyter Notebook option in the image below.
Just click on the Launch button and the notebook will open on the localhost page, as we can see below.
We can click on the New button and select Python 3. We are now ready to use the Jupyter Notebook.
## 3. NumPy Data Types
NumPy provides a wider range of numeric data types than Python. The additional data types in NumPy are designed for numerical calculations. Here is the list of NumPy data types and the characters used to represent them:
### 3.1. i (integer)
The ‘i‘ is used to represent signed integer types. The number of bits used to store the integer depends on the machine. For example, on a 64-bit machine, the ‘i‘ is 64 bits wide, while on a 32-bit machine, it is 32 bits wide.
If we execute the below code, the output will differ on various machines. On my machine, the output is: ‘int32‘, which indicates that the array is of a 32-bit signed integer type. On other machines, the output may be int64 or vary. Here, the .dtype attribute is used to print the data type.
import numpy as np
arr = np.array([1, 2, 3])
print(arr.dtype) # Prints 'int64'
We can also explicitly specify the data type when creating an array. For example, the following code will create an array of 64-bit integers.
import numpy as np
arr = np.array([1, 2, 3], dtype=np.int32)
print(arr.dtype) # Prints 'int32'
### 3.2. b (boolean)
The ‘b‘ represents Boolean values which can either be True or False. Let’s create an array of Boolean values and print its data type.
import numpy as np
arr = np.array([True, False, True])
print(arr.dtype) # Prints 'bool'
### 3.3. u (unsigned integer)
The ‘u‘ represents unsigned integer data types. Unsigned integers can only store positive values. The size of the unsigned integer depends on the machine. For example, the below code will create a numpy array of 8-bit unsigned integers.
import numpy as np
arr = np.array([1, 2, 3], dtype=np.uint8)
print(arr.dtype) # Prints 'uint8'
If we try to include the negative values, then we will get an error.
import numpy as np
arr = np.array([1, 2, 3, -4], dtype=np.uint8)
print(arr.dtype) # Error: DeprecationWarning: NumPy will stop allowing conversion of out-of-bound Python integers to integer arrays. The conversion of -1 to uint8 will fail in the future.
### 3.4. f (float)
The ‘f‘ represents floating point numbers. The precision of floating point numbers depends on the platform. It may be 16-bit floating point numbers, 32-bit, and 64-bit. For example, if we run the below code, the result will vary. It may be 32-bit, 64-bit, etc. Like in my case, it’s float64.
import numpy as np
arr = np.array([1.1, 2.2, 3.3])
print(arr.dtype) # Prints 'float64'
You can also create floating point numbers of specific bits by mentioning their data type, as shown below, we are creating an array of 32-bit floating point numbers.
import numpy as np
arr = np.array([1.1, 2.2, 3.3], dtype="float32")
print(arr.dtype) # Prints 'float32'
### 3.5. c (complex float)
The ‘c‘ represents complex numbers and is often denoted as complex64 or complex128. Complex numbers have both real and imaginary parts. In complex64, both real and imaginary part is represented using a 32-bit floating point number, and in complex128, each part is represented using a 64-bit floating point number.
We can access the real and imaginary parts using the numpy built-in functions: .real and .imag. Here is an example:
import numpy as np
array1 = np.array([1 + 2j, 3 - 4j], dtype=np.complex64)
array2 = np.array([1.5 + 2.5j, -3.5 - 4j], dtype=np.complex128)
print(array1)
print(array2)
print("Real part of Array 1:", array1.real)
print("Imaginary part of Array 1:", array1.imag)
The program output:
[1.+2.j 3.-4.j]
[ 1.5+2.5j -3.5-4.j ]
Real part of Array 1: [1. 3.]
Imaginary part of Array 1: [ 2. -4.]
### 3.6. m (timedelta64)
The ‘m‘ represents time delta which means time durations or intervals. It also allows us to perform arithmetic operations on time intervals such as adding different time durations.
In the following code, we create four-time delta objects that show day, hour, minute, and second.
import numpy as np
time_day = np.timedelta64(4, 'D')
time_hr = np.timedelta64(7, 'h')
time_min = np.timedelta64(60, 'm')
time_sec = np.timedelta64(120, 's')
print(time_day, time_hr, time_min, time_sec) # Prints '4 days 7 hours 60 minutes 120 seconds'
### 3.7. M (datetime64)
The ‘M‘ represents date and time. When we create a datetime64 object, It holds the year, month, day, hour, minute, second, and even fractions of a second. For example, consider the following datetime:
np.datetime64('2023-08-18T12:30:45.500')
In this case, datetime holds the date: ‘August 18th, 2023, and time: 12:30:00.50‘. We can then perform various operations with this datetime object, such as comparing it to other date-times, calculating time intervals, and more.
In the below code, we have calculated the duration by subtracting the start time from the end time.
import numpy as np
start = np.datetime64('2023-08-19T11:00:00')
end = np.datetime64('2023-08-21T16:40:00')
event_duration = end - start
print(event_duration) # Prints '193200 seconds'
## 4. Creating Arrays with NumPy
NumPy arrays are similar to Python lists, except lists can store elements of different data types whereas all of the elements in a NumPy array should be homogeneous.
There are several ways to create NumPy arrays. We will explore some of these methods below.
### 4.1. Creating an Empty Array
An empty array allocates memory for the array elements without initializing them to any particular value. It’s crucial to recognize that the array’s elements are uninitialized and may retain previous memory values.
We can create an empty array using the np.empty() function. In the following example, the shape (3, 4) specifies that the array should have 3 rows and 4 columns
import numpy as np
empty_array = np.empty((3, 4)) # Creates a 3x4 empty array
### 4.2. Creating N-d Arrays
NumPy provides following inbuilt methods to create arrays:
• array()
• arrange()
• zeros()
• ones()
• linespace()
Let us learn how to use these methods in brief.
#### 4.2.1. numpy.array()
The numpy.array() function creates an array from any iterable object, such as a list, tuple, or range. For example, the following code creates an array from a list:
import numpy as np
list1 = [1, 2, 3, 4]
array1 = np.array(list1)
print(array1) # Prints: [1 2 3 4]
We can also create multi-dimensional arrays by providing nested lists or tuples:
# Create a 2D NumPy array from a nested Python list
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my_2d_array = np.array(matrix)
# Create a 3D NumPy array from nested Python lists
cube = [[[1, 2], [3, 4]], [[5, 6], [7, 8]]]
my_3d_array = np.array(cube)
#### 4.2.2. numpy.arrange()
The numpy.arrange() creates one-dimensional arrays of evenly spaced numbers. It takes three arguments:
• start: The starting value of the array.
• stop: The ending value of the array, is not included.
• step: The step size between elements in the array.
# Create an array of integers from 0 to 9 (exclusive)
arr1 = np.arange(10)
print(arr1) # Output: [0 1 2 3 4 5 6 7 8 9]
# Create an array of even integers from 2 to 10 (exclusive)
arr2 = np.arange(2, 10, 2)
print(arr2) # Output: [2 4 6 8]
# Create an array of floating-point numbers from 0.0 to 1.0 (exclusive) with a step of 0.1
arr3 = np.arange(0.0, 1.0, 0.1)
print(arr3) # Output: [0. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9]
We can use arrange() with other NumPy functions to create multi-dimensional arrays if desired. For example, we are using reshape() to convert a single-dimension array to 2-D array.
# Create a one-dimensional array of 12 elements
arr1d = np.arange(12)
# Reshape it into a 3x4 two-dimensional array
arr2d = arr1d.reshape(3, 4)
print(arr2d)
# Output
[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
#### 4.2.3. numpy.zeros()
The numpy.zeros() creates an array of zeros. It takes one argument: shape (size of the array).
array3 = np.zeros(8)
print(array3) # Output: [0. 0. 0. 0. 0. 0. 0. 0.]
To create multi-dimensional arrays, we can specify the shape by passing a tuple containing the desired dimensions.
# Create a 2D array with dimensions 3x4 filled with zeros
zeros_2d = np.zeros((3, 4))
# Create a 3D array with dimensions 2x3x2 filled with zeros
zeros_3d = np.zeros((2, 3, 2))
# Create a 4D array with dimensions 2x3x2x2 filled with zeros
zeros_4d = np.zeros((2, 3, 2, 2))
#### 4.2.4. numpy.ones()
The numpy.ones() creates an array of ones. It takes one argument: shape (size of the array).
array4 = np.ones(8)
print(array4) # Output: [1. 1. 1. 1. 1. 1. 1. 1.]
Similar to zeros(), we can specify the dimensions for creating a multi-dimensional arrays.
# Create a 2D array with dimensions 3x4 filled with ones
ones_2d = np.ones((3, 4))
# Create a 3D array with dimensions 2x3x2 filled with ones
ones_3d = np.ones((2, 3, 2))
# Create a 4D array with dimensions 2x3x2x2 filled with ones
ones_4d = np.ones((2, 3, 2, 2))
#### 4.2.5. numpy.linspace()
The numpy.linspace() creates an array of evenly spaced numbers over a specified interval. The linspace() function takes four arguments:
• start: The starting value of the array.
• stop: The ending value of the array, inclusive.
• num: The number of elements in the array.
• endpoint: Whether to include the stop value in the array.
array5 = np.linspace(0, 1, 5)
print(array5) # Output: [0. 0.2 0.4 0.6 0.8]
We can use the reshape() for converting an array created with linespace() to N-dimensional array.
# Create a 2D array with 4 rows and 3 columns, with values evenly spaced from 0 to 1 (inclusive)
arr_1d = np.linspace(0, 1, 12) # 12 values
arr_nd = arr_1d.reshape(4, 3) # Reshape into a 2D array
NumPy provides functions like numpy.loadtxt(), numpy.genfromtxt(), and numpy.load() for reading data from text files, CSV files, and NumPy binary files, respectively.
The numpy.load() function loads arrays from ‘.npy’ files (NumPy Binary Files), previously saved with its own binary format using numpy.save().
import numpy as np
arr = np.array([1, 2, 3, 4])
np.save('my_array.npy', arr)
print(array) # Output: [1, 2, 3, 4]
The numpy.genfromtxt() or numpy.loadtxt() functions load arrays from text files or CSV files, generally, previously saved using numpy.savetxt(). It takes the following arguments:
• filename: The name of the file to load.
• delimiter: The delimiter is used to separate the values in the file.
• dtype: The data type of the values in the file.
• skip_header: The number of lines to skip at the top of the file.
• comments: A character to identify comment lines in the file.
import numpy as np
arr = np.array([1, 2, 3, 4])
np.savetxt('my_array.txt', arr)
array = np.genfromtxt("my_array.txt", delimiter=",")
print(array) # Output: [1, 2, 3, 4]
## 5. Constants and Attributes
Constants are predefined values that can be used without having to define them first. Attributes are properties that can be accessed using the dot notation.
### 5.1. Constants
NumPy provides several important mathematical constants that can be accessed for various calculations. Here are some of the most commonly used constants:
### 5.2. Attributes
NumPy arrays have several attributes that provide information about the array’s properties. Here are some of the most important attributes:
import numpy as np
print(np.pi) # 3.141592653589793
print(np.e) # 2.718281828459045
print(np.inf) # inf
print(np.nan) # nan
array = np.array([1, 2, 3])
print(array.dtype) # int32
print(array.shape) # (3,)
print(array.size) # 3
print(array.ndim) # 1
## 6. Working with NumPy Arrays
We can perform various operations like addition, subtraction, and multiplication, using statistical functions, comparing two arrays, performing matrix operations, set operations, and much more on NumPy arrays.
### 6.1. Add, Subtract, Multiply and Divide Arrays
We can perform arithmetic operations on NumPy arrays using the standard arithmetic operators (+, -, *, /). The results of these operations will be arrays of the same data type as the input arrays.
import numpy as np
array1 = np.array([1, 2, 3])
array2 = np.array([4, 5, 6])
add_res = array1 + array2
subtract_res = array1 - array2
multiply_res = array1 * array2
divide_res = array1 / array2
print(add_res, subtract_res, multiply_res, divide_res)
Here’s the output:
[5 7 9] [-3 -3 -3] [ 4 10 18] [0.25 0.4 0.5 ]
### 6.2. Statistical Functions
NumPy has a bunch of statistical functions that can summarize data. Here are some of them:
• mean(): Returns the mean of an array of values.
• median(): Returns the middle value in an array when it is sorted in increasing or decreasing order.
• min(): Returns the smallest value in an array.
• max(): Returns the largest value in an array.
• std(): Returns the standard deviation of the values in an array.
• var(): Returns the variance of the values in an array.
import numpy as np
data = np.array([11, 13, 17, 20, 23])
mini = np.min(data)
maxi = np.max(data)
mean = np.mean(data)
median = np.median(data)
std_dev = np.std(data)
variance = np.var(data)
print(mini)
print(maxi)
print(mean)
print(median)
print(std_dev)
print(variance)
Here’s the output:
11
23
16.8
17.0
4.4
19.36
### 6.3. Comparing Two Arrays
We can use the comparison operators (==, !=, <, >, <=, >=) to compare two NumPy arrays element-wise. The results of these comparisons will be Boolean arrays.
import numpy as np
arr1 = np.array([2, 3, 4])
arr2 = np.array([4, 3, 2])
more_than = arr1 > arr2
equal_to = arr1 == arr2
print(more_than)
print(equal_to)
Here’s the result:
[False False True]
[False True False]
### 6.4. Manipulating Strings Stored in Arrays
NumPy provides a set of string functions that can be used to manipulate strings stored in arrays. Some functions include:
• lower(): Converts all the characters in a string to lowercase.
• upper(): Converts all the characters in a string to uppercase.
• str_len(): To find the length of all the strings.
import numpy as np
names = np.array(['Alice', 'Bob', 'Charlie'])
uppercase = np.char.upper(names)
lowercase = np.char.lower(names)
length = np.char.str_len(names)
print(uppercase)
print(lowercase)
print(length)
Here’s the result:
['ALICE' 'BOB' 'CHARLIE']
['alice' 'bob' 'charlie']
[5 3 7]
### 6.5. Matrix Operations
NumPy has lots of functions for working with matrix arrays. Some of these functions are:
• dot(): calculates a matrix’s dot product.
• transpose(): transposes a matrix.
• inverse(): calculates a matrix’s inverse.
import numpy as np
matrix1 = np.array([[1, 2], [3, 4]])
matrix2 = np.array([[5, 6], [7, 8]])
result_dot = np.dot(matrix1, matrix2)
print("Dot Product:")
print(result_dot)
transpose_matrix = matrix1.T
print("Transposed Matrix:")
print(transpose_matrix)
inverse_matrix = np.linalg.inv(matrix1)
print("Inverse Matrix:")
print(inverse_matrix)
Here’s the result:
Dot Product:
[[19 22]
[43 50]]
Transposed Matrix:
[[1 3]
[2 4]]
Inverse Matrix:
[[-2. 1. ]
[ 1.5 -0.5]]
### 6.6. Set Operations
We can do set operations on arrays as well. These include:
• union(): returns the union of two arrays.
• intersection(): returns the intersection of two arrays.
• difference(): returns the difference between two arrays.
import numpy as np
set1 = np.array([1, 2, 3, 4])
set2 = np.array([3, 4, 5, 6])
union = np.union1d(set1, set2)
intersection = np.intersect1d(set1, set2)
difference = np.setdiff1d(set1, set2)
print(union)
print(intersection)
print(difference)
Here’s the result:
[1 2 3 4 5 6]
[3 4]
[1 2]
### 6.7. Vectorization
A vectorization involves applying a function to each element of an array. Consider the following example where we apply the square root and sin function on each element of the array.
import numpy as np
arr = np.array([4, 9, 16, 25])
result_sqrt = np.sqrt(arr)
result_sin = np.sin(arr)
print("\nElement-wise Square Root:")
print(result_sqrt)
print("\nElement-wise Sine:")
print(result_sin)
Here’s the result:
Element-wise Square Root:
[2. 3. 4. 5.]
Element-wise Sine:
[-0.7568025 0.41211849 -0.28790332 -0.13235175]
## 7. Error Handling
NumPy provides several functions that can be used to handle errors, including:
• try/except: handles errors that occur in the code.
• assert: checks for conditions that should never be true.
• isnan(): checks if a value is NaN (Not a Number).
import numpy as np
# Assert Statement
x = np.array([1, 2, 3])
y = np.array([1, 2, 4])
try:
assert len(x) == len(y), "Arrays must have the same length"
except AssertionError as e:
print("Assertion Error:", e)
else:
print("Lenghts are same")
# isnan() Function
z = np.array([1.0, np.nan, 3.0, np.nan, 5.0])
is_nan = np.isnan(z)
print("\nOriginal Array:")
print(z)
print(is_nan)
Here’s the result:
Lenghts are same
Original Array:
[ 1. nan 3. nan 5.]
[False True False True False]
## 8. Data Visualization
Next, we will dive into data visualization using NumPy.
First, we need to install matplotlib and import it. We should also open a Jupyter notebook so we can run all the code and see the actual visualization. NumPy has a number of cool ways to show data visually, such as line plots, scatter plots, bar graphs, and histograms. Visualizing data helps us quickly understand large data sets.
### 8.1. Line Plot
In NumPy, a line plot displays data as a series of points connected by a line. The plot() function is used to line plot the data, and it takes two arguments: the x-coordinates and the y-coordinates.
Let’s see an example.
import numpy as np
import matplotlib.pyplot as plt
fruit = np.array(["Apple", "Banana", "Orange", "Grapes", "Mango", "Strawberry"])
weight = np.array([150, 120, 180, 85, 200, 50])
plt.plot(fruit, weight)
plt.show()
Here’s the result:
In this case, we’ve used plot() to plot the data. The x and y coordinates are set according to the fruit and weight arrays.
### 8.2. Scatter Plot
The scatter plot displays data as a collection of points. Use the scatter() function to plot the data.
import numpy as np
import matplotlib.pyplot as plt
fruit = np.array(["Apple", "Banana", "Orange", "Grapes", "Mango", "Strawberry"])
weight = np.array([150, 120, 180, 85, 200, 50])
plt.scatter(fruit, weight)
plt.show()
Here’s the result:
So, in this example, we used the scatter() function to plot the data points we had. We just passed the fruit and weight as the x and y coordinates. But, in a scatter plot, we can also pass the c and s arguments to set the color and size of the points. For example,
import numpy as np
import matplotlib.pyplot as plt
fruit = np.array(["Apple", "Banana", "Orange", "Grapes", "Mango", "Strawberry"])
weight = np.array([150, 120, 180, 85, 200, 50])
colors = np.array([1, 2, 3, 4, 5, 6])
sizes = np.array([21, 41, 61, 81, 101, 121])
plt.scatter(fruit, weight, c=colors, s=sizes)
plt.show()
Here’s the result:
### 8.3. Bar Graph
Bar graphs are like rectangular boxes that show data. NumPy has bar() function that we can use to plot data in a bar graph.
For example,
import numpy as np
import matplotlib.pyplot as plt
fruit = np.array(["Apple", "Banana", "Orange", "Grapes", "Mango", "Strawberry"])
weight = np.array([150, 120, 180, 85, 200, 50])
plt.bar(fruit, weight)
plt.title('Bar Graph')
plt.show()
Here’s the result:
Here, we have used the bar() function to plot the bar graph and pass two arrays, fruit and weight, as its arguments.
### 8.4. Histogram
NumPy uses hist() to create histograms. Here’s an example.
import numpy as np
import matplotlib.pyplot as plt
weight = np.array([0.6, 1.8, 2.2, 2.5])
plt.hist(weight)
plt.show()
Here’s the result:
## 9. Advantages of NumPy
Let’s discuss some of the great advantages of NumPy:
• NumPy arrays use less memory. NumPy’s arrays are more compact in size than Python lists.
• The speed is also great. NumPy arrays perform computations faster than Python lists. The NumPy library uses the BLAS (Basic Linear Algebra Subroutines) library as its backend.
• In comparison with Python lists, NumPy is more efficient and faster at performing mathematical calculations on arrays and matrices.
• It is open source and all features can be accessed for free.
• In Numpy arrays, there are various functions, methods, and variables, which simplify the computation of matrices.
## 10. Limitations of NumPy
Apart from advantages, there are also some limitations.
• NumPy can have a steep learning curve, especially for beginners who are not familiar with array programming concepts.
• NumPy arrays can consume more memory than Python lists because they store additional metadata and type information with each element. This can lead to memory problems, especially in systems with limited memory.
• NumPy arrays do not have a built-in way to represent missing values (NaN). This can be a problem for data analysis tasks requiring missing values handling.
• NumPy arrays require all elements to be of the same data type. This can limit their use for handling data structures that contain different types of data.
## 11. Conclusion
In this Python tutorial, we have discussed how to get started with NumPy. We took a look at the data types of NumPy, and how to use NumPy for data visualization, followed by the advantages and limitations of NumPy.
Happy Learning!
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# Number Theory Warmups: Level 3 Challenges
One of the seven goblets above is made of real gold. If you start counting at A and wind back and forth while counting (A, B, C, D, E, F, G, F, E, D, ...), then the golden goblet would be the $1000^\text{th}$ one that you count.
Which one is the golden goblet?
$\Huge {\color{#3D99F6}9}^{{\color{#20A900}8}^{{\color{#D61F06}7}^{{\color{#624F41}6} ^{\color{magenta}5}}}}$
What are the last two digits when this integer fully expanded out?
Find the sum of all positive integers $\displaystyle n$, such that $\displaystyle \dfrac{(n+1)^2}{n+7}$ is an integer.
Find the sum of all prime numbers $p$ that divides $\underset{\text{the digit 1 is repeated }p \text{ times}}{\underbrace{111111111111111\cdots 1}}$.
There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.
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## 1st Edition
Chapman and Hall/CRC
376 pages
Hardback: 9781584882039
pub: 2000-07-21
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### Description
Boolean algebras have historically played a special role in the development of the theory of general or "universal" algebraic systems, providing important links between algebra and analysis, set theory, mathematical logic, and computer science. It is not surprising then that focusing on specific properties of Boolean algebras has lead to new directions in universal algebra.
In the first unified study of polynomial completeness, Polynomial Completeness in Algebraic Systems focuses on and systematically extends another specific property of Boolean algebras: the property of affine completeness. The authors present full proof that all affine complete varieties are congruence distributive and that they are finitely generated if and only if they can be presented using only a finite number of basic operations. In addition to these important findings, the authors describe the different relationships between the properties of lattices of equivalence relations and the systems of functions compatible with them.
An introductory chapter surveys the appropriate background material, exercises in each chapter allow readers to test their understanding, and open problems offer new research possibilities. Thus Polynomial Completeness in Algebraic Systems constitutes an accessible, coherent presentation of this rich topic valuable to both researchers and graduate students in general algebraic systems.
### Reviews
"This book gives a thorough, systematic treatment of various notions of polynomial completeness … the book is overdue as a reference for universal algebraists."
Mathematical Reviews, 2003a
ALGEBRAS, LATTICES, AND VARIETIES
Algebras, Languages, Clones, Varieties
Congruence Properties
CHARACTERIZATIONS OF EQUIVALENCE LATTICES
Introduction
Arithmeticity
Compatible Function Lifting
PRIMALITY AND GENERALIZATIONS
Primality and Functional Completeness
Near Unanimity Varieties
Arithmetical Varieties
Generalizations of Primality
Categorical Equivalence
AFFINE COMPLETE VARIETIES
Introduction and Instructive Examples
General properties
Varieties with a Finite Residual Bound
Locally Finite Affine Complete Varieties
POLYNOMIAL COMPLETENESS IN SPECIAL VARIETIES
Strictly Locally Affine Complete Algebras
Modules
Lattices
Algebras Based on Distributive Lattices
Semilattices
Miscellaneous Results | 539 | 2,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-09 | longest | en | 0.863766 |
http://math.stackexchange.com/questions/133794/existence-of-infinite-number-of-infinite-cardinals-st | 1,466,975,446,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395560.14/warc/CC-MAIN-20160624154955-00016-ip-10-164-35-72.ec2.internal.warc.gz | 192,797,120 | 18,216 | # Existence of infinite number of infinite cardinals st:
Prove the existence of infinite number of infinite cardinals
1) $\alpha$, such that $\alpha<\alpha^\aleph$
2) $\beta$, such that $\beta=\beta^\aleph$
-
You might want to look at meta.math.stackexchange.com/questions/1803/… for some tips on asking questions here (even if this isn't homework). – Gerry Myerson Apr 19 '12 at 6:36
The first problem is quite a bit harder than the second.
For $1$), use König's theorem. We still then need to show that, for example, there are infinitely many cardinals of cofinality say $\omega$, but that part is not hard. For suppose we start at the infinite cardinal $\kappa_0$. Let $\kappa_1=2^{\kappa_0}$, $\kappa_2=2^{\kappa_1}$, and so on, and let $\kappa_\omega=\bigcup \kappa_n$. Then $\kappa_\omega$ has a right cofinality, and König's theorem applies. For the next one, start at $\kappa_\omega$.
For $2$), we can use for $\beta$ anything of shape say $\kappa^\aleph$, where $\kappa$ is an infinite cardinal.
-
Actually just taking $\alpha=\kappa^{+\omega}$ would ensure that $\alpha$ has countable cofinality. – Asaf Karagila Apr 19 '12 at 18:55
Yes, I am taking (possibly) larger hops than necessary. – André Nicolas Apr 19 '12 at 18:56
Using $\aleph$ as a general cardinal might be ambiguous (there are places where it is used particularly for $2^{\aleph_0}$), the answer remains the same regardless to the intended use of $\aleph$.
1. Recall that for every infinite $\kappa$ we have $\kappa<\kappa^{\operatorname{cf}(\kappa)}$. Simply show that there are infinitely many cardinals whose cofinality is $\aleph$. You can show that there exists a sequence $\alpha_0<\alpha_0^\aleph<\alpha_1<\ldots$ by starting the construction of $\alpha_{n+1}$ from the construction of $\alpha_n^\aleph$.
2. Cardinal exponentiation has the property $\left(\kappa^\lambda\right)^\mu=\kappa^{\lambda\cdot\mu}$. Take $\alpha$ from the previous part and take $\beta=\alpha^\aleph$, now we have $\beta^\aleph=\alpha^{\aleph\cdot\aleph}=\alpha^\aleph=\beta$.
Note that for the second part you don't really need the first part, however it is easy to show that there are infinitely many $\beta$ if you use the fact that $\alpha_n^\aleph\neq\alpha_k^\aleph$ for $n\neq k$.
-
If anyone is interested, there's a short proof of the OP's question in the case of $\aleph_{0}$ at groups.google.com/group/sci.math/msg/324c39aa393fa801 Like Asaf Karagila's argument, it shows slightly more than what was asked by showing the existence of arbitrarily large such cardinals, and not just infinitely many such cardinals. – Dave L. Renfro Apr 19 '12 at 15:30 | 752 | 2,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2016-26 | latest | en | 0.835123 |
https://www.smore.com/tx921 | 1,638,196,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00345.warc.gz | 1,107,102,078 | 16,348 | ## Weekly Objectives
Reading- I can determine the central idea and supporting details of nonfiction texts.
Writing- I can compose an expository composition.
Math-I can identify the difference between multiplication and division word problems. I can use strategies to solve multi-step word problems in multiplication and division.
Science-I can identify the importance of conservation of our natural resources. I can come up with ways to reduce, reuse, and recycle.
Social Studies- I can identify traits of an everyday hero.
## Homework/Tests
Reading- Read each day for at least 20 minutes. We are studying nonfiction right now, so reading any type of nonfiction text would be great!
Vocabulary Unit 7 Quiz on Thursday
We are starting a new program called Lexia Core 5. Students can log in through Clever and practice at home. The activities are specially curated for each student in order to address his or her personal needs, and students need to spend at least 40 minutes a week on the program between school and home.
Writing- Spelling Quiz on Friday!
Math-Practice Multiplication Facts 3x a week for 10 minutes!
Math Quiz Friday, January 22nd.
We will review all week to be prepared for our quiz.
Science- Unit 8 Formation of Soil and Natural Resources Assessment will be on January 27th. Study guide to begin reviewing concepts:
Social Studies- None
## Water Bottles
Please remember to send a water bottle with your child, so he or she will have something to drink throughout the day and can refill it as needed. We don't have access to our regular water fountains due to Covid.
## Vocabulary Website to Practice at Home
Click on the green book. We are working on unit 7.
## MULTIPLICATION PRACTICE AT HOME: PRACTICE, PRACTICE, PRACTICE!
A big piece of 3rd grade math is to master multiplication facts up to X9. This will help them succeed in both 3rd and 4th grade. This is a huge goal, but can easily be achieved with practice at school and at home.
Have your child set a schedule to practice 2-3 times a week for 10 minutes a day for mastery.
They can also focus on a few facts a month to break the goal into chunks. Here is an example schedule:
December: 0, 1, 2, and 3
January: 5 and 10
February: 4 and 6
March: 7 and 8
April: 9
May: All Facts Fluently
Below are some ways to practice at home:
• Your child can sign on to xtramath.org at home and practice.
• Old-fashioned flash cards (send your child's teacher an email if you need some)
• Create chants and rhymes to help memorize (example-https://www.lewispalmer.org/cms/lib/CO01900635/Centricity/Domain/874/MultiplicationRhymes.docx.pdf)
• Play Multiplication War at-home (directions-https://www.multiplication.com/our-blog/jen-wieber/multiplication-war-card-game-0)
• Listen to Multiplication Songs (example-https://www.youtube.com/playlist?list=PLWphMREEQDrjzT4i4YKwkG93UUsM_X3Hp
Here is a website that has songs, apps, and hands-on tools for at-home practice as well:
## Please subscribe to our Edlio pages.
We will send an update every Friday. Please subscribe to our Edlio pages to get our updates. | 745 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.945142 |
https://www.physicsforums.com/threads/uniform-convergence-and-derivatives-question.615830/ | 1,508,219,910,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820927.48/warc/CC-MAIN-20171017052945-20171017072945-00200.warc.gz | 1,236,809,347 | 16,161 | # Uniform convergence and derivatives question
1. Jun 22, 2012
### Boorglar
In Spivak's Calculus, there is a theorem relating the derivative of the limit of the sequence {fn} with the limit of the sequence {fn'}.
What I don't like about the theorem is the huge amount of assumptions required:
" Suppose that {fn} is a sequence of functions which are differentiable on [a,b], with integrable derivatives fn', and that {fn} converges (pointwise) to f. Suppose, moreover, that {fn'} converges uniformly on [a,b] to some continuous function g. Then f is differentiable and f'(x) = lim n-->infinity fn'(x). "
Are really EACH of these assumptions necessary for this to be true? Are there counterexamples for any combination of missing hypotheses? With all these assumptions the proof is quite easy, and I suspect this might be the reason, but in this case, how many of these assumptions can we get rid of?
I've seen the counterexample of fn = sqrt(x^2+1/n^2), which converges uniformly to |x|, which is not differentiable. And also fn = 1/n*sin(n^2 x) which converges to 0 but the derivatives of fn do not always converge.
But what about counterexamples involving non-integrable derivatives, non-uniform convergence to a continuous g, or uniform convergence to a function g which is not continuous? And doesn't uniform convergence of the derivatives imply at least pointwise convergence of the functions? etc, etc... I think you get my point (no pun intended)...
Last edited: Jun 22, 2012
2. Jun 22, 2012
### micromass
Staff Emeritus
That seems a bit stronger than it needs to be. Here is a more general theorem:
Let $(f_n:[a,b]\rightarrow \mathbb{R})$ a sequence of functions such that
1) $f_n$ is continuous on [a,b].
2) $f_n$ is differentiable on ]a,b[.
3) There exists an $x_0\in [a,b]$ such that $f_n(x_0)$ converges.
4) The sequence $(f_n\vert_{]a,b[})_n$ converges uniformly.
Then
1) $(f_n)_n$ is uniformly convergent
2) The uniform limit f is differentiable on ]a,b[
3) $f_n^\prime(x)\rightarrow f^\prime(x)$ for all x in ]a,b[
Note that in complex analysis, if we use complex differentiability, then this statement simplifies even more!!
3. Jun 23, 2012
### Boorglar
Thanks for this reply! I thought about it a lot and I found counterexamples when uniform convergence of {fn'} is not assumed, which means that this condition is in fact essential. Also, I see how we could simplify the theorem to your version, but I still have a problem:
Basically, if {fn} and {fn'} both converge uniformly on [a,b], I showed that the theorem is true. Now I need to show that uniform convergence of {fn'} together with convergence of {fn(x0)} implies uniform convergence of {fn}. This step seems difficult, and I could only prove it by assuming integrable fn'. The reason is that I can't make any link between fn' and fn if fn' is not integrable, since the FTC does not apply anymore...
Do you know how this step is proven for non integrable fn' ? (It's already quite hard to think of fn' which are not integrable, and they probably never appear in practice, but I simply want to know, out of curiosity and satisfaction).
4. Jun 23, 2012
### micromass
Staff Emeritus
Apply that $f_n(x_0)$ is Cauchy to find
$$\|f_p(x_0)-f_q(x_0)\|\leq \frac{\varepsilon}{2}$$
Apply uniform convergence of $f^\prime_n$ to find
$$\sup_{y\in ]a,b[}{\|f_p^\prime(y)-f_q^\prime(y)\|}\leq \frac{\varepsilon}{2(b-a)}$$
Apply the mean-value theorem to get
$$\begin{eqnarray*} \|f_p(x)-f_q(x)\| & \leq & \|(f_p-f_q)(x)-(f_p-f_q)(x_0)\|+\|(f_p-f_q)(x_0)\|\\ & \leq & \sup_{y\in ]a,b[}{\|f_p^\prime(y)-f_q^\prime(y)\|} +\|(f_p-f_q)(x_0)\|\\ & \leq & \varepsilon \end{eqnarray*}$$
So $f_n$ is a uniform Cauchy sequence and thus uniformly convergent by completeness.
5. Jun 25, 2012
### Skrew
You need uniform convergence of f'n(x) and to make it easier restrict f'n(x) to be continuous.
As an example f_k(x) = (1\k)sin(x*k) converges uniformly on R to 0 but it's derivative cos(xk) doesn't converge.
In the reals you don't have any restrictions on the derivative of the function based on the max/min values the function takes.
In complex analysis you have the cauchy estimate and the cauchy integral formual, which lets you show that if a sequence of holomorphic functions converges to a function g, then g is holomorphic and the derivative of the sequence converges to g'. | 1,217 | 4,367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-43 | longest | en | 0.908523 |
https://socratic.org/questions/an-object-is-at-rest-at-2-1-6-and-constantly-accelerates-at-a-rate-of-1-4-m-s-2- | 1,623,669,557,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00579.warc.gz | 461,422,455 | 6,624 | # An object is at rest at (2 ,1 ,6 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (6 ,4 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.
Jul 20, 2017
The time is $= 6.39 s$
#### Explanation:
The distance is
$s = \sqrt{{\left(6 - 2\right)}^{2} + {\left(4 - 1\right)}^{2} + {\left(5 - 6\right)}^{2}}$
$= \sqrt{16 + 9 + 1}$
$= \sqrt{26} m$
The initial velocity is $u = 0 m {s}^{-} 1$
The acceleration is $= \frac{1}{4} m {s}^{-} 2$
We apply the equation of motion
$s = u t + \frac{1}{2} a {t}^{2}$
$t = \sqrt{\frac{2 s}{a}}$
$t = \sqrt{\frac{2 \cdot \sqrt{26}}{\frac{1}{4}}}$
$t = 6.39 s$
Jul 20, 2017
$6.387$ seconds
#### Explanation:
The distance between $\left(2 , 1 , 6\right)$ and 6,4,5) is
$\sqrt{{\left(6 - 2\right)}^{2} + {\left(4 - 1\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{16 + 9 + 1} = \sqrt{26}$, and it is in meters.
As distance $S = \frac{1}{2} a {t}^{2}$, where $a$ is accelaration and $t$ is time taken, then
$t = \sqrt{\frac{2 S}{a}} = \sqrt{\frac{2 \sqrt{26}}{\frac{1}{4}}}$
= $\sqrt{8 \sqrt{26}} = 2 \sqrt{2 \times 5.09902} = 2 \sqrt{10.19804} = 6.387$ | 504 | 1,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 19, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.706724 |
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