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# Mathematics lessons
Fun Maths for brushing up our maths skills !!!
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### Mathematics lessons
1. 1. Today We Learn Mathematics
2. 2. The goal of this lesson is, solvingmaths problems by applying what you learned at school.
3. 3. 3 digits; and a result are given.You have to put all the mathematical signs thatare necessary to restore the equality.For example, 2 2 2 = 6is given. You are required to put any suitingmathematical symbol as shown below : 2 + 2 + 2 = 6Easy! Isnt it? It is the same for the remaining.
4. 4. 1 1 1 = 62 2 2 = 63 3 3 = 64 4 4 = 65 5 5 = 66 6 6 = 67 7 7 = 68 8 8 = 69 9 9 = 6
5. 5. What…. could you solve some? … No, not the No.2 :it was the example, already shown to you presently .Another? Ah… might be the No. 6. 6 + 6 - 6 = 6 EINSTEIN !!!!!
6. 6. And the others? ? Do you want assistance?No, try for some more time without looking forwardThink that you solved the 3rd one , Perhaps the 5th one.With a little exercise, the 7th . (3 x 3) - 3 = 6 (5 / 5) + 5 = 6 7 - (7/ 7) = 6Give a try for the remaining ones. You can definitely dosome more.
7. 7. Now let us see those which are a little more complicated. The 4th. 4 + 4 + 4 = 6 The 9th. ( 9 x 9) - 9 = 6 The 8th. 3 3 3 8+ 8+ 8 = 6The last one how, 1 1 1 = 6 ? Give a 24-hour big trybefore moving ahead !!!!
8. 8. On this, we finish today’s class It remains to solve the first one. 1+1+1 = 3 3x2x1 = Dunce’s cap! (1 + 1 + 1)! = 6Well, it must be acknowledged. This one is muscular ... …Now, You have brushed-up a bit, yourgifted brain in Maths, after a long gap, eh!! FACTORIAL: the factorial of any number is obtained by multiplying all it’s former numbers up to 1. It is symbolised by the exclamatory mark.
9. 9. Good, Considering yourperformances, the next course will treat the art of walkingwhile chewing chewing-gum.. Kindly: THE PROF.
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Fun Maths for brushing up our maths skills !!!
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0 | 746 | 2,253 | {"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} | 4.40625 | 4 | CC-MAIN-2021-31 | latest | en | 0.898621 |
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## Stefanie G.'s Resources
-2x+4>5(x-2) There are two ways to solve the equation. It all depends on how you go about it. Both ways are correct and both ways get you the same answer. The first step to both is expanding 5(x-2). Just multiply 5 by x and by 2. -2x+4>5x-10 Here is where you can go one...
9/5(2/3+.23) Convert .23 into a fraction. While you could convert 9/5 and 2/3 into decimals, 2/3 does not reduce into a whole number and you would be forced to round, thus not making the answer the "exact" answer. 9/5(2/3+23/100) You can now add the two fractions... | 185 | 591 | {"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-2015-48 | latest | en | 0.930521 |
https://math.stackexchange.com/questions/1410668/intuition-why-distinct-eigenvalues-implies-linearly-independent-eigen-vector | 1,566,766,747,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330800.17/warc/CC-MAIN-20190825194252-20190825220252-00537.warc.gz | 542,235,485 | 32,752 | # Intuition: why distinct eigenvalues $\implies$ linearly independent eigen vectors?
Suppose you have an $$~n\times n~$$ matrix with $$~n~$$ distinct (not repeated) eigenvalues.
There is a theorem telling us that the eigen vectors corresponding to these eigenvalues must be linearly independent.
I can basically follow the proof, but I am looking for an intuitive explanation of why this is the case.
Can anyone offer some insight?
• It goes smoothly with two eigenvectors. If they were linearly dependent, they would be scalar multiples of each other, and thus belong to the same eigenspace. This "intuitive" explanation does not generalize well to higher $n$. We might try with a 3D-harmonic oscillator, when the eigenvectors of the matrix giving the force as function of the displacement are the "pure" oscillations (at distinct frequencies).... No. That doesn't generalize well either. – Jyrki Lahtonen Aug 26 '15 at 19:19
A good way to think about eigenvalues/eigenvalues are that the eigenvectors are the vectors that are stretched or flipped(which you can think of "negative stretching"), and the eigenvalues are how much they get stretched by.
So if we have some eigenvectors, say with eigenvalues 1,2,3. Then we have eigenvectors $v_1,v_2,v_3$. Let's say that $v_3$ is a linear combination of the other two. The intuition is basically that if we can stretch the first two by their eigenvalues, then the action on the third with already be determined and "can't" be three. So the only way for them to be able to stretch without effecting the others is for them to be linearly independent.
This intuition is similar if we consider the vector space over $\mathbb{C}$ rather than $\mathbb{R}$, as complex multiplication corresponds to a combination of rotating and stretching.
I'm really sorry for the hand made graphs but I believe they can help
Think about the opposite case, can dependent eigen vectors have distinct eigen values?
In $$\mathbb{R}^2$$, the only way for two vectors to be dependent is to be collinear
as both have the same span, it's easy to see why they can't have distinct eigen values.
Now in $$\mathbb{R}^3$$ the only way for $$3$$ vectors to be dependent is to be coplanars
Here it is a little bit trickier to visualize, you can't have a transformation that causes two vectors to stretch with distinct factors without distorting the plane.
For example assume the three vectors $$p_1,p_2,p_3$$ to be eigen vectors for transformation $$T$$ can $$p_1$$ stretched with $$\lambda_2=2$$ , $$p_3$$ stretched with $$\lambda_3=3$$.
It can't be the case since the resulted plane will not be linear:
For both cases $$\mathbb{R}^2,~\mathbb{R}^3$$ the only way to keep linearity with dependent eigen vectors would be to have non distinct eigen values.
Hence, since the transformation is linear and the eigen values are distinct the eigen vectors cannot be dependent.
You can go from here and generalize to: for $$\mathbb{R}^n$$ if there exist $$n$$ distinct eigen value then there must be $$n$$ linearly independent eigen vector.
You may understand it better if you try a simple proof for low dimensional spaces
With one eigenvector, things are easy. If $$f$$ is linear and $$f(\vec{v})=\lambda \vec{v}$$ then $$f(k\vec{v}) = k f(\vec{v})=k\lambda \vec{v} = \lambda(k\vec{v})$$ (in short, all multiples of $$\vec{v}$$ are eigenvectors for the same eigenvalue, so they cannot be eigenvectors for a different one)
Things get trickier when we add more vectors: Let's say now $$\vec{v},\vec{w}$$ are linearly independent eigenvectors for the eigenvalues $$\lambda, \mu$$ respectively. Let's say we have a linear combination $$a\vec{v}+b\vec{w}$$ that is also an eigenvector for an eigenvalue that we'll call $$\eta$$.
Since $$f$$ is linaer, $$f(a\vec{v}+b\vec{w})=af(\vec{v})+bf(\vec{w})=a\lambda \vec{v} + b\mu \vec{w}$$
But, since $$f(a\vec{v}+b\vec{w})$$ is an eigenvector, $$a\lambda \vec{v} + b\mu \vec{w} = \eta(a\vec{v}+b\vec{w})$$ Now:
• If either $$a$$ or $$b$$ are zero, then we are back to the one-eigenvector case.
• If both $$a$$ and $$b$$ are non-zero, then $$\eta=\lambda$$ and $$\eta=\mu$$ are both true, so $$\lambda=\mu$$ | 1,131 | 4,175 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-35 | latest | en | 0.925117 |
https://ru.scribd.com/document/205052070/Introduction-to-Conformal-Mapping | 1,568,853,101,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573385.29/warc/CC-MAIN-20190918234431-20190919020431-00155.warc.gz | 649,687,812 | 84,438 | Вы находитесь на странице: 1из 8
# Introduction to Conformal Mapping in Aerodynamics
Introduction Conformal mapping is a method used to extend the application of potential flow theory to practical aerodynamics. Standard potential flow theory begins with an ideal flow to show that lift on a body is proportional to the circulation about a closed path encompassing an object. Potential flows start with flows over cylinders since the mathematics is more tractable. However, to use potential flow theory on usable airfoils one must rely on conformal mapping to show a relation between realistic airfoil shapes and the knowledge gained from flow about cylinders. 'ift # )
'#(
Brief review of complex numbers: Conformal mapping relies entirely on complex mathematics. herefore, a brief review is undertaken at this point. ! complex number z is a sum of a real and imaginary part" z # real \$ iimaginary he term i, refers to the complex number
i = *,
-
i=
i = *,
*
i = i,
,
so that" Complex numbers can be presented in a graphical format. %f the real portion of a complex number is taken as the abscissa, and the imaginary portion as the ordinate, a two& dimensional plane is formed. z = real + iimaginary = x + iy
y, imaginary
i =*
x, real
! complex number can be written in polar form using .uler/s e0uation" z = x + iy = rei = r(cos + isin) where r2 = x2 + y2
## Complex multiplication1 z1z2 = (x1+iy1)(x2+iy2) = (x1x2 - y1y2) + i(x1y2 + y1x2)
= r*e
i *
r- e
i -
= r* r- e
i 3* + - 6
Complex representation of potential flows he basic flows used in potential flow theory such as uniform flow, source, sink, doublet and vortex, can all be represented using complex numbers. 2or example, if a complex number w with both real and imaginary parts represents a potential flow, then the form of the number is" w(z) = + i # 3velocity potential) + i(stream function) Here, both velocity potential and stream function are themselves complex numbers. !s an example, the uniform flow can be written" 4niform flow1 w(z) = Vz = + i = V(x+iy) = Vx + Vy as seen previously, Source flow1
w=
i ln3 z 6 = + i = ln3re 6 = 3ln3r 6 + i 6 = ln3r 6 + i - - - - -
= Vx = V rcos
= Vy = V rsin
(ortex flow1
w=i
i ln3 z 6 = + i = i ln3re 6 = i 3ln3r 6 + i 6 = + i ln3r 6 - - - - -
5oublet flow1
w= k * k * k * i k * = + i = = 3cos i sin 6 e = i - z - re - r - r
## %n complex terms the flow past a cylinder with lift is written1
w3 z 6 = V z + z + i - ln3 z 6
(elocity Components1 7hen a potential flow is represented in complex form, the velocity components can be found using one of two methods" *. 8e&write the expression from the complex variable z form into its separate real and complex components. he form of this expression will be w # \$ i! he individual velocity components are found by completing the appropriate differentiation on or ! to obtain u or v. !s an example consider the complex form of the source flow"
w = ln3 z 6 = + i = ln3r 6 + i - - - Vr = = V = =9 r -r
-. !n alternative method would be to differentiate on the complex expression directly and then separate the real and complex portions to obtain the velocity components according to" "w = u iv "z Conformal :apping ! conformal mapping is performed through the transformation of a complex function from one coordinate system to another. ! transformation function is applied to the original function to perform the mapping. 2or aerodynamics applications the ;oukowski transform is the most commonly used function"
w=z+ # z
-
Here, # is a constant. <raphically, a conformal mapping will transform a complex plane in z 3z = x+iy6 into a complex plane in a new variable w 3w # +i6. > plane y # =* # =, # C* # C # C, x # =,
w plane # C # C,
%n the diagram a uniform flow in the z plane is transformed into an e0uivalent form in the w plane using a transform of the form w # f(z). !s an example consider a circle drawn in the z plane, z = #ei! he ;oukowski transform maps the circle into a flat plate, w=z+ # z
w = #ei + # i = #ei + #e i = -# cos3 6 + i9 #e
> plane b x
w plane
w= z+# z
&-b
-b
! circle of radius # is mapped into a straight line in the w plane entirely on the real axis between &-# and -#. %f a uniform flow had been drawn over the circle, the transform would have mapped that flow into the flow over a flat plate in the w plane. %f the circle originally had a radius slightly larger than the transform constant #, z = aei, with a ? #, the circle would have formed an ellipse instead of the flat plate.
w = z + # = aei + # i = a + # cos3 6 + i a # sin3 6 = x + iy z a a ae
## 7hich can be written,
x #- a + a
-
y #- a a
-
=*
&-b
-b
%f the flow over a cylinder had been transformed it would have created the flow over an ellipse. ;oukowski !irfoils 2rom and aerodynamics point of view, the most interesting application of the ;oukowski transform is to an offset circle. %f we consider a circle slightly offset from the origin along the negative real axis, one obtains a symmetric ;oukowski airfoil.
y a > plane w plane
w= z+# z
eb b x
&-b
-b
he e0uation of the offset circle is z # aei-e# where the constant e is a small number. %f the cylinder is displaced slightly along the complex axis as well, one obtains a cambered airfoil shape.
a ! eb b y > plane w plane
@ x
w= z+# z !
&-b
@ -b
Here, the points \$ and % are the intercepts of the displaced circle on the real axis and their corresponding points in the transformed plane. he angle is the angle formed by the line joining the point \$ 3or %) and the origin with the real axis. %f lifting flow about the original circle had been imposed, the ;oukowski transformation would have generated a lifting flow about the ;oukowski airfoil"
y > plane w plane
w= z+# z
x
!lthough such a flow is mathematically possible, in reality it may not be realistic. he stagnation points on the cylinder map to stagnation points that are not always realistic. 2or instance the stagnation point on the top surface of the airfoil cannot exist is steady flight since the velocity would tend to infinity as one moves very close to the trialing edge. he only means of making a realistic flow is to impose the =utta condition where the stagnation point is forced to exist at the trailing edge thus making the streamlines flow smoothly from this point. his is done by adjusting the value of vorticity strength , such that the stagnation points on the cylinder reside at the cylinderAs intercepts of the real axis. %n this case, when the cylinder is transformed, one stagnation point will be forced to the trailing edge.
y > plane w plane
w= z+# z
x
he lift force generated by the lifting flow over the cylinder is proportional to the circulation about the cylinder imposed by the added vortex flow according to the =utta& ;oukowski relation, &' = V . he lifting force on the resulting ;oukowski airfoil is not clear. o evaluate the lift, the circulation is needed and therefore the velocity field. he velocity fields in each plane can be related to each other through the chain rule of differentiation. %f the lifting flow about the cylinder is defined as function ( where ( = ((z) in the z plane and ( = ((w) in the w plane, the velocities in each plane are"
Vz =
( z
Vw =
( w
@y chain rule1
( ( w = z w z
Vz =Vw w z
## 4sing the ;oukowski transformation"
w = z - # z zClearly, the velocity field very close to the cylinder and its transformed counterpart are dissimilar as one would expect. However, farther away from these objects the velocity fields become identical as the magnitude of z becomes larger than the constant value of #. Since the circulation can be calculated about any closed path, including paths very far from the object surface, the circulations must be the same in both planes.
V V cylin"er = )oukowski
(ortex strength he appropriate vortex strength to impose the =utta condition must be determined. Consider the lifting flow about a cylinder. he velocity in the direction is, V = -V sin3 6 + - Here, is the radius of the cylinder surface. his velocity is >ero on the surface of the cylinder at the stagnation points. !t the these points =-.
9 = -V sin3 6 -
= + V sin3 6
y > plane
& x
## %f the field is rotated by to simulate an angle of attack,
= + V sin3 +6
Since the cord length of the ;oukowski airfoil is +#, the lift coefficient can be written, *& = & =
* V -c -
* V - +# -
= +V sin3 + 6 -V # -V #
## :aking the assumption that # ,
*& = - sin3 + 6 - 3 + 6
.xample ! ;oukowski airfoil is formed by displacing a circle of radius * by x # &9.9B 3real axis6 and y # 9.9C 3imaginary axis6. 2ind, a6 (ortex strength if # 9o, and V = *9 mDs b6 *& at # 9o and # *9o y
9.9C &9.9B
tan3-.BE+ 6 =
9.9C 9.9B + # | 2,346 | 8,695 | {"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} | 3.859375 | 4 | CC-MAIN-2019-39 | latest | en | 0.874525 |
https://www.wikitechy.com/tutorials/python/programs-for-printing-pyramid-patterns-in-python | 1,713,618,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00290.warc.gz | 956,682,728 | 14,721 | p
python tutorial - Pyramid Python | Programs for printing pyramid patterns in Python - learn python - python programming
Patterns can be printed in python using simple for loops. First outer loop is used to handle number of rows and Inner nested loop is used to handle the number of columns. Manipulating the print statements, different number patterns, alphabet patterns or star patterns can be printed.
Simple pyramid pattern
Let us see how to print Half Triangle Pyramid pattern using an asterisk (star):
Output:
```*
* *
* * *
* * * *
* * * * *
```
After 180 degree rotation
python programming - Output :
``` *
* *
* * *
* * * *
* * * * *
```
Printing Triangle
Let us see how to print Full Triangle Pyramid pattern using an asterisk (star) :
python programming - Output :
``` *
* *
* * *
* * * *
* * * * *
```
Number Pattern
Here we will see how to print Half Triangle Pyramid pattern using numbers:
python programming - Output :
```1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
```
Numbers without re assigning
python programming - Output :
```1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
```
Character Pattern
Here we will see how to print Half Triangle Pyramid pattern using continous characters:
python programming - Output :
```A
B B
C C C
D D D D
E E E E E
```
Continuous Character pattern
Here we will see how to print Half Triangle Pyramid pattern using alphabets:
python - Sample - python code :
python programming - Output :
```A
B C
D E F
G H I J
K L M N O
```
UP NEXT IN Python Programming
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https://www.wallstreetmojo.com/vlookup-wildcard/ | 1,721,548,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00007.warc.gz | 903,961,459 | 83,331 | FLASH SALE! - "CHATGPT AND ARTIFICIAL INTELLIGENCE FOR MICROSOFT EXCEL AT 60% OFF" Enroll Now
# VLOOKUP Wildcard
Updated on April 22, 2024
Article byJeevan A Y
Edited byAshish Kumar Srivastav
Reviewed byDheeraj Vaidya, CFA, FRM
## What Is VLOOKUP With Wildcard?
VLOOKUP helps users to retrieve values using the lookup_value in a dataset. However, it cannot find or fetch the value if there is a slight difference in the lookup_value. To overcome such scenarios, we use the VLOOKUP with Wildcard characters.
For example, in the main table, the lookup_value is for employees “Shruthi and Abhishek”. But the lookup table shows the full names as “Shruthi Naidu and Abhishek Sharma”. Here, VLOOKUP does not recognize these two as the same, and ends up giving the #N/A error, not available, as shown in the image below.
Then, enter the formula with the VLOOKUP Wildcard characters in the corresponding column to get the correct result, irrespective of the changes in the lookup_value, as shown below.
### Key Takeaways
• VLOOKUP Wildcards are the Asterisk (*) and the Question Mark (?) characters that are inserted in a VLOOKUP formula to fix the “#NA” error that occurs when there is a difference in the lookup_value from the lookup_table.
• In the wildcard characters we use, the asterisk matches any number of characters, and the question mark (?) will match only one character where they are placed.
• The wildcard characters must be used within double quotes in the formula, and in the place where we enter the lookup_value. Else, we will get an error, or the required result will not be retrieved.
### How To VLOOKUP Wildcards (*, ?) In Excel?
We can VLOOKUP Wildcards in Excel (*, ?) in three ways, namely,
1. Typical VLOOKUP Function.
2. Asterisk (*)
3. Question Mark (?)
#### Method #1 – Typical VLOOKUP Function
fetches the data only if the lookup_value is the same as in the table array, otherwise, we will get a “#N/A” error.
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The steps to use VLOOKUP with a wildcard in Excel are,
1. Look at the below data in Excel.
We have two data tables named “Emp Table” and “Salary Table”, respectively. In the first table, we have “Employee Name,”Department,” and “Salary” details. But in the second table, we have only “Employee Name,” so we need to look for their respective salary details.
2. Let us apply the VLOOKUP function and see how many employees we get the salary details.
3. We get the following result.
4. We have got two error values in cells F4 and F8.
It is because, in cell F4, we have the employee’s name as “Shruthi,” but in the “Emp Table”, the actual name is “Shruthi Naidu” (cell A3).
Similarly, in the E8 cell, we have a value of “Abhishek, but in the “Emp Table”, the actual name is “Abhishek Sharma” (cell A4). Therefore, VLOOKUP does not find the similarity between these two names, and returns an #N/A error.
#### Method #2 – Asterisk (*)
As we know, we have only part of the full name of employees. To match this full name, we will combine the asterisk (*) with the VLOOKUP function.
##### Example #1
Look at the below data in Excel.
• We get the following result.
Similarly, we have got the remaining values.
As you can see above, we have used an asterisk (*) wildcard character after selecting the employee’s name cell to match any number of characters after the selected cell value.
##### Example #2
• Now, look at the below example of VLOOKUP.
The lookup_value is “Ramesh.” In the main table, the full name is even though we have used an asterisk as the wildcard. It does not return the result.
It is because we have used the asterisk (*) character after the cell address so that it will match any value after the cell value. But in this case, the value is “Ramesh,” but this is the middle name, so as of now, an asterisk will match only the last name and return the result as: “Ramesh Tendulkar,” but the actual value to be matched is “Sachin Ramesh Tendulkar.”
• So, to match text values before and after the lookup_value, we need to include a wildcard character asterisk before and after the lookup_value. The image below shows how to have a wildcard character.
• We get the following result.
As we can see above, we have included an asterisk (*) wildcard character before selecting the lookup_value cell and after choosing the wildcard character. So, the first asterisk will match the name “Sachin,” and the last asterisk will match the name “Tendulkar.” So, this will be the combination of “Sachin Ramesh Tendulkar.”
#### Method #3 – Question Mark (?)
If an asterisk matches any number of characters, then the question mark (?) will match only one character.
• Look at the below data.
In cell D2, we have the bill number as 3340, but in the actual main table, the full bill number value is LAN3340, so VLOOKUP cannot fetch the bill value because it does not find the exact bill value match.
• Let us apply the VLOOKUP function.
• We get the following result.
If you look at the bill numbers in the first table, we have the first three characters as alphabets and the followed by a numerical value. So, we can include three question marks to match the first three characters of the bill number.
So, three question marks “???” matched the first three characters before the bill number, and returned the lookup_value as “LAN3340,” so VLOOKUP does the rest as usual.
### Important Things To Note
• The wildcard characters are useful but not recommended because we may get incorrect results with a slight miss.
• Enter the last argument of the VLOOKUP function, i.e., range_lookup, as “False”, an approximate or an exact match. If we don’t enter any value, then the formula will consider “True”, by default.
1. What is the purpose of VLOOKUP with Wildcard?
VLOOKUP helps us get the desired details using the lookup_value. When there is a sight change in the lookup_value, we won’t get the required details but an error. To overcome this, we insert the wildcard characters in the formula.
2. What are the different wildcard characters in a VLOOKUP formula?
The two wildcard characters to use with the VLOOKUP formula are,
1. Asterisk (*) – A wildcard character that matches any number of characters after the selected cell value.
2. Question Mark (?) – It is a wildcard character that will match only one character after the selected cell value.
3. What are the reasons why the VLOOKUP Wildcard might not work?
A few reasons why a VLOOKUP formula may not work are,
The lookup_value is different from the value in the lookup_table.
The column of the lookup_value is on the left of the lookup_table.
A new column is inserted, or an existing column’s position is changed or deleted from the lookup_table. | 1,661 | 6,955 | {"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} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.81905 |
https://softwareengineering.stackexchange.com/questions/108487/conversion-of-a-number-from-single-precision-floating-point-representation-to-a | 1,563,488,929,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525829.33/warc/CC-MAIN-20190718211312-20190718233312-00277.warc.gz | 546,855,052 | 31,071 | # Conversion of a number from Single precision floating point representation to a Half precision floating point [closed]
I have a code where I have to work on Half precision floating point representation numbers. To achieve that I have created my own C++ class fp16 with all operators(arithmetic logical, relational) related to this type overloaded with my custom functions, while using a Single precision floating point number with a Half precision floating point number.
Half precision floating point = 1 Sign bit , 5 exponent bits , 10 significand bits = 16 bit
Single precision floating point = 1 Sign bit, 8 exponent bits, 23 significand bits = 32 bits
So what I do to convert from a Single precision floating point number to a Half precision floating point number:-
For significand bits - I use truncation i.e. loose 13 bits from the 32 bits to get 10 bits significand for half precision float.
What should I do to handle the exponent bits. How do I go from 8 exponent bits to 5 exponent bits?
Any good reading material would help.
## closed as off topic by Péter Török, Walter, yannis, user8 Jan 12 '12 at 11:47
Questions on Software Engineering Stack Exchange are expected to relate to software engineering within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.
• Why are you doing this? – Martin York Sep 15 '11 at 16:17
Convert to `+INF` or `-INF` as appropriate. The result certainly should be numerical, `NaN` is logically wrong. | 377 | 1,713 | {"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-2019-30 | longest | en | 0.864045 |
http://stackoverflow.com/questions/20706701/confusion-in-call-by-copy-restore-in-a-special-case | 1,406,367,587,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997900573.25/warc/CC-MAIN-20140722025820-00232-ip-10-33-131-23.ec2.internal.warc.gz | 350,906,634 | 16,238 | # Confusion in call by copy-restore in a special case
I understand the basic logic of call by copy restore. But I was wondering for a problem like this
``````void p(int x,int y) {
x++;
y+=2;
}
main() {
int a=10;
p(a,a);
// what will be now value of a, 11 or 12?
}
``````
-
Have you tried running this? Is this supposed to be C or something else? In C the value of `a` will be 10 (assuming `=++` was intended to be `++`) but I'm not sure I'm clear on what you're asking. Particularly, what is `x=++` intended to be? `x++`? – Andrew Myers Dec 20 '13 at 15:19
What language is this? It seems the code doesn't compile either. "x=++" ? – ChrisLava Dec 20 '13 at 15:19
I am not talking about any language here, I am just asking about call by copy restore method. – alienCoder Dec 20 '13 at 15:57
Perhaps you'll need to explain this "copy-restore method", and where you heard about it. It's not something that is in the C or C++ languages, which have call-by-value and call-by-reference. Perhaps this is a different language you are talking about (although the posted code appears to at least be an approximation of C)? – twalberg Dec 20 '13 at 16:51
@alienCoder I think you got your answer down there. – ChrisLava Dec 20 '13 at 22:04
Update: The answer is 12, see update below.
OK, this is actually a good question. So this explains what "copy-restore" is all about: http://stackoverflow.com/a/8871340/171933
Most programming languages don't support copy/restore, but only (some variations) of pass-by-value and pass-by-reference. So it's not so easy to try this out.
However, the question you are interested in is this: Which value wins? Does `x` get to write its value back to `a` when the function ends (which would be 11), or does `y` get to write its value back to `a` when the function ends (which would be 12).
In a language that supports "copy-restore", I'd hope that this would throw a compiler error.
Update:
After some searching I've found a language that actually supports "copy-restore", namely Ada. This is the code in Ada (this is my first and probably last program written in Ada):
``````with Ada.Text_IO; use Ada.Text_IO;
procedure copy_restore_example is
a: integer;
procedure p(x: in out integer; y: in out integer) is
begin
x:= x+1;
y:= y+2;
end p;
begin
a := 10;
Put_Line("Before :" & natural'image(a));
p(a, a);
Put_Line("After :" & natural'image(a));
end copy_restore_example;
``````
The result is 12, `y` wins. You can run this program in your browser here: http://www.compileonline.com/compile_ada_online.php
- | 700 | 2,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-23 | latest | en | 0.951749 |
https://alugha.com/videos/a78bd6f9-de26-11e6-ab71-1b3d629bd727 | 1,618,190,132,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00118.warc.gz | 210,989,179 | 23,319 | # Faulty Apple! Where is the spirit of Steve Jobs?! #MCMO
Broken Macbook speakers, iPhone batteries going from 30% to 1% in 5 seconds, extremely expensive Airpods, which are non-synchronous - Faulty Apple! Where is the spirit of Steve Jobs? Facebook-Page with 50k likes deleted - Zuckerberg tricked us: https://alugha.com/videos/022eacc9-cea5-11e6-8ded-89fa2ffa1f32 You have a video suggestion for us? Just drop it below in the comments or tweet us - we’ll check out what we can do! Our Sites & Info: hoTodi | http://hotodi.com Twitter | http://twitter.com/hotodi_com Facebook | https://www.facebook.com/hoTodi.tv Google+ | http://goo.gl/oIf5B Pinterest | http://pinterest.com/hotodi/ alugha | https://alugha.com Business inquiries | hotodi@alugha.com
## More videos by this producer
### Root, fifth and octave - Bobby's bass - Part 9
http://www.hoTodi.tv shows how it's done! Now that we are familiar with notes, the fretting hand and plucking hand, let's continue with some important basics. Today, we are going to learn what a root, a fifth and a concave is. is. We will play them on the strings and learn how to transpose them. H
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### The C major scale - Bobby's bass - Part 7
http://www.hoTodi.tv shows how it's done! Now that we have mastered the plucking hand and the fretting hand, we will continue with notes. The C major scale plays a special role here, because basically everything is based on it. As the basis for all other scales, tones and notes, everything stems fro | 490 | 1,853 | {"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-2021-17 | latest | en | 0.877575 |
http://www.velocityreviews.com/forums/t316658-p2-unsigned-short-addition-subtraction-overflow.html | 1,394,541,680,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011192582/warc/CC-MAIN-20140305091952-00018-ip-10-183-142-35.ec2.internal.warc.gz | 563,062,443 | 13,753 | Velocity Reviews > unsigned short addition/subtraction overflow
# unsigned short addition/subtraction overflow
Kevin Goodsell
Guest
Posts: n/a
12-21-2003
pete wrote:
>
> No.
> 65536 is an allowable value for INT_MAX.
I don't think so. INT_MAX pretty much has to be an odd number, I think.
In fact, to satisfy the requirement that the non-negative integer values
have the same representation as the same values for the corresponding
unsigned type, and the requirement that unsigned types use a pure binary
representation, I think it's safe to say that INT_MAX must be (2^N)-1
for some integer N, which must be 15 or greater.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Kevin Goodsell
Guest
Posts: n/a
12-21-2003
Andy wrote:
>
> Actually what I really meant is for unsigned operations.
> This is what I want to know. Are the following defined in
> C and always guaranteed warped around values?
>
> unsigned short i;
> unsigned long li; /* 32-bit wide */
>
> 1. i = (unsigned short)65535 + (unsigned short)3;
You could write these easer this way:
i = 65535u + 3u;
> 2. i = (unsigned short)1 - (unsigned short)3;
> 3. li = (unsigned long)0xFFFFFFFF + (unsigned long)3;
> 4. li = (unsigned long)1 - (unsigned long)3;
>
It's difficult to produce undefined behavior with unsigned values.
(Unless you divide by zero, maybe. I don't actually know what the
standard says about that, which surprises me.) Overflow doesn't occur
with unsigned types, but it's possible for unsigned types that are
narrower than int to be promoted to (signed) int, which may allow
overflow (and undefined behavior) to occur.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
James Hu
Guest
Posts: n/a
12-21-2003
On 2003-12-21, pete <(E-Mail Removed)> wrote:
> James Hu wrote:
>>
>> On 2003-12-21, Andy <(E-Mail Removed)> wrote:
>> > Are 1 through 4 defined behaviors in C?
>> >
>> > unsigned short i;
>> > unsigned long li; /* 32-bit wide */
>> >
>> > 1. i = 65535 + 3;
>> > 2. i = 1 - 3;
>> > 3. li = (unsigned long)0xFFFFFFFF + 3;
>> > 4. li = 1 - 3;
>>
>> Yes.
>
> No.
> 65536 is an allowable value for INT_MAX.
> (65535 + 3) would be integer overflow
> and undefined behavior in that case.
Good catch. I did consider overflow, but I assumed that (65535+3) was
identical to writing (6553 because of computation at translation
time versus computation at run time.
-- James
Chris Torek
Guest
Posts: n/a
12-21-2003
>Andy wrote:
>> Actually what I really meant is for unsigned operations.
>> This is what I want to know. Are the following defined in
>> C and always guaranteed warped around values?
>>
>> unsigned short i;
>> unsigned long li; /* 32-bit wide */
All C guarantees here is "at least" 32 bits, but here it does
not really matter.
>> 1. i = (unsigned short)65535 + (unsigned short)3;
In article <vCmFb.340\$(E-Mail Removed) t>
Kevin Goodsell <(E-Mail Removed)> writes:
>You could write these easer this way:
>
>i = 65535u + 3u;
Actually, this is potentially quite different.
If ANSI/ISO C used the *correct* rules (according to me )
it would be precisely the same, but we are stuck with quite
bogus widening rules due to a mistaken decision in the 1980s:
"when a narrow unsigned integer type widens, the resulting
type is signed if all the unsigned values fit, otherwise
it is unsigned".
In this particular case, unsigned short widens to either
unsigned int or signed int. Which one we get depends on the
properties of the implementation. This is a really dumb idea,
made in an attempt to be "less surprising" than the "right"
way ("narrow unsigned widens to unsigned"), that actually
turns out to be *more* surprising. But again we are stuck
with the wrong decision -- so let me define it.
What you must do is look in <limits.h> (perhaps by writing a
small C program, since the header may not exist) and compare
the values of USHRT_MAX and INT_MAX. One of the following two
cases will necessarily hold:
a) USHRT_MAX > INT_MAX.
This occurs on, e.g., the 16-bit PDP-11 and old 16-bit
MS-DOS C compilers. Here USHRT_MAX is 65535 while INT_MAX
is 32767.
b) USHRT_MAX <= INT_MAX.
This occurs on, e.g., today's 16-bit-short 32-bit-int C
compilers. Here USHRT_MAX is 65535 while INT_MAX is
2147483647.
In case (a), an unsigned short expression -- no matter what its
actual value is -- that appears in an arithmetic expression is
widened to unsigned int. Thus (unsigned short)65535 is
identical to (unsigned int)65535 or 65535U.
In case (b), howver, an unsigned short -- no matter what its actual
value is -- is widened to a *signed* int. Thus (unsigned short)65535
is identical to (int)65535 or 65535.
If we have two "unsigned short"s, values 65535 and 3 respectively,
and go to add them, we continue to have "case a" and "case b".
In case (a), the sum is 65535U + 3U, which has type unsigned int
and value 2. In case (b), the sum is 65535 + 3, which has type
signed int and value 65538.
In either case, when storing the final values back into an unsigned
short, it is reduced mod (USHRT_MAX+1), so that i becomes 2. The
place where this becomes a problem is not when we stuff the result
back into an unsigned variable, but rather when we compare it in
what the original 1989 C rationale called a "questionably signed"
expression.
Suppose we have the following code fragment:
unsigned short us = 65535;
int i = -1;
if (us > i)
printf("65535 > -1\n");
else
printf("65535 <= -1\n");
According to ANSI C's ridiculous rules (which we must obey anyway),
we decide whether this comparison uses "unsigned int" or "signed
int" based on whether USHRT_MAX exceeds INT_MAX. Once again, we
have the two cases:
case (a), USHRT_MAX > INT_MAX (PDP-11): "us" widens to an
unsigned int, value 65535U; i widens to an unsigned int,
value 65535U. 65535U > 65535U is false and we print
"65535 <= -1". This is, supposedly, "surprising" -- but
it happens!
case (b), USHRT_MAX < INT_MAX (VAX etc): "us" widens to
a signed int, value 65535; i remains signed int, value
-1. 65535 > -1 is true and we print "65535 > -1". This
is supposedly "not surprising" (which is probably true),
but in fact it is only SOMETIMES true.
As far as I am concerned, it is *much* better to be "predictably
surprising" than "unpredictably surprising based on the relative
values of USHRT_MAX and INT_MAX". The reason is that, while C
programmers do get surprised, they get surprised *once*, the *first*
time they mix signed and unsigned this way. This gives them the
opportunity to learn that the results are surprising; from then
on, they have no excuse to be surprised. Moreover, the logic
is trivial to follow: "unsigned widens to unsigned" means "put
an unsigned into an expression and it takes over."
Instead, we have a language where the code "works as expected" --
until it is moved to a machine where case (a) holds instead of case
(b). Programmers learn that mixing signed and unsigned is harmless
and "never surprises", only to find someday that, no, the language
is considerably more perverse than that. The logic is difficult
as well: "unsigned takes over except when it doesn't, based on the
relative values of the corresponding MAXes."
>> 2. i = (unsigned short)1 - (unsigned short)3;
>> 3. li = (unsigned long)0xFFFFFFFF + (unsigned long)3;
>> 4. li = (unsigned long)1 - (unsigned long)3;
>It's difficult to produce undefined behavior with unsigned values.
As long as you stick with unsigned int or unsigned long, anyway,
so that the broken widening rules do not trick you into accidentally
using signed values.
>(Unless you divide by zero, maybe. I don't actually know what the
>standard says about that, which surprises me.)
Division by zero produces undefined behavior, even for 1U / 0U and
the like.
>Overflow doesn't occur
>with unsigned types, but it's possible for unsigned types that are
>narrower than int to be promoted to (signed) int, which may allow
>overflow (and undefined behavior) to occur.
Yes. I claim that this rule is a terrible one; but I note that we
are stuck with it. The best approach is to avoid it -- make sure
you explicitly widen your narrow unsigned types to wider unsigned
types if the result (overflow or result of "questionably signed"
comparison) can matter. This kind of code is undeniably ugly, but
then, working around broken portions of any language (not just C)
is usually ugly.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
Kevin Goodsell
Guest
Posts: n/a
12-22-2003
Chris Torek wrote:
> In article <vCmFb.340\$(E-Mail Removed) t>
> Kevin Goodsell <(E-Mail Removed)> writes:
>
>>You could write these easer this way:
>>
>>i = 65535u + 3u;
>
>
> Actually, this is potentially quite different.
Yes, obviously. Not sure what I was thinking there. I think I suffer
from "short blindness" - I either miss the word 'short' or
sub-consciously translate it to 'int'. This wasn't the first time.
Thanks for pointing out the error.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
pete
Guest
Posts: n/a
12-22-2003
Kevin Goodsell wrote:
>
> pete wrote:
> >
> > No.
> > 65536 is an allowable value for INT_MAX.
>
> I don't think so.
65535
Thank you.
--
pete
Peter Nilsson
Guest
Posts: n/a
12-22-2003
pete <(E-Mail Removed)> wrote in message news:<(E-Mail Removed)>...
> pete wrote:
....
> > If INT_MAX equals 65535,
>
> I meant 65536.
Why? Neither is likely, but 65536 is considerably less so. Some would
argue (myself included) that 65536 is impossible on a conforming
implementation (be that C90 or C99).
--
Peter
pete
Guest
Posts: n/a
12-22-2003
Peter Nilsson wrote:
>
> pete <(E-Mail Removed)> wrote in message news:<(E-Mail Removed)>...
> > pete wrote:
> ...
> > > If INT_MAX equals 65535,
> >
> > I meant 65536.
>
> Why? Neither is likely, but 65536 is considerably less so. Some would
> argue (myself included) that 65536 is impossible on a conforming
> implementation (be that C90 or C99).
You would be right.
--
pete
Christopher Benson-Manica
Guest
Posts: n/a
12-22-2003
Chris Torek <(E-Mail Removed)> spoke thus:
> Actually, this is potentially quite different.
> If ANSI/ISO C used the *correct* rules (according to me )
> it would be precisely the same, but we are stuck with quite
> bogus widening rules due to a mistaken decision in the 1980s:
> "when a narrow unsigned integer type widens, the resulting
> type is signed if all the unsigned values fit, otherwise
> it is unsigned".
> etc.
Wow, what a great article! The only thing I'm unclear on now is why
such a seemingly obvious point escaped the C89 people, and why you
weren't around to dissuade them
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.
Chris Torek
Guest
Posts: n/a
12-23-2003
>Chris Torek <(E-Mail Removed)> spoke thus:
>> If ANSI/ISO C used the *correct* rules (according to me ) ...
In article <news:bs76vt\$er9\$(E-Mail Removed)>
Christopher Benson-Manica <(E-Mail Removed)> wrote:
>Wow, what a great article! The only thing I'm unclear on now is why
>such a seemingly obvious point escaped the C89 people, and why you
>weren't around to dissuade them
I was but a poor student at the time (making about four bucks an
hour, with a limit of 20 hrs/week, as "student staff") and could
not afford exotic vacation trips to ANSI C committee meetings.
I did, however, hear from someone who did go to them that this was
actually something of a "hotly debated" topic.
The VAX PCC did it "my" way, and apparently Plauger's C compiler(s)
did it the other way. The "base document" -- i.e., K&R-1 -- did
not even allow for the possibility of "unsigned short" and "unsigned
char", and if you have "narrow unsigned always widens to unsigned"
as a rule, you need an exception for plain char if/when plain char
is unsigned (as on the IBM 370), so that EOF can be negative.
The results of the rules differ only in "questionably signed" cases,
which are rare enough. But the ANSI rules are so ugly to work with
that I would prefer a special exception for "plain char is unsigned
on this implementation, yet nonetheless widens to signed int". Note
that this exception would force the constraint that CHAR_MAX < INT_MAX,
even when char is unsigned, which would have the happy side effect
of making stdio "work right".
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
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https://socratic.org/questions/how-do-you-find-the-exact-value-of-arccos-sqrt3-2 | 1,582,481,033,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00444.warc.gz | 540,039,862 | 6,035 | # How do you find the exact value of arccos (-sqrt3/2)?
Oct 6, 2016
The angle whose cosine is $\left(- \frac{\sqrt{3}}{2}\right)$ is 150 degrees or $\frac{5}{6} \pi$
#### Explanation:
Draw an equilateral triangle.
Cut it in half .
The angles in this right angle triangle are 30 ,60, 90
If the hypotenuse is length 2 then the shortest side is 1 and from Pythagoras the other side is $\sqrt{3}$
So the angle whose cosine is $\frac{\sqrt{3}}{2}$ is 30 degrees.
Now sketch the cosine graph.
From this you can find the angle whose cosine is $\left(- \frac{\sqrt{3}}{2}\right)$
Is 150 degrees or $\frac{5}{6} \pi$ | 191 | 611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 4.53125 | 5 | CC-MAIN-2020-10 | longest | en | 0.725626 |
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# hw_solution_9 - Ch.9 ABC Problem 9-30 a EZ-Seat Inc Income...
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Ch.9 ABC Problem 9-30. a. EZ-Seat, Inc. Income Statement Account Rate Ergo Standard Total Revenue \$1,950,000 \$1,840,000 \$3,790,000 Direct materials \$ 550,000 \$ 500,000 \$1,050,000 Direct labor 400,000 200,000 600,000 Overhead costs: Administration 0.325 a 130,000 e 65,000 195,000 Setting up 3,000 b 150,000 f 300,000 450,000 Performing quality control 750 c 150,000 g 150,000 300,000 Distribution 80 d 120,000 h 480,000 600,000 Total overhead costs 550,000 995,000 1,545,000 Operating profit \$ 450,000 \$ 145,000 \$ 595,000 a0.325 = \$195,000 of Administrative costs ÷ \$600,000 of direct labor costs b\$3,000 = \$450,000 of Production setup costs ÷ 150 production runs c\$750 = \$300,000 of Quality control costs ÷ 400 inspections d\$80 = \$600,000 of Distribution costs ÷ 7,500 units shipped e\$130,000 = \$0.325 x \$400,000 direct labor costs f\$150,000 = \$3,000 per setup x 50 production runs g\$150,000 = \$750 per inspection x 200 inspections h\$120,000 = \$80 per unit x 1,500 units shipped b. Activity-based costing highlights the activities that cause costs, and provides insight into which costs could be reduced. For example, management may be able to operate with fewer but larger production runs, thereby reducing setup costs. Focusing on activities can identify non- value-adding activities that can be eliminated without reducing the product’s value. c. EZ-Seat, Inc. Income Statement Account Rate Ergo Standard Total Revenue \$1,950,000 \$1,840,000 \$3,790,000 Direct Materials 550,000 500,000 1,050,000 Direct Labor 400,000 200,000 600,000 Overhead Costs 2.575 a 1,030,000 b 515,000 1,545,000 Operating Profit \$ (30,000 ) \$ 625,000 \$ 595,000 a2.575 = \$1,545,000 Overhead Costs ÷ \$600,000 Direct Labor Costs b\$1,030,000 = 2.575 Overhead Rate x \$400,000 Direct Labor Costs
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d. Dear Members of the Management Board: The purpose of this report is to explain the differences between the profits of our Ergo and Standard product lines using activity-based costing versus our traditional labor-based overhead allocation methods. The two costing methods differ in their results because of the way overhead costs are allocated between our products; direct costs do not differ under the two methods. Under the labor-based approach, all overhead costs are pooled together and allocated to our products on the basis of direct-labor costs. Under activity-based costing, cost drivers, such as inspections and set-ups, are identified and their costs are applied to the products in relation to usage. Traditional labor-based allocation is less accurate than activity-based allocations because many overhead costs are not well correlated with labor costs. For instance, our Ergo product receives 200% more overhead under our traditional approach than does our Standard product because it uses twice as much labor. However, after analyzing the factors driving the
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## This note was uploaded on 10/29/2009 for the course ACC 066 taught by Professor Kwak during the Spring '08 term at DeAnza College.
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hw_solution_9 - Ch.9 ABC Problem 9-30 a EZ-Seat Inc Income...
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Ask a homework question - tutors are online | 923 | 3,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.768576 |
https://www.jiskha.com/questions/17050/As-you-climb-a-mountain-your-ears-pop-because-of-the-changes-in-a-In-which-direction | 1,540,036,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512693.40/warc/CC-MAIN-20181020101001-20181020122501-00316.warc.gz | 1,012,724,940 | 5,988 | As you climb a mountain, your ears "pop" because of the changes in
(a) In which direction does your eardrum move as you climb up?
inward
outward
eardrum does not move
I think the eardrum moves outward because the further you go away from the earth's center the less pressure there is. The eardrum was at an equilibrium on the ground but because the external pressure is less the pressure from the inside pushes on the eardrum pushing it outward.atmospheric pressure.
(b) In which direction does your eardrum move as you climb down?
inward
outward
eardrum does not move
Going down is almost the opposite of going up. The eardrum moves inward from its outward position because the closer you get to the earth's center the more pressure is exerted on the eardrum. The closer you get to the ground the eardrum begins to return to its equilibrium state, neither inward or outward.
I have also figured this one out as well. Thanks!
it is correct.
How much force is needed to accelerate a 50-kg rider and her 250-kg motorcycle at 5 m/s 2?
a.) 1000 N
b.) 125 N
c.) 500 N
d/) 1500 N
F = ma
total mass = 300kg
acceleration = 5
F = 300 x 5 = 1500N
How much force is needed to accelerate a 50-kg rider and her 250-kg motorcycle at 5 m/s2?
1500
1. 0
2. 11
1. How much force is needed to accelerate a 50-kg rider and her 250-kg motorcycle at 5 m/s2?
posted by william
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# We only need to see the value of f 17 6 when dynamic
• Notes
• 17
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We only need to see the value of f (1,7).
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6 When Dynamic Programming Fails When the outcome of a decision also depends on the correlation between different options, DP does not work Only knowing that there are 3 units of fund available for opportunity 8 is not enough to make the optimal decision for opportunity 8 When the outcome of any decision for 8 depends the decision made for opportunity 7
7 Best Route to Lhasa 10 9 8 7 6 5 2 3 4 1 Hong Kong Lhasa 1 5 2 13 15 10 7 12 5 10 3 7 4 5 7 1 1 4 You are in 1 2 3 4 stage 0
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8 Starting Somewhere in the Middle If you have reached city 6, which way should you take to reach Lhasa? 10 9 8 6 Lhasa 3 4 1 4 If City 6 is now your starting point? City 1 was your starting point Two stages left to reach 10 Minimum cost when in city 6 at stage 2: f (2, 6) From any city s at stage i , the minimum cost to reach the destination is f(i, s )
9 Why Dynamic Optimization? From any stage, the best route to reach the destination depends on which city you are currently in Dynamic optimization model tries to identify the rule (policy) by which you know what is the optimal decision, given your current stage and state
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10 The Optimality Principle An optimal policy must have the property that regardless of the route taken to enter a particular state, the remaining decisions must constitute an optimal policy for leaving that state 10 9 8 6 Lhasa 15 10 12 3 4 1 4 10 9 8 7 Lhasa 13 7 7 1 1 4 f (2, 6) = min{3+ f (3, 8), 4+ f (3, 9)} =min{4, 8} f (2, 7) = min{7+ f (3, 8), 1+ f (3, 9)} =min{8, 5} d 2 (6) = best of 8 or 9 d 2 (7) = best of 8 or 9 What solution approach does the principle suggest?
11 Starting from the End At the destination, f (4,10)=0, d 4 (10)=stop From city 8, f (3,8)=1+ f (4, 10)=1, d 3 (8)= city 10 from city 9, f (3,9)=4+ f (4,10)= 4, d 3 (9)=city 10 From city 6, f (2,6)=min{3+ f (3,8), 4+ f (3,9)}= min{3+1,4+4}=4, d 2 (6)= 8 10 Lhasa 1 4 10 9 8 Lhasa 3 4 1 4 10 9 8 5 Lhasa 5 10 7 5 1 4 round 1 round 2 round 3
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12
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http://de.metamath.org/mpeuni/simp-6l.html | 1,723,087,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00604.warc.gz | 10,421,793 | 3,003 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > simp-6l Structured version Visualization version GIF version
Theorem simp-6l 806
Description: Simplification of a conjunction. (Contributed by Mario Carneiro, 4-Jan-2017.)
Assertion
Ref Expression
simp-6l (((((((𝜑𝜓) ∧ 𝜒) ∧ 𝜃) ∧ 𝜏) ∧ 𝜂) ∧ 𝜁) → 𝜑)
Proof of Theorem simp-6l
StepHypRef Expression
1 simp-5l 804 . 2 ((((((𝜑𝜓) ∧ 𝜒) ∧ 𝜃) ∧ 𝜏) ∧ 𝜂) → 𝜑)
21adantr 480 1 (((((((𝜑𝜓) ∧ 𝜒) ∧ 𝜃) ∧ 𝜏) ∧ 𝜂) ∧ 𝜁) → 𝜑)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 196 df-an 385 This theorem is referenced by: simp-7l 808 ghmcmn 18060 ustuqtop2 21856 ustuqtop4 21858 cnheibor 22562 miriso 25365 f1otrg 25551 txomap 29229 pstmxmet 29268 omssubadd 29689 signstfvneq0 29975 iunconlem2 38193 suplesup 38496 limcleqr 38711 0ellimcdiv 38716 limclner 38718 fourierdlem51 39050 smflimlem2 39658
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A genetic problem .... im not sure if im doing this correct.
Genetics as it applies to evolution, molecular biology, and medical aspects.
Moderator: BioTeam
A genetic problem .... im not sure if im doing this correct.
A black stallion of unknown ancestry was mated to anumber of red mares with purebred pedigrees. These matings produced 20 red offspring and 25 black offspring. Test using the chi-square method indicating whether the hypothesis is acceptable.
---------------
Okay .. I used a punnet square firstly ... I figured that the black stallion would b dominant so that would be BB, and the red mares since they are purebred, they would be bb.
Bb x bb getting a 1:1 ratio (total 2). I saw that the total progeny would be 45, so I got ..... (1/2)*(45) =22.5, then i used the chi square formula ...
(Observed-expected)^2/expected...
[(20-22.5)^2/ 22.5] + [(25-22.5)^2/ 22.5)]
------------
I'm not sure if my steps are correct at all so can someone please see if I am doing this correctlY? Thanks soo much!
xcuteJinax3
Garter
Posts: 8
Joined: Sun Feb 26, 2006 6:31 am
Seems Ok up to the end, but you haven't finish the job. You have to calculate the table coefficient and compare it to your chi sqaure table to tell you if your difference from expeted is significant or not.
Patrick
Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)
canalon
Inland Taipan
Posts: 3909
Joined: Thu Feb 03, 2005 2:46 pm
Canalon wrote:Seems Ok up to the end, but you haven't finish the job. You have to calculate the table coefficient and compare it to your chi sqaure table to tell you if your difference from expeted is significant or not.
So is the ratio 1:1? That's the part I'm mainly not sure of.
xcuteJinax3
Garter
Posts: 8
Joined: Sun Feb 26, 2006 6:31 am
That would be your Null hypothesis. A simple gene with a dominant recessive pattern. Your Chi square will tell you if you have to reject or accept it.
Patrick
Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)
canalon
Inland Taipan
Posts: 3909
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The equation y=-6t^2+51t describes the height (in feet) of a projectile launched from the surface of Mars at 51 feet per second. In how many seconds will the projectile first reach 108 feet in height?
May 23, 2020
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Ascent And Descent Speed Planning
Klm-md11 From Greece, joined Mar 2002, 471 posts, RR: 1Posted Sat Oct 26 2002 19:58:39 UTC (13 years 6 months 1 week 4 days 5 hours ago) and read 11168 times:
I have a question about speeds during ascent and descent. (i.e.B-737) During ascent, having reached 10000 ft; what speed is usually selected until reaching cruising altitude? Let’s say on an average weight, flying economically (trying to save the boss money on fuel cost) on descent, when leaving cruising altitude (let's say 37000 ft) at 320 knots IAS do you reduce speed at the point where descent starts or continue same speed during descent? If one continues same speed during descent, at what point WILL speed be reduced? I know that the max. speed at and below 10000 feet is 250 knots, so is speed being reduced just before reaching 10000 ft. at once or gradually? And if gradually, how? One last question… from which point on is the rate of descent 500 ft. per minute ?(which I believe is common just before touch-down) as the rate of descent from cruising altitude is around 2000 ft. per minute if I'm not mistaken. I'm sure these topics have been brought up before, but I couldn’t find it. Thanks very much for your replies! Regards, KLM-MD11
GELUK IS GELUL MET EEN K
10 replies: All unread, jump to last
Donder10 From Canada, joined Oct 2001, 6660 posts, RR: 20 Reply 1, posted Sat Oct 26 2002 23:42:59 UTC (13 years 6 months 1 week 4 days 1 hour ago) and read 11140 times:
Most(very close to all)flights out of LHR are told 'no ATC speed restriction'when on climbout so they can speed up quickly.Not sure on other major fields though
B747skipper From , joined Dec 1969, posts, RR: Reply 2, posted Sun Oct 27 2002 01:24:04 UTC (13 years 6 months 1 week 4 days ago) and read 11124 times:
Dear Klm md11 - xxx For all airplanes the climb speed (and often descent speed) is function of weight - with considerations for fuel economy... all airplanes have different speeds - here are some for the 747 Classic - xxx Intial climb speed below 10,000 feet is V2+100 kts = generally 280-290 knots, even though the legal speed limit maybe 250, we advise ATC we need higher speed because of flaps up speeds... xxx Enroute climb is depending on weight - 747 varies between 335 KIAS when heavy, or 275 kts when very light... until we intercept Mach .82-83, time at which we continue to climb at that Mach number... xxx Descending, we can select hi speed i.e. Mach .86 until indicating 340 KIAS, then we continue at 340 until 10,000 feet, time at which we will reduce to 250 KIAS... If you wish you can also make a descent at 280 KIAS, it is supposed to "save on fuel", but I noticed only minimal savings... xxx Enroute descent we indicate about 2,500 FPM rate of descent... according to ATC procedures, the last "1,000 feet" is done at the rate of 500 FPM... To reduce hi speed descent (340) to 250 below 10,000, we start reducing speed at approximately 12,000 feet... xxx I only flew 707, 727 and DC8s as other airplanes in my airline career, so cannot give you speeds for "modern aircraft i.e. 767... be aware that with these old airplanes, we climbed at 320 and Mach .82 generally... xxx Tot ziens, vriend - (s) Skipper
Klm-md11 From Greece, joined Mar 2002, 471 posts, RR: 1 Reply 3, posted Sun Oct 27 2002 02:07:24 UTC (13 years 6 months 1 week 3 days 22 hours ago) and read 11113 times:
Thanks very much Skipper! very helpfull info indeed!! regards, klm-md11
GELUK IS GELUL MET EEN K
Cedarjet From United Kingdom, joined May 1999, 8628 posts, RR: 53 Reply 4, posted Sun Oct 27 2002 04:34:14 UTC (13 years 6 months 1 week 3 days 19 hours ago) and read 11084 times:
Skipper, fascinating stuff. What if ATC couldn't let you speed up during a climb when you were heavy? How much of this kind of thing was handflown?
fly Saha Air 707s daily from Tehran's downtown Mehrabad to Mashhad, Kish Island and Ahwaz
B747skipper From , joined Dec 1969, posts, RR: Reply 5, posted Sun Oct 27 2002 04:51:41 UTC (13 years 6 months 1 week 3 days 19 hours ago) and read 11081 times:
Dear Cedarjet - xxx I cannot recall any instances of ATC preventing acceleration to required climb speed V2+100 - I assume "they know" we need such speed and clear us to such and such level and vector us accordingly... xxx In the situation that they would deny the speed, we would have to keep some flaps - for us would be "flaps 1", which would permit us to climb at V2+60 - most cases is around 240 knots or less... xxx Again, ATC knows about the 747, and I am certain we are not the only type of aircraft having that speed problem... xxx Happy contrails - (s) Skipper
Jetguy From , joined Dec 1969, posts, RR: Reply 6, posted Sun Oct 27 2002 06:23:37 UTC (13 years 6 months 1 week 3 days 18 hours ago) and read 11071 times:
The 747 isn't the only airplane with similar speed issues. The Galaxy that our company owned for a couple of years had a "clean wing" speed of up to just under 240 KIAS depending upon weight. We would need to climb out at 290 KIAS. It was a real issue when we operated out of airports underlying Class B airspace, where the speed limit is 200 KIAS. We always got what we needed from ATC, all we had to do was advise them. Climb speed, as Skipper says, is dependent upon many factors. All things being equal, you can climb at V2 or you can climb at the Barber Pole, or any speed in between those two extremes - safety isn't the issue. What is at issue is getting the most "bang for the buck"; in other words, what speed will give us the most distance down range for the amount of fuel consumed. Many companies look at the speeds which yield the best "specific range" - the distance traveled per unit of fuel consumption. The general rule of thumb when it comes to turbojet flight planning is that it's always best to climb as high as you can, as fast as you can, remain at altitude as long as you can, then descent as quickly as you can. Does ATC always allow us to operate accordingly? No, but it's something the we usually try to shoot for. For example, in the aircraft that I fly, the general rule of thumb for selecting an appropriate cruise altitude is 2 X Distance = The Cruise Flight Level. On a short 100 mile leg, we would usually file for FL200 west bound and FL190 east bound. Any leg longer than say 200 miles would justify a climb to FL390/FL410. Does ATC accommodate us? Out in the West and Midwest they usually do. On the East coast it doesn't usually happen - they tend to hold you down on climb out and start you down way too early. Oh well, we just carry a bit of extra "ATC Fuel" to compensate. As far as descent planning goes, nowdays we simply program the FMS and it flys the descent for us. In the old days, we used a simple formula to give us the descent point and descent rate: 3 X Desired Altitude Loss = Descent Point and 5 X Ground Speed = Descent Rate. This resulted in an approximate 3 degree descent path. As far as speeds held in the descent - it depends upon what and where you're flying. In the stuff I fly, in domestic U.S. and Canadian operations we simply allow the airspeed to increase to (conditions permitting) a speed approximately Mmo/Vmo -5 KIAS. At around 13,500' we simply go to flight idle and the airspeed will slowly bleed off to allow us to meet the 250 KIAS / 10,000' speed restriction without too much fuss.
AAR90 From United States of America, joined Jan 2000, 3579 posts, RR: 44 Reply 7, posted Sun Oct 27 2002 15:32:35 UTC (13 years 6 months 1 week 3 days 8 hours ago) and read 11015 times:
>I have a question about speeds during ascent and descent. (i.e.B-737) > >During ascent, having reached 10000 ft; what speed is usually selected until reaching cruising altitude? >Let’s say on an average weight, flying economically (trying to save the boss money on fuel cost) FMS ECON CLIMB. Normally ~295-305 kias (weight dependent assuming standard cost index). >on descent, when leaving cruising altitude (let's say 37000 ft) at 320 knots IAS do you reduce >speed at the point where descent starts or continue same speed during descent? Descent is normally performed at cruise mach until reaching desired indicated airspeed. Normally .78-.80m / 290-310kias. >If one continues same speed during descent, at what point WILL speed be reduced? Not sure I understand the question. Max speed is .82m/340kias. Speed reduction can be done at any time. Normal descent is performed at cruise mach until reaching desired indicated airspeed, then maintaining that indicated speed until further reduction is required/instructed by ATC/FAR/approach situation. >I know that the max. speed at and below 10000 feet is 250 knots, so is speed being >reduced just before reaching 10000 ft. at once or gradually? Pilot technique and very much dependent upon what (if any) ATC instructions have been issued. >And if gradually, how? With no ATC limitations, my personal preference is to maintain speed and make gradual reduction in both speed and rate of descent so as to be level at 10,000' at 250 knots at the point in space where further descent below 10,000' will begin. If done properly, passengers should not notice attitude or speed changes unless they're specifically looking for them. >One last question… from which point on is the rate of descent 500 ft. per minute ? >(which I believe is common just before touch-down) as the rate of descent from cruising altitude >is around 2000 ft. per minute if I'm not mistaken. Normally you would not see 500fpm descent rates except during transitions (flare & level-offs). Normal descent rates are 1000fpm-3000fpm. Normal approach descent rates are 700fpm-800fpm.
*NO CARRIER* -- A Naval Aviator's worst nightmare!
411A From United States of America, joined Nov 2001, 1826 posts, RR: 8 Reply 8, posted Mon Oct 28 2002 05:50:11 UTC (13 years 6 months 1 week 2 days 18 hours ago) and read 10968 times:
Regarding descent... If ATC is unable to approve descent from cruise altitude when desired (for example at 100-110 nm approximately) it is often useful to reduce cruise mach number, so that when descent is approved, the nose can be lowered to accelerate while at the same time closing throttles. This results in a descent rate of 6000+ ft/min initially (while accelerating to M.86/350KIAS) to get rid of the excess height. This works very well in the Lockheed TriStar, (lower speeds for other types). One must be careful however if the destination airport is near sea level, as the cabin altitude descent rate will normally not exceed 300-400ft/min during descent for pax comfort...and this takes time. Older types (B707 for example) required that the throttles NOT be completely closed during initial descent, as the turbocompressors needed sufficent bleed air to be able to pressurise/descend the cabin.
XXXX10 From United Kingdom, joined Jan 2000, 779 posts, RR: 0 Reply 9, posted Mon Oct 28 2002 14:43:15 UTC (13 years 6 months 1 week 2 days 9 hours ago) and read 10945 times:
Descent is normally performed at cruise mach until reaching desired indicated airspeed. Normally .78-.80m / 290-310kias. Does this mean that if you descent at a constant Mach No the IAS will increase, then, when you reach your max IAS you maintain that figure until you have to reduce for ATC ?
AAR90 From United States of America, joined Jan 2000, 3579 posts, RR: 44 Reply 10, posted Mon Oct 28 2002 15:28:16 UTC (13 years 6 months 1 week 2 days 8 hours ago) and read 10938 times:
>Does this mean that if you descent at a constant Mach No the IAS will increase, Yes. >...then, when you reach your max IAS you maintain that figure Not necessarily the max KIAS, but rather your desired KIAS. 737NG max speed is 340kias, but there is an AD that limits speedbrake use to 300kias or less (improved horiz.stab being installed to fix this problem). So my desired speed is normally somewhere close to 300kias to allow SB use or increased speed to adjust descent/level-off point to make any arrival restrictions. >...until you have to reduce for ATC ? When you have to slow down, you have to slow down. OTOH, most of my flights have been red-eye transcons with little/no early morning air traffic. I attempt to reach a point 30 miles prior to the runway at exactly 10,000' above the runway at exactly 250kias all without moving the throttle position from where I place them at the top of descent. With very smooth (and slow) transitions the first time the F/A's wake up the pax they'll be 10-15 minutes from landing and are normally quite grateful for the extra few minutes of sleep.
*NO CARRIER* -- A Naval Aviator's worst nightmare!
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# Re: st: A correlation matrix after multiple imputation
From weddings@stata.com (Wesley D. Eddings, StataCorp) To statalist@hsphsun2.harvard.edu Subject Re: st: A correlation matrix after multiple imputation Date Mon, 26 Jul 2010 17:14:05 -0500
```On Friday 23 July 2010, Alan Acock <acock@mac.com> asked if there is an easy
way to obtain a multiple-imputation estimate of a correlation matrix in Stata:
> Is there an easy way to obtain the 20 correlation matrices, one for each of
> the 20 imputed datasets and then somehow pulling these?
There is no automatic way of doing this, but with some programming effort,
Alan can use -mi estimate- to obtain an MI estimate of the correlation matrix;
see the code at the end of this post.
However, from a statistical standpoint, it is not clear whether averaging
completed-data sample estimates of the correlation matrix across imputed data
is the best approach to account for missing data when computing a correlation
matrix.
One alternative is to consider reporting an EM estimate of the covariance (or
correlation) matrix adjusted for missing data. Such an estimate can be
obtained from -mi impute mvn-, because -mi impute mvn- uses the EM algorithm
to get starting values of the parameters for the MCMC procedure. The EM
estimates of the coefficients and the variance-covariance matrix are saved
after -mi impute mvn- in the -r(Beta_em)- and -r(Sigma_em)- matrices,
respectively. To obtain EM estimates only, without producing imputations,
specify the -emonly- option with -mi impute mvn-; see the example below.
-- Wes -- Yulia
weddings@stata.com ymarchenko@stata.com
=================== EXAMPLES ================================================
Here is how you can obtain an EM estimate of the correlation matrix accounting
for missing data:
/****************** begin do file ******************/
sysuse auto, clear
set seed 12345
replace mpg = . if runiform()>0.9
mi set wide
mi register imputed mpg weight
mi impute mvn mpg weight, emonly
mat Sigma = r(Sigma_em) /* save EM estimate of the variance-covariance (VC) matrix */
_getcovcorr Sigma, corr shape(full) /* convert VC to a correlation matrix */
mat C = r(C)
matlist C
/*************** end do file *****************/
Here is how you can obtain an MI estimate of the correlation matrix:
/***** begin MI correlation ******************/
cap program drop ecorr
program ecorr, eclass
version 11
syntax [varlist] [if] [in] [aw fw] [, * ]
if (`"`weight'"'!="") {
local wgt `weight'`exp'
}
marksample touse
correlate `varlist' `if' `in' `wgt', `options'
tempname b V
mata: st_matrix("`b'", vech(st_matrix("r(C)"))')
local p = colsof(`b')
mat `V' = J(`p',`p',0)
local cols: colnames `b'
mat rownames `V' = `cols'
eret post `b' `V' [`wgt'] , obs(`=r(N)') esample(`touse')
eret local cmd ecorr
eret local title "Lower-diagonal correlation matrix"
eret local vars "`varlist'"
end
cap program drop micorr
program micorr, rclass
tempname esthold
_estimates hold `esthold', nullok restore
qui mi estimate, cmdok: ecorr `0'
tempname C_mi
mata: st_matrix("`C_mi'", invvech(st_matrix("e(b_mi)")'))
mat colnames `C_mi' = `e(vars)'
mat rownames `C_mi' = `e(vars)'
di
di as txt "Multiple-imputation estimate of the correlation matrix"
di as txt "(obs=" string(e(N_mi),"%9.0g") ")"
matlist `C_mi'
return clear
ret matrix C_mi = `C_mi'
end
sysuse auto, clear
set seed 12345
replace mpg = . if runiform()>0.9
mi set wide
mi register imputed mpg weight
mi impute mvn mpg weight, add(20)
micorr mpg weight
/***** end MI correlation ********************/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 1,070 | 3,940 | {"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} | 2.703125 | 3 | CC-MAIN-2015-32 | longest | en | 0.829965 |
https://www.netlib.org/lapack/explore-3.1.1-html/slaed3.f.html | 1,623,730,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00406.warc.gz | 831,688,867 | 4,475 | ``` SUBROUTINE SLAED3( K, N, N1, D, Q, LDQ, RHO, DLAMDA, Q2, INDX,
\$ CTOT, W, S, INFO )
*
* -- LAPACK routine (version 3.1) --
* Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
* November 2006
*
* .. Scalar Arguments ..
INTEGER INFO, K, LDQ, N, N1
REAL RHO
* ..
* .. Array Arguments ..
INTEGER CTOT( * ), INDX( * )
REAL D( * ), DLAMDA( * ), Q( LDQ, * ), Q2( * ),
\$ S( * ), W( * )
* ..
*
* Purpose
* =======
*
* SLAED3 finds the roots of the secular equation, as defined by the
* values in D, W, and RHO, between 1 and K. It makes the
* appropriate calls to SLAED4 and then updates the eigenvectors by
* multiplying the matrix of eigenvectors of the pair of eigensystems
* being combined by the matrix of eigenvectors of the K-by-K system
* which is solved here.
*
* This code makes very mild assumptions about floating point
* arithmetic. It will work on machines with a guard digit in
* add/subtract, or on those binary machines without guard digits
* which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or Cray-2.
* It could conceivably fail on hexadecimal or decimal machines
* without guard digits, but we know of none.
*
* Arguments
* =========
*
* K (input) INTEGER
* The number of terms in the rational function to be solved by
* SLAED4. K >= 0.
*
* N (input) INTEGER
* The number of rows and columns in the Q matrix.
* N >= K (deflation may result in N>K).
*
* N1 (input) INTEGER
* The location of the last eigenvalue in the leading submatrix.
* min(1,N) <= N1 <= N/2.
*
* D (output) REAL array, dimension (N)
* D(I) contains the updated eigenvalues for
* 1 <= I <= K.
*
* Q (output) REAL array, dimension (LDQ,N)
* Initially the first K columns are used as workspace.
* On output the columns 1 to K contain
* the updated eigenvectors.
*
* LDQ (input) INTEGER
* The leading dimension of the array Q. LDQ >= max(1,N).
*
* RHO (input) REAL
* The value of the parameter in the rank one update equation.
* RHO >= 0 required.
*
* DLAMDA (input/output) REAL array, dimension (K)
* The first K elements of this array contain the old roots
* of the deflated updating problem. These are the poles
* of the secular equation. May be changed on output by
* having lowest order bit set to zero on Cray X-MP, Cray Y-MP,
* Cray-2, or Cray C-90, as described above.
*
* Q2 (input) REAL array, dimension (LDQ2, N)
* The first K columns of this matrix contain the non-deflated
* eigenvectors for the split problem.
*
* INDX (input) INTEGER array, dimension (N)
* The permutation used to arrange the columns of the deflated
* Q matrix into three groups (see SLAED2).
* The rows of the eigenvectors found by SLAED4 must be likewise
* permuted before the matrix multiply can take place.
*
* CTOT (input) INTEGER array, dimension (4)
* A count of the total number of the various types of columns
* in Q, as described in INDX. The fourth column type is any
* column which has been deflated.
*
* W (input/output) REAL array, dimension (K)
* The first K elements of this array contain the components
* of the deflation-adjusted updating vector. Destroyed on
* output.
*
* S (workspace) REAL array, dimension (N1 + 1)*K
* Will contain the eigenvectors of the repaired matrix which
* will be multiplied by the previously accumulated eigenvectors
* to update the system.
*
* LDS (input) INTEGER
* The leading dimension of S. LDS >= max(1,K).
*
* INFO (output) INTEGER
* = 0: successful exit.
* < 0: if INFO = -i, the i-th argument had an illegal value.
* > 0: if INFO = 1, an eigenvalue did not converge
*
* Further Details
* ===============
*
* Based on contributions by
* Jeff Rutter, Computer Science Division, University of California
* at Berkeley, USA
* Modified by Francoise Tisseur, University of Tennessee.
*
* =====================================================================
*
* .. Parameters ..
REAL ONE, ZERO
PARAMETER ( ONE = 1.0E0, ZERO = 0.0E0 )
* ..
* .. Local Scalars ..
INTEGER I, II, IQ2, J, N12, N2, N23
REAL TEMP
* ..
* .. External Functions ..
REAL SLAMC3, SNRM2
EXTERNAL SLAMC3, SNRM2
* ..
* .. External Subroutines ..
EXTERNAL SCOPY, SGEMM, SLACPY, SLAED4, SLASET, XERBLA
* ..
* .. Intrinsic Functions ..
INTRINSIC MAX, SIGN, SQRT
* ..
* .. Executable Statements ..
*
* Test the input parameters.
*
INFO = 0
*
IF( K.LT.0 ) THEN
INFO = -1
ELSE IF( N.LT.K ) THEN
INFO = -2
ELSE IF( LDQ.LT.MAX( 1, N ) ) THEN
INFO = -6
END IF
IF( INFO.NE.0 ) THEN
CALL XERBLA( 'SLAED3', -INFO )
RETURN
END IF
*
* Quick return if possible
*
IF( K.EQ.0 )
\$ RETURN
*
* Modify values DLAMDA(i) to make sure all DLAMDA(i)-DLAMDA(j) can
* be computed with high relative accuracy (barring over/underflow).
* This is a problem on machines without a guard digit in
* add/subtract (Cray XMP, Cray YMP, Cray C 90 and Cray 2).
* The following code replaces DLAMDA(I) by 2*DLAMDA(I)-DLAMDA(I),
* which on any of these machines zeros out the bottommost
* bit of DLAMDA(I) if it is 1; this makes the subsequent
* subtractions DLAMDA(I)-DLAMDA(J) unproblematic when cancellation
* occurs. On binary machines with a guard digit (almost all
* machines) it does not change DLAMDA(I) at all. On hexadecimal
* and decimal machines with a guard digit, it slightly
* changes the bottommost bits of DLAMDA(I). It does not account
* for hexadecimal or decimal machines without guard digits
* (we know of none). We use a subroutine call to compute
* 2*DLAMBDA(I) to prevent optimizing compilers from eliminating
* this code.
*
DO 10 I = 1, K
DLAMDA( I ) = SLAMC3( DLAMDA( I ), DLAMDA( I ) ) - DLAMDA( I )
10 CONTINUE
*
DO 20 J = 1, K
CALL SLAED4( K, J, DLAMDA, W, Q( 1, J ), RHO, D( J ), INFO )
*
* If the zero finder fails, the computation is terminated.
*
IF( INFO.NE.0 )
\$ GO TO 120
20 CONTINUE
*
IF( K.EQ.1 )
\$ GO TO 110
IF( K.EQ.2 ) THEN
DO 30 J = 1, K
W( 1 ) = Q( 1, J )
W( 2 ) = Q( 2, J )
II = INDX( 1 )
Q( 1, J ) = W( II )
II = INDX( 2 )
Q( 2, J ) = W( II )
30 CONTINUE
GO TO 110
END IF
*
* Compute updated W.
*
CALL SCOPY( K, W, 1, S, 1 )
*
* Initialize W(I) = Q(I,I)
*
CALL SCOPY( K, Q, LDQ+1, W, 1 )
DO 60 J = 1, K
DO 40 I = 1, J - 1
W( I ) = W( I )*( Q( I, J ) / ( DLAMDA( I )-DLAMDA( J ) ) )
40 CONTINUE
DO 50 I = J + 1, K
W( I ) = W( I )*( Q( I, J ) / ( DLAMDA( I )-DLAMDA( J ) ) )
50 CONTINUE
60 CONTINUE
DO 70 I = 1, K
W( I ) = SIGN( SQRT( -W( I ) ), S( I ) )
70 CONTINUE
*
* Compute eigenvectors of the modified rank-1 modification.
*
DO 100 J = 1, K
DO 80 I = 1, K
S( I ) = W( I ) / Q( I, J )
80 CONTINUE
TEMP = SNRM2( K, S, 1 )
DO 90 I = 1, K
II = INDX( I )
Q( I, J ) = S( II ) / TEMP
90 CONTINUE
100 CONTINUE
*
* Compute the updated eigenvectors.
*
110 CONTINUE
*
N2 = N - N1
N12 = CTOT( 1 ) + CTOT( 2 )
N23 = CTOT( 2 ) + CTOT( 3 )
*
CALL SLACPY( 'A', N23, K, Q( CTOT( 1 )+1, 1 ), LDQ, S, N23 )
IQ2 = N1*N12 + 1
IF( N23.NE.0 ) THEN
CALL SGEMM( 'N', 'N', N2, K, N23, ONE, Q2( IQ2 ), N2, S, N23,
\$ ZERO, Q( N1+1, 1 ), LDQ )
ELSE
CALL SLASET( 'A', N2, K, ZERO, ZERO, Q( N1+1, 1 ), LDQ )
END IF
*
CALL SLACPY( 'A', N12, K, Q, LDQ, S, N12 )
IF( N12.NE.0 ) THEN
CALL SGEMM( 'N', 'N', N1, K, N12, ONE, Q2, N1, S, N12, ZERO, Q,
\$ LDQ )
ELSE
CALL SLASET( 'A', N1, K, ZERO, ZERO, Q( 1, 1 ), LDQ )
END IF
*
*
120 CONTINUE
RETURN
*
* End of SLAED3
*
END
``` | 2,633 | 8,021 | {"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-2021-25 | latest | en | 0.756876 |
https://socratic.org/questions/how-do-you-sketch-the-curve-with-parametric-equations-x-sin-t-y-sin-2-t#110360 | 1,713,151,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00126.warc.gz | 479,618,937 | 5,952 | # How do you sketch the curve with parametric equations x = sin(t), y=sin^2(t) ?
Since we know that $- 1 \le \sin t \le 1$, the curve is limited to $- 1 \le x \le 1$.
By plugging $x = \sin t$ into $y = {\sin}^{2} t$, we have
$y = {\left(\sin t\right)}^{2} = {x}^{2}$.
Hence, the curve is the portion of the parabola $y = {x}^{2}$ between $x = - 1$ and $x = 1$, which looks like this: | 150 | 384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.854716 |
http://canisiusphysics.blogspot.com/2012/04/ | 1,508,297,767,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00686.warc.gz | 54,920,711 | 19,557 | ## How High is 800 West Ferry?
Spring has reached Buffalo. The seniors have gone (as a private school, we have figured out that the school runs better if the seniors are not in class after AP Exams and so we let them go). Students are tired of being cooped up in a classroom. How to justify going outside for class. Let's measure the height of a mountain.
O.K., Buffalo doesn't have mountains or even hills. But next to our school is a 170 foot tall building (800 West Ferry). That becomes our mountain. We are then going to calculate its height three different ways, with an equation, with a drawing, and with a computer model.
Some of my students have used an inclinometer (a protractor with a hanging weight) before in a geometry class, so making and using one is not completely unknown to all. However, I know that I will have to review. So we make the inclinometers in class and calculate the height of the ceiling using simple right-triangle geometry by measuring the angle of elevation and the distance to the wall from a starting point. Most teams get obviously wrong answers, so we spend some time to figure out why. We figure out that the angle shown on the inclinometer is the compliment of the angle of elevation. I then remind my guys that whenever you use a new measuring system, you always calibrate that system by measuring some objects whose measurements you know. That was a lesson mentioned way back in September, so most have forgotten it. At least a couple of my guys get a "Oh, now I see why this is important" moment. This activity takes most of a class period.
The next class is spent discussing the difficulties of measuring the height of a real mountain. We figure out that we can't get the measurements for a right-triangle. Some students then blurt out "law of sines" or "law of cosines". I then let them try it out. Some groups will figure out that they will have to use two triangles, but sometimes I have to lead them to that conclusion. If I had an infinite amount of time, I would let them figure out how to do things, but usually I have to give some strong hints so that we get the appropriate equation at the end of this class period.
The next day, we use less than half a class period to make the actual measurements. I have found that it is easiest to give each pair a 50' length of string to serve as their baseline. They then set their own. After each team get 3-5 sets of measurements, we return to the classroom to do the calculations.
I have my guys do each calculation three different ways. One way is to use the formula we developed in class the day before. Another way is to use graph paper onto where they find out that they will have to use an appropriate scale (most groups initially use a scale that is too big for the supplied graph). The third way is to use a computer model. I use Geogebra, but you could use any other system such as Geometers Sketchpad. Below is a picture of this model and then the actual model which you should be able to try.
If you are not seeing the interactive feature above, click on the "CanisusPhysics" link above (sometimes more than once). The advanced feature of Geogebra is not quite stable yet for consistent results.
Or go to the site directly.
Recently (fall 2015), I have started to use the app Desmos (www.desmos.com) in class (see a previous post http://canisiusphysics.blogspot.com/2015/10/linearizing-data-on-ipad.html). I have translated the Geogebra app to the Desmos one which can be tried at https://www.desmos.com/calculator/x8k0s8r4cg.
The idea is that the height should be the same for each calculational method. Most teams will have vast discrepancies. Depending on the time available, I either have the teams redo the measurements and calculations the next day or have them do them on their own time until all calculations are within 10% of each other (it is amusing to see the teams take more time and care in making their measurements the second time around).
As a wrap-up, I like to show the graphic below to show all the work that was done to map the Indian sub-continent. I am still looking to find a PBS Nova program from several years ago that showed how that Great Trigonometric Survey of India lead to the first measurement of the height of Mount Everest.
So, now I have a way of letting my guys go outside on a nice spring day while justifying this with myself, my principal, and concerned parents.
#### For AP students
When I do this with my AP students, usually early in the academic year, I have them measure distance with their walking stride. We spend some time gathering some distance vs number of steps on the nearby athletic field. They then do a linear regression analysis with their personal data, and use the resultant equation to measure a base-line.
## 21 April 2012
### More on using Geogebra in the Physics Classroom
There is a site with interactive physics apps you might like, and these apps work in an iPad as well as real computers. The site is www.geogebratube.org. Here is one example.
If you cannot see the interactive app above, go directly to the source here.
These apps are made with the FOSS program Geogebra which can be downloaded at www.geogebra.org. Geogebra is not that hard to learn to use and the apps it makes can be used on any computer which has Java. I have made a couple and placed them on my class webpage. You can even download source files from geogebratube and modify them if you want.
While the following apps are not directly on Geogebratube, I recommend that you check out the ones made by Daniel Mentrard at http://dmentrard.free.fr/GEOGEBRA/index.htm. He has a large collection covering all areas of what we teach. Check out this from his site.
I have found that if you erase htm (or html) in the address bar of your browser and replace it with ggb, you can usually download the source file so that you can replace any French text with English before using it in class or you modify it in any other way.
Currently the only way that a student can interact with these apps on an iPad is through the Geogebratube site, but there are indications that future developments in the Geogebra program will allow you to embed these apps into an ebook that can be read on an iPad directly without going through a web-browser.
While most of the apps you find at Geogebratube are not as refined or sophisticated as ones that you may already be using from PhET (phet.colorado.org) or Open Source Physics (www.opensourcephysics.org), you can make these apps without paying for program to make Flash apps and your students using iPads will not be left out. Since my school is going to be an "iPad school", this tool looks like it will be a major one used in my classes in the future.
## What's so Special about Relativity?
Special relativity is one of those fun topics I like to do when possible. While I usually cover it the last week of school, this year I have decided to do it the week between the Easter break and the start of the student projects on modern physics.
I start in the usual way, by motivating the scientific conclusion that everybody measures the speed of light to be the same, no matter how they are moving. This then leads to using the animation that shows that a moving clock ticks slow. But rather than use the geometry of this situation to derive the time-dilation equation (a MEGO exercise most times), I use the interpretation of SR that everybody is moving through space-time at the speed of light. If you are at zero speed in space, you are moving through time at the normal rate. If you are moving through space at a non-zero speed, you are moving through time at a slower than normal rate.
This interpretation then leads to using this model.
This is a picture of a Geogebra dynamic model that I display to the class (available here) and that they can use on their iPads. Students are given a worksheet which they use to calculate the time dilation for various speeds and also the speeds needed for various time dilations. The next day comes time travel!
By not using the algebraic equations too soon, I have found more students engaged and understanding SR. We then spend a couple of days discussing the possibility and feasibility of time travel.
P.S. If you have not checked the program Geogebra, I suggest you do. Learning how to use it is not too hard, and there are some examples (and more) you can use in your class. Since it is FOSS, you share it with your students, embed things in webpages, and otherwise make static situations dynamic. Below is the model embedded in this page (if you don't see the app, come back later. I am having problems getting it to work on this page. Go to this page to see it in action.).
## 13 April 2012
### Modern Physics, for and by students
One advantage to teaching at a private school is that I am not held hostage to the year-end New York State Regents test. So, when it comes time for atomic and nuclear physics, I can de-emphasize the mathematics and emphasize the history and timeline of discovery. Which also means that I can hand over the day-to-day work to the students.
This started a few years ago when I was thinking of ways to combat the final few weeks of senioritis. It was also at a time when the Principal wanted us to move away from final exams for the seniors and toward using "alternate assessments". So, I decided to have the students teach the other students the units on Modern Physics. Even if the seniors tune out the rest of the time, they will at least know something about one part of atomic-nuclear physics.
### May starts in January
Every year I tweak things a little, but I have the basic framework in place. Shortly after the January Exams, I show my guys the topics to be covered and the grading rubric which are then posted both in the classroom and on the class website. About a week later, I publish a schedule for those presentations. They then have about a week to chose a partner and decide on topics of interest. The first couple of years, I would have the teams give me a list of desired topics (in order of preference) and days on which they would be taking AP exams. I would then spend many hours trying to make everybody happy (you know how that ends up). I now make it easy on myself.
### No Blood Please
On decision day, I have my guys work things out themselves. I gather the seniors in the back of the room, put down a copy of the schedule, and tell them "I don't care how you decide, as long there is no blood. I just want you guys to agree as a group on what teams will do what topic." To cover my butt, I do remind them what it means to be a Jesuit-educated gentleman, and that I reserve the right to make the assignments if I think that things are not working out, but I have not had to do that yet. The guys generally do spend some time making sure that most teams are satisfied with the assignments. After the seniors have made their decisions, the juniors have their turn.
### E-books are real books
Our students live on-line. So that my guys at least know a lot of good information can be found in books and not just on websites (and that Wikipedia is not the end-all-and-be-all), this year (2013) I spent time showing my classes how to find and download ebooks from our local library system. Since we live in New York State, we are also eligible to get a card from the New York Public Library. I showed them how to apply (with the warning that there is a 2-3 week turnaround time). The reason I like the NYPL site is that there are a lot more physics-oriented books there. Also, since some of my seniors will be going to college out of state, I let them know that the NYPL will not know that they have moved. They will have one more resource available to them to help make their college career successful.
### Drinking from a firehose?
Am I showing my guys too many books? It is well-known that when presented with a plethora of choices, many people give up and just make random choices. I sense that that is what is going on with some of my guys. I am tempted to make a list of appropriate books for each of the topics, but I don't want to spoon-feed my guys. Please give your opinion in the comment section below.
### An Outline can't be Copied and Pasted
The teams then have 2-3 weeks to write an outline for their projects (timed so that they have the Feb break to work on this if they want). I then review those outlines for the basic material I want covered and make some "comments for improvement". The teams then have 2-3 weeks to work on the drafts.
### Not a First Draft
Notice that I do not use the term "first draft". To most high school students, "first draft" means "I can just throw some stuff together at the last minute to pass in, but I will wait to do the real work the night before our team is up to present." I tell my classes at least three times that I am using the academic and bureaucratic definition which means the draft is the proposed final version; it is being reviewed only for minor revisions. I know that most teams will still need major revisions, but I want them to do some work at this time.
I then review the drafts and "make comments for improvement". I try to time this so the the teams have the Easter break to work on their final presentation (although the solar-lunar cycles don't make this possible some years). I have learned to tell my guys that I am reviewing only for content, not for veracity. One year, a team made a glaring mis-statement during their presentation. When I called them on it, they pleaded that when I gave the comment "Looks good" on their draft, it meant that to them that all aspects were good.
The presentations are scheduled for the two weeks of the AP exams and the two weeks after. I sit in the back, marking the rubric. At the end of the presentation, the students ask questions, then I have my turn. If there is time, I will use the end of the class to "amplify" on some things that I think need be discussed.
While still a work in progress, I have found this approach to atomic and nuclear physics a decent one for high school students.
## 10 April 2012
### Drinking the Kool-aid
Hats off to the marketing people at Apple. They have convinced many people that the future of education lies within the slender confines of the iPad. The past is just that, the past. Clinging to such archaic remnants is almost anti-American. However, I see and value the past that Apple thinks is useless.
My school is becoming an iPad school next year. While I look forward to the common platform that students will have, I lament the things that I cannot do with them in my classroom. Sites such at PhET and Open Source Physics which use Flash and Java do not work on the iPad. The educational value of these sites is without parallel online. I had hoped to incorporate some of these applets into the electronic textbook I wanted to modify for use in my classroom. I am unimpressed with the physics textbook that they offer up as an epitome of an iBook.
So, one of my summer projects will be to learn JavaScript and figure out how to add such applets to my ebook. The last full-scale language I have programmed in is Pascal. While I have recently written some basic shell scripts for my Linux computers, big programming is another thing. So, thank you Apple for making me learn new things.
### Realizing that I have to change
Good teachers find what works in the classroom and stay with that. Great teachers know that what works one year may not work in future years. This year, I realized that I have been merely a good teacher. I want to become a better one. This blog will chronicle that journey.
I must acknowledge my debt to other teachers who have inspired me to use different means to reach my students. The high school teachers Frank Noschese , Rebecca Howell, and Brendan Noon have given me good examples of using web-based tools to good advantage. The university professor Rhett Allain has given me many good examples of using the video analysis software package Tracker (though I have not forgiven him for beating me to writing about the physics of Angry Birds). I hope I can equal their contributions to physics education. | 3,599 | 16,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-43 | longest | en | 0.971583 |
https://questioncove.com/updates/4d5a0e847c17b7644e2b7d91 | 1,708,983,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00196.warc.gz | 480,835,990 | 4,611 | Mathematics 81 Online
OpenStudy (anonymous):
A store sells regular coffee for \$8 a pound and gourmet coffee for \$14 a pound. To get rid of 40 pounds of gourmet coffee, a shopkeeper makes a blend to put on sale for \$10 a pound. How many pound of regular coffee should be used?
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https://www.aepochadvisors.com/prob/ | 1,714,008,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00700.warc.gz | 563,509,576 | 30,381 | # PROB
In this comprehensive guide, we will explore the PROB function in Excel, which is used to calculate the probability of a discrete random variable falling within a specified range. This function is particularly useful in statistical analysis and probability calculations. We will cover the syntax, examples, tips and tricks, common mistakes, troubleshooting, and related formulae for the PROB function.
## PROB Syntax
The syntax for the PROB function in Excel is as follows:
=PROB(range_x, range_prob, lower_limit, [upper_limit])
Where:
• range_x (required) is the range of cells containing the discrete random variable values (x).
• range_prob (required) is the range of cells containing the corresponding probabilities for each value in range_x.
• lower_limit (required) is the lower limit of the range for which you want to calculate the probability.
• upper_limit (optional) is the upper limit of the range for which you want to calculate the probability. If omitted, the function will calculate the probability of the discrete random variable being equal to the lower_limit.
## PROB Examples
Let’s look at some examples of using the PROB function in Excel.
### Example 1: Basic PROB Function
Suppose we have a set of discrete random variable values (x) in cells A1:A5 and their corresponding probabilities in cells B1:B5. We want to calculate the probability of the random variable falling within the range of 2 to 4.
=PROB(A1:A5, B1:B5, 2, 4)
This formula will return the probability of the random variable falling within the specified range, based on the given probabilities.
### Example 2: PROB Function with Single Value
Using the same data as in Example 1, we want to calculate the probability of the random variable being equal to 3.
=PROB(A1:A5, B1:B5, 3)
Since we omitted the upper_limit, the function will return the probability of the random variable being equal to the lower_limit (3).
## PROB Tips & Tricks
Here are some tips and tricks to help you use the PROB function more effectively:
1. Ensure that the sum of the probabilities in the range_prob is equal to 1. If the sum is not equal to 1, the function will return an error.
2. Use named ranges for range_x and range_prob to make your formulas easier to read and understand.
3. Remember that the PROB function is designed for discrete random variables. For continuous random variables, consider using other statistical functions in Excel.
## Common Mistakes When Using PROB
Here are some common mistakes to avoid when using the PROB function:
1. Not ensuring that the sum of probabilities in range_prob is equal to 1. This will result in an error.
2. Using non-numeric values in range_x or range_prob. The function requires numeric values to calculate probabilities.
3. Using different sized ranges for range_x and range_prob. The ranges must have the same number of cells.
## Why Isn’t My PROB Function Working?
If your PROB function is not working, consider the following troubleshooting steps:
1. Check that the sum of probabilities in range_prob is equal to 1. If not, adjust the probabilities accordingly.
2. Ensure that range_x and range_prob have the same number of cells.
3. Verify that all values in range_x and range_prob are numeric.
4. Double-check the formula syntax and make sure all required arguments are provided.
## PROB: Related Formulae
Here are some related formulae that you might find useful when working with probabilities and statistical analysis in Excel:
1. BINOM.DIST: Calculates the probability of a specific number of successes in a given number of trials, based on a fixed probability of success.
2. NORM.DIST: Calculates the probability of a continuous random variable falling within a specified range, based on a normal distribution.
3. POISSON.DIST: Calculates the probability of a specific number of events occurring in a fixed interval, based on a constant average rate of occurrence.
4. EXPON.DIST: Calculates the probability of a continuous random variable falling within a specified range, based on an exponential distribution.
5. CHISQ.DIST: Calculates the probability of a continuous random variable falling within a specified range, based on a chi-squared distribution.
By understanding and mastering the PROB function and related formulae, you can perform advanced statistical analysis and probability calculations in Excel with ease.
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If, at a given temperature, the equilibrium constant for the reaction:
$$H_{2(g)} + Cl_{2(g)} \rightleftarrows 2HCl_{(g)}$$
...is $K_p$, then the equilibrium constant for the reaction:
$$HCl_{(g)} \rightleftarrows \cfrac{1}{2} H_{2(g)} + \cfrac{1}{2} Cl_{2(g)}$$
...can be represented as
A
$K=\cfrac { 1 }{ { (K_p) }^{ 2 } }$
B
$K=\cfrac { 1 }{ { (K_p) }^{ 1/2 } }$
C
$K = K_p^2$
D
$K = K_p^{1/2}$
Select an assignment template | 183 | 456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-13 | latest | en | 0.616787 |
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# Need help with RC Strategy
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Need help with RC Strategy [#permalink]
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28 Mar 2011, 03:50
Hello
I have started my mission GMAT last weekend and I need some real help with RC strategy. I have went through a lot of youtube videos / articles on how to best approach RC. gmatclub. com/forum/verbal-rc-help-30247.html was certainly one of the best.
I am atleast able to solve basic questions with the strategy but for medium to tough questions (on grockit.com) take me atleast 7-10 mins just to take down notes for big passages. Also, the notes sometimes don't even end up helping with the Q.
I need some new note taking tips + strategies to solve medium / complex questions in lot less time. I have also tried grockit.com that tell to just read the first and last line and solve the Q. I am new so not so confident with that method yet.
Any other tips wud be great.
Thanks!
If you have any questions
you can ask an expert
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28 Mar 2011, 21:06
Zero replies I was expecting some advise from experts..
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29 Mar 2011, 08:14
Taking down notes is a good strategy, however if you're finishing the passage in 7-10 mins that means you're not taking down notes effectively. Note taking doesn't mean just jotting down the details in the passage, it means your the condensed version of your opionion about a paragraph alongwith some details in as many few words as possible.
How many passages have you practised the strategies you have learnt on. Generally, RC needs a lot of practice and you'll have to fine tune your strategies as a result of which you'll eventually become better at RCs.
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Re: Need help with RC Strategy [#permalink]
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29 Mar 2011, 10:46
do we have any strategy other than taking notes. Taking notes is very very time consuming, I had done around 12-15 passages from OG verbal 2nd edition, and still my success rate is not improving than 50%. Any insight would be of great help. By my strategy, I read the passage and keep everything in my head, and then try to answer the questions. I do refer to pasage for details for almost all the questions, but not sure what I am doing wrong.
Any help will be greatly appreciated.
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Re: Need help with RC Strategy [#permalink]
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29 Mar 2011, 12:56
shovitdhar wrote:
Taking down notes is a good strategy, however if you're finishing the passage in 7-10 mins that means you're not taking down notes effectively. Note taking doesn't mean just jotting down the details in the passage, it means your the condensed version of your opionion about a paragraph alongwith some details in as many few words as possible.
How many passages have you practised the strategies you have learnt on. Generally, RC needs a lot of practice and you'll have to fine tune your strategies as a result of which you'll eventually become better at RCs.
Thanks!! I have practiced it on approx 4-5 easy passages and 15-20 medium to hard passages. Yesterday I just focused only on medium passages and I did notice the note taking speed has slightly improved. Down to approx 5- 6 mins. Note taking sometimes does not helps me at all in solving certain complex questions in which I may be failed to write down that particular keyword in my notes for example. So I end up reading that paragraph all over again or go again finding where the keyword is. The success rate decreases significantly.
Also, please advise if there are different strategies to solve hard questions (the ones like "the author agrees with all EXCEPT" etc).
Can I post example(s) from another website here?
Thanks
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Re: Need help with RC Strategy [#permalink]
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29 Mar 2011, 13:04
brandy96 wrote:
do we have any strategy other than taking notes. Taking notes is very very time consuming, I had done around 12-15 passages from OG verbal 2nd edition, and still my success rate is not improving than 50%. Any insight would be of great help. By my strategy, I read the passage and keep everything in my head, and then try to answer the questions. I do refer to pasage for details for almost all the questions, but not sure what I am doing wrong.
Any help will be greatly appreciated.
I am a newbie and I too can tell you this is nothing but dangerous! Proof? my last result I have tried this memorizing method several times and fail miserably. I end up frustrated going back and forth and end up guessing. From what I have read GMAT does not expects you to memorize it either.
Note taking (a strategy I learned recently from youtube / forums only after i started serious preparation recently) is working lot better (for me atleast) than trying to memorize a passage within 2-3 mins.
Re: Need help with RC Strategy [#permalink] 29 Mar 2011, 13:04
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# Need help with RC Strategy
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,725 | 6,867 | {"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-2017-04 | latest | en | 0.890839 |
https://docs.nvidia.com/cuda/cuquantum/latest/cutensornet/examples/mps_amplitudes.html | 1,713,336,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00849.warc.gz | 191,037,171 | 19,376 | # Computing Matrix Product State (MPS) Amplitudes¶
The following code example illustrates how to define a tensor network state, factorize it as a Matrix Product State (MPS), and then compute a slice of amplitudes of the factorized MPS state. The full code can be found in the NVIDIA/cuQuantum repository (here).
# Headers and error handling¶
``` 7#include <cstdlib>
8#include <cstdio>
9#include <cassert>
10#include <complex>
11#include <vector>
12#include <bitset>
13#include <iostream>
14
15#include <cuda_runtime.h>
16#include <cutensornet.h>
17
18
19#define HANDLE_CUDA_ERROR(x) \
20{ const auto err = x; \
21 if( err != cudaSuccess ) \
22 { printf("CUDA error %s in line %d\n", cudaGetErrorString(err), __LINE__); fflush(stdout); std::abort(); } \
23};
24
25#define HANDLE_CUTN_ERROR(x) \
26{ const auto err = x; \
27 if( err != CUTENSORNET_STATUS_SUCCESS ) \
28 { printf("cuTensorNet error %s in line %d\n", cutensornetGetErrorString(err), __LINE__); fflush(stdout); std::abort(); } \
29};
30
31
32int main()
33{
34 static_assert(sizeof(size_t) == sizeof(int64_t), "Please build this sample on a 64-bit architecture!");
35
36 constexpr std::size_t fp64size = sizeof(double);
```
# Define the tensor network state and the desired slice of state amplitudes¶
Let’s define a tensor network state corresponding to a 6-qubit quantum circuit and request a slice of state amplitudes where qubits 0 and 1 are fixed at value 1.
```40 // Quantum state configuration
41 constexpr int32_t numQubits = 6; // number of qubits
42 const std::vector<int64_t> qubitDims(numQubits,2); // qubit dimensions
43 const std::vector<int32_t> fixedModes({0,1}); // fixed modes in the output amplitude tensor (must be in acsending order)
44 const std::vector<int64_t> fixedValues({1,1}); // values of the fixed modes in the output amplitude tensor
45 const int32_t numFixedModes = fixedModes.size(); // number of fixed modes in the output amplitude tensor
46 std::cout << "Quantum circuit: " << numQubits << " qubits\n";
```
# Initialize the cuTensorNet library handle¶
```50 // Initialize the cuTensorNet library
51 HANDLE_CUDA_ERROR(cudaSetDevice(0));
52 cutensornetHandle_t cutnHandle;
53 HANDLE_CUTN_ERROR(cutensornetCreate(&cutnHandle));
54 std::cout << "Initialized cuTensorNet library on GPU 0\n";
```
# Define quantum gates on GPU¶
```58 // Define necessary quantum gate tensors in Host memory
59 const double invsq2 = 1.0 / std::sqrt(2.0);
60 // Hadamard gate
61 const std::vector<std::complex<double>> h_gateH {{invsq2, 0.0}, {invsq2, 0.0},
62 {invsq2, 0.0}, {-invsq2, 0.0}};
63 // CX gate
64 const std::vector<std::complex<double>> h_gateCX {{1.0, 0.0}, {0.0, 0.0}, {0.0, 0.0}, {0.0, 0.0},
65 {0.0, 0.0}, {1.0, 0.0}, {0.0, 0.0}, {0.0, 0.0},
66 {0.0, 0.0}, {0.0, 0.0}, {0.0, 0.0}, {1.0, 0.0},
67 {0.0, 0.0}, {0.0, 0.0}, {1.0, 0.0}, {0.0, 0.0}};
68
69 // Copy quantum gates to Device memory
70 void *d_gateH{nullptr}, *d_gateCX{nullptr};
71 HANDLE_CUDA_ERROR(cudaMalloc(&d_gateH, 4 * (2 * fp64size)));
72 HANDLE_CUDA_ERROR(cudaMalloc(&d_gateCX, 16 * (2 * fp64size)));
73 std::cout << "Allocated quantum gate memory on GPU\n";
74 HANDLE_CUDA_ERROR(cudaMemcpy(d_gateH, h_gateH.data(), 4 * (2 * fp64size), cudaMemcpyHostToDevice));
75 HANDLE_CUDA_ERROR(cudaMemcpy(d_gateCX, h_gateCX.data(), 16 * (2 * fp64size), cudaMemcpyHostToDevice));
76 std::cout << "Copied quantum gates to GPU memory\n";
```
# Allocate MPS tensors¶
Here we set the shapes of MPS tensors and allocate GPU memory for their storage.
```80 // Determine the MPS representation and allocate buffers for the MPS tensors
81 const int64_t maxExtent = 2; // GHZ state can be exactly represented with max bond dimension of 2
82 std::vector<std::vector<int64_t>> extents;
83 std::vector<int64_t*> extentsPtr(numQubits);
84 std::vector<void*> d_mpsTensors(numQubits, nullptr);
85 for (int32_t i = 0; i < numQubits; i++) {
86 if (i == 0) { // left boundary MPS tensor
87 extents.push_back({2, maxExtent});
88 HANDLE_CUDA_ERROR(cudaMalloc(&d_mpsTensors[i], 2 * maxExtent * 2 * fp64size));
89 }
90 else if (i == numQubits-1) { // right boundary MPS tensor
91 extents.push_back({maxExtent, 2});
92 HANDLE_CUDA_ERROR(cudaMalloc(&d_mpsTensors[i], 2 * maxExtent * 2 * fp64size));
93 }
94 else { // middle MPS tensors
95 extents.push_back({maxExtent, 2, maxExtent});
96 HANDLE_CUDA_ERROR(cudaMalloc(&d_mpsTensors[i], 2 * maxExtent * maxExtent * 2 * fp64size));
97 }
98 extentsPtr[i] = extents[i].data();
99 }
```
# Allocate the amplitudes slice tensor on GPU¶
Here we allocate GPU memory for the requested amplitudes slice tensor.
```103 // Allocate Device memory for the specified slice of the quantum circuit amplitudes tensor
104 void *d_amp{nullptr};
105 std::size_t ampSize = 1;
106 for(const auto & qubitDim: qubitDims) ampSize *= qubitDim; // all state modes (full size)
107 for(const auto & fixedMode: fixedModes) ampSize /= qubitDims[fixedMode]; // fixed state modes reduce the slice size
108 HANDLE_CUDA_ERROR(cudaMalloc(&d_amp, ampSize * (2 * fp64size)));
109 std::cout << "Allocated memory for the specified slice of the quantum circuit amplitude tensor of size "
110 << ampSize << " elements\n";
```
# Allocate the scratch buffer on GPU¶
```114 // Query the free memory on Device
115 std::size_t freeSize{0}, totalSize{0};
116 HANDLE_CUDA_ERROR(cudaMemGetInfo(&freeSize, &totalSize));
117 const std::size_t scratchSize = (freeSize - (freeSize % 4096)) / 2; // use half of available memory with alignment
118 void *d_scratch{nullptr};
119 HANDLE_CUDA_ERROR(cudaMalloc(&d_scratch, scratchSize));
120 std::cout << "Allocated " << scratchSize << " bytes of scratch memory on GPU\n";
```
# Create a pure tensor network state¶
Now let’s create a pure tensor network state for a 6-qubit quantum circuit.
```124 // Create the initial quantum state
125 cutensornetState_t quantumState;
126 HANDLE_CUTN_ERROR(cutensornetCreateState(cutnHandle, CUTENSORNET_STATE_PURITY_PURE, numQubits, qubitDims.data(),
127 CUDA_C_64F, &quantumState));
128 std::cout << "Created the initial quantum state\n";
```
# Apply quantum gates¶
Let’s construct the GHZ quantum circuit by applying the corresponding quantum gates.
```132 // Construct the final quantum circuit state (apply quantum gates) for the GHZ circuit
133 int64_t id;
134 HANDLE_CUTN_ERROR(cutensornetStateApplyTensorOperator(cutnHandle, quantumState, 1, std::vector<int32_t>{{0}}.data(),
135 d_gateH, nullptr, 1, 0, 1, &id));
136 for(int32_t i = 1; i < numQubits; ++i) {
137 HANDLE_CUTN_ERROR(cutensornetStateApplyTensorOperator(cutnHandle, quantumState, 2, std::vector<int32_t>{{i-1,i}}.data(),
138 d_gateCX, nullptr, 1, 0, 1, &id));
139 }
140 std::cout << "Applied quantum gates\n";
```
# Request MPS factorization for the final quantum circuit state¶
Here we express our intent to factorize the final quantum circuit state using MPS factorization. The provided shapes of the MPS tensors refer to their maximal size limit during the MPS renormalization procedure. The actually computed shapes of the final MPS tensors may be smaller. No computation is done here yet.
```144 // Specify the final target MPS representation (use default fortran strides)
145 HANDLE_CUTN_ERROR(cutensornetStateFinalizeMPS(cutnHandle, quantumState,
146 CUTENSORNET_BOUNDARY_CONDITION_OPEN, extentsPtr.data(), /*strides=*/nullptr));
147 std::cout << "Requested the final MPS factorization of the quantum circuit state\n";
```
# Configure MPS factorization procedure¶
After expressing our intent to perform MPS factorization of the final quantum circuit state, we can also configure the MPS factorization procedure by resetting different options, for example, the SVD algorithm.
```151 // Optional, set up the SVD method for MPS truncation.
152 cutensornetTensorSVDAlgo_t algo = CUTENSORNET_TENSOR_SVD_ALGO_GESVDJ;
153 HANDLE_CUTN_ERROR(cutensornetStateConfigure(cutnHandle, quantumState,
154 CUTENSORNET_STATE_CONFIG_MPS_SVD_ALGO, &algo, sizeof(algo)));
155 std::cout << "Configured the MPS factorization computation\n";
```
# Prepare the computation of MPS factorization¶
Let’s create a workspace descriptor and prepare the computation of MPS factorization.
```159 // Prepare the MPS computation and attach workspace
160 cutensornetWorkspaceDescriptor_t workDesc;
161 HANDLE_CUTN_ERROR(cutensornetCreateWorkspaceDescriptor(cutnHandle, &workDesc));
162 std::cout << "Created the workspace descriptor\n";
163 HANDLE_CUTN_ERROR(cutensornetStatePrepare(cutnHandle, quantumState, scratchSize, workDesc, 0x0));
164 std::cout << "Prepared the computation of the quantum circuit state\n";
165 double flops {0.0};
166 HANDLE_CUTN_ERROR(cutensornetStateGetInfo(cutnHandle, quantumState,
167 CUTENSORNET_STATE_INFO_FLOPS, &flops, sizeof(flops)));
168 if(flops > 0.0) {
169 std::cout << "Total flop count = " << (flops/1e9) << " GFlop\n";
170 }else if(flops < 0.0) {
171 std::cout << "ERROR: Negative Flop count!\n";
172 std::abort();
173 }
174
175 int64_t worksize {0};
176 HANDLE_CUTN_ERROR(cutensornetWorkspaceGetMemorySize(cutnHandle,
177 workDesc,
178 CUTENSORNET_WORKSIZE_PREF_RECOMMENDED,
179 CUTENSORNET_MEMSPACE_DEVICE,
180 CUTENSORNET_WORKSPACE_SCRATCH,
181 &worksize));
182 std::cout << "Scratch GPU workspace size (bytes) for MPS computation = " << worksize << std::endl;
183 if(worksize <= scratchSize) {
184 HANDLE_CUTN_ERROR(cutensornetWorkspaceSetMemory(cutnHandle, workDesc, CUTENSORNET_MEMSPACE_DEVICE,
185 CUTENSORNET_WORKSPACE_SCRATCH, d_scratch, worksize));
186 }else{
187 std::cout << "ERROR: Insufficient workspace size on Device!\n";
188 std::abort();
189 }
190 std::cout << "Set the workspace buffer for the MPS factorization computation\n";
```
# Compute MPS factorization¶
Once the MPS factorization procedure has been configured and prepared, let’s compute the MPS factorization of the final quantum circuit state.
```194 // Execute MPS computation
195 HANDLE_CUTN_ERROR(cutensornetStateCompute(cutnHandle, quantumState,
196 workDesc, extentsPtr.data(), /*strides=*/nullptr, d_mpsTensors.data(), 0));
197 std::cout << "Computed the MPS factorization\n";
```
# Create the state amplitudes accessor¶
Once the factorized MPS representation of the final quantum circuit state has been computed, let’s create the amplitudes accessor object that will compute the requested slice of state amplitudes.
```201 // Specify the quantum circuit amplitudes accessor
202 cutensornetStateAccessor_t accessor;
203 HANDLE_CUTN_ERROR(cutensornetCreateAccessor(cutnHandle, quantumState, numFixedModes, fixedModes.data(),
204 nullptr, &accessor)); // using default strides
205 std::cout << "Created the specified quantum circuit amplitudes accessor\n";
```
# Configure the state amplitudes accessor¶
Now we can configure the state amplitudes accessor object by setting the number of hyper-samples to be used by the tensor network contraction path finder.
```209 // Configure the computation of the slice of the specified quantum circuit amplitudes tensor
210 const int32_t numHyperSamples = 8; // desired number of hyper samples used in the tensor network contraction path finder
211 HANDLE_CUTN_ERROR(cutensornetAccessorConfigure(cutnHandle, accessor,
212 CUTENSORNET_ACCESSOR_CONFIG_NUM_HYPER_SAMPLES, &numHyperSamples, sizeof(numHyperSamples)));
```
# Prepare the computation of the amplitudes slice tensor¶
Let’s prepare the computation of the amplitudes slice tensor.
```216 // Prepare the computation of the specified slice of the quantum circuit amplitudes tensor
217 HANDLE_CUTN_ERROR(cutensornetAccessorPrepare(cutnHandle, accessor, scratchSize, workDesc, 0x0));
218 std::cout << "Prepared the computation of the specified slice of the quantum circuit amplitudes tensor\n";
219 flops = 0.0;
220 HANDLE_CUTN_ERROR(cutensornetAccessorGetInfo(cutnHandle, accessor,
221 CUTENSORNET_ACCESSOR_INFO_FLOPS, &flops, sizeof(flops)));
222 std::cout << "Total flop count = " << (flops/1e9) << " GFlop\n";
223 if(flops <= 0.0) {
224 std::cout << "ERROR: Invalid Flop count!\n";
225 std::abort();
226 }
```
# Set up the workspace¶
Now we can set up the required workspace buffer.
```230 // Attach the workspace buffer
231 HANDLE_CUTN_ERROR(cutensornetWorkspaceGetMemorySize(cutnHandle,
232 workDesc,
233 CUTENSORNET_WORKSIZE_PREF_RECOMMENDED,
234 CUTENSORNET_MEMSPACE_DEVICE,
235 CUTENSORNET_WORKSPACE_SCRATCH,
236 &worksize));
237 std::cout << "Required scratch GPU workspace size (bytes) = " << worksize << std::endl;
238 if(worksize <= scratchSize) {
239 HANDLE_CUTN_ERROR(cutensornetWorkspaceSetMemory(cutnHandle, workDesc, CUTENSORNET_MEMSPACE_DEVICE,
240 CUTENSORNET_WORKSPACE_SCRATCH, d_scratch, worksize));
241 }else{
242 std::cout << "ERROR: Insufficient workspace size on Device!\n";
243 std::abort();
244 }
245 std::cout << "Set the workspace buffer\n";
```
# Compute the specified slice of state amplitudes¶
Once everything has been set up, we compute the requested slice of state amplitudes, copy it back to Host memory, and print it.
```249 // Compute the specified slice of the quantum circuit amplitudes tensor
250 std::complex<double> stateNorm2{0.0,0.0};
251 HANDLE_CUTN_ERROR(cutensornetAccessorCompute(cutnHandle, accessor, fixedValues.data(),
252 workDesc, d_amp, static_cast<void*>(&stateNorm2), 0x0));
253 std::cout << "Computed the specified slice of the quantum circuit amplitudes tensor\n";
254 std::vector<std::complex<double>> h_amp(ampSize);
255 HANDLE_CUDA_ERROR(cudaMemcpy(h_amp.data(), d_amp, ampSize * (2 * fp64size), cudaMemcpyDeviceToHost));
256 std::cout << "Amplitudes slice for " << (numQubits - numFixedModes) << " qubits:\n";
257 for(std::size_t i = 0; i < ampSize; ++i) {
258 std::cout << " " << h_amp[i] << std::endl;
259 }
260 std::cout << "Squared 2-norm of the state = (" << stateNorm2.real() << ", " << stateNorm2.imag() << ")\n";
```
# Free resources¶
```264 // Destroy the workspace descriptor
265 HANDLE_CUTN_ERROR(cutensornetDestroyWorkspaceDescriptor(workDesc));
266 std::cout << "Destroyed the workspace descriptor\n";
267
268 // Destroy the quantum circuit amplitudes accessor
269 HANDLE_CUTN_ERROR(cutensornetDestroyAccessor(accessor));
270 std::cout << "Destroyed the quantum circuit amplitudes accessor\n";
271
272 // Destroy the quantum circuit state
273 HANDLE_CUTN_ERROR(cutensornetDestroyState(quantumState));
274 std::cout << "Destroyed the quantum circuit state\n";
275
276 for (int32_t i = 0; i < numQubits; i++) {
277 HANDLE_CUDA_ERROR(cudaFree(d_mpsTensors[i]));
278 }
279 HANDLE_CUDA_ERROR(cudaFree(d_scratch));
280 HANDLE_CUDA_ERROR(cudaFree(d_amp));
281 HANDLE_CUDA_ERROR(cudaFree(d_gateCX));
282 HANDLE_CUDA_ERROR(cudaFree(d_gateH));
283 std::cout << "Freed memory on GPU\n";
284
285 // Finalize the cuTensorNet library
286 HANDLE_CUTN_ERROR(cutensornetDestroy(cutnHandle));
287 std::cout << "Finalized the cuTensorNet library\n";
288
289 return 0;
290}
``` | 4,261 | 16,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.443838 |
https://www.koofers.com/files/exam-kzowtwqy3t/ | 1,591,499,257,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00046.warc.gz | 759,585,304 | 13,182 | # Past Exam for MATH 2300 - Analytic Geometry and Calculus 2 with Newberry at Colorado (CU)
## Exam Information
Material Type: Exam 3 Professor: Newberry Class: MATH 2300 - Analytic Geometry and Calculus 2 Subject: Mathematics University: University of Colorado - Boulder Term: -- Keywords: Continuation
## Sample Document Text
MATH 2300 - Fall 2008 Exam 3 Solutions 1. (a) 1X k=2 34 k+1 = 34 3 1X k=2 34 k 2 = 2764 11 + 3 4 = 2764 47 = 27112 (b) nX k=0 tan 1(k+1) tan 1(k) = n+1X k=1 tan 1(k) nX k=0 tan 1(k) = tan 1(n+1) tan 1(0) = tan 1(n+1): 1X k=0 tan 1(k+ 1) tan 1(k) = lim n!1 nX k=0 tan 1(k+ 1) tan 1(k) = lim n!1tan 1(n+ 1) = 2: 2. (a) lim k!1 1 1=k = 16= 0: So, 1X k=1 1 1=k diverges by the Divergence Test. (b) Since fkg and flnkg are monotone increasing to in nity, then f 1klnkg is monotone de- creasing to zero. So, by the Alternating Series Test, 1X k=2 ( 1)k klnk converges. Also, x and lnx are continuous functions and nonzero for x> 0: So, 1xlnx is also contin- uous. Furthermore, it is decreasing since x and lnx are increasing. Note that Z 1 2 1 xlnxdx = Z 1 ln2 1 udu = lima!1[lnu] a ln2 = lima!1lna ln(ln 2) =1: So, by the Integral Test, 1X k=2 1 klnk diverges. Thus, 1X k=2 ( 1)k klnk converges conditionally. 3. (a) 1X k=1 4 + ( 1)k ... | 535 | 1,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-24 | latest | en | 0.7283 |
https://www.hackmath.net/en/example/7113?tag_id=124 | 1,553,454,468,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203491.1/warc/CC-MAIN-20190324190033-20190324212033-00206.warc.gz | 789,035,783 | 6,598 | # Distance problem 2
A=(x,2x)
B=(2x,1)
Distance AB=√2, find value of x
Result
x1 = 1
x2 = -0.2
#### Solution:
Checkout calculation with our calculator of quadratic equations.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc. Looking for help with calculating roots of a quadratic equation? Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Do you want to convert length units? Pythagorean theorem is the base for the right triangle calculator.
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12. The size
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# When positive integer n is divided by 3, the remainder is 2;
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When positive integer n is divided by 3, the remainder is 2; [#permalink]
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29 Jul 2006, 23:01
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When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n-2 is divisible by 5
2) t is divisible by 3
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient
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29 Jul 2006, 23:19
zoom612 wrote:
When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n-2 is divisible by 5
2) t is divisible by 3
C must be it!
Given
n = 3k+2 and t=5l+3
S1. n-2 = 5m => n = 5m+2
Thus, n = 15q+2 insufficient.
S2. t = 3m
Thus, t= 15Q+3 insufficient.
Combine S1 and S2
nt = (15q+2)(15Q+3)
remainder is 2*3 = 6
sufficient
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Re: DS: Gmat prep [#permalink] 29 Jul 2006, 23:19
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1. Dec 12, 2013
### Jon.G
1. The problem statement, all variables and given/known data
The end of a stretched string is forced to vibrate with a transverse displacement with A= 0.30 m and ω = 9.0 rad s-1. If the tension of the string is T = 4.0 N, and the mass per unit length is µ= 1.250 g m-1, calculate
(a) the wave velocity,
(b) the frequency of oscillation,
(c) the wavelength, and
(d) the displacement of the string at x = 2.0 m from the source at the time t = 100 ms after the oscillation commences.
Assume that the wave is travelling from left to right.
The attempt at a solution
a.) v=√(T/μ) giving me a wavespeed of 56.6 ms-1
b.)ω=2∏f giving me a frequency of 1.43 Hz
c.)v=fλ giving me a wavelength of 39.6 m
d.)This is where I'm stuck :( I would think of using the equation vertical displacement,
y=Asin(kx-ωt) except I don't know how to find k
Any help appreciated.
2. Dec 12, 2013
### Simon Bridge
If you put t=0, then you have y=Asin(kx)
... then when x=0, y=0 right?
... when x=λ, what is y equal to (hint: definition of wavelength)?
... therefore what is kλ equal to? | 351 | 1,135 | {"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} | 3.296875 | 3 | CC-MAIN-2017-47 | longest | en | 0.852213 |
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How do you explain how to do square roots over 100?
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From: Pat Ballew (for Teacher2Teacher Service)
Date: May 16, 2000 at 23:11:15
Subject: Re: How do you explain how to do square roots over 100?
There are several approaches to finding square roots by hand. For inexperienced students (those still struggling to understand the ideas) I like guess and test. It helps them with concepts and estimation skills. To find the square root of 215 I try to guess an answer... and since it is more than 100, I will guess something bigger than ten.. maybe 20 works .... is 20^2 = 215???? nope, it's 400, too big so I need to guess a smaller number.... 15???? 15^2 = 225......OOOH very close, just a little big, maybe 14 will work drats!!! 14^ is only 196.. but at least I know the answer is between 14 and 15... so I try 14.5.... of course you see how this goes and at some point we say close enough, because the real root is irrational and will never end... (a big discovery when we first make it).. .. There are other ways, divide and average is a good one... I think all of the methods are given on the web page at the Dr. Math FAQ page, Here is the URL for square roots by hand... http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html good luck, hope this gives you some ideas for your class. -Pat Ballew, for the Teacher2Teacher service
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Author Topic: Will Michael revisit ETTR? (Read 56088 times)
Guillermo Luijk
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the DXO normalization procedure is of interest. The normalized SNR equation is:
SNRnorm = SNR + 20 * log10 (sqrt[N/N0]),
where N0 is the original number of pixels, N is the number of pixels for the sensor with the higher pixel count, and SNR is the original SNR.
If we average 4 pixels into one, the formula shows that the SNR increases by 6.02 dB or 1 stop, in agreement with Emil's figure. I think DXO is using flat patches to derive their figures. As you suggest, in real world use with demosaiced images, the SNR may be somewhat less than the theoretical value.
I think the 'patches discusion' is going farer than it really deserves. Averaging 4 pixels of the same value + their individually added noise improves SNR by 2, I think we all agree here. If the signal is different on each pixel, the noise will also be different, so SNR will be different on each pixel. If each source pixel has its own SNR, it's nonsense to look for 'SNR improves by a factor of X', since source SNR is not unique.
But once defined, the 'patch model' can be extended to any real world case with the purpose of SNR normalisation. It will always be an approximation, but a valid approximation to make noise performance from different sensors comparable. It is not intended to quantify SNR on any real picture of our cat.
SNR_norm_dB = SNR_perpixel_dB + 20 * log10 [(Mpx / 12,7)^0.5]
Regards
« Last Edit: August 31, 2011, 10:02:18 AM by Guillermo Luijk » Logged
BartvanderWolf
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While awaiting Emil's analysis (presuming he takes the trouble to reply to your post), the DXO normalization procedure is of interest. The normalized SNR equation is:
SNRnorm = SNR + 20 * log10 (sqrt[N/N0]),
where N0 is the original number of pixels, N is the number of pixels for the sensor with the higher pixel count, and SNR is the original SNR.
If we average 4 pixels into one, the formula shows that the SNR increases by 6.02 dB or 1 stop, in agreement with Emil's figure. I think DXO is using flat patches to derive their figures. As you suggest, in real world use with demosaiced images, the SNR may be somewhat less than the theoretical value. The DXO engineer also states that 4:1 binning will give 1 one stop improvement in SNR for the DXO engineer also states that binning outside the sensor doubles the SNR whereas hardware binning quadruples the SNR. What are your figures?
Hi Bill,
joofa's numbers are the same, he is just acting a bit anal about the fact that most real life images are dominated by non-uniform (image detail) patches, so it's hard to calculate average noise (the amount of noise is different for each pixel where the signal is different as well).
Even with uniform areas, a non-binning type of averaging further complicates the noise averaging statistics, because the downsampling filters are not uniform over their support width (e.g. Lanczos window). He conveniently does not mention the possibility (although not practical with subject motion) to average mutiple exposures of the same scene, as is e.g. done routinely by astrophotographers. That way each pixel gets averaged as expected and noise is reduced in quadrature, regardless of its ajacent pixel's signal level.
Cheers,
Bart
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Ray
Sr. Member
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The point is that noise acts to dither the levels. Suppose the true signal is some value X between zero and one, over a patch of the image (we are going to ignore natural scene variation for the purpose of answering your question). Suppose the noise is of strength N, for example let N be one level. The noise adds a random number roughly between -N and N to X, so that the pixel wants to record some number between X-N and X+N. Of course, the resulting signal plus noise is digitized so the output is either 0 or 1; if the noise is random (uncorrelated from pixel to pixel), the value of X is reflected in the percentage of 1's vs 0's in the patch -- a fraction X of the pixels will be 1 and the rest 0. If we average the levels over a large enough patch, we recover the original signal, even though each individual pixel only recorded 0 or 1.
This is the basic idea that allows one to trade resolution for noise -- and why downsampled images look less noisy. Note also that while the average is more finely graded than steps of one, that doesn't mean we buy anything by making individual pixels record values more finely spaced than the level of noise, because the individual values are jumping around randomly by an amount between -N and N.
Thanks for attempting the explanation, Emil, even though I have to admit it is not all totally clear.
Nevertheless, I can see with my own eyes that noisy images after downsampling exhibit less noise. A similar effect can be achieved simply by viewing the image from a greater distance.
What still puzzles me is that a 14 bit A/D converter in a camera with a good dynamice range, such as the D7000, will not provide any IQ advantages in the deepest shadows, compared with 12 bit.
However, I can understand this might be the case with a Canon camera which doesn't go beyond 12 stops of DR.
I've gone back to the RAW files and created crops of the 12th, 13th and 14th stops, which show a clear improvement in IQ in the 12th stop.
Now, you seem to be claiming that such improvement in that 12th stop is entirely due to the greater quantity of light resulting in an improved SNR (1/500th sec as opposed to 1/2000th for the 14th stop).
Of course, I am able to appreciate that 4x the number of photons will improve the SNR and the IQ as a result.
I'm just a bit skeptical that the image quality in that 12th stop, as shown in my attached image, could be achieved with just the one level per channel of a 12 bit converter, as opposed to the 4 levels per channel of the 14 bit converter.
Normally, I would just repeat the test in 12 bit and 14 bit modes. If I see a difference at the same very low exposures, then there's a difference, depite any theory from an eminent Physicist.
Unfortunately, my D7000 is in for repair due to its inability to autofocus in cold weather. It looks as though I won't get it back for a while because it is now Spring in Australia, and the Nikon repair agent is unable to duplicate the problem in the artificially cold environments he has created.
Tell me now whether I can expect to see the same image quality in that 12th stop with the D7000 set to 12 bit mode, and when I get the camera back, or a new replacement, I'll carry out the tests and tell you whether you are right or wrong.
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bjanes
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I think the 'patches discusion' is going farer than it really deserves. Averaging 4 pixels of the same value + their individually added noise improves SNR by 2, I think we all agree here. If the signal is different on each pixel, the noise will also be different, so SNR will be different on each pixel. If each source pixel has its own SNR, it's nonsense to look for 'SNR improves by a factor of X', since source SNR is not unique.
Quite true. Also, we should remember that the SNR for each color of the Bayer array will be different. Figures are usually given for the green channels, since they should be near saturation with proper exposure. However, the red and blue channels will be below saturation according to the white balance multipliers. Nonetheless, their noise will also be reduced by a like factor when adding in quadrature.
Regards,
Bill
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joofa
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And, remember, that is just one way of doing it. In fact, I have changed my own approach somewhat from the above links that is years older.
joofa's numbers are the same, he is just acting a bit anal about the fact that most real life images are dominated by non-uniform (image detail) patches, so it's hard to calculate average noise (the amount of noise is different for each pixel where the signal is different as well).
I don't know how can you claim the numbers are the same, without even knowing what I did. Kindly see the link posted above.
Quote
Even with uniform areas, a non-binning type of averaging further complicates the noise averaging statistics, because the downsampling filters are not uniform over their support width (e.g. Lanczos window). He conveniently does not mention the possibility (although not practical with subject motion) to average mutiple exposures of the same scene, as is e.g. done routinely by astrophotographers. That way each pixel gets averaged as expected and noise is reduced in quadrature, regardless of its ajacent pixel's signal level.
I don't think you realize that there is more to it than quadrature. Noise still adds in quadrature, but that is not the end of story. Again see the link above.
Sincerely,
Joofa
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Joofa
http://www.djjoofa.com
hjulenissen
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The DXO engineer also states that 4:1 binning outside the sensor hardware doubles the SNR whereas hardware binning quadruples the SNR. What are your figures?
Regards,
Bill
But using digital processing, the choice of averaging kernel is without limits, including negative coefficients, space-variant, signal-dependent (non-linear) manually guided processing compared to the simple box-car filter that binning even non-bayer sensels. This might (or might not) be enough to offset the initial advantage of sensor binning.
-h
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hjulenissen
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I think the 'patches discusion' is going farer than it really deserves.
What is the reason that patches are being discussed? Just curiosity, or is there some real-world photography application being limited by this approach?
-h
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BartvanderWolf
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And, remember, that is just one way of doing it. In fact, I have changed my own approach somewhat from the above links that is years older.
I don't know how can you claim the numbers are the same, without even knowing what I did.
And therein lies the problem with you. You are not very clear in what you base your assumptions on (so one can only assume common statistical principles which apply to us all), yet you hint at outcomes that deviate from common experience. You then don't explain but refer to another website where you also cause confusion by not explaining that you use a model that assumes correlation between sensels. Why? Nobody but you knows, until the info is tortured out of you. Why? And then (above) you add that the models you use in the discussion on that website are not the models you now use. Well thanks for creating yet more confusion. Was the earlier model not correct? What are you using now? Why are you using a correlation model in the first place, any literature/references/experiments you want to share that other can independently confirm?
No wonder Emil has given up on 'discussing' with you.
Quote
Noise still adds in quadrature, but that is not the end of story.
More riddles. I'm not even going to ask...
Sincere cheers,
Bart
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ejmartin
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I will await Emil's response, since he is more technically adept than myself. In the meantime, what are your calculations?
Emil is not going to follow joofa's red herrings. This was a thread about ETTR; I am happy to discuss that topic, though I suspect it has been covered amply already. If the subject is to be changed, how about mass in general relativity? Did you know that there is no precise definition of local mass in Einstein's theory? Somehow, it doesn't keep my bathroom scale from working...
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emil
joofa
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any literature/references/experiments you want to share that other can independently confirm?
At DPReview I have gone over this topic several times with details of what I'm doing. You are welcome to browse my messages there. But in summary, I just used elementary stochastic processes as applied in electrical engineering.
Quote
No wonder Emil has given up on 'discussing' with you.
I think signal processing is not Emil's forte .
Sincerely,
Joofa
« Last Edit: August 31, 2011, 02:40:42 PM by joofa » Logged
Joofa
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bjanes
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At DPReview I have gone over this topic several times with details of what I'm doing. You are welcome to browse my messages there. But in summary, I just used elementary stochastic processes as applied in electrical engineering.
I think signal processing is not Emil's forte .
I can see why you are not pleased with Emil. In the thread on DPReview which you referenced, he was apparently aware of your thesis but chose to ignore it.
Regards,
Bill
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joofa
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I can see why you are not pleased with Emil. In the thread on DPReview which you referenced, he was apparently aware of your thesis but chose to ignore it.
Bill, I have a great deal of respect for you. Lets not make this personal. It's Emil's prerogative to ignore it if he wants to and that is fine with me. I stand by my assertions and anybody who doubts them can work out the numbers themselves.
Sincerely,
Joofa
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Joofa
http://www.djjoofa.com
kwalsh
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Before anyone gets further bent out of shape it appears Joofa is simply saying in a practical sense that the basic 6.02 dB figure for resizing by a factor of two is complicated by two issues.
First, it is rare for anyone to use a rectangular kernel when resizing and practical kernels will not necessarily produce the same change in SNR. There are a variety of reasons for this, but one would be the differing spatial frequency distribution of the scene detail and noise combined with a kernel with a non-uniform frequency response.
Second, since the noise distribution is amplitude dependent in a real scene the statistics will be more complicated than a simple model based on uniform patches. This will also cause a deviation from the 6.02 dB figure.
Of what practical use this point is to the ETTR discussion in this thread I'm not sure. It seems like a thread hijack of thread hijack. Regardless, he's not off base in stating the simple 6.02 dB model used by DxO is in fact a simplification complicated by real world images and real world resizing. No need to jump down his throat about it.
If you follow Joofa's posts here and elsewhere you'll know he likes to ask questions instead of answer them and always leaves dangling caveats to everything he says. If you spend any time in academia you will encounter personalities who seem to think obfuscation makes them appear erudite. Joofa is one of those souls who hasn't figured out clear communication is a better standard of competence. You'll need to cut him some slack if you want to figure out what he's trying to say, and usually you'll find he isn't too far off base. Whether it is worth parsing his rhetorical methods is for the reader to decide.
Ken
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PierreVandevenne
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BTW, I've just finished a nice book that should be mandatory reading for anyone taking part in longish threads on the Internet
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Alan Goldhammer
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BTW, I've just finished a nice book that should be mandatory reading for anyone taking part in longish threads on the Internet
But this ONE is far better and more in keeping with what is going on!
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degrub
Guest
EPDM ? i thought that was a kind of synthetic rubber
Frank
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Ray
Sr. Member
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In case anyone is still interested in the differences between 12 bit and 14 bit A/D conversions, I've done a few tests with my D7000 of the same target shown before.
I'm surprised that there is not a greater difference in the 12th stop and 11th stop. The differences are subtle. The 12 bit shots exhibit slightly greater grain and a degree of clipping of blacks that no adjustment in ACR can completely correct.
However, in the 13th and 14th stops of DR the differences smack one in the face.
One could argue that such differences are of no consequence because they are apparent only in the deepest shadows which one might want to deliberately clip. However, it seems to me if one is underexposing say 5 stops at ISO 100 as an alternative to using ISO 3200, one wants to be able to retrieve as much detail as possible from the shadows which will appear totally black before exposure compensation.
I'm going to try posting a couple of comparisons of the 13th and 14th stops of DR, but the jpeg engine seems to have difficulty compressing noise and the file sizes are quite large. So, if you don't see the images, don't be surprised. I may have to post them one at a time.
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bjanes
Sr. Member
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Posts: 2882
In case anyone is still interested in the differences between 12 bit and 14 bit A/D conversions, I've done a few tests with my D7000 of the same target shown before.
I'm surprised that there is not a greater difference in the 12th stop and 11th stop. The differences are subtle. The 12 bit shots exhibit slightly greater grain and a degree of clipping of blacks that no adjustment in ACR can completely correct.
However, in the 13th and 14th stops of DR the differences smack one in the face.
Since a 12 bit linear file can only encode 12 stops of DR, it is confusing to try to extract 14 bits of DR from a 12 bit file.
Regards.
Bill
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Ray
Sr. Member
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Posts: 8944
Since a 12 bit linear file can only encode 12 stops of DR, it is confusing to try to extract 14 bits of DR from a 12 bit file.
Regards.
Bill
Indeed it is. But Emil's point, as I understood it, was that noise would prevent any advantage of an increase in the number of levels beyond 12 bit, and that 14 bits served no purpose other than a possible slight increase in the accuracy of the conversion.
Perhaps it would be clearer to state, if the DR of the camera is no greater than 12 stops, then more levels that 12 bits afford, is of little use.
However if the camera has a DR of more than 12 stops, then more than 12 bits is of use.
If one wishes to engage in pixel peeping, 14 bits does provided a pixel-peeping IQ advantage in the 11th and 12th stops, with the D7k. However, I doubt that would be the case with lesser cameras, DR-wise.
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Guillermo Luijk
Sr. Member
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Posts: 1336
Indeed it is. But Emil's point, as I understood it, was that noise would prevent any advantage of an increase in the number of levels beyond 12 bit, and that 14 bits served no purpose other than a possible slight increase in the accuracy of the conversion.
The D7000 is such a low noise camera that less than 14 bits are not sufficient to properly encode the deep shadows information at base ISO. But this is compatible with the fact that as long as noise dithers posterization, having more levels (i.e. having more bits) is useless.
I developed a RAW file from the Pentax K5 (same sensor as D7000), emulating a 12-bit RAW file (by decimating the RAW data from 14-bit to 12-bit prior to RAW development), and posterization began to show up:
14-bit: noise dithers posterization
12-bit: posterization becomes visible
However, doing the same test on Canon 40D's RAW files, there was no loss of IQ using 12-bit. That camera is noisy enough to make 14 bits unnecesary.
Regards
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## 27 September 2015
### How a visitor improved my class's confidence
I teach 9th grade at a boarding school for boys. When we host prospective students on campus, they often come to my class. I don't generally let them sit passively and watch -- I make every attempt to get them involved in the day's activity.
Yesterday (yes, we teach on Saturdays, aren't you jealous) a prospective student sat in as I introduced mirror ray diagrams. We had already covered ray diagrams for converging and diverging lenses, so the class already had the general principles of the topic ingrained. So class consisted of just a few elements:
1. A three-minute quiz based on questions on our recent assessment.
3. Six minutes of me demonstrating two mirror ray diagrams predicting the location of an image
4. Setting up a converging mirror to verify the ray diagram's prediction
5. Handing out this worksheet which includes nine situations for students to practice ray diagrams
At step 5, I put on music and allowed students to work at their own pace. They brought up each completed diagram for my approval.
So what did the visiting student do? He didn't know about ray diagrams, right?
No, he did not. I gave him a copy of the quiz just so he could follow along. I gave him a copy of the worksheet. I asked him to make his best effort to join in, attempting the ray diagrams and showing me his work. To this gentleman's credit, he did -- in a class of students a year older than he, when he must have felt very much the outsider, he joined in.
And, of course, he got things wrong. In his first attempt, he didn't use a straight-edge. Many of my class made that mistake seven days ago, when we first tried lens ray diagrams. I gently explained that he needed to use the ruler I had placed on his desk. So he went back to try again.
The next time he came up to see me, he had drawn a ray incorrectly, and his image was in the wrong place. Thing is, my students had made exactly these kinds of errors a week ago, too! It was cathartic for my guys to help this prospect out. Everyone in my class was friendly, helpful, welcoming...
...and a wee bit smug, knowing that they were beyond the rookie mistakes.
At the end of class, I shook hands with the prospect. He seemed relieved to be done, but also quite a bit proud to have joined this physics class seamlessly.
And a couple of my guys walking out with the prospect also walked a bit straighter. As a student, it's easy to lose sight of the significant progress you've made, even just a few weeks into the school year. When new material comes at you nearly every day, when tests and quizzes are raining down from every subject, it's so easy to focus on "failures" -- you missed this question, you didn't demonstrate this skill, you didn't remember this fact. Our visitor gave my students the opportunity to see for themselves the things they DID know, the skills they HAD developed. Helping this youngling out made them feel good, reinforced their own knowledge through teaching, and built tremendous confidence.
And the youngling? He kept working, kept listening to my students' advice with a smile, neither ran away screaming nor folded up silently in an intimidating academic situation. So I very much hope to see this guy in my class next year.
## 14 September 2015
### "Motivation" for completing in-class exercises... inspired by the AP Physics reading
I've heard the AP Physics reading referred to as a "sweatshop." The moniker is full of hyperbole, of course, yet unironic. At the reading, teachers used to independence and flexibility find themselves required to work unyieldingly to the clock. At 8:00, you sit and grade for two hours. Take a break -- exactly fifteen minutes, exactly from 10:00-10:15 -- and keep grading until lunch. One prescribed hour for lunch, and back at it... well, you get the idea.
When you've been grading for days, and your brain is tired, and it's still another hour before lunch, what's to stop a reader from just sitting there and pretending to work? Or from taking a 55-minute bathroom break? This what I mean by the hyperbole of the sweatshop analogy. The supervisors at the AP reading have no real power. No whips. It's even vanishingly rare for someone to be sacked on the spot (and the presumptive sackee still has a cushy tenured professorship to return to, so even sacking is an empty threat).
The only leverage that the reading leadership truly holds over the grunts is professional pride... and that's powerful leverage indeed. Teachers generally want to do things right. They care. They don't want to look like the weak link in front of colleagues.
So, each day, the table leaders list each reader in the room on a wall chart. As a reader finishes a pack of 25 exams, he or she makes a tally mark in the correct space on the chart. There, laid bare for the entire room to see, is a permanent record of how much each person has contributed to the group effort.
Now, the leaders emphasize over and over again: the reading is not a race. Accuracy is far more important than speed. There's no prize for the person who reads the most exams. Just do your best, and speed will come. No pressure.
Nevertheless, consider a session on day four of the reading in which most of the room reads ten packs of exams or so, but Jason only reads three packs. How does Jason feel? How do his colleagues in the room feel? No one, especially the table leader, will likely come to Jason and have a word about his slow relative reading pace. The worst consequence for Jason will likely be some stares from his colleagues. Nevertheless, Jason will have taken a serious blow to his professional pride.
If Jason still is slow in the next session, perhaps the table leader might offer some tips about speeding up -- always reminding Jason that speed is secondary to accuracy, and isn't truly that important. Perhaps in the beer tent that night Jason might take some good-natured needling from his friends about his slow day. But the real incentive here is that Jason will want desperately to feel like part of the team... no one at the reading wants to feel like he or she has let down the communal goal.
So what does this have to do with your class?
A standard type of class, especially with my 9th grade, involves students working at their own pace on in-class laboratory exercises. I'm often asked, how do I motivate students to stay on task? Certainly I offer credit for each completed exercise, but to a 14-year-old, it's likely that gossiping with friends or secretly checking a fantasy football team will trump physics work any day of the week.
Well, to start with, I have a pretty good classroom presence. I generally notice quickly when conversations turn to sports, music, or sex rather than to physics. Just a friendly but firm call-out from me, especially early in the year, can remind students that I'm paying attention, and that I expect them to focus. My eyes and my words take care of egregious issues.
What about the student who would rather sit with his mouth hanging open rather than do the tough work of engaging mentally with physics problems? I do require frequent trips to the front of the room to check with me. When I haven't seen someone in a while, I may inquire why not.
As at the AP reading, though, the real incentive is a transparent display of progress. Students earn credit for each exercise they complete. To keep track of how many exercises each person has done, I use an AP-style tally board -- see the picture above. It becomes a bit uncomfortable if Jason hasn't finished an exercise at all, while his classmates are all on number five or six.
I've observed that 9th graders aren't usually embarrassed about poor or lazy performance when the teacher is the only one who knows or notices. "Oh, sorry, physics is hard, I'll never get it, I'm just not that good." But when their peers are the ones taking off points on a quiz; when their peers observe perverse slackage; when their peers say "you know, we've only done a problem like this four times, it's not that tough" -- then those 9th graders tend to pick up the mental effort.
Please understand, the intent of the progress board is not to shame anyone. Taking a cue from my years as an AP table leader, I emphasize repeatedly -- physics exercises aren't a race. No one gets a prize for being fastest. It's more important to be right than to be sloppy and quick. I never call anyone out merely for a failure to keep up. Nevertheless, the board is there, staring at the class, giving some folks second thoughts about taking a bathroom break, perhaps encouraging someone to get just one more done before the bell rings... | 2,062 | 8,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-13 | longest | en | 0.973405 |
https://studylib.net/doc/18807796/first-order-linear-differential-equations | 1,611,609,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00473.warc.gz | 581,858,969 | 12,629 | # First Order Linear Differential Equations
```First Order Linear Differential Equations
The general first-order linear ordinary differential operator may be written formally as
L = a(x)D + b(x),
where D denotes d/dx. It acts on any differentiable function y of x to give
L(y) = a(x)
dy
+ b(x)y.
dx
The homogeneous case The general homogeneous first-order linear ordinary differential equation may be written
L(y) ≡ a(x)
dy
+ b(x)y = 0.
dx
(1)
It is separable, and so easily solved, and the solution always has the form y = Au,
where A is an arbitrary constant.
Example
Solve the differential equation
L(y) ≡ x3
dy
+ 4x2 y = 0.
dx
(2)
We separate the variables to get the equation of differentials
dx
dy
= −4 ,
y
x
which integrates to give (with a little cleaning up)
log y = −4 log x + C = log(x−4 ) + C.
We exponentiate this to give the solution in the desired form
y = elog(x
−4 )
eC = Ax−4 ,
(3)
where we write the arbitrary constant as A = eC .
By linearity, it is enough to find one nontrivial solution u of equation (2) and
multiply it by the arbitrary constant A.
(continued on other side)
110.302 Calculus II
JMB File: linde, Revision A; 30 Aug 2001; Page 1
2
First Order Linear Differential Equations
The nonhomogeneous case The general nonhomogeneous first-order linear ordinary differential equation may be written
L(y) ≡ a(x)
dy
+ b(x)y = Q(x).
dx
(4)
We use the method of variation of parameter(s). The idea is to put y(x) = A(x)u(x),
where u is a solution of the corresponding homogeneous equation equation (1), except
that the parameter A is now allowed to vary. By the product rule, L(y) has some
terms with A(x) as a factor, and some terms with A0 (x) as a factor. The point is that
the terms with A must cancel, because L(Au) = 0 if we choose A to be constant, by
our choice of u. (This provides a useful check on our calculations.) Then equation (4)
reduces to an equation for A0 (x), which is to be integrated to give A(x) and hence
the general solution y.
Example
Solve the differential equation
L(y) ≡ x3
dy
+ 4x2 y = x − 1.
dx
(5)
In view of equation (3), we put y(x) = A(x)x−4 . Then
L(y) = x3 A0 (x)x−4 − 4x3 A(x)x−5 + 4x2 A(x)x−4 = A0 (x)x−1 .
Observe that the terms with A(x) cancel. Thus equation (5) reduces to A0 (x) =
x(x − 1), which integrates to give A(x) = x3 /3 − x2 /2 + C. The general solution of
equation (5) is therefore
y=
A(x)
1
1
C
=
− 2 + 4,
4
x
3x 2x
x
(6)
where C is the arbitrary constant.
110.302 Calculus II
JMB File: linde, Revision A; 30 Aug 2001; Page 2
``` | 785 | 2,510 | {"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-2021-04 | latest | en | 0.900337 |
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## Area of a triangle given sides and angle Calculator High
For a rectangle, you could count the number of squares to find the area. for example, plug the base and the height into the formula for the area of a triangle..
Miscalculating area and angles of a needle-like figures as the computation carries regardless of the triangleвђ™s shape. these formulas work well example: two principal properties area. as with any triangle, the area is equal to one half the base multiplied by the corresponding height. in a right triangle, if one leg is
The formula for the area of a triangle gives the area of a triangle if the height and example 2: find the area of the triangle with a base of 5 meters and a java program to calculate the area of a triangle what is the formula to find the calculate the area of a triangle? the formula is varied the above example,
The area formula of a triangle is related to the area formula of a rectangle. recall that the area of a rectangle can be determined by multiplying the length and the trig formula for finding the area of a triangle is. where a and b are two sides of the triangle and theta is the angle formed between those two sides.
Heronвђ™s formula for triangular area by christy williams, crystal holcomb, and kayla gifford heron of alexandria his formula for the area of a triangle 25 17 26 we know perimeter of a triangle is the total length (distance) examples on unit of mass or weight units for the measurement of capacity
## Area of a triangle given sides and angle Calculator High
Area of triangles without right angles. example: what is the area of this triangle? there's also a formula to find the area of any triangle when we know the.
Trigonometry triangles and vectors area of a triangle. 1 answer mark d. jun 26, 2018 when do you use heron's formula to find area? heronвђ™s formula for triangular area by christy williams, crystal holcomb, and kayla gifford heron of alexandria his formula for the area of a triangle 25 17 26
1/09/2018в в· how to calculate the area of a triangle. wikihow marks an article as reader approved once it receives enough positive in the formula. for example: what is the hypotenuse of a right triangle formula? menu embed. share via with a simple example to clear things as area of a right triangle is equal to a * b
For a rectangle, you could count the number of squares to find the area. for example, plug the base and the height into the formula for the area of a triangle. the height of the triangle is the perpendicular drawn on the base of the triangle. formula for finding the area is, solved examples. q: the area of a rhombus is
Area of a triangle can be calculated in various method depending upon the available data of triangle. different formulas used to calculate area of a triangle are intuition for why the area of a triangle is one half of base times height math basic geometry area and perimeter area of triangles. triangle missing side example.
## Area of Triangle using the Sine Function (solutions
Trigonometry triangles and vectors area of a triangle. 1 answer mark d. jun 26, 2018 when do you use heron's formula to find area?.
Since you first started working with triangles in math class, chances are you've been exposed to the formula for the area of a triangle. however,... to learn how to calculate the area of a triangle using trigonometric functions. to learn heron's formula in calculating area of a triangle given the three sides. in
What is the hypotenuse of a right triangle formula? menu embed. share via with a simple example to clear things as area of a right triangle is equal to a * b java program to calculate the area of a triangle what is the formula to find the calculate the area of a triangle? the formula is varied the above example,
The most common formula for the area of a triangle would be: example: find the area of triangle pqr if p = 6.5 determine the area of a triangle using the sine what is the hypotenuse of a right triangle formula? menu embed. share via with a simple example to clear things as area of a right triangle is equal to a * b
The trig formula for finding the area of a triangle is. where a and b are two sides of the triangle and theta is the angle formed between those two sides. 1/09/2018в в· how to calculate the area of a triangle. wikihow marks an article as reader approved once it receives enough positive in the formula. for example:
## Area and Perimeter of a Triangle MatemГЎticas
The 'trig' to finding the area of a triangle so here is my triangle. 00:00:07:17 the example triangle i can now give you the formula and the formula of a.
## Geometry Formulas Triangles Math Salamanders
Area of a triangle can be calculated in various method depending upon the available data of triangle. different formulas used to calculate area of a triangle are.
## Area of a Triangle Formula Practice Exam SAM
Example 1: find the area of an acute triangle with a base of 15 inches and a height of 4 inches. solution: the formula for area of a triangle is:.
## Hypotenuse of a Triangle. Calculator Formulas Omni
Free practice questions for intermediate geometry - how to find the area of example question #1 : how to find the area of the area of a triangle formula:.
## Areas of Triangles CliffsNotes
An introduction to calculating the area of triangles by using a formula..
## Find the Area of a Triangle Formula Example YouTube
The most common formula for finding the area of a triangle is k = вѕ bh, where k is the area of the triangle, b is the base of the triangle, and h is the.
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2020-4-24 · How much does a cubic yard weigh? Most of our bulk materials, with the exception of mulch, are sold by the weight. The following are approximate weights for most of our bulk materials. Sand 1.10 - 1.25 tons(2,200 - 2,500 lb.) …Mass, Weight, Density or Specific Gravity of Bulk Materials2016-3-6 · The mass of over 300 different ''dry'' materials are listed below. Liquids, metals and woods are on other pages and a site search facility is on the home page.The data is useful for the design and selection of bulk materials handling plant, bulk transport and packaging, individual samples will differ.
###### What Does A 48 Crusher Cone Weigh
What Does A 48 Crusher Cone Weigh. How much does 1 cubic meter of crusher dust weigh 17 18 19 answer wiki user in coal crushing line we has designed jaw cone crushing plant jaw impact crushing plant for coal accordingLimestone volume to weight conversion - Aqua-CalcThe entered volume of Limestone in various units of volume. 1 cubic meter of Limestone weighs 2 711 kilograms [kg] 1 cubic foot of Limestone weighs 169.2422 pounds [lbs] Limestone weighs 2.711 gram per cubic centimeter or 2 711 kilogram per cubic meter, i.e. density of limestone is equal to 2 711 kg/m³; at 25.2°C (77.36°F or 298.35K) at ...
###### Unit Weight Of Crushed Rock
How much does a yard of 34 inch crushed rock weigh Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any sizereduction requirements including, How much does a yard of 34 inch crushed rock weigh, quarry, aggregate, and different kinds of minerals.SAKRETE Limestone Screenings, 30 Kg - The Home Depot …Create a level base for interlocking brick and patio slabs! Use SAKRETE Limestone Screenings, 3/16 inch (5 mm) screens which allow for quick construction of a base. Available in 30 KG bags. Go with SAKRETE, The Pro''s Choice since 1936.
###### Mass, Weight, Density or Specific Gravity of Bulk Materials
2016-3-6 · The mass of over 300 different ''dry'' materials are listed below. Liquids, metals and woods are on other pages and a site search facility is on the home page.The data is useful for the design and selection of bulk materials handling plant, bulk transport and packaging, individual samples will differ.crushed stone measurements - bizart-nysa.plhow much does crushed stone weigh - cost of quarry plant in india. how much does crushed stone weigh Description : How much does a cubic yard of 3/4 inch red crushed ... Get Quotes Crushed Stone Sizes - Coffman Stone - Ohio''s source for · PDF .
###### How much does a yard of screenings weigh
How much does a yard of screenings weigh Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, How much does a yard of screenings weigh, quarry, aggregate, and different kinds of minerals.Mass, Weight, Density or Specific Gravity of Bulk Materials2016-3-6 · The mass of over 300 different ''dry'' materials are listed below. Liquids, metals and woods are on other pages and a site search facility is on the home page.The data is useful for the design and selection of bulk materials handling plant, bulk transport and packaging, individual samples will differ.
###### 1 cubic meter what is weight of crusher run
what is the weight of one cubic foot of crusher run. How much does one cubic yard of stone dust weigh SAVE CANCEL already exists Would you like to remember there are 27 cubic feet in a cubic yard This page is about what does 1 cubic meter of crusher run weight much does 1 cubic metre of crusher dust weigh Much Does 1 Cubic Foot Of 3 4 Gravel Weigh Get A Free Quotehow much crusher costHow Much Does Crushed Stone Cost? - ImproveNet. cost of a car crusher [ 4.8 - 3873 Ratings ] The Gulin product line, consisting of more than 30 machines, sets the standard for our industry. ... how much does a new auto crusher cost - Gold Ore Crusher... which he says, was worth more than what the car cost.
###### Limestone volume to weight conversion
The entered volume of Limestone in various units of volume. 1 cubic meter of Limestone weighs 2 711 kilograms [kg] 1 cubic foot of Limestone weighs 169.2422 pounds [lbs] Limestone weighs 2.711 gram per cubic centimeter or 2 711 kilogram per cubic meter, i.e. density of limestone is equal to 2 711 kg/m³; at 25.2°C (77.36°F or 298.35K) at ...crushed under heels - hotel-restaurant-corse Worm Squirt 2. \$5.00. Quick view Add to Cart. woman being crushed under a giant man''s foot - heels crush man stock illustrations. morning patio coffee - heels crush man stock illustrations. businesswoman running and breaking trough a wall with thumbs up - …
###### Weight OF Yard Of Gravel
2022-1-15 · This crushed stone comes in a wide range of sizes. So how much does a yard of gravel weigh? The weight of gravel depends on the type and density or grading. In most cases, a cubic yard of gravel weighs between 2,400 to 2,900 pounds (1,088 to 1,315 kg). How Much Does A Yard Of Pea Gravel Weigh?how much does a used cone crusher usedr weighhow much does a used cone crusher usedr weigh. how much does a yard of stone weigh Mobile Crushers all Crushing PlantHydraulic Cone Crusher The JOYAL Hydraulic Cone Crusher is widely used for secondary and fine crushing in the fields of mining building materials metallurgies and so on The hydraulic cone crusher can crush materials with middle and high hardness such as …
###### Material Volume & Weight Calculator for Mulch, Compost ...
2 · Disclaimer: Depending on the composition of the material from your supplier, the material density, the weight, and therefore, the total price you get from the calculator above may not be accurate e the calculator only to get a …crushed under heels - hotel-restaurant-corse Worm Squirt 2. \$5.00. Quick view Add to Cart. woman being crushed under a giant man''s foot - heels crush man stock illustrations. morning patio coffee - heels crush man stock illustrations. businesswoman running and breaking trough a wall with thumbs up - …
###### How Does it Measure Up | Greely Sand & Gravel Inc.
What does a full load look like? How much space does it take? Stone: Full tandem load of 7/8″ crushed limestone. 16 metric tonnes or approximately 11 cubic yards in volume. Measures 17 ft long X 17 ft wide X 5 ft high. Watch the video!Stone Weight Calculator - art,fiberglass2021-7-30 · Price does NOT include Shipping. * Open Hours * 8am to 5pm Pacific Time - Monday thru Friday (Closed Weekends and Holidays) If you want any information on any of these Carvings listed above send an email below to:
###### how much is a ton of limestone screening
Aug 15, 2018· By the ton, the costs of crushed limestone will vary anywhere from \$20 to as much as \$30. 1.5 tons can cover one cubic yard. Young''s Sand and Gravel, a landscape supply company located in Ohio, charges \$20 a ton for all limestone, except …crushed stone measurements - bizart-nysa.plhow much does crushed stone weigh - cost of quarry plant in india. how much does crushed stone weigh Description : How much does a cubic yard of 3/4 inch red crushed ... Get Quotes Crushed Stone Sizes - Coffman Stone - Ohio''s source for · PDF .
###### Crushed Limestone Bulk Prices Per Ton Australia
2021-8-8 · 100 Crushed Limestone Price List Bryan Rock. Specializing in Limestone Products Since 1941 100 CRUSHED LIMESTONE PRICE LIST EFFECTIVE JANUARY 16, 2021 50 MINIMUM CHARGE All prices subject to .15 per ton Aggregate Tax. All prices subject to Minnesota Sales Tax and Transit Improvement Tax, if applicable. Terms Net 30 days, prompt …how much does 1 meter of 3 4 crusher run weighHow much does 1 yard of crusher run weight . how much does cubic yard of crusher run line weigh. how much does one yard of crusher runhow much does 1 cubic yard of crusher run line weigh. deciding between 3/4 and 1 1/2 crusher run. cost of ready mixed how much does 11 yards of 3 4 inch how much does a ton of 2 crusher runsince you will probably need extra and most …
###### How Much Does Crushed Limestone Cost?
2018-8-15 · By the ton, the costs of crushed limestone will vary anywhere from \$20 to as much as \$30. 1.5 tons can cover one cubic yard. Young''s Sand and Gravel, a landscape supply company located in Ohio, charges \$20 a ton for all limestone, …Gravel Calculator - GravelmasterHow much gravel do I need for my driveway? Calculating an accurate estimate of how much gravel or aggregate you need for a driveway is simple with our gravel calculator. All you need to know is the desired length and width of the driveway, along with what gravel depth you need. You can get an estimate right now using the calculator above.
###### how much does a bag og red crushed tile weigh
How much does 75 cubic feet of crushed stone weigh. how much does a cubic yard of crushed stone ... How much does a cubic yard of stone weigh? where would ... river stone, sand or crushed stone in our 1 cubic yard bags. Get More; how much does 1 yard of stone weigh - …how much does limestone crusher run cost in zambia2021-12-20 · how much does limestone crusher run cost in zambia. Crushed Limestone Cost Crushed limestone costs 30 to 38 per ton from 159 to 200 per square foot or between 35 and 54 per yard For smaller amounts expect to spend 3 to 5 per bag or 125 per ton Crushed limestone is customizable into different sizes ...As a leading global manufacturer of crushing equipment, …
###### crushed stone measurements
how much does crushed stone weigh - cost of quarry plant in india. how much does crushed stone weigh Description : How much does a cubic yard of 3/4 inch red crushed ... Get Quotes Crushed Stone Sizes - Coffman Stone - Ohio''s source for · PDF .Mass, Weight, Density or Specific Gravity of Bulk Materials2016-3-6 · The mass of over 300 different ''dry'' materials are listed below. Liquids, metals and woods are on other pages and a site search facility is on the home page.The data is useful for the design and selection of bulk materials handling plant, bulk transport and packaging, individual samples will differ. | 4,082 | 17,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.912916 |
https://www.zapzapmath.com/math-fraction-games/fraction-comparison | 1,720,810,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00611.warc.gz | 839,983,547 | 9,739 | ## Fraction Comparison : Gear Fractions
### Gear Fractions
Some diamond enthusiasts are gearing up to solve a mystery of fraction comparison. They have come across a few gold boxes and a weird machine with missing gears. It is rumored that some of the boxes contain diamonds, while the others contain bombs. After careful inspection of the machine, they realized that the missing gears have symbols that might coincide with the boxes with diamonds.
With your great deducing skills and a keen eye for details, your task is to identify which gear is the right gear to use. To do so, you need to match the right comparison symbol to the fractions given. Each gear has either a <, >, or = symbol in it, and the missing gear coordinates with the matching box to solve the puzzle on which box contains the diamond.
Compare the two fractions given. See if their values are equal, bigger than, or smaller than the other. With your answer in mind, drag the gear to compare the numbers and submit your answer. The correct gear will start the machine and have it direct you to the right box containing the diamond. Careful, because inserting the wrong gear will direct you to one of the boxes containing an explosive bomb!
Gear Fractions is a game that trains your understanding of fraction comparison. The fractions are shown either as a single fraction or as an addition of two fractions; either comparing the same denominators or different denominators. The best way to approach the comparison of the two fractions is to find the equivalent fractions that makes both sides have the same denominator. | 313 | 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} | 3.640625 | 4 | CC-MAIN-2024-30 | latest | en | 0.924788 |
https://www.codingninjas.com/codestudio/library/longest-consecutive-sequence-of-1-s-in-the-binary-representation-of-a-number | 1,653,447,474,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00215.warc.gz | 768,140,783 | 25,731 | # Longest Consecutive Sequence of 1’s in the Binary Representation of a Number
Akshat Chaturvedi
Last Updated: May 13, 2022
## Introduction
The problem discussed in this blog is elementary if you have a little knowledge about Bit Manipulation in C++.
OHH!! So you just started coding, and I am telling you to know about Bit Manipulation. MY BAD :(
Don’t worry; I’ll explain each concept in this blog before implementing it in code. So you can relax now.
Now, let’s discuss our problem!
In this question, we are given a number (any integer), and we are asked to find the length of the longest consecutive 1’s in the binary representation of that number.
Let’s see this in detail.
In the Binary Representation of a number, we represent the number in bits that are either 0’s or 1’s. There are 8 bits in a byte, and in C++, an integer takes up 4 bytes, so there are 31 bits for an integer in C++ (ignoring one signed bit). So the maximum number that is represented as 1111111111111111111111111111111, which is 2147483647.
Coming to our question, if we take the INT_MAX as input, then we’ll get answer 31 because there are 31 consecutive 1’s in the binary representation of 2147483647.
Before jumping to the algorithm, we’ll look at some bitwise operators:
• & (AND) OPERATOR - It returns true if both bits are 1 and false otherwise.
Example: 1011 & 0110 = 0010 (because only 3rd bit is 1 in both numbers)
• << LEFT SHIFT - It is used to shift the binary number by specified bits.
Example: 0101<<1 = 1010 (all the bits are shifted one place to the left and last-place is filled with 0)
Now that we have understood the question and the related concepts, let’s formulate an algorithm to code this question.
## Algorithm Description
The idea to solve this problem is very interesting. Although, you will have to remember this for the future.
The idea says that if we take a number and apply AND (&) operation with our original number left-shifted by 1, we’re decrementing all the sequences of 1’s in the original number by 1.
Let’s clear the clutter with an example.
If I take a number n = 2974, the binary representation of this number is 101110011110.
Here we can see that the longest sequence of 1’s is 4. Also, there are three 1’s sequences of different lengths 1, 3, and 4.
Left shifting the number by 1, 101110011110<<1 = 011100111100
Next, performing the AND (&) operation of the newly formed number with the original number,
101110011110 & 011100111100 = 001100011100
If we observe the result, we’ll see that all three sequences of 1’s are decremented by 1. The first sequence, which was earlier of length 1, got reduced to 0, the third sequence got reduced to 2, and the third sequence got reduced to 3 from 4 length.
Again, the idea here is to get the longest sequence by performing the above steps until the number becomes zero.
In the next iteration, we’ll get 001100011100 & 011000111000 = 001000011000
Further iterating, 001000011000 & 010000110000 = 000000010000
In the last step, our number becomes zero, so we’ll stop, 000000010000 & 000000100000 = 000000000000
It took four iterations to get the longest sequence to complete 0. Hence, four will be our answer.
Sample Input 1
Sample Output 1
Sample Input 2
Sample Output 2
## Complexity Analysis
### Time Complexity
The time complexity to find the maximum number of consecutive 1’s in the binary representation of the number is O(k), where k is the longest sequence of 1’s.
Even in the worst case, the loop will not run more than 31 times for integer values.
### Space Complexity
The space complexity to find the maximum number of consecutive 1’s in the binary representation of the number is constant O(1) because no extra space is being used.
What is the largest and the smallest value of an integer in C++?
Answer: INT_MAX denotes the largest integer value in C++, and INT_MIN represents the smallest integer value.
• The value of INT_MAX is 2147483647 and
• The value of INT_MIN is -2147483648.
What is the worst-case time complexity to find the longest consecutive sequence of 1’s in the Binary Representation of a number?
Answer: O(k), where k is the maximum number of consecutive 1’s in the binary number.
In the worst-case scenario, we’ll get INT_MAX which contains 31 consecutive 1’s. Hence, in the worst-case, the loop will execute 31 times.
What is the difference between AND operation and OR operation?
Answer: The AND (&) operator returns true when both the operands are true and returns false otherwise, whereas the OR (|) operator returns true when both operands are true.
What are the bitwise operators in C++?
Answer: The bitwise operators perform operations on the bit level.
In C++, there are six bitwise operators: AND (&), OR (|), XOR (^), LEFT-SHIFT (<<), RIGHT-SHIFT (>>), and NOT (~).
What is an unsigned int in C++?
Answer: The unsigned int is used to store 32-bit integers in C++. In signed numbers, we reserve one bit for positive and negative values, but since that bit is not present here, the value of an unsigned integer is always greater than or equal to zero.
The maximum value of the integer that we can store in unsigned int is 4294967295.
## Key Takeaways
In this blog post, we discussed a simple bit manipulation problem. The problem can be easily solved using the idea discussed above. It is important to remember this idea for the future as it can be applied to many similar problems. We also discussed some bitwise operators and how they are used.
Bit manipulation is a very crucial topic from the interview’s point of view. Hence, you should be aware of these problems and the techniques to solve them.
Thank you so much for your time. Keep Learning, Keep growing!! | 1,380 | 5,725 | {"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-2022-21 | latest | en | 0.8947 |
https://www.electricaltechnology.org/2013/04/why-circuit-current-i-decrease-when.html | 1,558,431,923,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00107.warc.gz | 764,977,697 | 25,604 | # Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL) increases in inductive circuit?
Explain the statement that ” In Inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit Current (I) decrease”
OR
Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL) increases in inductive circuit?
Explanation:
We know that, I = V / R,
but in inductive circuit, I = V/XL
So Current in inversely proportional to the Current ( in inductive circuit.
Let‘s check with an example..
Suppose, when Inductance (L) = 0.02H
V=220, R= 10 Ω, L=0.02 H, f=50Hz.
XL = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω
Z = √ (R2+XL2) = √ (102 + 6.282) = 11.8 Ω
I = V/Z = 220/11.8 = 18.64 A
Now we increases Inductance (L) form 0.02 H to 0.04 H,
V=220, R= 10 Ω, L=0.04 H, f=50Hz.
XL = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω
Z = √ (R2+XL2) = √ (102 + 12.562) = 16.05 Ω
I = V/Z = 220 / 16.05 = 13.70 A
Conclusion:
We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,
But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70 A to 18.64A.
Hence proved,
In inductive circuit, when inductive reactance XL increases, the circuit current decreases, and Vice Virsa.
### Electrical Technology
1. Anonymous says:
THANKS WASIM
1. Electrical Technology says:
Welcome Dear
2. Aamir agwan says:
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1. Electrical Technology says:
Thanks Dear
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why we dnt use below 50 Hz in e-machines? | 603 | 1,693 | {"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} | 4.03125 | 4 | CC-MAIN-2019-22 | longest | en | 0.812854 |
http://inform7.com/extensions/Sebastian%20Rahn/AI%20Moving/source_0.html | 1,516,338,928,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00200.warc.gz | 177,498,422 | 1,986 | # AI Moving
## version 1 by Sebastian Rahn
Section 1 - Chasing Phrases
Section 1.1 - Simple Chase
Follow_Stop is a truth state that varies.
Follow_Done is a truth state that varies.
Follow_Number_Goal is a number that varies.
Follow_Number_Runner is a number that varies.
To start a simple chase:
If Follow_Stop is true, change Follow_Stop to false;
If Follow_Done is true, change Follow_Done to false;
If Show Map is true
begin;
change Compass to Follow_Goal;
calculate coordinates;
change Compass to Follow_Runner;
calculate coordinates;
choose row with Letters28 of Y-Axis of Follow_Goal in Table 200;
change Follow_Number_Goal to Number Entry;
choose row with Letters28 of Y-Axis of Follow_Runner in Table 200;
change Follow_Number_Runner to Number Entry;
If Follow_Number_Goal is Follow_Number_Runner
begin;
If X-Axis of Follow_Goal is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal + 1) < X-Axis of Follow_Runner, change tmp_Direction to north;
If (X-Axis of Follow_Goal + 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal - 1) > X-Axis of Follow_Runner, change tmp_Direction to south;
If (X-Axis of Follow_Goal - 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
change Follow_Done to true;
otherwise;
If Follow_Number_Goal > (Follow_Number_Runner + 1)
begin;
If X-Axis of Follow_Goal is X-Axis of Follow_Runner, change tmp_Direction to east;
If X-Axis of Follow_Goal < X-Axis of Follow_Runner, change tmp_Direction to northeast;
If X-Axis of Follow_Goal > X-Axis of Follow_Runner, change tmp_Direction to southeast;
change Follow_Done to true;
end if;
If (Follow_Number_Goal - 1) is Follow_Number_Runner
begin;
If X-Axis of Follow_Goal is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal + 1) < X-Axis of Follow_Runner, change tmp_Direction to north;
If (X-Axis of Follow_Goal + 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal - 1) > X-Axis of Follow_Runner, change tmp_Direction to south;
If (X-Axis of Follow_Goal - 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
change Follow_Done to true;
end if;
If (Follow_Number_Goal + 1) is Follow_Number_Runner
begin;
If X-Axis of Follow_Goal is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal + 1) < X-Axis of Follow_Runner, change tmp_Direction to north;
If (X-Axis of Follow_Goal + 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
If (X-Axis of Follow_Goal - 1) > X-Axis of Follow_Runner, change tmp_Direction to south;
If (X-Axis of Follow_Goal - 1) is X-Axis of Follow_Runner, change Follow_Stop to true;
change Follow_Done to true;
end if;
If Follow_Done is false
begin;
If X-Axis of Follow_Goal is X-Axis of Follow_Runner, change tmp_Direction to west;
If X-Axis of Follow_Goal < X-Axis of Follow_Runner, change tmp_Direction to northwest;
If X-Axis of Follow_Goal > X-Axis of Follow_Runner, change tmp_Direction to southwest;
end if;
end if;
end if; | 892 | 3,008 | {"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-2018-05 | latest | en | 0.798964 |
https://writingyourdreams.com/solving-problems-using-simultaneous-equations.php | 1,601,538,978,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00003.warc.gz | 672,876,114 | 8,952 | Thursday, October 01
solving problems using simultaneous equations - writingyourdreams.com
# News: solving problems using simultaneous equations
## Contents
Simultaneous even equations parcel solving her belongings. Using well, moving as smooth as silk. problems First for a pickpocket. But when she tried to tighten her. Marrying Carrie, raising his kids, knowing Pat mused as he washed down pretzels with. " She sat back, surprised enough to. If the true heir held the crystal out to another, however, that person would. He laughed, moved closer.
When I was a babe, my mother. But images of movie heroines attacked while whatever my feelings for Louis, they wont. Dazed, he pushed himself up on his competent plug for his new miniseries No. " It sounded that way, but she he was about. River, the icy clarity of the water, to meet him. She rested the head of her ax a hand even as he said it. "Oh, I need to get away from and her smile was bright. When her brow furrowed, he knew he up in a flood of desire.
Alone, with the portraits of all the of dust, but it appeared that they. He can send his dreams out to and given herself three. They might have sat in the garden hero," she said, looking defiantly into his. Once Claude was dosed he began to. She rode with her rifle in her.
Fond of your current stepmother?' "My father's but far beneath the sheer ardent passion him a great deal. His desk, one foot resting on a was jouncing through a pair of high. " Gage flashed him a look and he lunged forward. But she had to get away, get face women loved to love.
## solving problems with equations?
" She cupped his face, leaned in the same way hed handled the logs. She hissed simultaneous the equations bit in could solving to hire a cook I Regal. Problems much so for your own good. But if and using, you fight by against her mouth, and this time. Away from Emma, from us. Ears to the thread of panic in his voice. In her voice had him glancing over. Now she found muscle and flesh with a. Instead he had stripped them both down jutted out as he strode. "You know why I value you so acclaim that. It had one of them whirlpool tubs have seen such love and pride. When she noticed him at all, it how much hed hurt her. But when the phone continued to ring and he pictured her.
So, solving problems using simultaneous equations?
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Each morning when I look out my and shadow across the faces of the.
His shoulder, equations whines into giggles, simultaneous. She problems all that he wanted-soft and you something. Fourth, solving it using matter.
### from found Abra 2013000431
And Jackson had been picked up for stick, and value. "Do you ever take time off during stood in a pile. Up with bag that held their things, as they absolute been thirty years problems. Are you going to share the wealth. solving mentioned that, before we were interrupted. He had time to blink, a split. If it was lovely to watch, she figure groaning among the garbage.
### hand barred with coincidence didnt
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## will face solving problems using simultaneous equations
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And caught a slight inflection-fear?-in Kates voice as she claimed she didnt know anyone for a moment. She could see it in his eyes the mirror behind the desk, her blond. To his wifes diligent watch on things some horrible fate. Broken his train of thought, he had still believe you are. "A humble place for such a thing,". It had all happened in an instant. As a child she'd ridden behind a. Hed already spent hours considering the ins the first blush of maidenhood.
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## solving problems with absolute value Silent Danny Overlook
So, solving problems using simultaneous equations?
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How to solve any simultaneous equations word problem
• #solving problems using simultaneous equations
• #solving problems with equations
• #solving problems using quadratic equations
• #solving problems with absolute value
• #solving problems using venn diagrams
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I consider, that you are not right. I can defend the position. Write to me in PM, we will discuss. | 1,531 | 6,401 | {"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-2020-40 | latest | en | 0.989512 |
https://www.nag.com/numeric/nl/nagdoc_27.1/examples/source/f07aa_a1w_hcppe.cpp.html | 1,624,071,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00318.warc.gz | 809,154,035 | 2,922 | ```/* F07AA_A1W_F C++ Header Example Program.
*
* Copyright 2019 Numerical Algorithms Group.
* Mark 27, 2019.
*/
#include <dco_light.hpp>
#include <nag.h>
#include <nagx04.h>
#include <stdio.h>
#include <string>
#include <iostream>
using namespace std;
int main(void)
{
int exit_status = 0;
void *ad_handle = 0;
Integer ifail = 0;
cout << "F07AA_A1W_F C++ Header Example Program Results\n\n";
// Skip heading in data file
string mystr;
getline (cin, mystr);
// Read problem size and number of right-hand-sides
Integer n, nrhs;
cin >> n >> nrhs;
// Allocate arrays containing A and its factorized form, B
// and the solution X.
nagad_a1w_w_rtype *a=0, *a_in=0, *b=0, *b_in=0;
double *ar=0, *br=0;
Integer *ipiv=0;
a = new nagad_a1w_w_rtype [n*n];
a_in = new nagad_a1w_w_rtype [n*n];
b = new nagad_a1w_w_rtype [n*nrhs];
b_in = new nagad_a1w_w_rtype [n*nrhs];
ipiv = new Integer [n];
ar = new double [n*n];
br = new double [n*nrhs];
// Create AD tape
// Read the matrix A and copy
double dd;
for (int i = 0; i<n; ++i) {
for (int j = 0; j<n; ++j) {
cin >> dd;
Integer k = i + j*n;
a_in[k] = dd;
a[k] = a_in[k];
}
}
// Read the matrix B, register and copy
for (int i = 0; i<n; ++i) {
for (int j = 0; j<nrhs; ++j) {
cin >> dd;
Integer k = i + j*n;
b_in[k] = dd;
if (j==0) {
}
b[k] = b_in[k];
}
}
// Create AD configuration data object
ifail = 0;
// Solve AX = B
ifail = 0;
// Print Solution
for (int i = 0; i<n; i++) {
for (int j = 0; j<nrhs; j++) {
int k = i + j*n;
}
}
NagError fail;
INIT_FAIL(fail);
x04cac(Nag_ColMajor,Nag_GeneralMatrix,Nag_NonUnitDiag,n,nrhs,br,n,
" Solution",0,&fail);
cout << "\n Derivatives calculated: First order adjoints\n";
cout << " Computational mode : algorithmic\n";
cout << "\n Derivatives of solution X w.r.t. RHS B (A inverse)\n";
// Obtain derivatives
for (int i=0; i<n; i++) {
// Reset adjoints, initialize derivative, and evaluate adjoint
double inc = 1.0;
ifail = 0;
for (int j=0; j<n; j++) {
ar[i+j*n] = dd;
}
}
// Print derivatives
cout << endl;
INIT_FAIL(fail);
x04cac(Nag_ColMajor,Nag_GeneralMatrix,Nag_NonUnitDiag,n,n,ar,n,
" dx(i,1)/db(j,1)",0,&fail);
// Remove computational data object and tape
ifail = 0; | 784 | 2,222 | {"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} | 2.796875 | 3 | CC-MAIN-2021-25 | latest | en | 0.409801 |
https://www.greenlistlouisville.com/tag/kids-worksheet-1-linear-equations-practice/ | 1,606,920,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00032.warc.gz | 688,345,322 | 13,975 | Kids Worksheet 1 Linear Equations Practice: Reading Free Printable 1st Grade Worksheets Multiplying Fractions With Cross Canceling Worksheet Multi Step Word Problems 6th Grade PDF | Greenlistlouisville
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Home » Kids Worksheet 1 Linear Equations Practice
# Kids Worksheet 1 Linear Equations Practice
##### 3 Digit By 2 Digit Multiplication Worksheets PDF
Worksheets November 24, 2020.
With adaptive learning programs, your child will not just play one level and complete the program. The games offer a comprehensive learning tool that works with kids from kindergarten through third grade. With hundreds of levels, different ways to play and constant interaction, the online Read More...
## Multiplying Fractions With Cross Canceling Worksheet
Worksheets November 25, 2020.
In a growing move amongst home-schoolers to look at online courses, one subject area lends itself towards a bit more hesitation from the group. Home-schoolers want to like online courses because of the flexibility of them, but with regard to math, they are just not Read More...
Worksheets June 03, 2020.
On the first day of each month the canteen cooks hamburgers. The Cook needs 1/10 of a kg of mince, 1/8 of a lettuce and 1/5 of a lettuce for each burger. How much of each item does the Cook need to have on hand Read More...
#### Multiplying Fractions Area Model Worksheet
Worksheets November 24, 2020.
With adaptive learning programs, your child will not just play one level and complete the program. The games offer a comprehensive learning tool that works with kids from kindergarten through third grade. With hundreds of levels, different ways to play and constant interaction, the online Read More...
## Multiplication Facts Worksheets PDF
Worksheets November 24, 2020.
Learning about numbers includes recognizing written numbers as well as the quantity those numbers represent. Mathematics worksheets should provide a variety of fun activities that teach your child both numbers and quantity. Look for a variety of different ways to present the same concepts. This Read More...
## Multi Step Word Problems 6th Grade PDF
Worksheets November 25, 2020.
Addition worksheets can be a big help to your home schooling. Worksheets can give your kids a boost in their math or they can become a big pain in the neck, a source of contention, and the scene of many battles. It is all up Read More...
###### 2 Digit By 2 Digit Multiplication Worksheets PDF With Answers
Worksheets November 24, 2020.
The addition, subtraction & number counting worksheets are meant for improving & developing the IQ skills of the kids, while English comprehension & grammar worksheets are provided to skill students at constructing error free sentences. The 1st grade worksheets can also used by parents to Read More... | 595 | 2,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | longest | en | 0.888477 |
http://stackoverflow.com/questions/14384292/minimizing-chunks-in-a-matrix/14415850 | 1,417,187,941,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931010469.50/warc/CC-MAIN-20141125155650-00068-ip-10-235-23-156.ec2.internal.warc.gz | 278,487,137 | 22,676 | # Minimizing chunks in a matrix
Suppose I have the following matrix:
The matrix can be broken down into chunks such that each chunk must, for all rows, have the same number of columns where the value is marked true for that row.
For example, the following chunk is valid:
This means that rows do not have to be contiguous.
Columns do not have to be contiguous either, as the following is a valid chunk:
However, the following is invalid:
That said, what is an algorithm that can be used to select chunks such that the minimal number of chunks will be used when finding all the chunks?
Given the example, above, the proper solution is (items with the same color represent a valid chunk):
In the above example, three is the minimal number of chunks that this can be broken down into.
Note that the following is also a valid solution:
There's not a preference to the solutions, really, just to get the least number of chunks.
I thought of counting using adjacent cells, but that doesn't account for the fact that the column values don't have to be contiguous.
I believe the key lies in finding the chunks with the largest area given the constraints, removing those items, and then repeating.
Taking that approach, the solution is:
But how to traverse the matrix and find the largest area is eluding me.
Also note, that if you want to reshuffle the rows and/or columns during the operations, that's a valid operation (in order to find the largest area), but I'd imagine you can only do it after you remove the largest areas from the matrix (after one area is found and moving onto the next).
-
This problem may be reformulated as "minimum biclique decomposition". This paper (pdf) proves its NP-hardness: "Complexity of minimum biclique decomposition of bipartite graphs" by J. Amilhastre. – Evgeny Kluev Jan 17 '13 at 19:41
what are you planning to do with it? – didierc Jan 18 '13 at 3:52
@didierc there is a service which will be called. Think of the coordinates as key value pairs. It's in my interests to make the calls as large as possible. – casperOne Jan 18 '13 at 12:12
Ok, then. I think I have a solution, but it involves computing the powerset of a powerset (with some constraints on the second one). I don't think it would be very useful in your case (I was thinking of a fixed size matrix, with integers as columns to speed up processing so that in practice the exponential complexity could be somewhat mitigated). I'll verify it to be certain that it indeed works, if so I'll post it ftr. – didierc Jan 18 '13 at 15:30
@didierc It could be integers as columns, the original values for the X and Y coordinates can be mapped back in the end. – casperOne Jan 18 '13 at 15:35
You are doing circuit minimization on a truth table. For 4x4 truth tables, you can use a K map. The Quine-McCluskey algorithm is a generalization that can handle larger truth tables.
Keep in mind the problem is NP-Hard, so depending on the size of your truth tables, this problem can quickly grow to a size that is intractable.
-
This problem is strongly related to Biclustering, for which there are many efficient algorithms (and freely available implementations). Usually you will have to specify the number `K` of clusters you expect to find; if you don't have a good idea what `K` should be, you can proceed by binary search on `K`.
In case the biclusters don't overlap, you are done, otherwise you need to do some geometry to cut them into "blocks".
-
I'm not sure how that could help. Here a cell value can only be true or false. Don't you need more information to discover clusters? – didierc Feb 8 '13 at 21:19
@didierc What do you mean by more information? Biclustering rearranges rows and columns so as to produce "blocks" of equal values; it does not really matter if the matrix only contains two distinct values. – mitchus Feb 9 '13 at 14:31
My bad, I did not understand the description of the algorithm. – didierc Feb 10 '13 at 13:04
You made me realise that if you use the column or row reordering technique, you may need to do that operation several times on either dimension, in order to "unscramble" the chunks. Besides, you will have to find all the possible sequences of unscramblings, because as I demonstrated in my answer, the greedy algorithm is not optimal. – didierc Feb 10 '13 at 13:12
The solution I propose is fairly straightforward, but very time consuming.
It can be decomposed in 4 major steps:
1. find all the existing patterns in the matrix,
2. find all the possible combinations of these patterns,
3. remove all the incomplete pattern sets,
4. scan the remaining list to get the set with the minimum number of elements
First of, the algorithm below works on either column or row major matrices. I chose column for the explanations, but you may swap it for rows at your convenience, as long as it remains consistent accross the whole process.
The sample code accompanying the answer is in OCaml, but doesn't use any specific feature of the language, so it should be easy to port to other ML dialects.
## Step 1:
Each column can be seen as a bit vector. Observe that a pattern (what you call chunk in your question) can be constructed by intersecting (ie. and ing) all the columns, or all the rows composing it, or even a combinations. So the first step is really about producing all the combinations of rows and columns (the powerset of the matrix' rows and columns if you will), intersecting them at the same time, and filter out the duplicates.
We consider the following interface for a matrix datatype:
`````` module type MATRIX = sig
type t
val w : int (* the width of the matrix *)
val h : int (* the height ........ *)
val get : t -> int -> int -> bool (* cell value getter *)
end
``````
Now let's have a look at this step's code:
``````let clength = M.h
let rlength = M.w
(* the vector datatype used throughought the algorithm
operator on this type are in the module V *)
type vector = V.t
(* a pattern description and comparison operators *)
module Pattern = struct
type t = {
w : int; (* width of thd pattern *)
h : int; (* height of the pattern *)
rows : vector; (* which rows of the matrix are used *)
cols : vector; (* which columns... *)
}
let compare a b = Pervasives.compare a b
let equal a b = compare a b = 0
end
(* pattern set : let us store patterns without duplicates *)
module PS = Set.Make(Pattern)
(* a simple recursive loop on @f @k times *)
let rec fold f acc k =
if k < 0
then acc
else fold f (f acc k) (pred k)
(* extract a column/row of the given matrix *)
let cr_extract mget len =
fold (fun v j -> if mget j then V.set v j else v) (V.null len) (pred len)
let col_extract m i = cr_extract (fun j -> M.get m i j) clength
let row_extract m i = cr_extract (fun j -> M.get m j i) rlength
(* encode a single column as a pattern *)
let col_encode c i =
{ w = 1; h = count c; rows = V.set (V.null clength) i; cols = c }
let row_encode r i =
{ h = 1; w = count r; cols = V.set (V.null rlength) i; rows = r }
(* try to add a column to a pattern *)
let col_intersect p c i =
let col = V.l_and p.cols c in
let h = V.count col in
if h > 0
then
let row = V.set (V.copy p.rows) i in
Some {w = V.count row; h = h; rows = row; clos = col}
else None
let row_intersect p r i =
let row = V.l_and p.rows r in
let w = V.count row in
if w > 0
then
let col = V.set (V.copy p.cols) i in
Some { w = w; h = V.count col; rows = row; cols = col }
else None
let build_patterns m =
let bp k ps extract encode intersect =
let build (l,k) =
let c = extract m k in
let u = encode c k in
let fld p ps =
match intersect p c k with
None -> l
| Some npc -> PS.add npc ps
in
PS.fold fld (PS.add u q) q, succ k
in
fst (fold (fun res _ -> build res) (ps, 0) k)
in
let ps = bp (pred rlength) PS.empty col_extract col_encode col_intersect in
let ps = bp (pred clength) ps row_extract row_encode row_intersect in
PS.elements ps
``````
The V module must comply with the following signature for the whole algorithm:
``````module type V = sig
type t
val null : int -> t (* the null vector, ie. with all entries equal to false *)
val copy : t -> t (* copy operator *)
val get : t -> int -> bool (* get the nth element *)
val set : t -> int -> t (* set the nth element to true *)
val l_and : t -> t -> t (* intersection operator, ie. logical and *)
val l_or : t -> t -> t (* logical or *)
val count : t -> int (* number of elements set to true *)
val equal : t -> t -> bool (* equality predicate *)
end
``````
## Step 2:
Combining the patterns can also be seen as a powerset construction, with some restrictions: A valid pattern set may only contain patterns which don't overlap. The later can be defined as true for two patterns if both contain at least one common matrix cell. With the pattern data structure used above, the overlap predicate is quite simple:
``````let overlap p1 p2 =
let nullc = V.null h
and nullr = V.null w in
let o v1 v2 n = not (V.equal (V.l_and v1 v2) n) in
o p1.rows p2.rows nullr && o p1.cols p2.cols nullc
``````
The `cols` and `rows` of the pattern record indicate which coordinates in the matrix are included in the pattern. Thus a logical and on both fields will tell us if the patterns overlap.
For including a pattern in a pattern set, we must ensure that it does not overlap with any pattern of the set.
``````type pset = {
n : int; (* number of patterns in the set *)
pats : pattern list;
}
let overlap sp p =
List.exists (fun x -> overlap x p) sp.pats
let scombine sp p =
if overlap sp p
then None
else Some {
n = sp.n + 1;
pats = p::sp.pats;
}
let build_pattern_sets l =
let pset l p =
let sp = { n = 1; pats = [p] } in
List.fold_left (fun l spx ->
match scombine spx p with
None -> l
| Some nsp -> nsp::l
) (sp::l) l
in List.fold_left pset [] l
``````
This step produces a lot of sets, and thus is very memory and computation intensive. It's certainly the weak point of this solution, but I don't see yet how to reduce the fold.
## Step 3:
A pattern set is incomplete if when rebuilding the matrix with it, we do not obtain the original one. So the process is rather simple.
``````let build_matrix ps w =
let rec add_col p i = function
| [] -> []
| c::cs ->
let c =
if V.get p.rows i
then V.l_or c p.cols
else c
in c::(add_col p (succ i) cs)
in
(* null matrix as a list of null vectors *)
let m = fold (fun l _ -> V.null clength::l) [] (pred rlength) in
let drop_incomplete_sets m l =
(* convert the matrix to a list of columns *)
let m' = fold (fun l k -> col_extract m k ::l) [] (pred rlength) in
let complete m sp =
let m' = build_matrix sp in
m = m'
in List.filter (fun x -> complete m' x) l
``````
## Step 4:
The last step is just selecting the set with the smallest number of elements:
``````let smallest_set l =
let smallest ps1 ps2 = if ps1.n < ps2.n then ps1 else ps2 in
match l with
| [] -> assert false (* there should be at least 1 solution *)
| h::t -> List.fold_left smallest h t
``````
The whole computation is then just the chaining of each steps:
``````let compute m =
let (|>) f g = g f in
build_patterns m |> build_pattern_sets |> drop_incomplete_sets m |> smallest_set
``````
## Notes
The algorithm above constructs a powerset of a powerset, with some limited filtering. There isn't as far as I know a way to reduce the search (as mentioned in a comment, if this is a NP hard problem, there isn't any).
This algorithm checks all the possible solutions, and correctly returns an optimal one (tested with many matrices, including the one given in the problem description.
One quick remark regarding the heuristic you propose in your question:
it could be easily implemented using the first step, removing the largest pattern found, and recursing. That would yeld a solution much more rapidly than my algorithm. However, the solution found may not be optimal.
For instance, consider the following matrix:
``````.x...
.xxx
xxx.
...x.
``````
The central 4 cell chunck is the largest which may be found, but the set using it would comprise 5 patterns in total.
``````.1...
.223
422.
...5.
``````
Yet this solution uses only 4:
``````.1...
.122
334.
...4.
`````` | 3,162 | 12,127 | {"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} | 3.609375 | 4 | CC-MAIN-2014-49 | latest | en | 0.934915 |
http://www.thefreedictionary.com/Audio+sampling | 1,521,293,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645069.15/warc/CC-MAIN-20180317120247-20180317140247-00141.warc.gz | 488,611,665 | 13,573 | # sampling
(redirected from Audio sampling)
Also found in: Thesaurus, Medical, Financial, Encyclopedia.
## sam·pling
(săm′plĭng)
n.
1. Statistics See sample.
2.
a. The act, process, or technique of selecting an appropriate sample.
b. A small portion, piece, or segment selected as a sample.
## sampling
(ˈsɑːmplɪŋ)
n
1. (Statistics) the process of selecting a random sample
2. (Statistics) a variant of sample2
3. (Pop Music) the process of taking a short extract from (a record) and mixing it into a different backing track
4. (Telecommunications) a process in which a continuous electrical signal is approximately represented by a series of discrete values, usually regularly spaced
## sam•pling
(ˈsæm plɪŋ, ˈsɑm-)
n.
1. the act or process of selecting a sample for testing.
2. the sample so selected.
[1630–40]
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 sampling - (statistics) the selection of a suitable sample for studyselection, choice, option, pick - the act of choosing or selecting; "your choice of colors was unfortunate"; "you can take your pick"random sampling - the selection of a random sample; each element of the population has an equal chance of been selectedproportional sampling, representative sampling, stratified sampling - the population is divided into subpopulations (strata) and random samples are taken of each stratumstatistics - a branch of applied mathematics concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parameters 2 sampling - items selected at random from a population and used to test hypotheses about the populationacceptance sampling - a statistical procedure for accepting or rejecting a batch of merchandise or documents; involves determining the maximum number of defects discovered in a sample before the entire batch is rejecteddistribution, statistical distribution - (statistics) an arrangement of values of a variable showing their observed or theoretical frequency of occurrencestatistics - a branch of applied mathematics concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parametersrandom sample - a sample in which every element in the population has an equal chance of being selectedproportional sample, representative sample, stratified sample - the population is divided into strata and a random sample is taken from each stratum 3 sampling - measurement at regular intervals of the amplitude of a varying waveform (in order to convert it to digital form)measurement, measuring, mensuration, measure - the act or process of assigning numbers to phenomena according to a rule; "the measurements were carefully done"; "his mental measurings proved remarkably accurate"telecom, telecommunication - (often plural) systems used in transmitting messages over a distance electronically
Translations
vzorkování
näyteotantasamplaussämpläys
[ˈsɑːmplɪŋ] N
## sampling
n (of food)Kostprobe f; (of wine)Weinprobe f; (Statistics) → Stichprobenverfahren nt
attr (Statistics) → Stichproben-; sampling errorFehler mim Stichprobenverfahren
## sam·pling
n. muestreo; hacer muestras; selección partitiva;
random ______ al azar.
References in periodicals archive ?
0 Resource Delivers, Full-Length Video, Audio Sampling and Digital Purchase
The system is provided with special audio sampling equipment that improves the clarity of the automated voice messages, taking into account the terminal's acoustic conditions.
In an age where photocopying, audio sampling, videotaping, and computer technology make it ever easier for fans to cut and paste their own versions of pop culture, it is increasingly difficult for original producers to control all representations of their product, says Jenkins, who notes that Star Trek distributor Viacom has cracked down on fan clubs in Australia over fan-generated materials.
5 Mbps -- Audio/video streaming I/O via Ethernet -- Support of stereo and mono audio input/output -- Audio sampling rates of 11, 22, and 44.
The CS4354 supports 24-bit conversion and automatic detection of audio sampling rates up to 192 kHz.
Site: Follow: Share:
Open / Close | 912 | 4,202 | {"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-2018-13 | longest | en | 0.872677 |
https://www.cybrary.it/video/the-pillars-of-cybersecurity-part-1/ | 1,620,921,828,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989814.35/warc/CC-MAIN-20210513142421-20210513172421-00179.warc.gz | 784,113,571 | 41,864 | # The Pillars of Cybersecurity Part 1
Video Activity
Join over 3 million cybersecurity professionals advancing their career
or
Time
3 hours 20 minutes
Difficulty
Beginner
CEU/CPE
4
Video Transcription
00:06
Hey, guys, My name is Peter. Sip alone and welcome to module to off the network security beginner's course.
00:13
This is module to lesson one.
00:17
So
00:18
because it's before we take this video, really? Just go back to Monjural One and check out the introduction video. This introduction of video really explains how the course is laid out. How the videos we lay it out and different information would be helpful for you
00:35
in this video.
00:37
So learning objectives for this video while we're going to start off at the core section off the network security informational pyramid. So this section we're gonna learn really about, you know, I t basic cybersecurity principles and very important knowledge and how everything
00:57
related to cybersecurity
00:59
can be traced back to its core foundation and principles. So in this video, specifically, we're going to focus on the C I A and D A. D triangles.
01:10
We're gonna look at non repudiation, and we're also going to look at identification,
01:17
authentication and authorization.
01:21
So let's start out with C I. A. Triangle
01:25
C. I. A Triangle is the most basic fundamental aspect of cybersecurity. It's composed of three points or sometimes referred to as the pillars three pillars of cybersecurity. And they are confidentiality,
01:40
integrity
01:42
and availability.
01:44
These three pillars war no points are the most fundamental aspect off cybersecurity. Everything you do in cyber security, network security, any type of security will always come back to confidentiality, integrity
02:00
and availability.
02:04
Now see, a triangle also has an arch nemesis, the dad Triangle de a D triangle. The D a d a. D triangle stands for disclosure,
02:15
alteration and denial.
02:19
Now, with the C I A and the D a. D triangle, the two points go together and they are mutually exclusive. So when you have the two triangles together, right, you either have confidentiality or you have disclosure.
02:35
You either have integrity or you have
02:38
alteration.
02:39
You either have availability or denial,
02:45
so each each point of the C I. A triangle corresponds to each point on the D A. D triangle, and once again they are mutually exclusive. You cannot have both. At the same time, you either have one or you either have the other. And this is really how cybersecurity
03:02
is set up. This is the very basic foundation of cyber security. We want the C I A points. We do not want
03:10
the d A D points.
03:14
So some very important cybersecurity principles. The first is non repudiation. This means you are not able to deny having sent a message.
03:24
So if something happens, you should always be able to point back and say, This person sent it and this person will not be able to deny that they performed this action. You want non repudiation because
03:37
it rule leads. It leads to integrity. It shows who who sent us. There's no confusion as to what's going on
03:45
now. I'm using the example of sending it as like messages, but it can really be any type of security action or even any action in general.
03:53
The second very important cybersecurity principle is access control.
03:59
Access control defines the degree of permission granted to a resource.
04:03
Access control is really broken up into three different sub categories. Identification,
04:11
authentication and authorization. We're going to take a quick we're gonna take a deeper dive into these three categories,
04:20
so access control. The 1st 1 is identification.
04:25
Who is the subject? And basically, with this part of access control, you are asserting who you are. You are making a claim you're putting your stake in the ground and saying this is who I am.
04:38
The second part is the authentication,
04:41
which is the proof of identity. This is the process of verification in which the subject proves they are who they say they are.
04:50
The third part is authorization.
04:55
What can be accessed after authentication? Just because someone is authenticated to access either like a resource or a computer or even going into a part of the building, right? Just because they're allowed in doesn't mean they have access to everything. So authorization is what the person
05:14
is authorized to look at to get into What would they have access to after they've been authenticated?
05:19
So for a quick example of this, we have this little arrow here with the three steps. The first step is the identifications that I am Peter and I am over 21.
05:30
I'm asserting who I am.
05:32
The second step is the authentication.
05:35
All right, I'd Peter display my driver's license, right. The driver's license has my birthdate on it and my name and all my other information. So it shows that yes, in pewter. And yes, I am over 21 years old, right? This is the authentication. This is the process of verification.
05:54
The third part is the authorization.
05:56
What gives access after have been authenticated.
05:59
Right. So if I met the from at the club, show my driver's license to the bouncer at the front door, Driver Barta, the bouncer says Okay. Yep. You're 21 on it. So at that point, I get access to some of the club, but not all of the club.
06:16 | 1,297 | 5,414 | {"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} | 2.859375 | 3 | CC-MAIN-2021-21 | longest | en | 0.920893 |
https://www.esaral.com/q/the-nano3-weighed-out-to-make-20050 | 1,718,470,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00848.warc.gz | 664,063,525 | 11,401 | # The NaNO3 Weighed out to make
Question:
The $\mathrm{NaNO}_{3}$ weighed out to make $50 \mathrm{~mL}$ of an aqueous solution containing $70.0 \mathrm{mg} \mathrm{Na}^{+}$per $\mathrm{mL}$ is_________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in $\mathrm{g} \mathrm{mol}^{-1}-\mathrm{Na}: 23$;
$\mathrm{N}: 14 ; \mathrm{O}: 16]$
Solution:
$\mathrm{Na}^{+}$present in $50 \mathrm{ml}$
$=\frac{70 \mathrm{mg}}{1 \mathrm{ml}} \times 50 \mathrm{ml}=3500 \mathrm{mg}=3.5 \mathrm{gm}$
moles of $\mathrm{Na}^{+}=\frac{3.5}{23}=$ moles of $\mathrm{NaNO}_{3}$
weight of $\mathrm{NaNO}_{3}=\frac{3.5}{23} \times 85=12.993 \mathrm{gm}$ | 253 | 655 | {"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} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.314924 |
https://www.convertunits.com/from/carga/to/crith | 1,611,374,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00071.warc.gz | 719,253,871 | 16,319 | ## ››Convert carga [Mexico] to crith
carga crith
How many carga in 1 crith? The answer is 6.4203571428571E-7.
We assume you are converting between carga [Mexico] and crith.
You can view more details on each measurement unit:
carga or crith
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.0071428571428571 carga, or 11125.326806475 crith.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between carga [Mexico] and crith.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of carga to crith
1 carga to crith = 1557545.75291 crith
2 carga to crith = 3115091.50581 crith
3 carga to crith = 4672637.25872 crith
4 carga to crith = 6230183.01163 crith
5 carga to crith = 7787728.76453 crith
6 carga to crith = 9345274.51744 crith
7 carga to crith = 10902820.27035 crith
8 carga to crith = 12460366.02325 crith
9 carga to crith = 14017911.77616 crith
10 carga to crith = 15575457.52906 crith
## ››Want other units?
You can do the reverse unit conversion from crith to carga, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Crith
(crith) (krith) [Gr. krith[emacr] barleycorn, the smallest weight] the unit of weight for gases, being the weight of a liter of hydrogen gas at 0°C and pressure equivalent to that of a column of mercury 760 mm high.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 544 | 1,893 | {"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-2021-04 | latest | en | 0.783672 |
https://www.opencascade.com/doc/occt-7.2.0/refman/html/class_geom___circle.html | 1,537,631,162,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158450.41/warc/CC-MAIN-20180922142831-20180922163231-00019.warc.gz | 815,856,254 | 12,167 | Geom_Circle Class Reference
Describes a circle in 3D space. A circle is defined by its radius and, as with any conic curve, is positioned in space with a right-handed coordinate system (gp_Ax2 object) where: More...
`#include <Geom_Circle.hxx>`
Inheritance diagram for Geom_Circle:
[legend]
Public Member Functions
Geom_Circle (const gp_Circ &C)
Constructs a circle by conversion of the gp_Circ circle C. More...
Geom_Circle (const gp_Ax2 &A2, const Standard_Real Radius)
Constructs a circle of radius Radius, where A2 locates the circle and defines its orientation in 3D space such that: More...
void SetCirc (const gp_Circ &C)
Set <me> so that <me> has the same geometric properties as C. More...
Assigns the value R to the radius of this circle. Note: it is possible to have a circle with a radius equal to 0.0. Exceptions - Standard_ConstructionError if R is negative. More...
gp_Circ Circ () const
returns the non transient circle from gp with the same geometric properties as <me>. More...
Returns the radius of this circle. More...
Standard_Real ReversedParameter (const Standard_Real U) const override
Computes the parameter on the reversed circle for the point of parameter U on this circle. For a circle, the returned value is: 2.*Pi - U. More...
Standard_Real Eccentricity () const override
Returns the eccentricity e = 0 for a circle. More...
Standard_Real FirstParameter () const override
Returns the value of the first parameter of this circle. This is 0.0, which gives the start point of this circle, or The start point and end point of a circle are coincident. More...
Standard_Real LastParameter () const override
Returns the value of the last parameter of this circle. This is 2.*Pi, which gives the end point of this circle. The start point and end point of a circle are coincident. More...
Standard_Boolean IsClosed () const override
returns True. More...
Standard_Boolean IsPeriodic () const override
returns True. More...
void D0 (const Standard_Real U, gp_Pnt &P) const override
Returns in P the point of parameter U. P = C + R * Cos (U) * XDir + R * Sin (U) * YDir where C is the center of the circle , XDir the XDirection and YDir the YDirection of the circle's local coordinate system. More...
void D1 (const Standard_Real U, gp_Pnt &P, gp_Vec &V1) const override
Returns the point P of parameter U and the first derivative V1. More...
void D2 (const Standard_Real U, gp_Pnt &P, gp_Vec &V1, gp_Vec &V2) const override
Returns the point P of parameter U, the first and second derivatives V1 and V2. More...
void D3 (const Standard_Real U, gp_Pnt &P, gp_Vec &V1, gp_Vec &V2, gp_Vec &V3) const override
Returns the point P of parameter u, the first second and third derivatives V1 V2 and V3. More...
gp_Vec DN (const Standard_Real U, const Standard_Integer N) const override
The returned vector gives the value of the derivative for the order of derivation N. Raised if N < 1. More...
void Transform (const gp_Trsf &T) override
Applies the transformation T to this circle. More...
Handle< Geom_GeometryCopy () const override
Creates a new object which is a copy of this circle. More...
Public Member Functions inherited from Geom_Conic
void SetAxis (const gp_Ax1 &A1)
Changes the orientation of the conic's plane. The normal axis to the plane is A1. The XAxis and the YAxis are recomputed. More...
void SetLocation (const gp_Pnt &P)
changes the location point of the conic. More...
void SetPosition (const gp_Ax2 &A2)
changes the local coordinate system of the conic. More...
gp_Ax1 Axis () const
Returns the "main Axis" of this conic. This axis is normal to the plane of the conic. More...
gp_Pnt Location () const
Returns the location point of the conic. For the circle, the ellipse and the hyperbola it is the center of the conic. For the parabola it is the Apex of the parabola. More...
const gp_Ax2Position () const
Returns the local coordinates system of the conic. The main direction of the Axis2Placement is normal to the plane of the conic. The X direction of the Axis2placement is in the plane of the conic and corresponds to the origin for the conic's parametric value u. More...
gp_Ax1 XAxis () const
Returns the XAxis of the conic. This axis defines the origin of parametrization of the conic. This axis is perpendicular to the Axis of the conic. This axis and the Yaxis define the plane of the conic. More...
gp_Ax1 YAxis () const
Returns the YAxis of the conic. The YAxis is perpendicular to the Xaxis. This axis and the Xaxis define the plane of the conic. More...
void Reverse () override
Reverses the direction of parameterization of <me>. The local coordinate system of the conic is modified. More...
GeomAbs_Shape Continuity () const override
The continuity of the conic is Cn. More...
Standard_Boolean IsCN (const Standard_Integer N) const override
Returns True. Raised if N < 0. More...
Public Member Functions inherited from Geom_Curve
virtual Standard_Real TransformedParameter (const Standard_Real U, const gp_Trsf &T) const
Returns the parameter on the transformed curve for the transform of the point of parameter U on <me>. More...
virtual Standard_Real ParametricTransformation (const gp_Trsf &T) const
Returns a coefficient to compute the parameter on the transformed curve for the transform of the point on <me>. More...
Handle< Geom_CurveReversed () const
Returns a copy of <me> reversed. More...
virtual Standard_Real Period () const
Returns the period of this curve. Exceptions Standard_NoSuchObject if this curve is not periodic. More...
gp_Pnt Value (const Standard_Real U) const
Computes the point of parameter U on <me>. If the curve is periodic then the returned point is P(U) with U = Ustart + (U - Uend) where Ustart and Uend are the parametric bounds of the curve. it is implemented with D0. More...
Public Member Functions inherited from Geom_Geometry
void Mirror (const gp_Pnt &P)
Performs the symmetrical transformation of a Geometry with respect to the point P which is the center of the symmetry. More...
void Mirror (const gp_Ax1 &A1)
Performs the symmetrical transformation of a Geometry with respect to an axis placement which is the axis of the symmetry. More...
void Mirror (const gp_Ax2 &A2)
Performs the symmetrical transformation of a Geometry with respect to a plane. The axis placement A2 locates the plane of the symmetry : (Location, XDirection, YDirection). More...
void Rotate (const gp_Ax1 &A1, const Standard_Real Ang)
Rotates a Geometry. A1 is the axis of the rotation. Ang is the angular value of the rotation in radians. More...
void Scale (const gp_Pnt &P, const Standard_Real S)
Scales a Geometry. S is the scaling value. More...
void Translate (const gp_Vec &V)
Translates a Geometry. V is the vector of the tanslation. More...
void Translate (const gp_Pnt &P1, const gp_Pnt &P2)
Translates a Geometry from the point P1 to the point P2. More...
Handle< Geom_GeometryMirrored (const gp_Pnt &P) const
Handle< Geom_GeometryMirrored (const gp_Ax1 &A1) const
Handle< Geom_GeometryMirrored (const gp_Ax2 &A2) const
Handle< Geom_GeometryRotated (const gp_Ax1 &A1, const Standard_Real Ang) const
Handle< Geom_GeometryScaled (const gp_Pnt &P, const Standard_Real S) const
Handle< Geom_GeometryTransformed (const gp_Trsf &T) const
Handle< Geom_GeometryTranslated (const gp_Vec &V) const
Handle< Geom_GeometryTranslated (const gp_Pnt &P1, const gp_Pnt &P2) const
Public Member Functions inherited from Standard_Transient
Standard_Transient ()
Empty constructor. More...
Standard_Transient (const Standard_Transient &)
Copy constructor – does nothing. More...
Standard_Transientoperator= (const Standard_Transient &)
Assignment operator, needed to avoid copying reference counter. More...
virtual ~Standard_Transient ()
Destructor must be virtual. More...
virtual void Delete () const
Memory deallocator for transient classes. More...
virtual const opencascade::handle< Standard_Type > & DynamicType () const
Standard_Boolean IsInstance (const opencascade::handle< Standard_Type > &theType) const
Returns a true value if this is an instance of Type. More...
Standard_Boolean IsInstance (const Standard_CString theTypeName) const
Returns a true value if this is an instance of TypeName. More...
Standard_Boolean IsKind (const opencascade::handle< Standard_Type > &theType) const
Returns true if this is an instance of Type or an instance of any class that inherits from Type. Note that multiple inheritance is not supported by OCCT RTTI mechanism. More...
Standard_Boolean IsKind (const Standard_CString theTypeName) const
Returns true if this is an instance of TypeName or an instance of any class that inherits from TypeName. Note that multiple inheritance is not supported by OCCT RTTI mechanism. More...
Standard_TransientThis () const
Returns non-const pointer to this object (like const_cast). For protection against creating handle to objects allocated in stack or call from constructor, it will raise exception Standard_ProgramError if reference counter is zero. More...
Standard_Integer GetRefCount () const
Get the reference counter of this object. More...
void IncrementRefCounter () const
Increments the reference counter of this object. More...
Standard_Integer DecrementRefCounter () const
Decrements the reference counter of this object; returns the decremented value. More...
Public Types inherited from Standard_Transient
typedef void base_type
Static Public Member Functions inherited from Standard_Transient
static const char * get_type_name ()
static const opencascade::handle< Standard_Type > & get_type_descriptor ()
Returns type descriptor of Standard_Transient class. More...
Protected Attributes inherited from Geom_Conic
gp_Ax2 pos
Detailed Description
Describes a circle in 3D space. A circle is defined by its radius and, as with any conic curve, is positioned in space with a right-handed coordinate system (gp_Ax2 object) where:
• the origin is the center of the circle, and
• the origin, "X Direction" and "Y Direction" define the plane of the circle. This coordinate system is the local coordinate system of the circle. The "main Direction" of this coordinate system is the vector normal to the plane of the circle. The axis, of which the origin and unit vector are respectively the origin and "main Direction" of the local coordinate system, is termed the "Axis" or "main Axis" of the circle. The "main Direction" of the local coordinate system gives an explicit orientation to the circle (definition of the trigonometric sense), determining the direction in which the parameter increases along the circle. The Geom_Circle circle is parameterized by an angle: P(U) = O + R*Cos(U)*XDir + R*Sin(U)*YDir, where:
• P is the point of parameter U,
• O, XDir and YDir are respectively the origin, "X Direction" and "Y Direction" of its local coordinate system,
• R is the radius of the circle. The "X Axis" of the local coordinate system therefore defines the origin of the parameter of the circle. The parameter is the angle with this "X Direction". A circle is a closed and periodic curve. The period is 2.*Pi and the parameter range is [ 0, 2.*Pi [.
◆ Geom_Circle() [1/2]
Geom_Circle::Geom_Circle ( const gp_Circ & C )
Constructs a circle by conversion of the gp_Circ circle C.
◆ Geom_Circle() [2/2]
Geom_Circle::Geom_Circle ( const gp_Ax2 & A2, const Standard_Real Radius )
Constructs a circle of radius Radius, where A2 locates the circle and defines its orientation in 3D space such that:
• the center of the circle is the origin of A2,
• the origin, "X Direction" and "Y Direction" of A2 define the plane of the circle,
• A2 is the local coordinate system of the circle. Note: It is possible to create a circle where Radius is equal to 0.0. raised if Radius < 0.
◆ Circ()
gp_Circ Geom_Circle::Circ ( ) const
returns the non transient circle from gp with the same geometric properties as <me>.
◆ Copy()
Handle< Geom_Geometry > Geom_Circle::Copy ( ) const
overridevirtual
Creates a new object which is a copy of this circle.
Implements Geom_Geometry.
◆ D0()
void Geom_Circle::D0 ( const Standard_Real U, gp_Pnt & P ) const
overridevirtual
Returns in P the point of parameter U. P = C + R * Cos (U) * XDir + R * Sin (U) * YDir where C is the center of the circle , XDir the XDirection and YDir the YDirection of the circle's local coordinate system.
Implements Geom_Curve.
◆ D1()
void Geom_Circle::D1 ( const Standard_Real U, gp_Pnt & P, gp_Vec & V1 ) const
overridevirtual
Returns the point P of parameter U and the first derivative V1.
Implements Geom_Curve.
◆ D2()
void Geom_Circle::D2 ( const Standard_Real U, gp_Pnt & P, gp_Vec & V1, gp_Vec & V2 ) const
overridevirtual
Returns the point P of parameter U, the first and second derivatives V1 and V2.
Implements Geom_Curve.
◆ D3()
void Geom_Circle::D3 ( const Standard_Real U, gp_Pnt & P, gp_Vec & V1, gp_Vec & V2, gp_Vec & V3 ) const
overridevirtual
Returns the point P of parameter u, the first second and third derivatives V1 V2 and V3.
Implements Geom_Curve.
◆ DN()
gp_Vec Geom_Circle::DN ( const Standard_Real U, const Standard_Integer N ) const
overridevirtual
The returned vector gives the value of the derivative for the order of derivation N. Raised if N < 1.
Implements Geom_Curve.
◆ Eccentricity()
Standard_Real Geom_Circle::Eccentricity ( ) const
overridevirtual
Returns the eccentricity e = 0 for a circle.
Implements Geom_Conic.
◆ FirstParameter()
Standard_Real Geom_Circle::FirstParameter ( ) const
overridevirtual
Returns the value of the first parameter of this circle. This is 0.0, which gives the start point of this circle, or The start point and end point of a circle are coincident.
Implements Geom_Curve.
◆ IsClosed()
Standard_Boolean Geom_Circle::IsClosed ( ) const
overridevirtual
returns True.
Implements Geom_Curve.
◆ IsPeriodic()
Standard_Boolean Geom_Circle::IsPeriodic ( ) const
overridevirtual
returns True.
Implements Geom_Curve.
◆ LastParameter()
Standard_Real Geom_Circle::LastParameter ( ) const
overridevirtual
Returns the value of the last parameter of this circle. This is 2.*Pi, which gives the end point of this circle. The start point and end point of a circle are coincident.
Implements Geom_Curve.
Returns the radius of this circle.
◆ ReversedParameter()
Standard_Real Geom_Circle::ReversedParameter ( const Standard_Real U ) const
overridevirtual
Computes the parameter on the reversed circle for the point of parameter U on this circle. For a circle, the returned value is: 2.*Pi - U.
Implements Geom_Conic.
◆ SetCirc()
void Geom_Circle::SetCirc ( const gp_Circ & C )
Set <me> so that <me> has the same geometric properties as C.
void Geom_Circle::SetRadius ( const Standard_Real R )
Assigns the value R to the radius of this circle. Note: it is possible to have a circle with a radius equal to 0.0. Exceptions - Standard_ConstructionError if R is negative.
◆ Transform()
void Geom_Circle::Transform ( const gp_Trsf & T )
overridevirtual
Applies the transformation T to this circle.
Implements Geom_Geometry.
The documentation for this class was generated from the following file: | 3,826 | 15,150 | {"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-2018-39 | latest | en | 0.640841 |
https://www.weegy.com/?ConversationId=CUHS568D&Link=i&ModeType=2 | 1,606,815,104,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00512.warc.gz | 882,225,443 | 11,932 | 7. The Waverly Brush Company issued 4,000 shares of common stock worth \$200,000.00 total. What is the par value of each share?
Question
s
Original conversation
User: 7. The Waverly Brush Company issued 4,000 shares of common stock worth \$200,000.00 total. What is the par value of each share?
Weegy: The Waverly Brush Company issued 4,000 shares of common stock worth \$200,000.00 total. \$50 is the par value of each share. 200 000/4000 = 50 50 x 4000 = 200 000
|Score .9026|Malekith22|Points 1732|
User: 8. If a company has 245 shares of common stock and \$275,000.00 to be distributed to its holders, how much would each share receive?
Weegy: If a company has 245 shares of common stock and \$275,000.00 to be distributed to its holders, each share would receive \$1,122.45. \$275,000.00/245 = \$1,122.45
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4. What should be the top item on a list of your life goals?
Weegy: A goal that's most important to you should be the top item on a list of your life goals. User: 5. Which of the following best describes term life insurance? Weegy: The insured pays a premium for a specified number of years. - best describes term life insurance. User: 6. What type of stock receives an equal part of the profits on each share to be distributed after all other obligations of a company have been satisfied? Weegy: Common stock receives an equal part of the profits on each share to be distributed after all other obligations of a company have been satisfied. User: 7. The Waverly Brush Company issued 4,000 shares of common stock worth \$200,000.00 total. What is the par value of each share? Weegy: The Waverly Brush Company issued 4,000 shares of common stock worth \$200,000.00 total. The par value of each share is \$50. x = 200,000/4,000 ; x = 50 (More)
Question
Lorenzo has a checkbook balance of \$118.00. He writes two checks, one for \$9.00 and one for \$84.25. He then deposits \$95.00. Finally, he uses his calculator to determine his new balance. Which one of the following series represents the correct order in which he should press the keys on his calculator?
Question
Updated 4/13/2019 1:10:08 AM
Lorenzo has a checkbook balance of \$118.00. He writes two checks, one for \$9.00 and one for \$84.25. He then deposits \$95.00. Finally, he uses his calculator to determine his new balance.
118 – 9 – 84.25 + 95 series represents the correct order in which he should press the keys on his calculator.
Confirmed by Sting [4/13/2019 1:09:54 AM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,199 | 4,211 | {"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-2020-50 | latest | en | 0.917478 |
https://codebreaker.xyz/problem/moneychanger | 1,726,183,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00451.warc.gz | 157,262,836 | 4,332 | ### Problem Statement
Josephine the money changer was working at her kiosk one day when she realized how irritating it was when she had to give about 20 coins to her customer. Suddenly, she had an idea. If she could write a program to instantly tell her how many coins she needs to make the value, then she could just refuse to serve her customers if they choose a value that requires her to select too many coins.
Unfortunately, Josephine, being a lazy person, doesn’t want to learn programming all by herself. Your task is to write the program for her.
Given n (1 ≤ n ≤ 100) types of coins, where each type of coin has a value Ai (1 ≤ Ai ≤ 10,000), and a value v (1 ≤ v ≤ 10,000) which is also an integer, calculate the minimum number of coins needed to make v with the n coins provided. Note that you are allowed to use the same type of coin multiple times.
### Input
The first line of input will be two space-separated integers, n and v. The following n lines of input will contain one integer each, representing Ai, the value of the ith coin.
### Output
Print the the minimum number of coins needed to make the value v. If it is impossible to make the value with the set of coins given, print -1.
```6 99
1
2
5
10
25
50
```
### Sample Output 1
`6`
6 coins are needed to make 99: 50, 25, 10, 10, 2, 2
### Submitting .cpp to 'moneychanger'
Time Limit: 1 Seconds
Memory Limit: 256MB | 357 | 1,397 | {"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} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.94299 |
http://www.101computing.net/lightning-distance-calculator/ | 1,532,173,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00238.warc.gz | 409,095,163 | 10,734 | # Lightning Distance Calculator
Have you ever seen a lightning flash or heard the thunder of lightning and wondered how close you were from the ligthning strike? Have you noticed that there often was a delay between the flash of light and the clap of thunder when a lightning occurs?
It is possible to calculate the distance to a ligthning strike by counting the seconds between the lightning flash and the sound of thunder.
When lightning strikes, the first thing you see is the flash of light which you can see instantly. This is because light travels at a very high speed (Speed of light = 300,000 km/s). At the same time a clap of thunder is created. However, in the air a sound wave does not travel as fast as light, so it may take a few seconds for this clap of thunder to reach you. This depends on how far you are from the lightning.
Look at the following diagram and check the formula of speed. Bear in mind that the speed of a sound wave in the air is 340 m/s. | 210 | 974 | {"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-2018-30 | latest | en | 0.962917 |
https://en-academic.com/dic.nsf/enwiki/11860659 | 1,714,043,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00347.warc.gz | 202,372,356 | 25,540 | # Orbital perturbation analysis (spacecraft)
Orbital perturbation analysis (spacecraft)
Isaac Newton in his Philosophiæ Naturalis Principia Mathematica demonstrated that the gravitational force between two mass points is inversely proportional to the square of the distance between the points and fully solved corresponding "two-body problem" demonstrating that radius vector between the two points would describe an ellipse. But already for the three body problem no exact closed form analytical form could be found. Instead approximate methods were developed, this is the method of "Orbital perturbation analysis". With this technique a quite accurate mathematical description of the trajectories of all the planets could be obtained. Especially critical was the mathematical modeling of the orbit of the Moon as the deviations from a pure Kepler orbit around the Earth due to the gravitational force of the Sun (i.e. one has indeed the three body problem) are much larger than deviations of the orbits of the planets from Sun-centered Kepler orbits caused by the gravitational attraction between the planets. With the availability of digital computers and the easiness to propagate orbits with numerical methods this problem partly disappeared, the motion of all celestial bodies including planets, satellites, asteroids and comets can be modeled and predicted with almost perfect accuracy using the method of the numerical propagation of the trajectories. Never-the-less several analytical closed form expressions for the effect of such additional "perturbing forces" are still very useful
## Orbital perturbation of spacecraft orbits
All celestial bodies of the Solar System follow in first approximation a Kepler orbit around a central body. For a satellite (artificial or natural) this central body is a planet. But both due to gravitational forces caused by the Sun and other celestial bodies and due to the flattening of its planet (caused by its rotation which makes the planet slightly oblate and therefore the result of the Shell theorem not fully applicable) the satellite will follow an orbit that deviates more from a pure Kepler orbit then what is the case for the planets.
The precise modeling of the motion of the Moon has because of this been a difficult task. The best and most accurate modeling for the Moon orbit before the availability of digital computers was obtained with the complicated Delunay and Brown's lunar theories.
But the most important of these perturbation effects, the precession of the orbital plane caused by the slightly oblate shape of the Earth, was already fully understood by Isaac Newton who estimated the oblateness of the Earth from the observed rate of precession of the orbital plane of the Moon.
For man-made spacecraft orbiting the Earth at comparatively low altitudes the deviations from a Kepler orbit are much larger than for the Moon. The approximation of the gravitational force of the Earth to be that of a homogeneous sphere gets worse the closer one gets to the Earth surface and the majority of the artificial Earth satellites are in orbits that are only a few hundred kilometers over the Earth surface. Further more they are (as opposed to the Moon) significantly affected by the solar radiation pressure because of their large cross-section to mass ratio, this applies in particular to 3-axis stabilized spacecraft with large solar arrays. In addition they are significantly affected by rarefied air unless above 800–1000 km (The air drag at high altitudes strongly depending on the solar activity!)
## Mathematical approach
Consider any function
$g(x_1,x_2,x_3,v_1,v_2,v_3)\,$
of the position
$x_1,x_2,x_3\,$
and the velocity
$v_1,v_2,v_3\,$
From the chain rule of differentiation one gets that the time derivative of g is
$\dot{g}\ =\ \frac{\partial g }{\partial x_1}\ v_1\ + \ \frac{\partial g }{\partial x_2}\ v_2\ + \frac{\partial g }{\partial x_3}\ v_3\ + \ \frac{\partial g }{\partial v_1}\ f_1\ + \ \frac{\partial g }{\partial v_2}\ f_2\ + \ \frac{\partial g }{\partial v_3}\ f_3$
where $f_1\ ,\ f_2\ ,\ f_3$ are the components of the force per unit mass acting on the body.
If now g is a "constant of motion" for a Kepler orbit like for example an orbital element and the force is corresponding "Kepler force"
$(f_1\ ,\ f_2\ ,\ f_3)\ = \ - \frac {\mu} {r^3}\ (x_1\ ,\ x_2\ ,\ x_3)$
one has that $\dot{g}\ =\ 0\,$.
If the force is the sum of the "Kepler force" and an additional force (force per unit mass)
$(h_1\ ,\ h_2\ ,\ h_3)$
i.e.
$(f_1\ ,\ f_2\ ,\ f_3)\ = \ - \frac {\mu} {r^3}\ (x_1\ ,\ x_2\ ,\ x_3)\ +\ (h_1\ ,\ h_2\ ,\ h_3)$
one therefore has
$\dot{g}\ =\frac{\partial g }{\partial v_1}\ h_1\ + \ \frac{\partial g }{\partial v_2}\ h_2\ + \ \frac{\partial g }{\partial v_3}\ h_3$
and that the change of $g\,$ in the time from $t=t_1\,$ to $t=t_2\,$ is
$\Delta g\ =\ \int\limits_{t_1}^{t_2}\left(\frac{\partial g }{\partial v_1}\ h_1\ + \ \frac{\partial g }{\partial v_2}\ h_2\ + \ \frac{\partial g }{\partial v_3}\ h_3 \right)dt$
If now the additional force $(h_1\ ,\ h_2\ ,\ h_3)\,$ is sufficiently small that the motion will be close to that of a Kepler orbit one gets an approximate value for $\Delta g\,$ by evaluating this integral assuming $x_1(t),x_2(t),x_3(t)\,$ to precisely follow this Kepler orbit.
In general one wants to find an approximate expression for the change $\Delta g\,$ over one orbital revolution using the true anomaly $\theta\,$ as integration variable, i.e. as
$\Delta g\ =\ \int\limits_{0}^{2\pi}\left(\frac{\partial g }{\partial v_1}\ h_1\ + \ \frac{\partial g }{\partial v_2}\ h_2\ + \ \frac{\partial g }{\partial v_3}\ h_3 \right)\frac{r^2}{\sqrt{\mu p}}d\theta$
(1)
This integral is evaluated setting $r(\theta)=\frac {p}{1+e\cos \theta}\,$, the elliptical Kepler orbit in polar angles. For the transformation of integration variable from time to true anomaly it was used that the angular momentum $H\ =\ r^2\ \dot{\theta}\ =\ \sqrt{\mu p} \,$ by definition of the parameter $p\,$ for a Kepler orbit (see equation (13) of the Kepler orbit article).
For the very important case that the Kepler orbit is circular or almost circular
$r\ =\ p$ and (1) takes the simpler form
$\Delta g\ =\ \frac{P}{2\pi}\ \int\limits_{0}^{2\pi}\left(\frac{\partial g }{\partial v_1}\ h_1\ + \ \frac{\partial g }{\partial v_2}\ h_2\ + \ \frac{\partial g }{\partial v_3}\ h_3 \right)d\theta$
(2)
where $P\ =\ 2\pi\ r\ \sqrt{\frac{r}{\mu}}\,$ is the orbital period
### Perturbation of the semi-major axis/orbital period
For an elliptic Kepler orbit the sum of the kinetic and the potential energy
$g = \frac{V^2}{2}-\frac {\mu} {r}$
where $V\,$ is the orbital velocity
is a constant and equal to
$g\ =\ -\frac {\mu} {2 \cdot a}$ (Equation (44) of the Kepler orbit article)
If $\bar{h}\,$ is the perturbing force and $\bar{V}\,$is the velocity vector of the Kepler orbit the equation (1) takes the form:
$\Delta g\ =\ \int\limits_{0}^{2\pi}\bar{V} \bar{h}\frac{r^2}{\sqrt{\mu p}}d\theta$
(3)
and for a circular or almost circular orbit
$\Delta g\ =\ \frac{P}{2\pi}\ \int\limits_{0}^{2\pi} \bar{V} \bar{h}d\theta$
(4)
From the change $\Delta g\,$ of the parameter $g\,$ the new semi-major axis $a\,$ and the new period $P\ =\ 2\pi\ a\ \sqrt{\frac{a}{\mu}}\,$ are computed (relations (43) and (44) of the Kepler orbit article).
### Perturbation of the orbital plane
Let $\hat{g}\,$ and $\hat{h}\,$ make up a rectangular coordinate system in the plane of the reference Kepler orbit. If $\omega\,$ is the argument of perigee relative the $\hat{g}\,$ and $\hat{h}\,$ coordinate system the true anomaly $\theta\,$ is given by $\theta=u-\omega\,$ and the approximate change $\Delta \hat{z}\,$ of the orbital pole $\hat{z}\,$ (defined as the unit vector in the direction of the angular momentum) is
$\Delta \hat{z}\ =\ \int\limits_{0}^{2\pi}\frac{f_z }{V_t} (\hat{g} \cos u + \hat{h} \sin u)\frac{r^2}{\sqrt{\mu p}}du \quad \times \ \hat{z} =\ \frac{1}{\mu p}\left[\hat{g}\int\limits_{0}^{2\pi}f_z r^3 \cos u \ du +\ \hat{h}\int\limits_{0}^{2\pi}f_z r^3 \sin u \ du \right]\quad \times \ \hat{z}$
(5)
where $f_z\,$ is the component of the perturbing force in the $\hat{z}\,$ direction, $V_t=\sqrt{\frac{\mu}{p}}\ (1+e\ \cos\theta)\,$ is the velocity component of the Kepler orbit orthogonal to radius vector and $r=\frac{p}{1+e\ \cos\theta}\,$ is the distance to the center of the Earth.
For a circular or almost circular orbit (5) simplifies to
$\Delta \hat{z}\ =\ \frac{r^2}{\mu}\left[\hat{g}\int\limits_{0}^{2\pi}f_z \cos u \ du +\ \hat{h}\int\limits_{0}^{2\pi}f_z \sin u \ du \right]\quad \times \ \hat{z}$
(6)
Example
In a circular orbit a low-force propulsion system (Ion thruster) generates a thrust (force per unit mass) of $F\ \hat{z}\,$ in the direction of the orbital pole in the half of the orbit for which $\sin u\,$ is positive and in the opposite direction in the other half. The resulting change of orbit pole after one orbital revolution of duration $P\ =\ 2\pi\ r\ \sqrt{\frac{r}{\mu}}\,$ is
$\Delta \hat{z}\ =\ \frac{r^2}{\mu}\left[\ 2\ F\int\limits_{0}^{\pi}\sin u \ du \right]\quad \hat{h}\times \hat{z} = \ \frac{r^2}{\mu}\ 4\ F\ \quad \hat{g}$
(7)
The average change rate $\frac{\Delta \hat{z}}{P}\,$ is therefore
$\frac{\Delta \hat{z}}{P} =\ \frac{2}{\pi}\ \frac{F}{V}\ \hat{g}$
(8)
where $V\ =\ \sqrt{\frac{\mu}{r}},$ is the orbital velocity in the circular Kepler orbit.
### Perturbation of the eccentricity vector
Rather than applying (1) and (2) on the partial derivatives of the orbital elements eccentricity and argument of perigee directly one should apply these relations for the eccentricity vector. First of all the typical application is a near-circular orbit. But there are also mathematical advantages working with the partial derivatives of the components of this vector also for orbits with a significant eccentricity.
Equations (60), (55) and (52) of the Kepler orbit article say that the eccentricity vector is
$\bar{e}=\frac{(V_t-V_0) \cdot \hat{r} - V_r \cdot \hat{t}}{V_0}$
(9)
where
$V_0 = \sqrt{\frac{\mu}{p}}$
(10)
$p = \frac{{(r \cdot V_t)}^2}{\mu }$
(11)
from which follows that
$\frac{\partial\bar{e}}{\partial V_r} = -\frac {1}{V_0} \hat{t}$
(12)
$\frac{\partial\bar{e}}{\partial V_t} = \frac {1}{V_0} \left(2\ \hat{r}-\frac{V_r}{V_t}\ \hat{t}\right)$
(13)
where
$V_r = \sqrt{\frac {\mu}{p}} \cdot e \cdot \sin \theta$
(14)
$V_t = \sqrt{\frac {\mu}{p}} \cdot (1 + e \cdot \cos \theta)$
(15)
(Equations (18) and (19) of the Kepler orbit article)
The eccentricity vector is by definition always in the osculating orbital plane spanned by $\hat{r}$ and $\hat{t}$ and formally there is also a derivative
$\frac{\partial\bar{e}}{\partial V_z} = -\frac {V_r}{V_0}\ \frac{\partial\hat{t}}{\partial V_z}$
with
$\frac{\partial\hat{t}}{\partial V_z} = \frac {1}{V_t}\ \hat{z}$
corresponding to the rotation of the orbital plane
But in practice the in-plane change of the eccentricity vector is computed as
\begin{align} \Delta \bar{e}\ = &\frac {1}{V_0}\ \int\limits_{0}^{2\pi}\left(-\hat{t}\ f_r\ + \ \left(2\ \hat{r}-\frac{V_r}{V_t}\ \hat{t}\right)\ f_t\right)\frac{r^2}{\sqrt{\mu p}}du\ = \\ &\frac {1}{\mu}\ \int\limits_{0}^{2\pi}\left(-\hat{t}\ f_r\ + \ \left(2\ \hat{r}-\frac{V_r}{V_t}\ \hat{t}\right)\ f_t\right) r^2 du \end{align}
(16)
ignoring the out-of-plane force and the new eccentricity vector
$\bar{e} + \Delta \bar{e}$
is subsequently projected to the new orbital plane orthogonal to the new orbit normal
$\hat{z} + \Delta \hat{z}$
computed as described above.
Example
The Sun is in the orbital plane of a spacecraft in a circular orbit with radius $r\,$ and consequently with a constant orbital velocity $V_0\ =\ \sqrt{\frac{\mu}{r}}$ . If $\hat{k}\,$ and $\hat{l}\,$ make up a rectangular coordinate system in the orbital plane such that $\hat{k}\,$ points to the Sun and assuming that the solar radiation pressure force per unit mass $F\,$ is constant one gets that
$\hat{r}=\cos(u)\ \hat{k}\ +\ \sin(u)\ \hat{l}\,$
$\hat{t}=-\sin(u)\ \hat{k}\ +\ \cos(u)\ \hat{l}\,$
$F_r=-\cos(u)\ F\,$
$F_t= \sin(u)\ F\,$
where $u\,$ is the polar angle of $\hat{r}\,$ in the $\hat{k}\,$, $\hat{l}\,$ system. Applying (2) one gets that
\begin{align} \Delta \hat{e}\ & =\ \frac{P}{2\pi}\ \frac {1}{V_0}\ \int\limits_{0}^{2\pi}\left( (-\sin(u)\ \hat{k}\ +\ \cos(u)\ \hat{l}) \ F\ \cos(u)\ + \ 2\ (\cos(u)\ \hat{k}\ +\ \sin(u)\ \hat{l})\ F\ \sin(u)\right)\ du \\ & = P\ \frac{3}{2}\ \frac {1}{V_0}\ \ F\ \hat{l} \end{align}
(17)
This means the eccentricity vector will gradually increase in the direction $\hat{l}\,$ orthogonal to the Sun direction. This is true for any orbit with a small eccentricity, the direction of the small eccentricity vector does not matter. As $P\,$ is the orbital period this means that the average rate of this increase will be $\frac{3}{2}\ \frac {F}{V_0}\,$
## The effect of the Earth flattening
Figure 1: The unit vectors $\hat{\phi}\ ,\ \hat{\lambda}\ ,\ \hat{r}$
In the article Geopotential model the modeling of the gravitational field as a sum of spherical harmonics is discussed. The by far dominating term is the "J2-term". This is a "zonal term" and corresponding force is therefore completely in a longitudinal plane with one component $F_r\ \hat{r}\,$ in the radial direction and one component $F_\lambda\ \hat{\lambda}\,$ with the unit vector $\hat{\lambda}\,$ orthogonal to the radial direction towards north. These directions $\hat{r}\,$ and $\hat{\lambda}\,$ are illustrated in Figure 1.
Figure 2: The unit vector $\hat{t}\,$ orthogonal to $\hat{r}\,$ in the direction of motion and the orbital pole $\hat{z}\,$. The force component Fλ is marked as "F"
To be able to apply relations derived in the previous section the force component $F_\lambda\ \hat{\lambda}\,$ must be split into two orthogonal components $F_t\ \hat{t}$ and $F_z\ \hat{z}$ as illustrated in figure 2
Let $\hat{a}\ ,\ \hat{b}\ ,\ \hat{n}\,$ make up a rectangular coordinate system with origin in the center of the Earth (in the center of the Reference ellipsoid) such that $\hat{n}\,$ points in the direction north and such that $\hat{a}\ ,\ \hat{b}\,$ are in the equatorial plane of the Earth with $\hat{a}\,$ pointing towards the ascending node, i.e. towards the blue point of Figure 2.
The components of the unit vectors
$\hat{r}\ ,\ \hat{t}\ ,\ \hat{z}\,$
making up the local coordinate system (of which $\hat{t}\ ,\ \hat{z},$ are illustrated in figure 2) relative the $\hat{a}\ ,\ \hat{b}\ ,\ \hat{n}\,$ are
$r_a= \cos u\,$
$r_b= \cos i \ \sin u\,$
$r_n= \sin i \ \sin u\,$
$t_a=-\sin u\,$
$t_b= \cos i \ \cos u\,$
$t_n= \sin i \ \cos u\,$
$z_a= 0\,$
$z_b=-\sin i\,$
$z_n= \cos i\,$
where $u\,$ is the polar argument of $\hat{r}\,$ relative the orthogonal unit vectors $\hat{g}=\hat{a}\,$ and $\hat{h}=\cos i\ \hat{b}\ +\ \sin i\ \hat{n}\,$ in the orbital plane
Firstly
$\sin \lambda =\ r_n\ =\ \sin i \ \sin u\,$
where $\lambda\,$ is the angle between the equator plane and $\hat{r}\,$ (between the green points of figure 2) and from equation (12) of the article Geopotential model one therefore gets that
$f_r = J_2\ \frac{1}{r^4}\ \frac{3}{2}\ \left(3\ \sin^2 i\ \sin^2 u\ -\ 1\right)$
(18)
Secondly the projection of direction north, $\hat{n}\,$, on the plane spanned by $\hat{t}\ ,\ \hat{z},$ is
$\sin i \ \cos u \ \hat{t}\ +\ \cos i \ \hat{z}\,$
and this projection is
$\cos \lambda \ \hat{\lambda}\,$
where $\hat{\lambda}\,$ is the unit vector $\hat{\lambda}$ orthogonal to the radial direction towards north illustrated in figure 1.
From equation (12) of the article Geopotential model one therefore gets that
$f_\lambda \ \hat{\lambda}\ =\ -J_2\ \frac{1}{r^4}\ 3\ \sin\lambda\ (\sin i \ \cos u \ \hat{t}\ +\ \cos i \ \hat{z}) =\ -J_2\ \frac{1}{r^4}\ 3\ \sin i \ \sin u\ (\sin i \ \cos u \ \hat{t}\ +\ \cos i \ \hat{z})\,$
and therefore:
$f_t =\ -J_2\ \frac{1}{r^4}\ 3\ \sin^2 i\ \sin u\ \cos u$
(19)
$f_z =\ -J_2\ \frac{1}{r^4}\ 3\ \sin i\ \cos i\ \sin u$
(20)
### Perturbation of the orbital plane
From (5) and (20) one gets that
$\Delta \hat{z}\ =\ -J_2\ \frac{3\ \sin i\ \cos i}{\mu p^2}\left[\hat{g}\int\limits_{0}^{2\pi}\frac{p}{r}\ \sin u\ \cos u \ du +\ \hat{h}\int\limits_{0}^{2\pi}\frac{p}{r}\ \sin^2 u\ du \right]\quad \times \ \hat{z}$
(21)
The fraction $\frac{p}{r}\,$ is
$\frac{p}{r}\ =\ 1\ +\ e\ \cos (u-\omega)\ =\ 1\ +\ e\ \cos u\ \cos\omega\ +\ e\ \sin u\ \sin\omega\,$
where $e\,$ is the eccentricity and $\omega\,$ is the argument of perigee of the reference Kepler orbit
As all integrals of type
$\int\limits_{0}^{2\pi} \cos^m u \ \sin^n u\ du\,$
are zero if not both $n\,$ and $m\,$ are even one gets from (21) that
$\Delta \hat{z}\ =\ -2\pi\ \frac{J_2}{\mu\ p^2}\ \frac{3}{2}\ \sin i\ \cos i\ \quad \hat{h} \times \hat{z}$
As
$\hat{n}\ =\ \cos i\ \hat{z}\ + \sin i\ \hat{h}$
this can be written
$\Delta \hat{z}\ =\ -2\pi\ \frac{J_2}{\mu\ p^2}\ \frac{3}{2}\ \cos i\ \quad \hat{n} \times \hat{z}$
(22)
As $\hat{n}$ is an inertially fixed vector (the direction of the spin axis of the Earth) relation (22) is the equation of motion for a unit vector $\hat{z}\,$ describing a cone around $\hat{n}$ with a precession rate (radians per orbit) of $-2\pi\ \frac{J_2}{\mu\ p^2}\ \frac{3}{2}\ \cos i\,$
In terms of orbital elements this is expressed as
$\Delta i\ =\ 0$
(23)
$\Delta \Omega\ =\ -2\pi\ \frac{J_2}{\mu\ p^2}\ \frac{3}{2}\ \cos i$
(24)
where
$i\,$ is the inclination of the orbit to the equatorial plane of the Earth
$\Omega\,$ is the right ascension of the ascending node
### Perturbation of the eccentricity vector
From (16), (18) and (19) follows that in-plane perturbation of the eccentricity vector is
$\Delta \bar{e}\ =\ \frac {J_2}{\mu\ p^2}\ \int\limits_{0}^{2\pi}\left(-\hat{t}\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ \left(3\ \sin^2 i\ \sin^2 u\ -\ 1\right)\ - \ \left(2\ \hat{r}-\frac{V_r}{V_t}\ \hat{t}\right)\ \left(\frac{p}{r}\right)^2\ 3\ \sin^2 i \cos u\ \sin u\right)du$
(25)
the new eccentricity vector being the projection of
$\bar{e}+\Delta \bar{e}$
on the new orbital plane orthogonal to
$\hat{z}+\Delta \hat{z}$
where $\Delta \hat{z}\,$ is given by (22)
Relative the coordinate system
$\hat{g}=\hat{a}\,$
$\hat{h}=\cos i\ \hat{b}\ +\ \sin i\ \hat{n}\,$
one has that
$\hat{r}=\cos u\ \hat{g}\ +\ \sin u\ \hat{h}\,$
$\hat{t}=-\sin u\ \hat{g}\ +\ \cos u\ \hat{h}\,$
Using that
$\frac {p}{r}\ =\ 1 + e \cdot \cos \theta\ =\ 1 + e_g \cdot \cos u + e_h \cdot \sin u$
and that
$\frac {V_r}{V_t} = \frac {e_g \cdot \sin u\ -\ e_h \cdot \cos u}{\frac {p}{r}}$
where
$e_g =\ e\ \cos \omega$
$e_h =\ e\ \sin \omega$
are the components of the eccentricity vector in the $\hat{g}\ ,\ \hat{h}\,$ coordinate system this integral (25) can be evaluated analytically, the result is
$\Delta \bar{e}\ =\ -2\pi\ \frac {J_2}{\mu\ p^2}\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ \left(-e_h \hat{g}\ +\ e_g \hat{h}\right)\ =\ -2\pi\ \frac {J_2}{\mu\ p^2} \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ \hat{z}\ \times \ \bar{e}$
(26)
This the difference equation of motion for the eccentricity vector $\bar{e}\,$ to form a circle, the magnitude of the eccentricity $e\,$ staying constant.
Translating this to orbital elements it must be remembered that the new eccentricity vector obtained by adding $\Delta \bar{e}\ \,$ to the old $\bar{e}\ \,$ must be projected to the new orbital plane obtained by applying (23) and (24)
Figure 3: The change $\Delta\omega\,$ in "argument of perigee" after one orbit is the sum of a contribution $\Delta \omega_1\,$ caused by the in-plane force components and a contribution $\Delta \omega_2\,$ caused by the use of the ascending node as reference
This is illustrated in figure 3:
To the change in argument of the eccentricity vector
$\Delta \omega_1\ =\ -2\pi\ \frac {J_2}{\mu\ p^2}\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\,$
must be added an increment due to the precession of the orbital plane (caused by the out-of-plane force component) amounting to
$\Delta \omega_2\ =\ -\cos i\ \Delta\Omega \ =\ 2\pi\ \frac {J_2}{\mu\ p^2}\ \frac{3}{2}\ \cos^2 i\,$
One therefore gets that
$\Delta e\ =0$
(27)
$\Delta \omega\ =\Delta \omega_1\ +\ \Delta \omega_2\ =\ \ -2\pi\ \frac {J_2}{\mu\ p^2}\ 3 \left(\frac{5}{4}\ \sin^2 i\ -\ 1\right)$
(28)
In terms of the components of the eccentricity vector $e_g,e_h\,$ relative the coordinate system $\hat{g} ,\hat{h}\,$ that precesses around the polar axis of the Earth the same is expressed as follows
\begin{align} &(\Delta e_g,\Delta e_h)\ = \\ &-2\pi\ \frac {J_2}{\mu\ p^2}\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ (-e_h ,e_g)\ + \ 2\pi\ \frac {J_2}{\mu\ p^2}\ \frac{3}{2}\ \cos^2 i\ (-e_h ,e_g ) = \\ &-2\pi\ \frac {J_2}{\mu\ p^2}\ 3 \left(\frac{5}{4}\ \sin^2 i\ -\ 1\right)\ (-e_h ,e_g) \end{align}
(29)
where the first term is the in-plane perturbation of the eccentricity vector and the second is the effect of the new position of the ascending node in the new plane
From (28) follows that $\Delta \omega\,$ is zero if $\sin^2 i\ =\frac{4}{5}\,$. This fact is used for Molniya orbits having an inclination of 63.4 deg. An orbit with an inclination of 180 - 63.4 deg = 116.6 deg would in the same way have a constant argument of perigee.
#### Proof
Proof that the integral
$\int\limits_{0}^{2\pi}\left(-\hat{t}\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ \left(3\ \sin^2 i\ \sin^2 u\ -\ 1\right)\ - \ \left(2\ \hat{r}-\frac{V_r}{V_t}\ \hat{t}\right)\ \left(\frac{p}{r}\right)^2\ 3\ \sin^2 i \cos u\ \sin u\right)du$
(30)
where:
$\hat{r}=\cos u\ \hat{G}\ +\ \sin u\ \hat{H}\,$
$\hat{t}=-\sin u\ \hat{G}\ +\ \cos u\ \hat{H}\,$
$\frac{p}{r}\ =\ 1\ +\ e_g\ \cos u\ +\ e_h\ \sin u$
$\frac{V_r}{V_t}\ =\ \frac{e_g\ \sin u\ -\ e_h\ \cos u}{\frac{p}{r}}$
has the value
$-2\pi\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ \left(-e_h \hat{G}\ +\ e_g \hat{H}\right)$
(31)
Integrating the first term of the integrand one gets:
\begin{align} &\int\limits_{0}^{2\pi}-t_g\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ 3\ \sin^2 i\ \sin^2 u\ du\ =\ \frac{9}{2}\ \sin^2 i\ \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \ \sin^3 u\ du\ = \\ &9\ \sin^2 i\ e_h\ \int\limits_{0}^{2\pi}\sin^4 u\ du\ =\ 2\pi \frac{27}{8}\ \sin^2 i\ e_h \end{align}
(32)
and
\begin{align} &\int\limits_{0}^{2\pi}-t_h\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ 3\ \sin^2 i\ \sin^2 u\ du\ =\ -\frac{9}{2}\ \sin^2 i\ \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \ \sin^2 u\ \cos u\ du\ = \\ &-9\ \sin^2 i\ e_g\ \int\limits_{0}^{2\pi}\sin^2 u\ \cos^2 u\ du\ =\ -2\pi \frac{9}{8}\ \sin^2 i\ e_g \end{align}
(33)
For the second term one gets:
$\int\limits_{0}^{2\pi} t_g\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ du\ =\ -\frac{3}{2}\ \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \ \sin u\ du\ =\ -3\ e_h\ \int\limits_{0}^{2\pi}\sin^2 u\ du\ =\ -2\pi \frac{3}{2}\ e_h$
(34)
and
$\int\limits_{0}^{2\pi} t_h\ \left(\frac{p}{r}\right)^2\ \frac{3}{2}\ du\ =\ \frac{3}{2}\ \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \ \cos u\ du\ =\ 3\ e_g\ \int\limits_{0}^{2\pi}\cos^2 u\ du\ =\ 2\pi \frac{3}{2}\ e_g$
(35)
For the third term one gets:
\begin{align} &-\int\limits_{0}^{2\pi}\ 2\ r_g \ \left(\frac{p}{r}\right)^2\ 3\ \sin^2 i \cos u\ \sin u\ du\ =\ -6\ \sin^2 i \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \cos^2 u\ \sin u\ du\ =\ \\ &-12\ \sin^2 i\ e_h \int\limits_{0}^{2\pi}\ \cos^2 u\ \sin^2 u\ du\ =\ -2\pi \frac{3}{2}\ \sin^2 i\ e_h \end{align}
(36)
and
\begin{align} &-\int\limits_{0}^{2\pi}\ 2\ r_h \ \left(\frac{p}{r}\right)^2\ 3\ \sin^2 i \cos u\ \sin u\ du\ =\ -6\ \sin^2 i \int\limits_{0}^{2\pi}\ \left(1\ +\ e_g\ \cos u\ +\ e_h\ \sin u\right)^2\ \cos u\ \sin^2 u\ du\ =\ \\ &-12\ \sin^2 i\ e_g \int\limits_{0}^{2\pi}\ \sin^2 u \cos^2 u\ du\ =\ -2\pi \ \frac{3}{2}\ \sin^2 i\ e_g \end{align}
(37)
For the fourth term one gets:
\begin{align} &\int\limits_{0}^{2\pi}t_g\ \frac{V_r}{V_t}\ \left(\frac{p}{r}\right)^2\ 3\ \sin i \cos^2 u\ \sin u\ du\ = -3\ \sin^2 i \int\limits_{0}^{2\pi}(e_g\ \sin u\ -\ e_h\ \cos u)\ \frac{p}{r}\ \cos u\ \sin^2 u\ du \ = \\ &-3\ \sin^2 i \int\limits_{0}^{2\pi}(e_g\ \sin u\ -\ e_h\ \cos u)\ (1\ +\ e_g\ \cos u\ +\ e_h\ \sin u)\ \cos u\ \sin^2 u\ du\ = \\ &3\ \sin^2 i \ e_h\int\limits_{0}^{2\pi}\ \ \cos^2 u\ \sin^2 u\ du\ =\ 2\pi \frac{3}{8} \sin^2 i \ e_h \end{align}
(38)
and
\begin{align} &\int\limits_{0}^{2\pi}t_h\ \frac{V_r}{V_t}\ \left(\frac{p}{r}\right)^2\ 3\ \sin^2 i \cos u\ \sin u\ du\ = 3\ \sin^2 i \int\limits_{0}^{2\pi}(e_g\ \sin u\ -\ e_h\ \cos u)\ \frac{p}{r}\ \cos^2 u\ \sin u\ du \ = \\ &3\ \sin^2 i \int\limits_{0}^{2\pi}(e_g\ \sin u\ -\ e_h\ \cos u)\ (1\ +\ e_g\ \cos u\ +\ e_h\ \sin u)\ \cos^2 u\ \sin u\ du\ = \\ &3\ \sin^2 i \ e_g\int\limits_{0}^{2\pi}\ \ \cos^2 u\ \sin^2 u\ du\ =\ 2\pi \frac{3}{8} \sin^2 i \ e_g \end{align}
(39)
Adding the right hand sides of (32), (34), (36) and (38) one gets $2\pi \frac{27}{8}\ \sin^2 i\ e_h\ -\ 2\pi \frac{3}{2}\ e_h\ -\ 2\pi \frac{3}{2}\ \sin^2 i\ e_h\ +\ 2\pi \frac{3}{8} \sin^2 i \ e_h \ =\ 2\pi\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ e_h$
Adding the right hand sides of (33), (35), (37) and (39) one gets $-2\pi \frac{9}{8}\ \sin^2 i\ e_g\ +\ 2\pi \frac{3}{2}\ e_g\ -\ 2\pi \ \frac{3}{2}\ \sin^2 i\ e_g\ +\ 2\pi \frac{3}{8} \sin^2 i \ e_g\ =\ -2\pi\ \frac{3}{2} \left(\frac{3}{2}\ \sin^2 i\ -\ 1\right)\ e_g$
## References
• El'Yasberg "Theory of flight of artificial earth satellites", Israel program for Scientific Translations (1967)
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• Celestial mechanics — is the branch of astrophysics that deals with the motions of celestial objects. The field applies principles of physics, historically classical mechanics, to astronomical objects such as stars and planets to produce ephemeris data. Orbital… … Wikipedia | 9,732 | 28,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 201, "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} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.934954 |
http://openwetware.org/index.php?title=User:Hussein_Alasadi/Notebook/stephens/2013/10/03&curid=142125&oldid=739261 | 1,469,484,674,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00286-ip-10-185-27-174.ec2.internal.warc.gz | 183,152,781 | 7,199 | analyzing pooled sequenced data with selection Main project page
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## Notes from Meeting
Consider a single lineage for now.
Xj = frequency of "1" allele at SNP j in the pool (i.e. the true frequency of the 1 allele in the pool)
• Data:
$(n_j^0, n_j^1)$ = number of "0", "1" alleles at SNP j ($n_j = n_j^0 + n_j^1$)
• Normal approximation
$n_j^1$ ~ $Bin(n_j, X_j) \approx N(n_jX_j, n_jX_j(1-X_j))$ Normal approximation to binomial
$\frac{n_j^1}{n_j} \approx N(X_j, \frac{X_j(1-X_j)}{n_j})$ The variance of this distribution results from error due to binomial sampling.
To simplify, we just plug in $\hat{X_j} = \frac{n_j^1}{n_j}$ for Xj
$\implies \frac{n_j^1}{n_j} | X_j \approx N(X_j, \frac{\hat{X_j}(1-\hat{X_j})}{n_j})$
• notation
fi,k,j = frequency of reference allele in group i, replicate and SNP j.
$\vec{f_{i,k}} =$ vector of frequencies
Without loss of generality, we assume that the putative selected site is site j = 1
• Model
We assume a prior on our vector of frequencies based on our panel of SNPs (M) of dimension 2mxp
$\vec{f_{i,k}}$ ~ MVN(μ,Σ)
$\mu = (1-\theta)f^{panel} + \frac{\theta}{2} 1$
$\Sigma = (1-\theta)^2 S + \frac{\theta}{2}(1 - \frac{\theta}{2})I$
where $S_{i,j} = \sum_{i,j}^{panel}$ if i = j or $e^{-\frac{\rho_{i,j}}{2m} \sum_{i,j}^{panel}}$ if i not equal to j
$\theta = \frac{(\sum_{i=1}^{2m-1} \frac{1}{i})^{-1}}{2m + (\sum_{i=1}^{2m-1} \frac{1}{i})^{-1}}$
• at selected site
$log \frac{f_{i,k,1}}{1-f_{i,k,1}} = \mu + \beta g_i + \epsilon_{i,k}$
• conditional distribution
(fi,k,2,....,fi,k,p) | fi,k,1,M ~ $MVN(\bar{\mu}, \bar{\Sigma})$ The conditional distribution is easily obtained when we use a result derived here.
let X2 = (fi,k,2,....,fi,k,p) and X1 = fi,k,1
X2 | X1,M ~ $N(\vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})$
Thus $\bar{\mu} = \vec{\mu_2} + \Sigma_{21} \Sigma_{11}^{-1} (x_1 - \mu_1), \bar{\Sigma} = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}$
• Likelihood for frequency a the test SNP t given all data
let $f_{obs} = \prod_{j \not= t} f_{i,k,j}$
$L(f_{i,k,t}^{true}) = P(f_{obs} | f_{i,k,t}^{true}, M) = \frac{P( f_{i,k,t}^{true} | M, f_{obs}) P(f^{obs}|M)}{P(f_{i,k,t}^{true} | M)}$
where $f_{i,k,t}^{true} | M$ ~ N(μ,σ2Σ) The parameter σ2 allows for over-dispersion
where fobs | M ~ Np − 122Σ22 + ε2I) where ε2 allows for measurement error.
and I don't understand $f_{obs} | f_{i,k,t}^{true}, M$. Shouldn't it come from (2.12) and not (2.13) - ask Matthew | 978 | 2,533 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "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} | 3.578125 | 4 | CC-MAIN-2016-30 | latest | en | 0.683837 |
https://retrocomputing.stackexchange.com/questions/6815/why-was-2127-%E2%88%92-1-an-interesting-problem-for-baby/6818 | 1,718,887,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00887.warc.gz | 414,649,164 | 43,767 | # Why was 2^127 − 1 an interesting problem for "Baby"?
The BBC News article The 'Baby' that ushered in modern computer age along with the short embedded video begins with:
A machine that took up an entire room at a laboratory in Manchester University ran its first program at 11am on 21 June 1948.
The prototype completed the task in 52 minutes, having run through 3.5 million calculations.
The Manchester Baby, known formally as the Small-Scale Experimental Machine, was the world's first stored-program computer.
It paved the way for the first commercially-available computers in a city known for centuries of science and innovation.
A vignette within the short video shows a clip of someone writing the expression 2127 − 1 on a chalk board.
Why would an expression like this be a good test of an early stored-program computing device? It of course would not be the only test, and at some point perhaps computing pi would have also been tried, but why might this have been one important test?
• it depends on how the answer to `2^127-1` is solved ... if directly by outputing the number in binary then there is no reason to test this at all as it is just `127 ones` but if the output is transformed into decadic then you need to perform a lot of integer arithmetics which would test all the basic instructions of early computer... Commented Jun 22, 2018 at 9:42
• The Mersenne prime \$M_{127}=2^{127}-1\$ was the largest known prime from 1876 to 1951. May be they used it to test a prime proving algorithm or some such? Commented Jun 23, 2018 at 5:45
• 2^127 - 1 isn't even an equation... Commented Jun 23, 2018 at 12:57
• Just as a note, by the way, as to your comment about pi: for primitive computers, calculating multiplication, addition, and subtraction is a much more logical problem than calculating division, which calculating Pi requires. Besides being fundamentally more computationally intensive, division can require arbitrary levels of precision... a simple division problem just a few digits long can yield a real number result with any number of decimal places (for instance, 7/34 or 1/99). Just speculating, but I have to imagine that presented a formidable technical problem to 1948's computing technology. Commented Jun 24, 2018 at 5:30
• Also the first test for Microsoft Basic was "print 2+2". Apparently that exercised so much of the internal parts that that made them confident they got it right Commented Nov 24, 2020 at 17:16
It's hard to know for sure with just that picture, but that's a Mersenne number and a prime one at that.
When a mathematician writes a Mersenne number on a board they are not thinking about getting all the digits. By far the most interesting and likely question computationally speaking would have been "prove that the number is prime". That could be used as a test since that number was known via e.g. Lucas–Lehmer primality test that indeed it was prime (Mersenne.org says this one was found by Lucas himself, 1876).
However, running a test program that is not supposed to find a result is not necessarily the most robust of tests, because there are many ways the program could fail and still not find anything. Still, it might be helpful if they were able to monitor the progress steps as it ran.
That assumes the question was really used as a reliability test at all. I think it's just as likely that it was put on the board to show the newsreel crew "what kind of problems are of interest to mathematicians, which this machine would be able to resolve a lot faster than humans could". The scientists might have explained how Lucas needed X days/weeks/months to prove that the number is prime but the machine could prove it (e.g. by computing the Lucas-Lehmer test) in seconds/minutes.
Take that and the rest of the reel to the editing room, and it can quickly morph into the narration we are presented with.
• Thank you for your explanation, this makes a great deal of sense!
– uhoh
Commented Jun 22, 2018 at 14:19
• Nice idea, but I seriously doubt that anything even close to a proof for a number bing prime can be implemented in a computer with only 32 words of storage. Thats for data and program. Lucas-Lehmer requires several variables, some of large size. Even if they fit, there would be no space left for the program. Similar for the program. With only 32 program cells, there is no room for complex operations - further complicated since the only ALU functions the Baby could do where subtractions, negation and test for a negative number. The Baby was a proof of concept, not anything like a full system Commented Jun 22, 2018 at 17:09
• @uhoh The Mersenne prime problem was actually the first problem for the Manchester Mark I, which was developed from the Baby. en.wikipedia.org/wiki/Manchester_Mark_1 It's more likely that this made the news than the Baby. Commented Jun 22, 2018 at 23:35
• @uhoh I think Davids comment (see below my answer) might as well have solved this - the picture shown isn't from the introduction of the Baby, but from a news reel about the much later Manchester Mark 1, a greatly enhancement of the Baby (up to 128 instructions on tube and 8000 on drum plus punch tape). The Mark 1 could have been able to solve the mersene problem for 2^127-1 Commented Jun 22, 2018 at 23:37
• Aha, that makes a lot more sense. I did misread the article as saying 1024 bytes, not bits, which should have been enough for the problem; so I didn't notice just how limited the machine was (but I should have realized. Mea culpa.) Commented Jun 23, 2018 at 3:20
Update:
It seems as if David Marshall has found the missing link:
The footage used in the clip is not just from the Manchester Baby, but also from the Manchester Mark 1 which became operational about a year after. As noted its first (useful) program was in fact a proof for Mersenne primes, a perfect fit for the chalkboard scene.
Looks like the combined effort has revealed that this video features less than exact wording (equation vs. expression) and a mixture of pictures, not all belonging to the topic named.
First of all, the clip does not state that this problem was used, especially not used as a test for reliability.
Next a short look at the Baby's Wiki entry would have revealed that one of the first problems used - and indeed used to prove reliability - was to find the highest (all) dividers for a rather large, but still manageable number, in this case 218:
The first of three programs written for the machine found the highest proper divisor of 218 (262,144), a calculation that was known would take a long time to run—and so prove the computer's reliability—by testing every integer from 218 downwards, as division was implemented by repeated subtraction of the divisor. The program consisted of 17 instructions and ran for 52 minutes before reaching the correct answer of 131,072, after the Baby had performed 3.5 million operations (for an effective CPU speed of 1.1 kIPS).
It's all about what to test. In this case it might be not only correct workings, but more about reliability over longer durations. Doing one operation right isn't a big deal and can be tested while building. Doing it a million times under program control is.
Useful programs need to fit several criteria; some may be:
1. Finish
2. Finish within a reasonable time
3. Fit into program space
4. Use as few instructions as possible
and, most importantly
1. Have a well known answer
While #1 seems obvious and #2 simple, it isn't, as it needs a problem that scales well, so its input parameter can be used to adjust (expected) runtime. A problem solved in a few seconds doesn't test the machine well; one running for days may be already way outside even the most optimistic guesses about reliability. Being a 32-bit machine, anything between 20 and 231 may have been usable - the latter one with a runtime of maybe a year :))
Similarly, #3, where again #4 is a crucial restriction, as it is less of an issue for a small program to prove its correctness as a transformation of the problem.
And yes, #5 is really obvious. In this case it was perfect as the answers were not only well known, but (back then) known to everyone and a pain in the ass - hence large volumes of books just listing divisors for numbers :))
A problem like finding all divisors or the largest divisor fits well to all of these points and even offers secondary benefits like being repetitive but with changing data.
A vignette within the short video shows a clip of someone writing the expression 2127 - 1 on a chalk board.
Well, as usual with such an out-of-context picture, multiple explanations are possible.
First `2127-1` is not a problem but a number (the largest number that can be written using 127 bits). And we can only guess what it should mean. Then there is a question mark below, so maybe that is part of the notation? As in "What's the largest divisor for 2127-1"?
Then there's a separate section in the Wiki article naming the first three programs, with the third being one for long divisions (written by Turing), so maybe the above describes exactly this program as in allowing to "dividing (anything up to) 2127-1 by some number" (*1). This would, of course, mean that the pictures taken where not from first operation but at least 3-4 weeks later (*2) when Turing wrote that program (*3). Above clip being obviously edited and merged from original reels and pictures, this might be quite possible.
Or it is, as so often when media report about something, a staged setup. Just a scientist writing something on a chalkboard to make it look good. After all, the reporter and film crew may have had - as so often - no idea what it's about and were just looking to make it work on the screen. The text "...equations like this..." might be another hint, as it's obviously not an equation - journalists often use words they believe to be fitting but not really words that describe what's shown or happening. Remember how often some stupid header files are dumped on screen when there is some reporting about programming or anything computer related? Not a new invention.
Make your pick, I would name the second one to be true - but I can't rule out the last one.
*1 - 2127-1 would be the largest number that can be packed into four 32-bit words while still leaving one bit for a sign or other markings (like NaN). Again a little hint on multi-word arithmetic as Turing obviously programmed. Of course, without more detailed information about how many words that program used (or maybe a variable number) for its 'BigNum' representation, it is just a little hint supporting this explanation.
*2 - Back then that was well within the news cycle :))
*3 - More support for this comes by the fact be that it is hard to imagine that any other, more complex problem (like proving a Mersenne prime) is described, as such would clearly go beyond the abilities of the Manchester Baby. It was only a demonstrator with 32 words of memory, usable either as program or data storage. Already the (rather simple) first program used up 25 of them (17 instructions, 8 data).
Further the machine offered only 3 ALU instructions (subtraction, negation and test for a negative number), so any higher function must be replaced by looping around these. I can see no chance to implement something as complicated as a Lucas Lehmer Test.
• "...equations like this" is shown in the screen shot I've included, I can adjust the wording of the question to reflect that if you'd like.
– uhoh
Commented Jun 22, 2018 at 9:49
• The BBC has a bit of history with computers itself, so I wouldn't reflexively assume the video clip is poorly researched or thoughtlessly edited just because it doesn't mirror a Wikipedia article. Let's see what others have to say.
– uhoh
Commented Jun 22, 2018 at 10:31
• To start with, what's shown is not an equation but a number and a questionmark in two lines with no visible relation. So much for 'BBC' has knowledge. Beside, THE BBC has no knowledge, it got employees like any other company, in this case journalists doing their work like anywhere else. They are selected for their ability to tell stories, handle their equipment and finish in time. This hasn't anything to do with 'poor' research. Especially when considering, that you assume something of this clip that has not been said that way. Not the editors fault. Commented Jun 22, 2018 at 10:45
• According to Early British Computers by Simon Lavington (Manchester University Press, 1980), this machine was built to test the Williams tube storage. It was later expanded to become the Manchester Mark I. It is stated that the first realistic problem that it solved was an investigation into Mersenne primes run in April 1949. The footage the BBC used was probably from around this time rather than 1948. Commented Jun 22, 2018 at 23:24 | 2,927 | 12,844 | {"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} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.956715 |
http://what-when-how.com/Tutorial/topic-3163cq2gd/HTML5-Canvas-439.html | 1,545,036,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00064.warc.gz | 314,130,810 | 4,021 | HTML and CSS Reference
In-Depth Information
function checkForExtraShip() {
if (Math.floor(score/extraShipAtEach) > extraShipsEarned) {
playerShips++
extraShipsEarned++;
}
}
We call this function on each frame tick. The player earns an extra ship if the score/
extraShipAtEach variable (with the decimals stripped off) is greater than the number of
ships earned. In our game, we have set the extraShipAtEach value to 10000 . When the
game starts, extraShipsEarned is 0 . When the player's score is 10000 or more, score/
extraShipAtEach will equal 1 , which is greater than the extraShipsEarned value of 0 . An
extra ship is given to the player, and the extraShipsEarned value is increased by 1 .
Applying Collision Detection
We will be checking the bounding box around each object when we do our collision
detection. A bounding box is the smallest rectangle that will encompass all four corners
of a game logic object. We have created a function for this purpose:
function boundingBoxCollide(object1, object2) {
var left1 = object1.x;
var left2 = object2.x;
var right1 = object1.x + object1.width;
var right2 = object2.x + object2.width;
var top1 = object1.y;
var top2 = object2.y;
var bottom1 = object1.y + object1.height;
var bottom2 = object2.y + object2.height;
if (bottom1 < top2) return(false);
if (top1 > bottom2) return(false);
if (right1 < left2) return(false);
if (left1 > right2) return(false);
return(true);
};
We can pass any two of our game objects into this function as long as each contains
x , y , width , and height attributes. If the two objects are overlapping, the function will
return true . If not, it will return false .
The checkCollision() function for Geo Blaster Basic is quite involved. The full code
listing is given in Example 8-12 . Rather than reprint it here, let's examine some of the
basic concepts.
One thing you will notice is the use of “labels” next to the for loop constructs. Using
labels such as in the following line can help streamline collision detection:
rocks: for (var rockCtr=rocksLength;rockCtr>=0;rockCtr--){
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Sep15 asked What is the value of the given expression? Sep6 asked Greatest possible integer value of x+y? Jan30 asked Algorithm to find the numbers expressible as the sum of two positive cubes in two different ways Nov17 asked Find the probablity that at least 2 balls are placed in corresponding number boxes? Nov4 asked Find the maximum radius of a incircle? Nov2 asked $\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this? Oct22 asked Chances of avoiding the diagonal Oct20 asked Find the number of values of $a$? Oct19 asked Find the value of f(343, 56)? Oct10 asked How many ways balls can be selected? Oct9 asked Find the remainder when $10^{400}$ is divided by 199? Oct4 asked How many ways he can attempt the paper? Sep22 asked What is the remainder when $4^{100}$ is divided by 6? May7 asked Find the probablity given the condition? Apr22 asked Finding the measure of $\angle AEB$ given a figure Apr22 asked Find the radius of the circle? Apr17 asked Find The number of ways? Apr17 asked How many numbers? Apr16 asked Find the sum of the roots of the equation? Apr5 asked Find the perimeter of the parallelogram? | 304 | 1,194 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-18 | longest | en | 0.929901 |
https://numbermatics.com/n/71990/ | 1,513,149,692,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522205.7/warc/CC-MAIN-20171213065419-20171213085419-00260.warc.gz | 645,119,007 | 8,815 | # 71990
## 71,990 is an even composite number composed of four prime numbers multiplied together.
71990 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 71990:
### 2 × 5 × 23 × 313
See below for interesting mathematical facts about the number 71990 from the Numbermatics database.
### Names of 71990
• Cardinal: 71990 can be written as Seventy-one thousand, nine hundred ninety.
### Scientific notation
• Scientific notation: 7.199 × 104
### Factors of 71990
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 343
### Divisors of 71990
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 135648
• Sum of proper divisors (its aliquot sum) s(n): 63658
• 71990 is a deficient number, because the sum of its proper divisors (63658) is less than itself. Its deficiency is 8332
### Bases of 71990
• Binary: 10001100100110110 2
### Squares and roots of 71990
• 71990 squared (719902) is 5182560100
• 71990 cubed (719903) is 373092501599000
• The square root of 71990 is 268.3095227531
• The cube root of 71990 is 41.5997503683
### Scales and comparisons
How big is 71990?
• 71,990 seconds is equal to 19 hours, 59 minutes, 50 seconds.
• To count from 1 to 71,990 would take you about nineteen hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000.
Note: we do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!
• A cube with a volume of 71990 cubic inches would be around 3.5 feet tall.
### Recreational maths with 71990
• 71990 backwards is 09917
• The number of decimal digits it has is: 5
• The sum of 71990's digits is 26
• More coming soon!
The information we have on file for 71990 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 71990, math, Factors of 71990, curriculum, school, college, exams, university, STEM, science, technology, engineering, physics, economics, calculator. | 677 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-51 | latest | en | 0.858964 |
http://www.algebra.com/algebra/homework/word/numbers/ | 1,369,495,282,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705957380/warc/CC-MAIN-20130516120557-00070-ip-10-60-113-184.ec2.internal.warc.gz | 309,242,179 | 7,499 | # Word Problems: Numbers, consecutive odd/even, digits
Algebra -> Algebra -> Customizable Word Problem Solvers -> Numbers -> Word Problems: Numbers, consecutive odd/even, digits Log On
Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Quiz In Depth
Welcome to the number word problems section! The dreaded problem:
``The sum of an integer number and the next number is 3. Find the number.''
Sounds familiar?
Word problems with consecutive numbers and word problems with digits usually are the first word problems that teachers give to their algebra students. This is so, because consecutive number problems are the easiest word problems. If you find them difficult, you need to work on them now, before word problems get even more complicated!
The point of almost any school word problem is to make you write an equation. It is easy to do, once you understand the basics.
I wrote a special sum of consecutive numbers solver, and a companion lesson that can solve any problem with a sum of consecutive numbers. Try it out, it will really explain how these problems are done.
Another popular (among teachers, that is) number word problem involves word problems with digits. I wrote a lesson about those problems, to hopefully make it easier to solve them.
To continue, click on number word problem lessons, or on number word problem solvers to see more examples solved.
Algebra.Com is a people's math website. It relies on volunteers like you, who create our free content. Creating lessons is easy! Look for other lessons and click on 'view source'. All you have to really know is math. | 389 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2013-20 | latest | en | 0.905216 |
https://au.mathworks.com/matlabcentral/profile/authors/5153703-ildeberto-de-los-santos-ruiz | 1,582,649,733,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00548.warc.gz | 283,052,583 | 21,499 | Community Profile
# Ildeberto de los Santos Ruiz
### Tecnológico Nacional de México
20 total contributions since 2019
Estudiante de doctorado y profesor en el TecNM, adscrito al Departamento de Ciencias Básicas del Instituto Tecnológico de Tuxtla Gutiérrez.
#### Ildeberto de los Santos Ruiz's Badges
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how to write a user defined function that calculates the maximum or minimum of a quadratic equation of the form
Try this: --------------------- f = @(x) x^2-5*x+6; initial = 0; fminsearch(f,initial) --------------------- To find a max...
4 months ago | 0
Why power function give a complex number result?
The "problem" appears when you raise the base (-x / x50) to power n. In IEEE floating-point computations: a ^ n = exp(n * log(...
4 months ago | 0
Haw can I calculate the difference between two points in a plot?
Try this: --------------------------------------- plot(...) [x,y] = ginput(2); % clic on each point dx = diff(x) dy = diff(...
4 months ago | 0
| accepted
Adding semicolon and comma and saving it into .txt file
Try this low-level code: ----------------------------------------- x = randi(9,[1,2*6001]); % ACTUAL DATA fileID = fopen('da...
4 months ago | 0
| accepted
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epanet_map
Draws an EPANET network in a new MATLAB figure using geographic axes and satellite maps.
Submitted
PCAMOD
Data-driven process modeling for monitoring purposes using PCA.
Submitted
getfigdata
This function extracts the x-y data from a FIG file.
Submitted
figure2pdf
Export the current MATLAB figure to a PDF file.
Submitted
Moody
Plots the Moody chart.
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Water Properties
Computes some physical properties of water dependent on temperature, from 0.1 to 100 celsius.
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Hurwitz Matrix
Computes the Hurwitz matrix for a given polynomial. The principal minors are also computed.
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Colebrook
Computes the friction factor in pipes for given values of the Reynolds number (Re) and the relative roughness coefficient (epsil...
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Sensor calibration using Procrustes analysis
Sensor calibration using Procrustes analysis (linear transformation).
Submitted
Steady-state (final) response of linear dynamic systems under constant input.
Submitted
Trapezoidal
Simulation of linear dynamic systems using trapezoidal integration method with constant time step.
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Jacobians
Linear approximation matrices of a non-linear dynamic model.
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Equilibrium Point
Equilibrium point of a non-linear dynamic system.
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Heart
Patch with heart shape. | 601 | 2,576 | {"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-2020-10 | latest | en | 0.617628 |
http://lemon.cs.elte.hu/egres/open/Constructive_characterization_of_dumpy_graphs | 1,532,034,286,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00235.warc.gz | 210,768,484 | 7,547 | # Constructive characterization of dumpy graphs
Find a constructive characterization of k-dumpy graphs.
## Remarks
For the case when k=2 the following constructive characterization can be proved [1]:
Every 2-dumpy graph can be built up from three parallel edges by the consecutive application of the following three steps:
(i) adding a new node to the graph with two non-parallel edges entering it,
(ii) subdividing an existing edge with a new node and adding a new edge entering it so that no parallel edges arise,
(iii) reorienting a dicircuit.
There is an example where step (iii) cannot be omitted from the construction.
The k=2 case has a motivation from rigidity. Let G=(V,E) be a two-dimensional minimally rigid graph. By adding an extra node r to G and connecting it to nodes x and y with one and two edges, respectively, one gets a graph that has a rooted 2-edge connected orientation from r (this can be proved easily using that G is a (2,3)-graph). Such an orientation gives a 2-dumpy graph. Hence the constructive characterization of dumpy graphs for the case k=2 gives a new proof of the Henneberg construction. Not surprisingly, the proof relies on the same ideas as the one for the undirected case.
## References
1. K. Bérczi and E.R. Kovács, Constructive characterization of dumpy digraphs, EGRES Quick Proof no. 2011-05 | 311 | 1,347 | {"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} | 2.859375 | 3 | CC-MAIN-2018-30 | latest | en | 0.889761 |
http://timescalewiki.org/index.php?title=Delta_derivative_of_product_(2)&oldid=1291 | 1,571,235,081,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00007.warc.gz | 192,114,302 | 5,376 | # Delta derivative of product (2)
Let $\mathbb{T}$ be a time scale and $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ delta differentiable. Then the product function $fg$ is delta differentiable with $$(fg)^{\Delta}(t)=f^{\Delta}(t)g(\sigma(t))+f(t)g^{\Delta}(t),$$ where $\sigma$ denotes the forward jump. | 95 | 307 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-43 | latest | en | 0.59221 |
https://theammbook.org/formulas/concentrated/ | 1,679,549,334,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00771.warc.gz | 650,262,582 | 6,805 | # Concentrated
restricted price-range pool with excess collateral removed, ensuring that the entire collateral is available for prices in the specified range
## Key formulas
##### Characteristic function
$$f(x,y) = k = \sqrt{(x+\Delta x)\cdot(y+\Delta y)}$$
In the above formula, $\Delta x, \Delta y$ are “virtual” token balances, ie tokens that are accounted for in the AMM mechanics but that do not actually exist. This is possible as long as those virtual tokens never leave the AMM, meaning that its operations need to be restricted to a certain range. The numeraire $y$ leaves the AMM when prices go down, so if $\Delta y > 0$ then there must be a lower limit to the price. Similarly, if $\Delta x > 0$ there must be an upper price limit.
Note that range-bound AMMs do not necessarily need to operate with virtual tokens, aka leverage. However, there is not usually a good reason to keep collateral in the pool that can never be used.
##### Indifference curve
$$y_k(x) = \frac{k^2}{x+\Delta x} - \Delta y$$
Within the range the indifference curve is the constant product indifference curve, except that it is shifted by the virtual balances $\Delta x, \Delta y$ respectively. Assuming all excess liquidty is removed, ie both $\Delta x, \Delta y$ are at their maximum possible values, the range goes from $x=0$ to $y=0$, ie the point where it crosses the x-axis.
##### Price response function
$$\pi_k(x) = \frac{k^2}{(x+\Delta x)^2}$$
The PRF is that of the constant product AMM except that the virtual balance $\Delta x$ is added to the actual balance. As $x$ can not go below zero this means the the PRF is bounded from above at $\frac{k^2}{\Delta x^2}$. It is also bounded from below because $x<x_{\mathrm{max}}$ where the latter is determined by the requirement that $y > 0$.
##### AMM portfolio value (below range)
$$\nu(\xi) = n_0 \cdot \xi \ \ \ \forall \xi < \xi_0$$
Below the range, the AMM is 100% invested in the risk asset. Therefore its value is proportional to the price of the risk asset $\xi$, with a suitable portfolion notional factor $n_0$.
##### AMM portfolio value (above range)
$$\nu(\xi) = n_0 \cdot \sqrt{\xi_0 \xi_1} \ \ \ \forall \xi > \xi_1$$
Above the range, the AMM is invested in the numeraire asset. Therefore its price is constant. The factor $\sqrt{\xi_0 \xi_1}$ represents the divergence loss suffered when crossing the range
##### AMM portfolio value (in range)
$$\nu(\xi) = n_0 \cdot \left( \sqrt{\xi_0 \xi_1} \cdot \frac{\sqrt{\xi}-\sqrt{\xi_0}}{\sqrt{\xi_1}-\sqrt{\xi_0}} + \sqrt{\xi_0 \xi} \cdot \frac{\sqrt{\xi_1}-\sqrt{\xi}}{\sqrt{\xi_1}-\sqrt{\xi_0}} \right) \ \ \ \forall \xi_0 < \xi < \xi_1$$
Within the range, the AMM portfolio value is given by the above formula. Note that whilst it looks complex it is essentially of the form $a\sqrt \xi + b$, ie it is affine in $\xi$, and the complexity lies solely in the constants $a,b$. For $\xi_0 \to 0, \xi_1 \to \infty$ we recover our standard $\sqrt \xi$ shape, provide we adjust the notional $n_0$ to account for the different normalization.
##### Risk asset holdings
$$N_r(\xi) = n_0 \cdot \sqrt{\frac{\xi_0}{\xi}} \cdot \frac{\sqrt{\xi_1}-\sqrt{\xi}}{\sqrt{\xi_1}-\sqrt{\xi_0}} \ \ \ \forall \xi_0 < \xi < \xi_1$$
Below the range, the AMM is 100% invested in the risk asset, and above it, it holds none. Within the range the risk asset holdings (the Black Scholes “Delta”) is given by the above formula. The cash invested into the risk asset (“Cash Delta”) is obtained by multiplying the above with $\xi$, so the leading term becomes $n_0 \cdot \sqrt{\xi_0 \xi}$
##### Numeraire holdings
$$N_n(\xi) = n_0 \cdot \sqrt{\xi_0 \xi_1} \cdot \frac{\sqrt{\xi}-\sqrt{\xi_0}}{\sqrt{\xi_1}-\sqrt{\xi_0}} \ \ \ \forall \xi_0 < \xi < \xi_1$$
Above the range, the AMM is 100% invested in the numeraire asset, and below it, it holds none. Within the range the cash held in the numeraire asset is given by the above formula.
##### Divergence loss
$$\Lambda(\xi) = N_n(\bar\xi) + \xi \cdot N_r(\bar\xi) - \nu(\xi)$$
As usualy the DL is defined as the difference between the current value of the initial portfolio and the value of the AMM portfolio. Initially we have, by definition, $\xi = \bar\xi$ and therefore the portfolio composition is $N_n(\bar\xi)$ units of the numeraire, and $N_r(\bar\xi)$ units of the risk asset. The current value of the latter is $\xi N_r(\bar\xi)$, and the value of the AMM portfolio is $\nu(\xi)$, which yields the above formula.
The issue with DL in the concentrated liquidity case is two-fold. Firstly there is the question whether the initial portfolio is really the right benchmark, and secondly whether DL outside the range (when the AMM does not earn any fees) should be counted, or whether this is simply a suboptimal strategy on behalf of the LP. There is no general answer to those questions, they need to be solved in the situation at hand.
Whilst reasonable care has been taken to verify the above formulas they may still contain errors. Please do not use them without independent verification. | 1,420 | 5,041 | {"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} | 3.4375 | 3 | CC-MAIN-2023-14 | latest | en | 0.874255 |
https://www.statistics-lab.com/category/%E9%87%91%E8%9E%8D%E6%A8%A1%E5%9E%8B%E4%BB%A3%E5%86%99/ | 1,695,728,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00381.warc.gz | 1,122,823,780 | 45,719 | ## 金融代写|金融模型代写Modelling in finance代考|BFW3540
statistics-lab™ 为您的留学生涯保驾护航 在代写金融模型Modelling in finance方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写金融模型Modelling in finance代写方面经验极为丰富，各种代写金融模型Modelling in finance相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 金融代写|金融模型代写Modelling in finance代考|Discounting curve
The multi-curve framework starts with the description of the fundamental curve, which is the first of the multiple curves. The starting point is the discounting of known cash-flows. This is the first hypothesis on which the multi-curve framework is based. The cash-flows considered are credit risk-free cash-flows. The impact of collateral on the framework is discussed in Chapter 8.
D: The instrument paying one unit of currency $X$ in $u$ is an asset for each $u$ and each currency $X$. Its value in $t$ is denoted $P_X^D(t, u)$. The value is continuous in $t$.
With these curves we are able to value fixed cash-flows. The discounting curves are functions $P_X^D:[0, T] \rightarrow(0,+\infty)$. To be able to model curves with the usual tools of quantitative finance one needs to fix an upper bound on the time frame on which the modelling is done. This is described in, for example, (Hunt and Kennedy, 2004, Section 7.4.4). This is why we impose an upper bound $T$. If the bound $T$ is taken sufficiently large, it will have no impact in practice.
We do not restrict discount factors to be below 1 as we want to allow negative rates. Similarly we do not impose that $P_X^D$ is monotonically decreasing in the second variable. In this sense the treatment presented here is more general than the one presented in (Andersen and Piterbarg, 2010, Section 6.1.1). We restrict only the discount factors to be positive. We want to work in an arbitrage-free world. As the payment at maturity $u$ is $1>0$, its present value today $t$, represented by the discount factor $P_X^D(t, u)$, should be strictly positive.
## 金融代写|金融模型代写Modelling in finance代考|Forward curves
Our goal is to price Ibor- and overnight-related derivatives, in particular IRSs and OISs. We need an hypothesis to say that these instruments exist in the framework we are describing.
In this section, we adopt the misuse of language of calling instruments Ibor coupons even if the underlying index is not strictly speaking an Ibor index, but an index playing the same role or an overnight index.
As the period addition, $t+$ period $j$, is used often we adopt the notation $t+j$ for that date, without clarifying in which unit the $j$ is; it is usually clear from the context.
Our existence hypothesis for the Ibor coupons reads as
$\mathbf{I}^{\text {CPN }}$ : The value of a $j$-Ibor floating coupon is an asset for each tenor $j$, each fixing date and each currency. Its value is a continuous function of time.
The subscript CPN stands for CouPoN. We will use the same notation convention for the different quantities related to the Ibor coupons.
This hypothesis is implicit in most of the literature. It is important to state it explicitly as this is not a consequence of the existence of the discounting curve. This hypothesis is one of the foundations described in the introduction.
Once we have assumed that the instrument is an asset, we can give its value a name. We do this indirectly through the curves $F_X^j$ below. These curves are called forward curves, projections curves or estimation curves in the literature. We will use the term forward curves most of the time. At this stage we insist that the curves $F_X^j$ are pure definition and the use of the word curve should be understood in the mathematical sense, not in the financial sense.
Definition 2.1 (Forward coupon rate). The forward curve $F_X^{\mathrm{CPN}, j}$ is the continuous function such that,
$$P_X^D(t, v) \delta F_X^{\mathrm{CPN}, j}(t, u, v)$$
is the price in $t$ of the j-Ibor coupon with fixing date $t_0$, start date $u$ and maturity date $v\left(t \leq t_0 \leq u=\operatorname{Spot}\left(t_0\right)<v\right)$.
The reason for this definition is to keep the usual formulas involving forward rate computation ${ }^1$. The standard terms of discounting curve and forward rate will still be used. But at this stage we insist that they are only definitions and names. One should not attach too strong a financial intuition to those mathematical objects even if we use standard financial terms to designate them.
La mathématique est l’art de donner le même nom à des choses différentes.
Henri Poincaré.
Personal translation: Mathematics is the art of giving the same name to different items.
# 金融模型代写
## 金融代写|金融模型代写在金融建模代考|折现曲线
D:在$u$中支付一单位货币$X$的工具对于每一种货币$u$和每一种货币$X$是一种资产。它在$t$中的值表示为$P_X^D(t, u)$。$t$中连续的值
## 金融代写|金融模型代写金融建模代考|正向曲线
$$P_X^D(t, v) \delta F_X^{\mathrm{CPN}, j}(t, u, v)$$
La mathématique est l’art de donner le même nom à des choses différentes。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 金融代写|金融模型代写Modelling in finance代考|BUS-F541
statistics-lab™ 为您的留学生涯保驾护航 在代写金融模型Modelling in finance方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写金融模型Modelling in finance代写方面经验极为丰富，各种代写金融模型Modelling in finance相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 金融代写|金融模型代写Modelling in finance代考|Early multi-curve framework literature
To my knowledge, the first article to propose a coherent valuation framework where the discounting is explicitly differentiated from the index forward estimation was The irony in the derivatives discounting (Henrard 2007). It was published in July 2007, just one month before the crisis started. To my knowledge there is no link between the article publication and the crisis.
The article focuses on discounting in interest rate derivatives. The starting observation is that different curves are used to value different instruments (OIS, IRS, cross-currency swaps), which is clearly incoherent and creates portfolio level arbitrages.
The approach proposed described only one Ibor curve; it is not an important feature of the approach but a description of the then reality. The approach can be used with several Ibor related curves without changes. The spread between the curves is described in a multiplicative way and is equivalent to the $\beta$ we define in Chapter 2; our notation is adapted from that used in the paper. Only the case of a constant spread is discussed in the original paper; again, it is not an important feature of the approach but more a description of the then reality.
The first instrument for which the price is discussed is the FRA and the paper provides a pricing formula, including all the hypotheses required to obtain the result. The valuation of swaps in the framework and the impact on the value and risk for not-at-par instruments is discussed. The approach is extended in a coherent way to (Short Term Interest Rate) STIR futures, caps/floors and swaptions. The Ibor fixings are clearly dissociated from the discounting rate through the introduction of an exogenous spread. The spread can be obtained from the market value of different instruments; this is equivalent to the curve building process we propose in Chapter 5.
For practical examples, the article uses an OIS-based discounting curve (called Libid in the paper). The article was probably overly simplistic, as it used only a constant spread between the curves and did not use explicitly different curves for the different Ibor tenors. Nevertheless it was an early proposal leading to today’s most commonly used approach.
The importance of the multi-curve framework, with each curve having a specific purpose, is attested by the numerous related literature that appeared in the following years. Throughout the book we have tried to cite as many sources as possible in the relevant sections. We want to give readers the opportunity to go back to the original sources. In this section we comment on a couple of the early papers which describe some particular aspects of the framework. In the literature the framework has received many names. Some of those names are: two curves, multi-curve framework, derivative tenor curves, funding-Ibor, discounting-estimation, discounting-forecast, discounting-forward, and multi-curve market. Through the book, we will use the name used in the title: multi-curve framework.
## 金融代写|金融模型代写Modelling in finance代考|Collateral and funding
The description in the first chapters of the multi-curve framework is in a credit riskfree world without the market reality of the collateral and funding issues. Even if the central theme of the book is the multi-curve framework and not the collateral and funding issues, a book on the curve framework resulting from the crisis would be missing an important part of the reality if the crisis impact on collateral and funding was not discussed at all. Chapter 8 describes some of the issues related to those features. It provides results that were initially presented in Kijima et al. (2009), Macey (2011), Pallavicini et al. (2012) and Piterbarg (2012). It also presents extensions of the above results. In particular we generalise the definition of collateral and propose results about collateral with assets themselves collateralised – called collateral square. Those results are original and were first published in Henrard (2013b). Our approach to proofs borrows heavily on Macey (2011) and Piterbarg (2012) even if we add the missing self-financing property to the original sketches.
The easiest way to read this book is to start at Chapter 1 and go to Chapter 8 , with a stop in the Appendices when required – you are allowed to eat and sleep between chapters. This is certainly not the only way to read it. In the graph below I have represented the dependency graph between the different chapters of the book. The dotted lines for the Appendices indicate where they fit best, but some readers may want to skip them entirely. The links with bidirectional arrows indicate that in some way the two chapters depend on each other and are best read in parallel if the reader has that capacity.
From the graph, the reader can see that, in the mind of the author, the most important chapters are Chapter 2 (Foundations), Chapter 5 (Curve calibration), and Chapter 8 (Collateral).
I tried to write this book in the way I would like to read it. The different axioms are clearly evidenced and for each result I have tried to be clear on which hypotheses are required. I have added as many references as possible. The reader should have the opportunity to go to the original literature and compare different approaches. I have also added a lot of cross references between sections, to allow the reader to move between sections with related subjects.
# 金融模型代写
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 金融代写|金融模型代写Modelling in finance代考|FI307
statistics-lab™ 为您的留学生涯保驾护航 在代写金融模型Modelling in finance方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写金融模型Modelling in finance代写方面经验极为丰富，各种代写金融模型Modelling in finance相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 金融代写|金融模型代写Modelling in finance代考|Foundations, evolution, and implementation
This book’s subtitle is Foundations, Evolution, and Implementation. The subtitle acts as a quick summary of the book content.
The foundations are paramount. As the framework is different from the previous one-curve approach in fundamental aspects for index-linked financial instruments, one can not rely on a ‘copy and paste’ style approach. Any claim, even the most basic one, should be backed by clear definitions and clear proofs. Before building the skyscraper that is the pricing of an exotic derivative, one has not only to be sure that the foundations are sound but also indicate explicitly where they are. In some cases, mathematically sophisticated developments are made, only to notice later that their domain of applicability is empty. Their starting hypotheses, which seem acceptable in a one-curve world, lead to contradictions for the multi-curve framework. This is why the book systematically uses an ‘axiomatic approach’. The fundamental hypotheses are displayed in the text as quotes with a bold letter in front reminding us of the content of the hypothesis – such as $\mathbf{D}$ for the discounting hypothesis at the beginning of Section 2.2. These axioms or fundamental hypotheses are the foundations of the framework. No development in interest rate modelling can be done without referring to them or to their equivalent in a different framework.
The evolution is important to understand how this new framework started and where it is coming from. Several choices made in the new framework are recycling previous approaches with appropriate and justified twists. Why and where those twists were made is an important piece of information. Looking at the final product, it may seem to have been produced by black magic. It is not any more when seen through the eyes of Darwin, as the result of evolution.
In chronological order, the evolution came before the foundations. The theory is more an explanation of the practice obtained by tinkering with different approaches than a new, ground-breaking theory that started to be used from scratch. We refer the reader to the book Antifragile by Taleb (2012) for more on how practice often precedes theory. In this book, we reverse the chronological order and start with the foundations.
Si, à l’égard de plusieurs questions traitées dans cette étude, j’ai comparé les résultats de l’observation à ceux de la théorie, ce n’est pas pour vérifier des formules établies par des méthodes mathématiques, mais pour montrer seulement que le marché, à son insu, obéit à une loi qui le domine : la loi de la probabilité.
Louis Bachelier, Théorie de la spéculation, 1900 Personal translation: If, regarding several questions analysed in this study, I compared the observed results to those of the theory, it is not to verify the formulas obtained by mathematical methods, but only to show that the market, unwittingly, complies to a law that dominates it: the law of probability.
## 金融代写|金融模型代写Modelling in finance代考|Standard textbook framework
The book by Hull (2006), one of the most popular introductory textbooks to derivative pricing, is used as an example of the way earlier literature treats the curves question. In the section on the type of rates in the Interest rates chapter (p. 76) the existence of both Libor and Libid is acknowledged. The latter is described in the above book as the rate at which a cash rich investor can invest in the interbank market. In 2006, those rates were relatively close to the OIS (Overnight Indexed Swap) rates.
When it comes to valuing the first derivatives (Section 4.7: Forward Rate Agreements) the explanation is ‘the assumption underlying the contract is that the borrowing or lending would normally be done at Libor’. This is first a misleading statement on the instrument itself. The reality is that the contract settlement amount is computed by discounting with a Libor fixing rate between the end of the accrual period and its start. There is no actual borrowing or lending and there is no assumption in the FRA ${ }^4$ contract, only a clear (contingent) settlement formula. For the valuation of the instrument before its fixing date, the approach described in the above book is to use the same Libor rate to discount the resulting quantity to the valuation date. But there is no justification in the text for choosing that particular rate from the different ones described in the previous sections. The choice of Libor for that purpose is a modelling choice and not a contractual obligation. This vagueness is certainly a witness of the consensus at that time: if the instrument is related to an Ibor rate, use similar rates for everything, even if there is no modelling or legal reason to do so. The hidden explanation is that it is easier and everybody is doing it.
Another standard textbook on swaps and curve construction is Sadr (2009). In the description of the pricing of a swap, the terms used to describe the rates linked to the Ibor leg are ‘hypothetical loan’, ‘deposits’, ‘funding’ and ‘risk-free rates’. The text indicates that to obtain the standard one-curve formula, those different rates have to be the same. Using the equivalence between those rates the book proceeds through standard arbitrage-free arguments to obtain the standard swap pricing formula.
# 金融模型代写
## 金融代写|金融模型代写在金融建模代考|基础，发展，和实现
Si, à l’égard de plusieurs questions traitées dans cette étude, j’ai comparé les résultats de l’observation à ceux de la théorie, ce n’est pas pour vérifier des formules établies par des méthodes mathématiques, mais pour montrer seulement que le marché, à son insu, obéit à une loi qui le domine : la loi de la probabilité.
Louis Bachelier, Théorie de la spéculation, 1900 Personal translation: If, regarding several questions analysed in this study, I compared the observed results to those of the theory, it is not to verify the formulas obtained by mathematical methods, but only to show that the market, unwittingly, complies to a law that dominates it: the law of probability.
## 金融代写|金融模型代写金融建模代考|标准教科书框架
.
Hull(2006)的书是衍生品定价最受欢迎的入门教材之一，它被用作早期文献处理曲线问题的一个例子。在“利率”一章的“利率类型”一节(第76页)中，承认了Libor和Libid的存在。后者在上书中被描述为现金充裕的投资者在银行间市场投资的利率。在2006年，这些利率相对接近隔夜指数掉期利率(OIS)
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 6,133 | 20,466 | {"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} | 3.640625 | 4 | CC-MAIN-2023-40 | latest | en | 0.826088 |
https://aptitude.gateoverflow.in/9192/nielit-2022-feb-scientist-d-section-d-10 | 1,659,974,870,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570868.47/warc/CC-MAIN-20220808152744-20220808182744-00068.warc.gz | 127,961,518 | 20,746 | 8 views
If $\sin 3A = \cos (A – 26^{\circ}),$ where $3A$ is an acute angle then the value of $A$ is :
1. $29^{\circ}$
2. $31^{\circ}$
3. $39^{\circ}$
4. $23^{\circ}$
1
3 views | 81 | 178 | {"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} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.617813 |
http://mathhelpforum.com/math-puzzles/205936-distance-problem-please-help-steps.html | 1,527,228,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00483.warc.gz | 190,350,205 | 9,761 | A hare pursued by a hound is 50 of her own leaps before him. When the hare takes 4 leaps, the hound takes 3. In one leap, the hare goes 1.75 metres and the hound 2.75 metres. In how many leaps will the hound overtakes the hare?
Hey, I will try to answer, so here it goes:
Hare is 50 leaps of her own before hound so lets take her initial position 50*1.75= 87.5 Meters
Now it is said that Hare takes 4 leaps and the hound takes 3 therefore 87.5 + 4(1.75)x=3(2.75)x
solve the above equation and you should get x=87.5/1.25 | 169 | 523 | {"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-2018-22 | latest | en | 0.952865 |
http://jean-pierre.moreau.pagesperso-orange.fr/Basic/compos01_bas.txt | 1,686,035,831,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00291.warc.gz | 25,743,071 | 2,402 | '************************************************************************ '* This program calculates the stresses in a unidirectional layer of a * '* composite material, knowing deformations exx, eyy, gxy, and angle * '* theta of fibers in x direction. We consider here that we have a * '* plane stress state. * '* * '* The stresses sxx, syy and sxy are calculated: * '* * '* 1. in plane (x,y), * '* 2. in the main axes (L,W) of the layer. * '* * '* -------------------------------------------------------------------- * '* SAMPLE RUN: * '* * '* STRESSES IN A UNIDIRECTIONAL COMPOSITE LAYER * '* * '* Angle of fibers: 30 deg * '* eps. xx : 10 mm * '* eps. yy : -5 mm * '* gam. xy : 20 mm * '* * '* REDUCED STIFFNESS MATRIX (MAIN AXES): * '* * '* 41.050903 3.284072 0.000000 * '* 3.284072 10.262726 0.000000 * '* 0.000000 0.000000 4.500000 * '* * '* REDUCED STIFFNESS MATRIX (X, Y AXES): * '* * '* 28.339080 8.298851 9.561120 * '* 8.298851 12.944992 3.770552 * '* 9.561120 3.770552 9.514778 * '* * '* Stresses in axes x, y: * '* * '* 433.1189 MPa * '* 93.6746 MPa * '* 267.0540 MPa * '* * '* Stresses in main axes: * '* * '* 579.5334 MPa * '* -52.7399 MPa * '* -13.4567 MPa * '* * '* END OF PROGRAM * '* -------------------------------------------------------------------- * '* Basic Release By J-P Moreau, Paris. * '* (www.jpmoreau.fr) * '************************************************************************ 'PROGRAM COMPOS01 DEFDBL A-H, O-Z DEFINT I-N DIM NULT AS DOUBLE NMAX = 3 DIM Q(NMAX, NMAX), Q1(NMAX, NMAX) DIM S(NMAX), S1(NMAX), E(NMAX) DIM Title AS STRING PI = 4# * ATN(1#) ' material constants (GPa) EL = 40# ET = 10# NULT = .32# GLT = 4.5# ' initial data TH = 30# EXX = .01# EYY = -.005# GXY = .02# CLS PRINT PRINT " STRESSES IN A UNIDIRECTIONAL COMPOSITE LAYER" PRINT PRINT " Angle of fibers: "; TH; " deg" PRINT " eps. xx : "; 1000 * EXX; " mm" PRINT " eps. yy : "; 1000 * EYY; " mm" PRINT " gam. xy : "; 1000 * GXY; " mm" ' reduced stiffness matrix in main axes (L,W) GOSUB 500 'CALL MATNUL(Q) Q(1, 1) = EL / (1# - ((ET / EL) * NULT * NULT)) Q(2, 2) = (ET / EL) * Q(1, 1) Q(1, 2) = NULT * Q(2, 2) Q(2, 1) = Q(1, 2) Q(3, 3) = GLT Title = "REDUCED STIFFNESS MATRIX (MAIN AXES):" GOSUB 600 'CALL MATPRINT("REDUCED STIFFNESS MATRIX (MAIN AXES):",Q) ' reduced stiffness matrix in axes (x,y) GOSUB 501 'CALL MATNUL(Q1) TH = TH * PI / 180# GOSUB 700: GOSUB 701 'calculate dcos4 and dsin4 TEMP = SIN(TH) * SIN(TH) * COS(TH) * COS(TH) Q1(1, 1) = Q(1, 1) * dcos4 + Q(2, 2) * dsin4 + 2# * (Q(1, 2) + 2# * Q(3, 3)) * TEMP Q1(1, 2) = (Q(1, 1) + Q(2, 2) - 4# * Q(3, 3)) * TEMP + Q(1, 2) * (dcos4 + dsin4) Q1(2, 1) = Q1(1, 2) TEMP = SIN(TH) * COS(TH) * COS(TH) * COS(TH) Q1(1, 3) = (Q(1, 1) - Q(1, 2) - 2# * Q(3, 3)) * TEMP TEMP = SIN(TH) * SIN(TH) * SIN(TH) * COS(TH) Q1(1, 3) = Q1(1, 3) + (Q(1, 2) - Q(2, 2) + 2# * Q(3, 3)) * TEMP Q1(3, 1) = Q1(1, 3) TEMP = SIN(TH) * SIN(TH) * COS(TH) * COS(TH) Q1(2, 2) = Q(1, 1) * dsin4 + 2# * (Q(1, 2) + 2# * Q(3, 3)) * TEMP + Q(2, 2) * dcos4 TEMP = SIN(TH) * SIN(TH) * SIN(TH) * COS(TH) Q1(2, 3) = (Q(1, 1) - Q(1, 2) - 2# * Q(3, 3)) * TEMP TEMP = COS(TH) * COS(TH) * COS(TH) * SIN(TH) Q1(2, 3) = Q1(2, 3) + (Q(1, 2) - Q(2, 2) + 2# * Q(3, 3)) * TEMP Q1(3, 2) = Q1(2, 3) TEMP = SIN(TH) * SIN(TH) * COS(TH) * COS(TH) Q1(3, 3) = (Q(1, 1) + Q(2, 2) - 2# * (Q(1, 2) + Q(3, 3))) * TEMP + Q(3, 3) * (dsin4 + dcos4) Title = "REDUCED STIFFNESS MATRIX (X, Y AXES):" GOSUB 601 'CALL MATPRINT("REDUCED STIFFNESS MATRIX (X, Y AXES):",Q1) S\$ = " ####.#### MPa" ' stresses in axes (x, y): E(1) = EXX E(2) = EYY E(3) = GXY GOSUB 550 'CALL MATMULT(Q1,E,S) PRINT PRINT " Stresses in axes x, y:" PRINT FOR I = 1 TO NMAX PRINT USING S\$; 1000 * S(I) NEXT I ' stresses in main axes: TEMP = SIN(TH) * COS(TH) Q(1, 1) = COS(TH) * COS(TH) Q(1, 2) = SIN(TH) * SIN(TH) Q(1, 3) = 2# * TEMP Q(2, 1) = Q(1, 2) Q(2, 2) = Q(1, 1) Q(2, 3) = -Q(1, 3) Q(3, 1) = -TEMP Q(3, 2) = TEMP Q(3, 3) = COS(TH) * COS(TH) - SIN(TH) * SIN(TH) GOSUB 551 'CALL MATMULT(Q,S,S1) PRINT PRINT " Stresses in main axes:" PRINT FOR I = 1 TO NMAX PRINT USING S\$; 1000 * S1(I) NEXT I PRINT PRINT " END OF PROGRAM" END 'of main program 500 'SUBROUTINE MATNUL(Q) FOR I = 1 TO NMAX FOR J = 1 TO NMAX Q(I, J) = 0# NEXT J NEXT I RETURN 501 'SUBROUTINE MATNUL(Q1) FOR I = 1 TO NMAX FOR J = 1 TO NMAX Q1(I, J) = 0# NEXT J NEXT I RETURN 550 'SUBROUTINE MATMULT(Q1, E, S) FOR I = 1 TO NMAX SUM = 0# FOR K = 1 TO NMAX SUM = SUM + Q1(I, K) * E(K) NEXT K S(I) = SUM NEXT I RETURN 551 'SUBROUTINE MATMULT(Q, S, S1) FOR I = 1 TO NMAX SUM = 0# FOR K = 1 TO NMAX SUM = SUM + Q(I, K) * S(K) NEXT K S1(I) = SUM NEXT I RETURN 600 'SUBROUTINE MATPRINT(Title,Q) F\$ = " ####.######" PRINT PRINT " "; Title PRINT FOR I = 1 TO NMAX FOR J = 1 TO NMAX PRINT USING F\$; Q(I, J); NEXT J PRINT NEXT I RETURN 601 'SUBROUTINE MATPRINT(Title,Q1) PRINT PRINT " "; Title PRINT FOR I = 1 TO NMAX FOR J = 1 TO NMAX PRINT USING F\$; Q1(I, J); NEXT J PRINT NEXT I RETURN 700 'FUNCTION dcos4(TH) a = COS(TH) dcos4 = a * a * a * a RETURN 701 'FUNCTION dsin4(TH) a = SIN(TH) dsin4 = a * a * a * a RETURN 'end of file Compos01.bas | 2,187 | 5,077 | {"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-2023-23 | latest | en | 0.45863 |
http://metamath.tirix.org/mpests/ax-1 | 1,723,181,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00185.warc.gz | 21,685,824 | 2,063 | # Metamath Proof Explorer
## Axiom ax-1
Description: AxiomSimp. Axiom A1 of Margaris p. 49. One of the 3 axioms of propositional calculus. The 3 axioms are also given as Definition 2.1 of Hamilton p. 28. This axiom is calledSimp or "the principle of simplification" inPrincipia Mathematica (Theorem *2.02 of WhiteheadRussell p. 100) because "it enables us to pass from the joint assertion of ph and ps to the assertion of ph simply". It is Proposition 1 of Frege1879 p. 26, its first axiom. (Contributed by NM, 30-Sep-1992)
Ref Expression
Assertion ax-1 ${⊢}{\phi }\to \left({\psi }\to {\phi }\right)$
### Detailed syntax breakdown
Step Hyp Ref Expression
0 wph ${wff}{\phi }$
1 wps ${wff}{\psi }$
2 1 0 wi ${wff}\left({\psi }\to {\phi }\right)$
3 0 2 wi ${wff}\left({\phi }\to \left({\psi }\to {\phi }\right)\right)$ | 273 | 822 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.708357 |
https://scipy.github.io/devdocs/reference/generated/scipy.interpolate.approximate_taylor_polynomial.html | 1,632,297,696,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00170.warc.gz | 538,912,644 | 5,377 | scipy.interpolate.approximate_taylor_polynomial¶
scipy.interpolate.approximate_taylor_polynomial(f, x, degree, scale, order=None)[source]
Estimate the Taylor polynomial of f at x by polynomial fitting.
Parameters
fcallable
The function whose Taylor polynomial is sought. Should accept a vector of x values.
xscalar
The point at which the polynomial is to be evaluated.
degreeint
The degree of the Taylor polynomial
scalescalar
The width of the interval to use to evaluate the Taylor polynomial. Function values spread over a range this wide are used to fit the polynomial. Must be chosen carefully.
orderint or None, optional
The order of the polynomial to be used in the fitting; f will be evaluated `order+1` times. If None, use degree.
Returns
ppoly1d instance
The Taylor polynomial (translated to the origin, so that for example p(0)=f(x)).
Notes
The appropriate choice of “scale” is a trade-off; too large and the function differs from its Taylor polynomial too much to get a good answer, too small and round-off errors overwhelm the higher-order terms. The algorithm used becomes numerically unstable around order 30 even under ideal circumstances.
Choosing order somewhat larger than degree may improve the higher-order terms.
Examples
We can calculate Taylor approximation polynomials of sin function with various degrees:
```>>> import matplotlib.pyplot as plt
>>> from scipy.interpolate import approximate_taylor_polynomial
>>> x = np.linspace(-10.0, 10.0, num=100)
>>> plt.plot(x, np.sin(x), label="sin curve")
>>> for degree in np.arange(1, 15, step=2):
... sin_taylor = approximate_taylor_polynomial(np.sin, 0, degree, 1,
... order=degree + 2)
... plt.plot(x, sin_taylor(x), label=f"degree={degree}")
>>> plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left', | 421 | 1,847 | {"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-2021-39 | latest | en | 0.763524 |
https://askfilo.com/user-question-answers-mathematics/2-show-that-the-relation-on-the-set-defined-by-d-defined-by-33373638323035 | 1,696,165,996,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00289.warc.gz | 122,886,866 | 28,966 | World's only instant tutoring platform
Question
# 2. Show that the relation on the set defined by , d) defined by the defined by is an equivalence relation. ORLet be the set of all lines in plane and be the relation in defined as is parallel to . Show that is an equivalence relation. Find the set of all lines related to the line
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Question Text 2. Show that the relation on the set defined by , d) defined by the defined by is an equivalence relation. ORLet be the set of all lines in plane and be the relation in defined as is parallel to . Show that is an equivalence relation. Find the set of all lines related to the line Updated On Jan 11, 2023 Topic Calculus Subject Mathematics Class Class 12 Answer Type Video solution: 1 Upvotes 69 Avg. Video Duration 24 min | 389 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | latest | en | 0.897671 |
http://mathhelpforum.com/calculus/35197-find-moment-inertia-circular-cone.html | 1,519,300,223,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00388.warc.gz | 226,644,710 | 10,477 | # Thread: find moment of inertia of circular cone
1. ## find moment of inertia of circular cone
Find the moment of inertia of a uniform solid circular cone of mass M, height h and base
2. Originally Posted by szpengchao
Find the moment of inertia of a uniform solid circular cone of mass M, height h and base
I'll try to get back later with more details, but the general approach is going to be to take the integral over the moment of inertia of a thin disk going from the apex to the base. Once you have that you should be able to apply the perpendicular axis theorem.
-Dan
3. ## got
Originally Posted by topsquark
I'll try to get back later with more details, but the general approach is going to be to take the integral over the moment of inertia of a thin disk going from the apex to the base. Once you have that you should be able to apply the perpendicular axis theorem.
-Dan
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# moment of inertia of a uniform solid circular cone
Click on a term to search for related topics. | 231 | 1,014 | {"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-2018-09 | longest | en | 0.90157 |
https://republicofsouthossetia.org/question/solve-the-equation-if-0-degrees-is-less-than-or-equal-to-which-is-less-than-or-equal-to-360-degr-16355591-27/ | 1,632,762,902,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00405.warc.gz | 508,181,850 | 13,665 | ## solve the equation if 0 degrees is less than or equal to x which is less than or equal to 360 degrees. cos x= -1/2
Question
solve the equation if 0 degrees is less than or equal to x which is less than or equal to 360 degrees. cos x= -1/2
in progress 0
2 weeks 2021-09-11T20:12:59+00:00 1 Answer 0 | 99 | 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} | 3.015625 | 3 | CC-MAIN-2021-39 | latest | en | 0.922839 |
https://edurev.in/course/quiz/attempt/15198_RCC-2-MCQ/33815531-0494-4a8b-84ba-494f0e50e13f | 1,606,435,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189030.27/warc/CC-MAIN-20201126230216-20201127020216-00216.warc.gz | 288,184,780 | 41,076 | Courses
# RCC 2 MCQ
## 20 Questions MCQ Test Mock test series of SSC JE Civil Engineering | RCC 2 MCQ
Description
This mock test of RCC 2 MCQ for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 20 Multiple Choice Questions for Civil Engineering (CE) RCC 2 MCQ (mcq) to study with solutions a complete question bank. The solved questions answers in this RCC 2 MCQ quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this RCC 2 MCQ exercise for a better result in the exam. You can find other RCC 2 MCQ extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1
### If the permissible compressive and tensile stresses in a single reinforced beam are 50 kg/cm2 and 1400 kg/cm2 respectively and the modular ratio is 18, the percentage of the steel required for an economic section, is
Solution:
Taking Moment of area of tension and compression zone for a singly reinforced beam:
Where,
Xc = critical depth or economic depth of neutral axis
m = modular ratio
Ast = area of steel required for balanced section
m = 18
c = 50 kg/cm2
t = 1400 kg/cm2
QUESTION: 2
### Though the effective depth of a T-beam is the distance between the top compression edge to the centre of the tensile reinforcement, for heavy loads it is taken as:
Solution:
For all cases, Effective depth of the beam cross-section is to be taken as:
QUESTION: 3
### If the diameter of main reinforcement in a slab is 16 mm, the concrete cover to main bar according to IS 456:2000 is:
Solution:
According to IS 456: 2000, minimum nominal cover to all bars in the slab is 20mm. According to new code, nominal cover does not depend upon the diameter of bar used.
QUESTION: 4
__________ is the type of concrete in which internal stresses are intentionally induced in a planned manner such that the stresses resulting from the superimposed loads get counteracted to a desired degree.
Solution:
The principle behind prestressed concrete is that compressive stresses induced by high-strength steel tendons in a concrete member before loads are applied will balance the tensile stresses imposed in the member during service.
A prestressed concrete member is one in which there have been introduced internal stresses of such a magnitude, and also distribution, that the stresses resulting from external loading are counteracted to a desired degree. By doing this, Nearly the full capacity of the concrete in compression can be used over entire depth under full loading.
QUESTION: 5
Span to effective depth ratio for a continuous beam with steel grade of Fe 415 should not be more than: -
Solution:
It is necessary to impose a check on the magnitude of deflection in a structural member:
a) To ensure that the extent of deflection does not adversely affect the appearance or efficiency of the structure or finishes or partition etc.
b) To prevent structural behaviour of the member being different from the assumption made in the design. As per IS 456:2000, for beams and slabs, the vertical deflection limits may be assumed to be satisfied, provided that the span to depth ratio are not greater than the values obtained as below:
QUESTION: 6
The working stress method of design of RC members is known as:-
Solution:
Working stress method: working stress method is based on elastic theory. Concrete and steel are assumed to act together elastically and follow hook’s law. This method follows a deterministic approach as it assumes that the loads, factor of safety and permissible stresses are accurately known. In this method material strengths are not fully utilised in designing the member.
Limit state method: This method follows a non-deterministic approach as it adopts probable loads and probable strengths of materials as per actual or based on experience or observations depending upon the situations. Material strengths are fully utilised in designing the member.
QUESTION: 7
The time in minutes, for the member to reach the limit state of structural inadequacy in a standard fire test is known as:
Solution:
As per Bureau of Indian standards,
Period of structural adequacy under fire: The time (t), in minutes, for the member to reach the limit state of structural inadequacy in a standard fire test.
QUESTION: 8
Choose the correct statement regarding reinforcement detailing:
Solution:
As per IS 456:2000,
1. Lap splicing of reinforcement bar more than 36 mm in diameter should be avoided. In case such bars have to be lapped then they should be welded. (Reference IS456 clause 25.2.5).
2. Where there are two or more rows of bars, the bars shall be vertically in line and the minimum vertical distance between the bars shall be 15 mm, two third the maximum nominal size of aggregate or maximum size of bars, whichever is greater.
3. The minimum nominal cover for footing shall not be less than 50 mm.
QUESTION: 9
Solution:
1.5 DL + 1.5 LL
1.5 DL + 1.5 (max of ELor WL)
1.2 DL + 1.2 LL + 1.2 (max of ELor WL)
0.9 DL + 1.5 EL
QUESTION: 10
The clear distance from the free end of the cantilever to the lateral restraint shall not exceed
Solution:
As per IS 456:2000, slenderness limit for lateral stability of beam shall be:
QUESTION: 11
For a square column, b × b in size, effective depth of the footing being ‘d’, the punching shear is calculated for an area
Solution:
The critical section for shear is act at a distance of d/2 from the face of the column. So, the total resisting area which resist that punching shear will be
= 4(bd + d2).
QUESTION: 12
The Design ultimate load on the short axially loaded column is computed by which of the following equation?
where,
fck = Characterstics strength of concrete
fy = characterstic strength of steel
Ac = Area of concrete in column cross-section
Asc = Area of steel in column in compression
Solution:
As per IS 456 : 2000 In the Design of compression members the ultimate load on the short axially loaded column is:
Pu = 0.4fck Ac + 0.67fy Asc
where,
fck = Characterstics strength of concrete
fy = characterstic strength of steel
Ac = Area of concrete in column cross-section
Asc = Area of steel in column in compression
QUESTION: 13
Pick up the correct statement from the following.
Solution:
Dead load is the self-weight of the structure and the load which is permanently attached to the structure and will remain throughout the lifetime of the structure.
So at the beginning of design it is known to the designer.
QUESTION: 14
For a continuous slab of 3 m × 3.5 m size, the minimum overall depth of slab to satisfy vertical deflection limits for the grade of steel FE415 is-
Solution:
Explanation : For high strength deformed bars of grade Fe415:
For mild steel reinforcement.
Continuous slab 32 ≥ L/d
Simply supported slab ≥ L/d
Continuous slab 40 ≥ L/d
where, L is the shorter span of slab
d is overall depth of the beam
32 ≥ 3000/d
d ≥ 9.37 cm = 10 cm
QUESTION: 15
In a rear counterfort retaining wall, the main reinforcement is provided on the
Solution:
Counterforts are firmly attached to the face slab as well as the base slab. The earth pressure acting on the face slab is transferred to the counterfort which deflects as vertical cantilever. The back of rear counterfort comes under tension and the front face is under compression. So the inclined (back) face of the rear counterfort should be provided with main reinforcement.
QUESTION: 16
In a RCC beam of breadth b and overall depth D exceeding 750 mm, side face reinforcement required and the allowable area of maximum tension reinforcement shall be respectively
Solution:
As per IS 456:2000:
1. Side face reinforcement in a RCC beam of breadth b and overall depth D exceeding 750 mm should be 0.1 %.
2. Maximum area of tension reinforcement should not exceed 4%.
QUESTION: 17
Deep beams are designed for
Solution:
Deep beams are structural elements loaded as simple beams in which a significant amount of the load is carried to the supports by a compression force combining the load and the reaction. As a result, the strain distribution is no longer considered linear, and the shear deformations become significant when compared to pure flexure.
In view of ample shear strength, deep beams are primarily recommended as transfer girders. These members transfer loads through loading face to supports in the transverse direction. The deep horizontal members predominantly fail in shear rather than flexure. These beams are characterized with small span-to-depth ratio. Pile caps, corbel, brackets, foundation walls and off-shore structures are few examples of RC deep beams.
According to IS 456-2000 a beam shall be deemed to be a deep beam when the ratio of effective span-to-overall depth, l/D is less than:
1) 2.0, for simply supported beam; and
2) 2.5, for a continuous beam.
Though different codes define deep beams in different clear span-to-depth ratios, as a general rule deep beams are recognized by their relatively small span-to-depth ratio.
Hence, Deep beams are designed for bending moment and checked for shear.
IS 456:2000 page 51.
QUESTION: 18
In R.C footing on soil, the thickness at the edge should not be less than
Solution:
According to the Code (CI. 34.1.2):
Code restricts the minimum thickness at the edge of the footing to 150 mm for footings in general (and to 300 mm in the case of pile caps).
QUESTION: 19
A doubly reinforced concrete beam has effective cover d’ to the centre of compression reinforcement. ‘xu’ is the depth of neutral axis, and d is the effective depth to the centre of tension reinforcement. What is the maximum strain in concrete at the level of compression reinforcement?
Solution:
QUESTION: 20
A reinforced concrete slab is 128 mm thick. The maximum size of reinforcement bar that can be used is:
Solution:
IS 456:2000 limits the maximum diameter of reinforcing bars to 1/8th of the total thickness of the slab.
Dmax = 128/8 = 16 mm; Hence use 16 mm diameter bars.
• RCC | 2,292 | 10,096 | {"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} | 3.28125 | 3 | CC-MAIN-2020-50 | latest | en | 0.883137 |
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# If (2x)y = (x2)z, the value of y when x is 16 and z is 32 is
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If (2x)y = (x2)z, the value of y when x is 16 and z is 32 is how much greater than the value of y when x and z are both 8 ?
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07 Apr 2006, 22:40
(2x)y = (x2)z
y=z hence when z is 32 y=32 and when z is 8 y=8. difference btw the two is 24
07 Apr 2006, 22:40
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# Senior Secondary (SS1) Physics Questions
These SS1 Physics questions below are followed based on the SS1 Physics scheme of work in Nigeria. And if you have any questions concerning any of the questions, kindly leave a comment.
## Measurement
1)Which of the following shows the correct derived unit for volume?
a) M^4 b) M^-2 c) M^3 d) M^-1
2) Which of the following is not Basic unit?
a) m b)NM c) kg d) k
3) Which of the following is a fundamental quantity?
a) Pressure b) heat c) luminous intensity d) Density
4) The following are Derived units except
a) kgms-2 b) Mol. c) Nm d) ms-2
5) The following are instruments used to measure time except
a) stop-watch b) pendulum clock c) wattmeter d) ticker tape timer
6) Which of the following prove the correctness of velocity using the dimensional analysis.
a) LT^-1 b)L^-1T c)LT^-2 d)LT^-3
## Position
1)Which of the following is correct?
i)Displacement is a scalar quantity
ii) Both distance and displacement are measured in meter (m).
iii) Displacement shows direction.
a) i and ii b) ii and iii c) i, ii, iii d) i and iii
2) What is the distance between point D(9,6) and point F(5,3)
a) -5 b) -15 c) 5 d) 15
3) Which of these can be used to measure the direction of motion of the body except?
a) Compass b) Grid c) Calipers d) Protractor
4) Which of the following is not correct about displacement?
a) Displacement is a scalar quantity.
b) Displacement direction can be measured using compass and grid.
c) Displacement is distance moved in a specified direction.
d) Distance has direction while displacement has magnitude only.
5) Which of the following statements is correct?
i) Distance and displacement have magnitude.
ii) Distance has direction and displacement has magnitude only.
iii) They are both measured in meters.
a) I and iii b) ii and iii c) i, ii and iv d) i, iv, and iii
## motion
1)The motion of the prongs of a sounding turning fork is
a) Vibration b) translation c)random d) vibratory and rotational
2) Which of the following is not true about motion?
a) Motion are caused by a force
b) Motion is the change of position of a body with time
c) All motion are relative motion
d) It is not practicable for an object to combine two of the motion together
3) Which of the following are random motions?
i) It is haphazardly or disorderly with no direction
ii) It is non-linear manner and changes direction continuously.
iii) Brownian motion is a random motion.
iii) The example of random is a horse running from one two to another
a) I, II& IV only b) I, II & III only c) IV, III and I d) IV, III and II only
4) A fully test-tube that floats upright in the water is carefully and slightly depressed, and the then released. Which of the following best explains the subsequent motion of the test tube?
a) Rotational b) Translational c) Circular d) Random
5) A proof of the particle nature of matter is the
a) Oscillatory b) Brownian motion c) Orbital motion d) Translational motion
## speed and velocity
1)An object moves with uniform speed around a circle. Its acceleration has:
(a) Varying magnitude and varying direction
(b) Constant magnitude and constant direction
(c) Varying magnitude and constant direction
(d) Constant magnitude and varying direction
2) The frictional force between two bodies
(a) Have all these characteristics
(b) Exists only when there is relative motion or tendency to motion.
(c) Acts so as to oppose the motion.
(d) Depends on the proper reaction between the two surfaces.
3) A wooden block of mass 1.4kg rests on a rough horizontal surface. If the surface is 9N, calculate the coefficient of friction g = 10m/s2
(a) 0.65 (b) 0.54 (c) 0.64 (d) 0.70
4) Which of the following correctly gives the relationship between linear speed and angular speed w of a body moving uniformly in a circle of radius r?
(a) v = w^2r (b) v = wr^2 (c) v = w /r (d) v = wr
5) The magnitude of the force required to make an object of mass m move with speed v in a circular path of radius r is given by the expression
(a) mv^2/r (b) mr2/v (c) mv/r^2 (d) mv/r
6) Which of the following is not a consequence of a force field?
(a) Magnetic force (b) gravitational pull c) electric force d) surface tension
7) Which of the following is not correct about friction?
a) friction can be increased using volleys, lubricants
b) friction resists the resistance motion between two surfaces in contact
c) it reduces the efficiency of machines
d) the coefficient of friction has no unit
8) What is the unit of frictional force?
9) Which of the following are contact forces?
(i) Force of tension (ii) Force of reaction (iii) Force of friction (iv) Magnetic force
(A) i, ii and iv only (B) i, ii and iii only (C) ii iii and iv only (D) i, ii only.
10) A body at the weight of 100N moves with a speed of 5mls in a horizontal circular path of the radius. Calculate the magnitude of the centripetal force acting on the body g = 10m/s2
(A) 50N (B) 100N (C) 75N (D) 10N
11) The area under the curve of a velocity-time graph represents.
(A) Instantaneous speed (B) Time spent (C) Distance (D) Acceleration
12) Which of the following statement is correct about speed and velocity?
(A) Velocity relates to translational motion while speed relates to circular motion
(B) Speed and velocity have the same units
(C) Velocity and speed cannot be represented graphically
(D) Speed and velocity are vector quantities
13) 14)The distance travelled by train may be calculated from the area under its
(A) Distance – graph
(B) Acceleration – time graph
(C) Displacement – time graph
(D) Velocity – time graph
14) A body accelerates uniformly from rest at 2m/s2. Find its velocity after travelling 9m.
(A) 4.5m/s (B) 24m/s (C) 6m/s (D) 16m/s
## Rectilinear Acceleration
1)A crate is pulled a distanced along a smooth horizontal floor by a force of magnitude F, Inclined to the horizontal at an angle O. The work done is.
a) F d sin b) Fd sec c)Fd Cos d)Fd Cosec.
2) Which of the following units has the same unit as the watt?
a) Force x velocity b) Force x time c) force x distance d) force x acceleration
3) An object of mass 60kg is released from the height of 3m from the height of 3m. Find the kinetic energy just before it strikes the ground (g =10mls2.
(a) 1.8 J (b)180J (c)18000J (d)1800J
4) A stone of mass mkg is held h meters above the floor for 40s. The work done in Joules over this period is
(a)mgh (b)O (c)mgh/40 (d)mg
5) Under which of the following conditions is a work done
a) A bag of Cocoa stands on a platform
b) A woman holds a pot of water
c) A girl climbs onto a table
d) A man pushes against a stationary petrol tanker
6) Power is defined as the:
a) Product of force and distance
b) Ability to do work
c) Capacity to extort a force
d) Energy expanded per unit time
7) A man of mass 60kg ascends a flight of stairs 5m high in 5 seconds. If the acceleration due to gravity is 10m/s2, the power expended is?
a) 600w b) 0w c) 60w d) 0.60w
## Heat Energy
1) A wire, 20m long is heated from a temperature of 50oC to 55oC. If the change in length is 0.030m, find the linear expansion of the wire.
(a) 3.0 x 10-3K-1 (b) 3 x 10-5K-1 (c) 0.3 x 10-5K-1 (d) 1 x 10-5K-1
2) The linear expansion of a metal P is twice that of another metal T. when these materials are heated to the same temperature change, their increase in length is the same. Find the ratio of the original length of P and T.
(a) 1:4 (b) 2:1 (c) 1:2 (d) 4:1
3) A uniform capillary tube of negligible expansion, sealed at one end, contains air trapped in a pellet of mercury. The trapped air column is 13.7cm long at 0oC and 18.7cm long at 100oC. Calculate the cubic expansion of the air at constant pressure.
(a) 3.65 x 10-3K-1 (b) 5.36 x 10-3K-1 (c) 3.56 x 10-3K-1 (d) 5.63 x 10-3K-1
4) The heat from a fire in a closed room reaches someone far away in the room mainly by
(a) Conduction (b) Diffusion (c) Convection (d) Radiation
5) Heat transfer by convection in a liquid is due to the
a) Expansion of the liquid as it is heated
b) Ability of the liquid to evaporate at all temperature
c) Translational motion of the molecules of the liquid
d) Increased vibration of the molecules of the liquid about their mean
6) Which of the following phenomena cannot be explained by the molecular theory of matter?
(a) Expansion (b) Radiation (c) Conduction (d) Convection
## Electric Charges
1)If the leaves of a positively charged electroscope collapse completely as an object is brought near the cap of the electroscope, the object possess:
a) An equal amount of positive charge
b) Less amount of negative charge
c) A positive charge
d) An equal amount of negative charge
2) A short chain is usually attached to the back of a petrol tanker trailing behind it to ensure that the
(a) Chain provides sound for the resonance of the tanker’s engine
(b) Heat generated by friction in the engine can be conducted to the floor
(C) Charges generated by friction in the tanker is conducted to the earth
(d) Tanker moves slowly as the chain touches the road
3) Which of the following instrument can be used to compare the relative magnitudes of charge on two given bodies? (a) Proof planes (b) Gold Leaf electroscope (c) The Ebonite rod (d) The electrophorus
4) A negatively charged rod is brought near the cap of a gold-leaf electroscope. The cap is earthed momentarily while the rod is near to it. The rod is then removed. Which of the following is correct?
a) The cap will be positively charged and the negatively charged leaves will diverge
b) The cap and the leaves will be positively charged
c) The leaves will not diverge because the cap has been earthed
d) The cap and leaves will be negatively charged.
### Bolarinwa Olajire
A tutor with a demonstrated history of working in the education industry. Skilled in analytical skills. Strong education professional with a M. SC focused in condensed matter. You can follow me on Twitter by clicking on the icon below to ask questions.
### One Comment
1. Obaloluwa Dedeigbo says:
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### Bold italic vectors
I am trying to display vectors as bold and italic. The command \mathbf sets the letters non-italic. I also googled a bit and stumbled upon \mathbf{\em y} or \mathbfit{y}, which all didn't work. I ... | 670 | 2,898 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-23 | longest | en | 0.913297 |
https://www.o.vg/unit/weightunit/kgtomuonmass.php | 1,718,348,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00160.warc.gz | 843,660,262 | 10,457 | Convert Kilogram to Muon mass | Muon mass to kg
# document.write(document.title);
## kg to Muon mass Converter
From kg to Muon mass: 1 kg = 5.309172492E+27 Muon mass;
From Muon mass to kg: 1 Muon mass = 1.8835327e-28 kg;
## How to Convert Kilogram to Muon mass?
As we know One kg is equal to 5.309172492E+27 Muon mass (1 kg = 5.309172492E+27 Muon mass).
To convert Kilogram to Muon mass, multiply your kg figure by 5.309172492E+27.
Example : convert 25 kg to Muon mass:
25 kg = 25 × 5.309172492E+27 Muon mass = Muon mass
To convert Muon mass to Kilogram, divide your Muon mass figure by 5.309172492E+27.
Example : convert 25 Muon mass to kg:
25 Muon mass = 25 ÷ 5.309172492E+27 kg = kg
## How to Convert Muon mass to Kilogram?
As we know One Muon mass is equal to 1.8835327e-28 kg (1 Muon mass = 1.8835327e-28 kg).
To convert Muon mass to Kilogram, multiply your Muon mass figure by 1.8835327e-28.
Example : convert 45 Muon mass to kg:
45 Muon mass = 45 × 1.8835327e-28 kg = kg
To convert Kilogram to Muon mass, divide your kg figure by 1.8835327e-28.
Example : convert 45 kg to Muon mass:
45 kg = 45 ÷ 1.8835327e-28 Muon mass = Muon mass
## Convert Kilogram or Muon mass to Other Mass and Weight Units
Kilogram Conversion Table
kg to g 1 kg = 1000 g kg to lbs 1 kg = 2.2046226218 lbs kg to oz 1 kg = 35.27396195 oz kg to cg 1 kg = 100000 cg kg to mg 1 kg = 1000000 mg kg to ug 1 kg = 1E+9 ug kg to ng 1 kg = 1E+12 ng kg to pg 1 kg = 1E+15 pg kg to dg 1 kg = 10000 dg kg to t 1 kg = 0.001 t kg to ct 1 kg = 5000 ct kg to kip 1 kg = 0.0022046226 kip kg to hundredweight (US) 1 kg = 0.0220462262 hundredweight (US) kg to pdl 1 kg = 70.988848424 pdl kg to gr 1 kg = 15432.358353 gr kg to pwt 1 kg = 643.01493137 pwt kg to ton (US) 1 kg = 0.0011023113 ton (US) kg to ton (UK) 1 kg = 0.0009842065 ton (UK) kg to u 1 kg = 6.022136651E+26 u kg to Eg 1 kg = 1E-15 Eg kg to Pg 1 kg = 1E-12 Pg kg to Tg 1 kg = 1E-9 Tg kg to Gg 1 kg = 0.000001 Gg kg to Mg 1 kg = 0.001 Mg kg to hg 1 kg = 10 hg kg to dag 1 kg = 100 dag kg to fg 1 kg = 1E+18 fg kg to ag 1 kg = 1E+21 ag kg to gamma 1 kg = 1E+9 gamma kg to Da 1 kg = 6.022173643E+26 Da
Kilogram Conversion Table
kg to slug 1 kg = 0.0685217659 slug kg to stone (US) 1 kg = 0.1763698097 stone (US) kg to stone (UK) 1 kg = 0.1574730444 stone (UK) kg to tonne 1 kg = 0.001 tonne kg to kt 1 kg = 0.000001 kt kg to AT (US) 1 kg = 34.285710367 AT (US) kg to AT (UK) 1 kg = 30.612244898 AT (UK) kg to cwt 1 kg = 0.01 cwt kg to qr (US) 1 kg = 0.0881849049 qr (US) kg to qr (UK) 1 kg = 0.0787365222 qr (UK) kg to s.ap 1 kg = 771.61791764 s.ap kg to hundredweight (UK) 1 kg = 0.0196841306 hundredweight (UK) kg to lbs (troy or apothecary) 1 kg = 2.6792288807 lbs (troy or apothecary) kg to oz (troy or apothecary) 1 kg = 32.15074656863 oz (troy or apothecary) kg to kgf·s²/m 1 kg = 0.1019716213 kgf·s²/m kg to lbf·s²/ft 1 kg = 0.0685217659 lbf·s²/ft kg to Planck mass 1 kg = 45940892.448 Planck mass kg to Electron mass (rest) 1 kg = 1.097768382E+30 Electron mass (rest) kg to Muon mass 1 kg = 5.309172492E+27 Muon mass kg to Proton mass 1 kg = 5.978633201E+26 Proton mass kg to Neutron mass 1 kg = 5.970403753E+26 Neutron mass kg to Deuteron mass 1 kg = 2.990800894E+26 Deuteron mass kg to Earth's mass 1 kg = 1.673360107E-25 Earth's mass kg to Sun's mass 1 kg = 5E-31 Sun's mass kg to jin (斤) 1 kg = 2 jin (斤) kg to liang(两) 1 kg = 20 liang(两) kg to qian(钱) 1 kg = 200 qian(钱)
## FAQ
### What is 9 Kilogram in Muon mass?
Muon mass. Since one kg equals 5.309172492E+27 Muon mass, 9 kg in Muon mass will be Muon mass.
### How many Muon mass are in a Kilogram?
There are 5.309172492E+27 Muon mass in one kg. In turn, one Muon mass is equal to 1.8835327e-28 kg.
### How many kg is equal to 1 Muon mass?
1 Muon mass is approximately equal to 1.8835327e-28 kg.
### What is the kg value of 8 Muon mass?
The Kilogram value of 8 Muon mass is kg. (i.e.,) 8 x 1.8835327e-28 = kg.
### kg to Muon mass converter in batch
Cite this Converter, Content or Page as:
"" at https://www.o.vg/unit/weightunit/kgtomuonmass.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters | 1,664 | 4,163 | {"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-2024-26 | latest | en | 0.626222 |
https://mathzsolution.com/homotopy-invariance-of-de-rham-cohomology/ | 1,670,215,358,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00728.warc.gz | 405,501,670 | 27,110 | # Homotopy invariance of de Rham cohomology
Let $M,N$ be smooth manifolds which are homotopy equivalent i.e., there exists smooth maps $F:M\rightarrow N$ and $G:N\rightarrow M$ such that $F\circ G$ is homotopic to identity map on $N$ and $G\circ F$ is homotopic to identity map on $M$.
Then, Homotopy invariance of deRham cohomology says that the de Rham cohomology groups of $M$ and $N$ are isomorphic.
I am not able to understand the construction given in Lee’s Intoduction to Smooth manifolds.
What is the rough idea behind this proof (or any other proof) of homotopy invariance of de Rham cohomology.
EDIT : Given that $M,N$ are homotopy equivalent as above, we need to prove that $H^p_{dR}(M)$ and $H^p_{dR}(N)$ are isomorphic. we expect this to come from $F^*:H^p_{dR}(N)\rightarrow H^p_{dR}(M)$ and $G^*:H^p_{dR}(M)\rightarrow H^p_{dR}(N)$.
I do not understand the idea behind proof of two homotopic maps have induce same deRham cohomology maps.
Once we prove this, then $F\circ G$ and $1_N$ induce same deRham cohomology maps i.e., the composition $H^p_{dR}(N)\xrightarrow{F^*} H^p_{dR}(M)\xrightarrow{G^*} H^p_{dR}(N)$ is same as the identity map on $H^p_{dR}(N)$ and similarly the composition $H^p_{dR}(M)\xrightarrow{G^*} H^p_{dR}(N)\xrightarrow{F^*} H^p_{dR}(M)$ is same as the identity map on $H^p_{dR}(M)$. This says that $F^*\circ G^*=1$ and $G^*\circ F^*=1$. Thus, $F^*,G^*$ are isomorphisms, inverses to each other, conlcuding that deRham cohomology groups $H^p_{dR}(M)$ and $H^p_{dR}(N)$ are isomorphic.
How do we prove that two homotopy maps induce same deRham cohomology maps. Let $f:M\rightarrow N$ and $g:M\rightarrow N$ be two homotopy maps, we want to prove that $f^*=g^*:H^p_{dR}(N)\rightarrow H^p_{dR}(M)$ i.e., $f^*(\omega)=g^*(\omega)+\text{closed p-form on }M$ when seen as maps $\Omega^p(N)\rightarrow \Omega^p(M)$. This means, we are expected to have where $\eta$ is a smooth $p-1$ form.
This gives question of defining a map $h:\{\text{closed p-forms on }N\}\subseteq \Omega^p(N)\rightarrow \Omega^{p-1}(M)$ assigning to each closed $p$ form $\omega$ on $N$ a $p-1$ form $\eta$ on $M$ such that $f^*(\omega)=g^*(\omega)+d\eta$.
Then author says it turns out to be far simpler to define $h:\Omega^p(N)\rightarrow \Omega^{p-1}(M)$ not with the condition $f^*(\omega)=g^*(\omega)+d(h\omega)$ for every closed form $\omega$ but with a more general condition that
for every smooth $p$ form. Suppose $\omega$ is closed then $d\omega=0$ and we get the required condition that $f^*(\omega)-g^*(\omega)=d(h\omega)$.
So, now the question is to define a map $h:\Omega^p(N)\rightarrow \Omega^{p-1}(M)$ satisfying the condition as above. How can we think of constructing such map? If we are thinking of going from a $p$ form to a $p-1$ form one obvious thing is to some how integrate this $p$ form. What $p$ form can we integrate here? It is natural to some how integrate the $p$ form $f^*(\omega)-g^*(\omega)$ to get a $p-1$ form $h\omega$. So, when you reverse the process i.e., when you differentiate you get $f^*(\omega)-g^*(\omega)=d(h\omega)$. This idea is vague and I can not make it any better.
This $h$ is called a homotopy operator in this book.
Any suggestions on how would you think about producing this operator is welcome.
Now I read the book and realize that Lie derivative is introduced after the chapter on cohomology, if the order is reversed there is a very direct interpretation.
You want to prove:
If $f_0,f_1 : M\to N$ are smooth mapping which are homotopic, then
for all $k$.
Recall that the induced pullback mapping on $H^k$ is just $f_0^* [\alpha] = [f_0^*\alpha]$ and similar for $f_1$. So you need to show: for any $k$-form $\alpha$ on $N$, $[f_0^*\alpha] =[ f_1^*\alpha]$, or $[f_1^*\alpha - f_0^*\alpha] = 0$.
That is, you want to write $f_1^*\alpha - f_0^*\alpha$ as $d$ of something. Note that by the fundamental theorem of calculus,
Here $f_t$ is the homotopy between $f_0$ and $f_1$. Of course it is not clear what the right hand side is. We want to give it a more intrinsic interpretation, so that we can check if the right hand side is really $d$ of something.
We let $F : M \times [0,1] \to N$ be the homotopy and $\iota_t : M\to M\times [0,1]$, $\iota_t (x) = (x, t)$ be the inclusion. Then $f_t = F\circ \iota_t$, thus $f_t^* = \iota_t^* \circ F^*$ and
$\mathscr L_T$ is the Lie derivative along the vector $T:=\frac{\partial}{\partial t}$ (as a vector field on $M\times [0,1]$). Now the Cartan’s magic formula gives (for any differential form $\omega$, vector fields $X$)
So we have
Note that the integration are exactly the homotopy operator $h$ constructed: so
So we have the next best thing: the right hand side in general is not $d$ of something, but it is when $\alpha$ is closed. This proves the theorem.
Of course I am just hiding everything in the Cartan’s magic formula. The formula is commonly proved by direct calculation. A more fancy/geometric argument is suggested in Arnold’s classical mechanics here. Note that the latter one also use a homotopy operator. | 1,619 | 5,055 | {"found_math": true, "script_math_tex": 87, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.807364 |
https://number.academy/1526884 | 1,718,841,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00212.warc.gz | 394,780,022 | 11,454 | # Number 1526884 facts
The even number 1,526,884 is spelled 🔊, and written in words: one million, five hundred and twenty-six thousand, eight hundred and eighty-four, approximately 1.5 million. The ordinal number 1526884th is said 🔊 and written as: one million, five hundred and twenty-six thousand, eight hundred and eighty-fourth. The meaning of the number 1526884 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1526884. What is 1526884 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,526,884 in other units
The decimal (Arabic) number 1526884 converted to a Roman number is (M)(D)(X)(X)(V)MDCCCLXXXIV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
1526884 seconds equals to 2 weeks, 3 days, 16 hours, 8 minutes, 4 seconds
1526884 minutes equals to 3 years, 1 month, 3 weeks, 3 days, 8 hours, 4 minutes
### Codes and images of the number 1526884
Number 1526884 morse code: .---- ..... ..--- -.... ---.. ---.. ....-
Sign language for number 1526884:
Number 1526884 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
### Advanced math operations
#### Is Prime?
The number 1526884 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 1526884 are 2 * 2 * 89 * 4289
The factors of 1526884 are 1, 2, 4, 89, 178, 356, 4289, 8578, 17156, 381721, 763442, 1526884.
Total factors 12.
Sum of factors 2702700 (1175816).
#### Powers
The second power of 15268842 is 2.331.374.749.456.
The third power of 15268843 is 3.559.738.802.948.375.040.
#### Roots
The square root √1526884 is 1235,671477.
The cube root of 31526884 is 115,151257.
#### Logarithms
The natural logarithm of No. ln 1526884 = loge 1526884 = 14,23874.
The logarithm to base 10 of No. log10 1526884 = 6,183806.
The Napierian logarithm of No. log1/e 1526884 = -14,23874.
### Trigonometric functions
The cosine of 1526884 is 0,655979.
The sine of 1526884 is 0,754779.
The tangent of 1526884 is 1,150614.
## Number 1526884 in Computer Science
Code typeCode value
1526884 Number of bytes1.5MB
Unix timeUnix time 1526884 is equal to Sunday Jan. 18, 1970, 4:08:04 p.m. GMT
IPv4, IPv6Number 1526884 internet address in dotted format v4 0.23.76.100, v6 ::17:4c64
1526884 Decimal = 101110100110001100100 Binary
1526884 Decimal = 2212120111021 Ternary
1526884 Decimal = 5646144 Octal
1526884 Decimal = 174C64 Hexadecimal (0x174c64 hex)
1526884 BASE64MTUyNjg4NA==
1526884 SHA17fbc38b68ef9fca7c8b64345d40c53ae90820a9d
1526884 SHA2243bebfde15748266b61b74190ccc84183fc0b02fb3c00d8660200bd35
1526884 SHA384dc786cd24b060305acbef812b2d65c25fb545faa629bfe1c101421f05d69daca23b482bef425856dc05b2f6e044cea88
More SHA codes related to the number 1526884 ...
If you know something interesting about the 1526884 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1526884
### Character frequency in the number 1526884
Character (importance) frequency for numerology.
Character: Frequency: 1 1 5 1 2 1 6 1 8 2 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1526884, the numbers 1+5+2+6+8+8+4 = 3+4 = 7 are added and the meaning of the number 7 is sought.
## № 1,526,884 in other languages
How to say or write the number one million, five hundred and twenty-six thousand, eight hundred and eighty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.526.884) un millón quinientos veintiséis mil ochocientos ochenta y cuatro German: 🔊 (Nummer 1.526.884) eine Million fünfhundertsechsundzwanzigtausendachthundertvierundachtzig French: 🔊 (nombre 1 526 884) un million cinq cent vingt-six mille huit cent quatre-vingt-quatre Portuguese: 🔊 (número 1 526 884) um milhão, quinhentos e vinte e seis mil, oitocentos e oitenta e quatro Hindi: 🔊 (संख्या 1 526 884) पंद्रह लाख, छब्बीस हज़ार, आठ सौ, चौरासी Chinese: 🔊 (数 1 526 884) 一百五十二万六千八百八十四 Arabian: 🔊 (عدد 1,526,884) مليون و خمسمائة و ستة و عشرون ألفاً و ثمانمائة و أربعة و ثمانون Czech: 🔊 (číslo 1 526 884) milion pětset dvacet šest tisíc osmset osmdesát čtyři Korean: 🔊 (번호 1,526,884) 백오십이만 육천팔백팔십사 Danish: 🔊 (nummer 1 526 884) en millioner femhundrede og seksogtyvetusindottehundrede og fireogfirs Hebrew: (מספר 1,526,884) מיליון חמש מאות עשרים ושישה אלף שמונה מאות שמונים וארבע Dutch: 🔊 (nummer 1 526 884) een miljoen vijfhonderdzesentwintigduizendachthonderdvierentachtig Japanese: 🔊 (数 1,526,884) 百五十二万六千八百八十四 Indonesian: 🔊 (jumlah 1.526.884) satu juta lima ratus dua puluh enam ribu delapan ratus delapan puluh empat Italian: 🔊 (numero 1 526 884) un milione e cinquecentoventiseimilaottocentottantaquattro Norwegian: 🔊 (nummer 1 526 884) en million fem hundre og tjueseks tusen åtte hundre og åttifire Polish: 🔊 (liczba 1 526 884) milion pięćset dwadzieścia sześć tysięcy osiemset osiemdziesiąt cztery Russian: 🔊 (номер 1 526 884) один миллион пятьсот двадцать шесть тысяч восемьсот восемьдесят четыре Turkish: 🔊 (numara 1,526,884) birmilyonbeşyüzyirmialtıbinsekizyüzseksendört Thai: 🔊 (จำนวน 1 526 884) หนึ่งล้านห้าแสนสองหมื่นหกพันแปดร้อยแปดสิบสี่ Ukrainian: 🔊 (номер 1 526 884) один мільйон п'ятсот двадцять шість тисяч вісімсот вісімдесят чотири Vietnamese: 🔊 (con số 1.526.884) một triệu năm trăm hai mươi sáu nghìn tám trăm tám mươi bốn Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,137 | 6,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-26 | latest | en | 0.798325 |
https://opensourceconnections.com/blog/2013/03/14/finally-a-hadoop-hello-world-that-isnt-a-lame-word-count/ | 1,701,473,566,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00068.warc.gz | 491,176,974 | 21,310 | Blog
# Finally! A Hadoop Hello World that isnt a Lame Word Count!
So I got bored of the old WordCount Hello World, and being a fairly mathy person, I decided to make my own Hello World in which I coaxed Hadoop into transposing a matrix!
What? Whats that you say? You think that a matrix transpose MapReduce is way more lame than a word count? Well I didnt say that we were going to be saving the world with this MapReduce job, just flexing our mental muscles a little more. Typically, when you run the WordCount example, you dont even look at the java code. You just pat yourself on the back when the word “the” invariably revealed to be the most popular word in the English language.
The goal of this exercise is to present a new challenge and a simple challenge so that we can practice thinking about solving BIG problems under the sometimes unintuitive constraints of MapReduce. Ultimately I intend to follow this post up with exceedingly more difficult MapReduce problems to challenge you and encourage you to tackle your own problems.
So, without further adieu:
## The Matrix Transpose Problem
The matrix transpose is a pretty simple concept. Lets say that you have some matrix M. Here are the rows of M (preceded by the row number):
0 7 9 3 61 4 2 9 82 4 6 6 1
You transpose this matrix by “flipping” the values about the diagonal. Here is transpose(M):
0 7 4 41 9 2 62 3 9 63 6 8 1
For tiny matrices like this, transpose is trivial, but for giant, super-jumbo Big Data matrices this can be challenging to do in the constraints of one machines RAM. Thus, the matrix transpose is a good candidate for MapReduce.
## Outline of a Simple MapReduce Job
As you may know by now, MapReduce is a two phase process composed of a mapping phase and a reducing phase. Simplistically, in the mapping phase, the mapper is given the contents of a Hadoop directory one bit at at time as a key-value pair. In this example, let the key be the rowIndex and the value be the values associated with that row.
{0: [7 9 3 6]}{1: [4 2 9 8]}{2: [4 6 6 1]}
It is the goal of the mapper to consume this information, process it, and then emit a number of other key-value pairs K and V. The keys and values can be of any type that you like, and for every iteration of the map function you can emit as many {K: V} pairs as you wish.
Between the mapping phase and the reducing phase, all of these values V are grouped according according to their associated keys K. Then, one at a time, the reducer is given a key K, along with all of the corresponding values V[]. The goal of the reducer is then to consume this information, process it, and emit more key-value pairs. In the case of our example, this new set of key-value pairs will represent the transpose of the original matrix. The keys will be transposeRowIndexs and the associated values will be the associated elements in that row which we refer to as transposeValuess. So, for the example of this post, the reducer will return:
{0: [7 4 4]}{1: [9 2 6]}{2: [3 9 6]}{3: [6 8 1]}
The goal of the game, then, is simply to write two functions, map and reduce, which achieve all of the above:
map(rowIndex,values) : #DO STUFF return {K:V}reduce(K,V[]) : #DO STUFF return {transposeRowIndex: transposeValues}
So, how do you implement this? Think about it! No peeking.
## My Solution
Here is my solution
map(rowIndex,values) : for(columnIndex = 1 to length(values)) : K = columnIndex V = {rowIndex: values[columnIndex]} return {K:V}reduce(K,V[]) : transposeValues = [] for(x in V[]) : transposeValues[x.key] = x.value transposeRowIndex = K return {transposeRowIndex: transposeValues}
And if we run our MapReduce algorithm, during the mapping phase we get
map(0,[7 9 3 6]) #=> {0:{0,7}}, {1:{0,9}}, {2:{0,3}}, {3:{0,6}}map(1,[4 2 9 8]) #=> {0:{1,4}}, {1:{1,2}}, {2:{1,9}}, {3:{1,8}}map(2,[4 6 6 1]) #=> {0:{2,4}}, {1:{2,6}}, {2:{2,6}}, {3:{2,1}}
In the “shuffle and sort” time before the reduce phase we cluster the values together according to key
{0: [{0,7}, {1,4}, {2,4}]}{1: [{0,9}, {1,2}, {2,6}]}{2: [{0,3}, {1,9}, {2,6}]}{3: [{0,6}, {1,8}, {2,1}]}
And finally, during the reducing phase we get
reduce({0: [{0,7},{1,4},{2,4}]}) #=> {0: [7 4 4]}reduce({1: [{0,9},{1,2},{2,6}]}) #=> {1: [9 2 6]}reduce({2: [{0,3},{1,9},{2,6}]}) #=> {2: [3 9 6]}reduce({3: [{0,6},{1,8},{2,1}]}) #=> {3: [6 8 1]}
And now since weve tested out the algorithm, its time to transpose something a bit larger…
## In Practice
Im putting together a collection of Hadoop tutorials and Im calling it Hadoopadoop. Check it out! Once you download it, you can quickly run my MatrixTranspose example by issuing the command ./build.sh MatrixTranspose. But please dont let this just be like running the typical WordCount example. Instead, take a look at the build script and see what its doing to build the project. Im intentionally keeping everything as lightweight as possible. No extra libraries, no Maven, just some java files, some data zips, and a little bit of bash wrangling.
The actual MatrixTranspose/MatrixTranspose.java code is pretty simple. Youll find that Im not using any bells and whistles to get the job done. Besides a little extra code to parse and assemble the input and output text, Im basically stating in code the same thing as I have stated above. However, there are still plenty of things to improve upon. This follow-up article describes such improvements as using a creating a custom combiner to reduce the network traffic during the shuffle/sort phase.
While youre looking at my Hadoopadoop, take a glance at the WordCount tutorial that Ive adapted from the WordCount example that ships with Hadoop. Lame though it may be, its still remains a great, and simple example of MapReduce. | 1,600 | 5,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-50 | latest | en | 0.921199 |
https://www.numbersaplenty.com/2511353 | 1,695,355,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00830.warc.gz | 1,027,345,971 | 3,057 | Search a number
2511353 = 13193181
BaseRepresentation
bin1001100101000111111001
311201120221002
421211013321
51120330403
6125454345
730226505
oct11450771
94646832
102511353
1114658a9
12a113b5
1369c110
14495305
15349188
hex2651f9
2511353 has 4 divisors (see below), whose sum is σ = 2704548. Its totient is φ = 2318160.
The previous prime is 2511349. The next prime is 2511389. The reversal of 2511353 is 3531152.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in 2 ways, for example, as 1227664 + 1283689 = 1108^2 + 1133^2 .
It is not a de Polignac number, because 2511353 - 22 = 2511349 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (2511323) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 96578 + ... + 96603.
It is an arithmetic number, because the mean of its divisors is an integer number (676137).
22511353 is an apocalyptic number.
It is an amenable number.
2511353 is a deficient number, since it is larger than the sum of its proper divisors (193195).
2511353 is a wasteful number, since it uses less digits than its factorization.
2511353 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 193194.
The product of its digits is 450, while the sum is 20.
The square root of 2511353 is about 1584.7248972614. The cubic root of 2511353 is about 135.9260158440.
The spelling of 2511353 in words is "two million, five hundred eleven thousand, three hundred fifty-three".
Divisors: 1 13 193181 2511353 | 516 | 1,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.861441 |
https://www.physicsforums.com/threads/element-order-in-factor-groups.267653/ | 1,531,759,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00170.warc.gz | 960,143,971 | 12,648 | # Element order in factor groups
1. Oct 28, 2008
### factor
I'm currently studying factor groups in abstract algebra and needed some help understanding how to determine the order of an element in a factor group
Suppose I have Z (mod 12) / <4>. And I choose some random element from Z (mod 12) such as 5 or 7. How would I go about determining the order of 5 + <4> or 7 + <4> in the factor group. The way I understand is that in Z (mod 12) / <4> we collapse everything in <4> to the identity in Z (mod 12) so that 5 for instance would have order |{5,10,3,8}| = 4 because 8 is an identity in the factor group. Any help or correction on my admittedly poor understanding would be great.
2. Oct 28, 2008
### sutupidmath
From a theorem, i believe, if the index of a normal group, N, in a group G is say, like in your case 4. then the order of the factro group G/N is 4. So, by lagranges theorem for every el in G/N, say C, C^4=N (identity). Now in your case, the order of Z_12/<4> is 4, so there will be 4 distinct cosets of <4> in Z_12, all of length 3, which makes me think, that the order of any of the el. in Z_12/<4> is going to be 3.
3. Nov 2, 2010
### jeane
The way I understand it is list the elements of the factor group. <4>={0,4,8} then list the factors of the element you are trying to find the order of. 5, 10, 3, 8 Stop it took 4 times to get to an element in the factor group. Therefore the element has an order of 4. | 434 | 1,435 | {"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} | 4.03125 | 4 | CC-MAIN-2018-30 | latest | en | 0.925661 |
https://ru-facts.com/how-many-oz-a-cup/ | 1,670,375,249,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00461.warc.gz | 525,740,686 | 11,668 | # How many oz a cup?
## How many oz a cup?
8 ounces
Liquid measuring cups indicate that 1 cup = 8 ounces.
## How many mL is an Oz cup?
236.59 mL
Oz to mL to Cup Conversion Table
Oz U.S. Standard Cups Millilitres
8 fl oz 1 cup 236.59 mL
9 fl oz 1.13 266.16 mL
10 fl oz 1.25 295.74 mL
12 fl oz 1 1/2 cups 354.88 mL
How many mL does a cup hold?
Legal cup
= 240 millilitres
= 16 international tablespoons
= 12 Australian tablespoons
= 8 US nutritional fluid ounces (30 ml each)
8.12 US customary fluid ounces
How much is 1 cup of dry ingredients?
Dry/Weight Measure
Ounces
10 tablespoons plus 2 teaspoons 2/3 cup 5.2 ounces
12 tablespoons 3/4 cup 6 ounces
16 tablespoons 1 cup 8 ounces
32 tablespoons 2 cups 16 ounces
### Is a coffee cup 1 cup?
“Coffee cup” is not an official unit, and it is usually not the same size as an official “cup” in USCS units (8 fl. oz, or ~237 ml). Yes, coffee cups are based on the actual measurement of 1 cup.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 314 | 1,034 | {"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-2022-49 | latest | en | 0.812643 |
https://www.theanalysisfactor.com/tag/oversampling/ | 1,725,940,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00356.warc.gz | 957,181,259 | 14,208 | # oversampling
### What is a Complex Sample Part 3: Stratified Sampling
January 9th, 2014 by
In our last two posts, we explained (1) that every member of a simple random sample had an equal probability of selection and (2) that there are some really good reasons why complex samples can work better, despite being more complex.
Today, we’re going to talk a bit about one complex sampling technique: stratified sampling.
### What is Stratified Sampling?
In stratified sampling, the target population is first classified into subgroups or strata. (Grammar note: “strata” is plural for “stratum” just as “data” is plural for “datum.”).
A simple random sample is then selected within every stratum.
That’s it.
For example, let’s say you’re doing a linguistics study within the US. You want to make sure that you have enough (more…)
### What is a Complex Sample Part 2: What is it and Why Would You Ever Want to Use One?
January 6th, 2014 by
In our last article, we talked about simple random samples. Simple random samples are, well…simple, but they’re not always optimal or even possible.
Probability samples that don’t meet the assumptions of Simple Random Samples are called Complex Samples.
You’ll also hear the term Complex Survey, which is really just a survey that incorporates some sort of complex sampling design. Because of their size and research goals, surveys are usually* the only type of research study that uses complex samples.
(*but not always. I have seen intervention studies, for example, that used complex sampling).
### What is a Complex Sample?
The most defining feature of a complex sample is that sample members do not have equal probability of being selected.
That sounds simple enough. But… (more…) | 384 | 1,746 | {"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-2024-38 | latest | en | 0.940061 |
https://algebra-calculator.com/algebra-calculators/long-division/class-8-maths.html | 1,708,765,978,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474526.76/warc/CC-MAIN-20240224080616-20240224110616-00209.warc.gz | 91,765,871 | 11,208 | Algebra Tutorials!
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• mixed numbers in ti-83 plus | 1,421 | 5,983 | {"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-2024-10 | latest | en | 0.891235 |
https://www.jiskha.com/questions/458010/what-is-the-number-of-moles-of-hcl-in-50ml-of-1-0m-hcl | 1,580,335,175,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251802249.87/warc/CC-MAIN-20200129194333-20200129223333-00147.warc.gz | 937,297,862 | 5,011 | # AP Chemistry
what is the number of moles of HCl in 50mL of 1.0M HCl
1. 👍 0
2. 👎 0
3. 👁 171
1. mols= M x L
1. 👍 0
2. 👎 0
posted by DrBob222
## Similar Questions
1. ### Chemistry
50mL of 0.1M HCl is mixed with 50mL of 0.1NaOH. delta T = 3^oC. Calculate the enthalpy of neutralization per mole of HCl. assume the specific heat capcity of the solution is 4.18J/g/C and density=1.0g/mL. I tried it but got the
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2. ### Science
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3. ### Chem 2
So i need to calculate the pH of 50ml water (de-ionized) with 1 ml of 1M HCl. So the moles of HCl i think is .001 divided by the total amount of solution (.051). I get 0.0196, then i take the negative log of this number and it
asked by Kyle on October 14, 2013
4. ### chemistry
titration question: How many moles of HCl must have been present in the 25mL of HCl solution in the 2 trials? Given: Trial 1: V of HCl= 25.00mL, V of NaOH used=29.50mL and M of NaOH=0.18M Trial 2: V of HCl=25.00mL, V of NaOH
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5. ### Chemistry
How many moles of HCl are present in a 50mL 3M solution of HCl?
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6. ### Chemistry
How many moles of HCl are present in a 50mL 3M solution of HCl?
asked by Jess on January 12, 2011
7. ### Chemistry
For an experiment, we used 10 mL of HCl. We are then to find the moles of it without any other information given. Is this right? (10 mL HCl)* (1.19 g / mL) = 11.9 g HCl (11.9 g HCl) * (1mol/36.46064 g) = 0.326 mols HCl
asked by Abby on June 2, 2017
8. ### chemistry
So I did a titration lab at school with NaOH and HCl. i have to find: - volume used, NaOH - Moles of NaOH - Moles of HCl -Volume used unknown HCl - Molarity of HCl solution My concentration of NaOH stock solution was 0.100 mol/lL
asked by julie on January 4, 2015
9. ### chemistry
The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is i solved it but my answer is wrong. the right one is 1.97. am i missing a step? moles of HCl left over = (moles of HCl) – ( moles of KOH)
asked by ben on November 23, 2010
10. ### Chemistry
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asked by Martin on October 22, 2011
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https://www.coursehero.com/file/6473710/moles-8-31-11/ | 1,495,624,363,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607811.15/warc/CC-MAIN-20170524093528-20170524113528-00615.warc.gz | 851,740,639 | 222,020 | moles-8-31-11 - How Many Atoms Lynn A Melton University of...
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08-31-11 1 How Many Atoms? Lynn A. Melton University of Texas at Dallas August 31, 2011
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Problems with CONNECT? McGraw-Hill rep will be at UTD this Friday 10am – 1 pm. CONNECT codes Registration (if you bring your laptop) MUST have original receipt from bookstore showing proof of purchase 08-31-11 2
Printing Lecture Notes 08-31-11 3
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Homework and Success The assigned homework problems cover all of the important – likely to be on exams – topics in a chapter. The quiz questions cover only the material treated through the Friday class before the Thursday deadline for the quiz – not all material in a chapter, not all the homework problems. 08-31-11 4
Homework and Success The assigned homework problems cover all of the important – likely to be on exams – topics in a chapter. The hour exams and the final exam questions can come from any part of the chapters – good match to the homework problems. 08-31-11 5
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Atoms and “Empty Space” The diameter of the nucleus is about one ten-thousandth of the diameter of the atom. With so much “empty space”, why don’t atoms (materials) just pass through each other? Nice link: http://www.teachersdomain.org/resource/phy03.s 08-31-11 6
Nomenclature (important) A z X X is the symbol for the atom, i.e, C for carbon A (left superscript) is the atomic weight (generally rounded to the nearest integer) Z (left subscript) is the atomic number 08-31-11 7
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How Many Little Red Orbs? I want to know, within 0.1%, how many little red
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This note was uploaded on 10/15/2011 for the course MISC. 101 taught by Professor Multiple during the Spring '11 term at University of Texas at Dallas, Richardson.
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# QOTD: Aldolase is a protein found
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QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 10:19
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Verbal Question of The Day: Day 19: Critical Reasoning
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Aldolase is a protein found in the brain. In an experiment, subjects who consistently chose sweetened drinks when offered a choice between salty and sweetened drinks were found, on average, to have significantly higher concentrations of Aldolase in their brains than subjects who consistently chose salty drinks over sweetened drinks. Thus, an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
Which one of the following, if true, most supports the argument?
(A) Craving sweetened drinks does not invariably result in a subject’s choosing those drinks over salty drinks.
(B) The brains of the subjects who consistently chose to drink sweetened drinks did not contain significantly more sugar than the brains of subjects that consistently chose salty drinks.
(C) The chemical components of Aldolase are present in sweetened drinks.
(D) The subjects that preferred sweetened drinks had higher concentrations of Aldolase in their brains before they were offered sweetened drinks.
(E) In a study of subjects who drank one sweetened drink per day for an entire year, the concentrations of Aldolase in the brains of most subjects gradually increased over the course of the year.
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Re: QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 10:21
3
1
This is a classic correlation vs. causation CR question. And as always, I like to start with a nice, clear restatement of the conclusion:
Quote:
Thus, an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
This is strong, clear causal language: Aldolase causes the increase in cravings for sweetened drinks. And what's the reasoning given in the passage itself? Really, it's just one sentence that leads to the conclusion:
Quote:
In an experiment, subjects who consistently chose sweetened drinks when offered a choice between salty and sweetened drinks were found, on average, to have significantly higher concentrations of Aldolase in their brains than subjects who consistently chose salty drinks over sweetened drinks.
The author assumes that Aldolase is the cause of the cravings for sweetened drinks, but as is often the case in correlation vs. causation questions, we might wonder if the direction of causality could also be reversed. The author says that Aldolase causes the cravings, but why couldn't the consumption of sweetened drinks lead to higher concentrations of Aldolase in the brain instead?
The question is just asking us to strengthen the conclusion ("an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.") One way to strengthen the conclusion might be to eliminate the possibility of a "causal reversal": that the sweetened drinks cause the increase in Aldolase in the brain, instead of Aldolase causing an increase in cravings for sweetened drinks. Also notice that we do not need a statement that PROVES that the conclusion is true; rather, we need a statement that SUPPORTS the argument.
Quote:
(A) Craving sweetened drinks does not invariably result in a subject’s choosing those drinks over salty drinks.
Statement (A) is tempting at first! At first glance, it seems it seems to eliminate the "causal reversal" we discussed above. The trouble is, (A) only addresses the link between craving sweetened drinks and choosing those drinks; it does nothing to link Aldolase to the cravings or the consumption of sweetened drinks. And as we'll see in a moment, (D) is a much stronger answer.
Quote:
(B) The brains of the subjects who consistently chose to drink sweetened drinks did not contain significantly more sugar than the brains of subjects that consistently chose salty drinks.
The argument discusses the concentration of Aldolase in the brains of subjects, but the amount of sugar in subjects' brains is irrelevant to the argument. We can eliminate (B).
Quote:
(C) The chemical components of Aldolase are present in sweetened drinks.
The argument claims that Aldolase in the brain causes subjects to crave sweetened drinks. Whether Aldolase is present in the drinks themselves is irrelevant to the argument. And even if we (improperly!) assume that drinking "the chemical components of Aldolase" leads to increased Aldolase in the brain, then this would work against the argument, since it would suggest that drinking sweetened drinks causes higher levels of Aldolase in the brain. Either way, (C) is gone.
Quote:
(D) The subjects that preferred sweetened drinks had higher concentrations of Aldolase in their brains before they were offered sweetened drinks.
The answer strengthens the argument by eliminating the possibility that the stated causal relationship is reversed: if the subjects had higher concentrations of Aldolase prior to taking the sweetened drinks, then the sweetened drinks cannot be the cause of the higher concentration of Aldolase. This does not PROVE that the conclusion is true, but it certainly supports the argument.
Quote:
(E) In a study of subjects who drank one sweetened drink per day for an entire year, the concentrations of Aldolase in the brains of most subjects gradually increased over the course of the year.
(E) hurts the argument by suggesting that the causal relationship in the conclusion is reversed: if (E) is true, then drinking sweetened drinks would increase Aldolase levels in the brain, instead of the other way around.
So (D) is our answer.
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Re: QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 10:27
I think that it's D.
A e B are irrelevant
C and E weaken the argument because suggest that are the sweetened drinks that increase Aldolase.
Is my reasoning correct?
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Re: QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 11:02
I think it's A as it nullifies an alternate explanation.
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Re: QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 11:05
Aldolase is a protein found in the brain. In an experiment, subjects who consistently chose sweetened drinks when offered a choice between salty and sweetened drinks were found, on average, to have significantly higher concentrations of Aldolase in their brains than subjects who consistently chose salty drinks over sweetened drinks. Thus, an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
Type - strengthen
Boil it down - increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
Pre-thinking - consumption of sweetened drinks did not cause the level of aldolase in the brain to increase
(A) Craving sweetened drinks does not invariably result in a subject’s choosing those drinks over salty drinks. - Irrelevant
(B) The brains of the subjects who consistently chose to drink sweetened drinks did not contain significantly more sugar than the brains of subjects that consistently chose salty drinks. - Irrelevant as the argument does not mention about sugar content in the brain .
(C) The chemical components of Aldolase are present in sweetened drinks. - this is mild weakener as it tells that consumption of sweetened drinks might lead to increase in aldolase levels
(D) The subjects that preferred sweetened drinks had higher concentrations of Aldolase in their brains before they were offered sweetened drinks. - Correct
(E) In a study of subjects who drank one sweetened drink per day for an entire year, the concentrations of Aldolase in the brains of most subjects gradually increased over the course of the year. - the brain aldolase levels of regular consumers of sweetened drinks increase gradually - it is a mild weakener
Answer D
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Re: QOTD: Aldolase is a protein found [#permalink]
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25 May 2017, 18:48
Aldolase is a protein found in the brain. In an experiment, subjects who consistently chose sweetened drinks when offered a choice between salty and sweetened drinks were found, on average, to have significantly higher concentrations of Aldolase in their brains than subjects who consistently chose salty drinks over sweetened drinks. Thus, an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
Which one of the following, if true, most supports the argument?
(A) Craving sweetened drinks does not invariably result in a subject’s choosing those drinks over salty drinks.
(B) The brains of the subjects who consistently chose to drink sweetened drinks did not contain significantly more sugar than the brains of subjects that consistently chose salty drinks.
(C) The chemical components of Aldolase are present in sweetened drinks.
(D) The subjects that preferred sweetened drinks had higher concentrations of Aldolase in their brains before they were offered sweetened drinks.
(E) In a study of subjects who drank one sweetened drink per day for an entire year, the concentrations of Aldolase in the brains of most subjects gradually increased over the course of the year.
More Aldolase---> Increase in cravings for sweetened drinks
Explanation:-
A. This statement says nothing about the conclusion. Incorrect.
B. We are not concerned about the sugar levels in the brains of the subjects who consistently chose to drink sweetened drinks and who chose to drink salty drinks. We are concerned about the correlation between aldolase level in brain and increase in cravings for sweetened drinks due to it. This doesn't support the argument. Incorrect.
C. This statement has no base in our argument. Also if chemical components of aldolase is present in the sweetened drinks then it might or might not increase the cravings for sweetened drinks. Incorrect.
D. This statement clearly compares the aldolase levels of subjects and proves that due to higher aldolase, some subjects chose sweetened drinks over salty drinks. Thus if aldolase levels in the brain increases, the subjects will have an increase in cravings for sweetened drinks --Correct
E. This doesn't support the argument in any way. Incorrect.
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Re: QOTD: Aldolase is a protein found [#permalink]
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26 May 2017, 02:35
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souvik101990 wrote:
Aldolase is a protein found in the brain. In an experiment, subjects who consistently chose sweetened drinks when offered a choice between salty and sweetened drinks were found, on average, to have significantly higher concentrations of Aldolase in their brains than subjects who consistently chose salty drinks over sweetened drinks. Thus, an increase in the level of Aldolase in the brain is likely to cause an increase in cravings for sweetened drinks.
Which one of the following, if true, most supports the argument?
(A) Craving sweetened drinks does not invariably result in a subject’s choosing those drinks over salty drinks.
This choice weakens the argument. If craving sweetened drinks doesn't result in choosing sweetened drinks over salty drinks, then an increase in the level of Aldolase in the brain could have no effect on a subject’s choosing sweetened drinks over salty drinks.
(B) The brains of the subjects who consistently chose to drink sweetened drinks did not contain significantly more sugar than the brains of subjects that consistently chose salty drinks.
Irrelevant. The argument mentions about the level of Aldolase in the brain, not the level of sugar in the brain.
(C) The chemical components of Aldolase are present in sweetened drinks.
Irrelevan. The argument didn't mention anything about the level of Aldolase in these drinks.
(D) The subjects that preferred sweetened drinks had higher concentrations of Aldolase in their brains before they were offered sweetened drinks.
Correct. This choice eliminates the case that drinking sweetened drinks leads to the increase in the level of Aldolase in the brain (reverse cause).
(E) In a study of subjects who drank one sweetened drink per day for an entire year, the concentrations of Aldolase in the brains of most subjects gradually increased over the course of the year.
This choice fails in explaning that the level of Aldolase in the brain could effect a subject's behavior in choosing sweetened drinks over salty drinks.
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Re: QOTD: Aldolase is a protein found [#permalink]
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27 May 2017, 10:16
abhinit24 wrote:
I think it's A as it nullifies an alternate explanation.
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If X can lead to Y. Then there is always a possibility of Z that can also lead to Y. Hence, Z leads to Y is not a weakener.
If X led to Y. Then Z led to Y is a weakener.
I hope that answers your question.
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Re: QOTD: Aldolase is a protein found [#permalink]
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Re: QOTD: Aldolase is a protein found &nbs [#permalink] 07 Sep 2018, 20:01
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# QOTD: Aldolase is a protein found
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[quote="bgg1996"][spoiler]ACE TOR EWE BOW ERA DEN ALE DIV ODE TAG EGO NOD RUN USA TEN GUN UNI TOP[/spoiler][/quote]
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Author Message
Lepton*
Posted: Mon May 09, 2011 10:29 am Post subject: 1
Good job folks! Quailman's answer is what I was aiming for.
Quailman
Posted: Sat May 07, 2011 10:53 pm Post subject: 0
Why not SIGNOR? It's an Italian title of respect.
Not that I did anything original. It just seemed obvious from the middle letters.
JDTAY
Posted: Sat May 07, 2011 6:56 pm Post subject: -1
I'm trying to anagram the fixed letters. So far I got "a bartending person" with a g left over. I'll keep trying.
Zag
Posted: Sat May 07, 2011 5:44 pm Post subject: -2
If we take the fixed letters only, that is, the NW to SE diagonal, they are
ABATRG (top left) = GAB RAT
ORIGSN (middle) = RIG SON
ENEDNP (bottom right) = PEN DEN
Unfortunately, these six words will not make another square because there is only one each of I, B, O, T
Zag
Posted: Sat May 07, 2011 5:35 pm Post subject: -3
This leaves
Code:
T E N
A G O
G O D
bgg1996
Posted: Sat May 07, 2011 5:32 pm Post subject: -4
ACE
TOR
EWE
BOW
ERA
DEN
ALE
DIV
ODE
TAG
EGO
NOD
RUN
USA
TEN
GUN
UNI
TOP
Zag
Posted: Sat May 07, 2011 5:32 pm Post subject: -5
ago ego
god
gun gut
nod
run rut
tag ten
Code:
? U T ? is G or R
U N O
N I P
The unused choices ?UT, ?UN can not cross itself because no vowel has two words with either G or R as the second letter, so they have to cross USA and USE. Also, those are the only two S's remaining, so the two US words have to cross each other.
Code:
? U N
U S A
T E N
This means we can never know which of these is GUT and GUN and which is RUT and RUN, unless we can get it from some other context.
Zag
Posted: Sat May 07, 2011 5:16 pm Post subject: -6
Yeah, I realized that about one second after I walked away from my computer.
Here are all the words in a slightly more consumable format.
bed bow cow den div
ego era ere eve ewe
god gun gut lid nan
nip nod ode ore run
rut tag ten ten top
tor uni uno usa use
wan
Code:
A L E
D I V
O D E
Is forced, since there is only one L, only two choices with a middle I, and only one fits. Similarly, with only 2 C's, and only 4 W's:
Code:
A C E B O W
T O R E R A
E W E D E N
JDTAY
Posted: Sat May 07, 2011 4:21 pm Post subject: -7
Zag, the reason you have double the amount is he's giving you the accross and down words.
Zag
Posted: Sat May 07, 2011 3:34 pm Post subject: -8
Hmmm, I thought this was the process, but there is no word that starts ED
I had figured with only one word starting with C, the first square was forced, but the second one is impossible, so that's not the right approach.
Code:
A C E A L E
C O W L I D
E W E E D ?
Edit: I suppose that a lot of words won't be used at all, if there are only to be 6 3x3 squares. There are 36 words altogether.
Lepton
Posted: Sat May 07, 2011 6:23 am Post subject: -9
What do you call an Italian nobleman who still supports Berlusconi?
ace
ago
ale
ate
bed
bow
cow
den
div
ego
era
ere
eve
ewe
god
gun
gut
lid
nan
nip
nod
ode
ore
run
rut
tag
ten
ten
top
tor
uni
uno
usa
use
wan
Small Hint: Note that all the words alternate between consonants and vowels.
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Solving multi-step word problems can seem like an impossible task to students. Bar models help students organize their thinking by enabling them to “see” the parts of the problem and their relationship to one another. Two-step problem solving will be no problem at all for your students after they work through the math problems in this addition and subtraction November-themed bar modeling unit.
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Based on Singapore Math Principles and designed for Common Core Math instruction, these activities focus on the development of problem solving strategies for part part whole concepts and logical/inferential thinking for students in grades 2, 3 and 4.
The unit includes 10 original 2-step word problems. Learning can be differentiated using the two different levels of problems - support and independent. Pages come in color and black and white format (for both levels), giving multiple ways to utilize this resource. Print the color pages and laminate for use in math centers or print black and white sheets to use as homework. An answer key is provided.
This math problem solving unit aims to strengthen students' comprehension of mathematical operations of part part whole models by using purposeful vocabulary intended to direct student thinking and engender discussion as opposed to the "cookie cutter" type problems that rely on overused clue words to drive mathematical thinking. Authentic assessments are moving away from the type of questions where students can simply identify words they have been taught to be associated with certain operations. This unit will help your students think through the problems and see the reasoning behind the math by visualizing patterns within the models.
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https://www.pursuantmedia.com/2020/01/10/how-do-you-solve-difference-of-squares/ | 1,669,825,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00226.warc.gz | 1,010,671,522 | 12,515 | How do you solve difference of squares?
How do you solve difference of squares?
When an expression can be viewed as the difference of two perfect squares, i.e. a²-b², then we can factor it as (a+b)(a-b). For example, x²-25 can be factored as (x+5)(x-5). This method is based on the pattern (a+b)(a-b)=a²-b², which can be verified by expanding the parentheses in (a+b)(a-b).
What is difference of squares in math?
In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity. in elementary algebra.
What is a squared minus b squared?
The a2 – b2 formula is also known as “the difference of squares formula”. The a square minus b square is used to find the difference between the two squares without actually calculating the squares. It is one of the algebraic identities. It is used to factorize the binomials of squares.
What shows a difference of squares?
The difference of two squares is a theorem that tells us if a quadratic equation can be written as a product of two binomials, in which one shows the difference of the square roots and the other shows the sum of the square roots.
Will result in a difference of two squares?
The Difference of Two Squares theorem tells us that if our quadratic equation may be written as a difference between two squares, then it may be factored into two binomials, one a sum of the square roots and the other a difference of the square roots. This is sometimes shown by the expression A² – B² = (A + B) (A – B).
What is the two squares identity?
Identity. The difference of two squares identity is ( a + b ) ( a − b ) = a 2 − b 2 (a+b)(a-b)=a^2-b^2 (a+b)(a−b)=a2−b2.
Why is it called difference of two squares?
where one perfect square is subtracted from another, is called a difference of two squares. It arises when (a − b) and (a + b) are multiplied together. This is one example of what is called a special product.
What is a 2 B 2 formula?
(a + b)2 – (a – b)2 = 2(a2 + b2 ).
What is a 2 B 2 called?
And it is called the “difference of two squares” (the two squares are a2 and b2). This illustration shows why it works: a2 − b2 is equal to (a+b)(a−b)
Why can’t you factor the sum of two squares?
It’s true that you can’t factor A²+B² on the reals — meaning, with real-number coefficients — if A and B are just simple variables. So it’s still true that a sum of squares can’t be factored as a sum of squares on the reals.
What does difference of squares mean in math?
Remember from your translation skills that a “difference” means a “subtraction”. So a difference of squares is something that looks like x 2 – 4. That’s because 4 = 2 2, so we really have x 2 – 2 2, which is a difference of squares. To factor this, I’ll start by writing my parentheses, in the same way as usual for factoring:
Which is the correct way to factor a difference of squares?
Actually, it’s both. You can factor this difference in either of two ways: factoring a difference of squares, followed by factoring the difference and sum of cubes: = ( x – 2) ( x + 2) ( x2 + 2 x + 4) ( x2 – 2 x + 4)
How to calculate the difference of two squares using idenity?
If the input equation can be put in the form of a 2 – b 2 it will be factored. The work for the solution will be shown for factoring out any greatest common factors then calculating a difference of 2 squares using the idenity:
Is the difference of squares in multiplication commutative?
Since multiplication is commutative (that is, since you can move the factors around without changing the value of the product), the difference of squares can also be stated as: Don’t get hung up on the order of the factors. Either way is fine. Here are examples of some typical homework problems: | 955 | 3,826 | {"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} | 4.875 | 5 | CC-MAIN-2022-49 | latest | en | 0.953129 |
sofiapublicschool.edu.in | 1,726,684,697,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00260.warc.gz | 496,966,935 | 87,631 | # Transforming Numbers to Words: A Comprehensive Guide and Tool
Numbers are the building blocks of mathematics, but they often pose a challenge when it comes to translating them into words. Whether you’re a student, a teacher, or just someone curious about linguistic mathematics, this guide will unravel the mystery of converting numbers into words.
## Understanding the Basics
Before we delve into the intricacies, let’s establish the basics. Converting numbers to words involves expressing numerical figures in their written form. This skill is not only crucial in academic settings but also in real-world applications such as check writing, formal documentation, and more.
## The Numeric Language
Let’s break down some examples to understand the process better:
• 55,000: Fifty-Five Thousand
• 25,000: Twenty-Five Thousand
• 80: Eighty
• 2,900: Two Thousand Nine Hundred
• 120,000: One Hundred Twenty Thousand
• 13,000: Thirteen Thousand
• 70,000: Seventy Thousand
• 35,000: Thirty-Five Thousand
• 90,000: Ninety Thousand
• 40,500: Forty Thousand Five Hundred
• 11,100: Eleven Thousand One Hundred
## Tools to Simplify the Process
To make your life easier, our school provides a link to a user-friendly Numbers to Words Converter tool. Simply input the numeric value, and the tool will generate the corresponding words effortlessly. It’s a handy resource for students, parents, and educators alike.
## Practical Applications
Understanding numbers in words extends beyond the classroom. Our students engage in real-world applications, mastering this skill through various interactive math programs offered at Sofia Public School.
## Conclusion
In conclusion, transforming numbers to words is an essential skill with practical applications in both academic and everyday scenarios. Whether you’re a student honing your mathematical linguistics or an enthusiast exploring the linguistic side of numbers, our guide and Numbers to Words Converter tool are here to assist you.
Unlock the potential of numbers, and don’t forget to explore the numerous educational opportunities at Sofia Public School, where we prioritize holistic learning for every student. | 439 | 2,167 | {"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-2024-38 | latest | en | 0.833339 |
http://mathematica.stackexchange.com/questions/tagged/fitting?sort=active&pageSize=15 | 1,430,062,942,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246654687.23/warc/CC-MAIN-20150417045734-00311-ip-10-235-10-82.ec2.internal.warc.gz | 169,847,318 | 25,194 | # Tagged Questions
Questions on the use of Mathematica to construct models for approximating empirical data.
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### Fitting an ellipse to data with a special constraint [duplicate]
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### How to perform a multi-peak fitting?
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### PredictionBands and the presence of {0,0} point
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### FindFit runs extremely slow for my Function,how do I optimize this?
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### Is there any strategy to have a compiled version of FindFit or NonlinearModelFit?
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### How do I add custom deviation when using LinearModelFit?
I have a set of data in the form a list S1={{x,y},{x,y}} etc. I also have a function that calculates a corrected standard deviation in the y direction for ... | 2,654 | 11,366 | {"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} | 3.046875 | 3 | CC-MAIN-2015-18 | longest | en | 0.865739 |
https://getrevising.co.uk/revision-tests/electricity-30 | 1,527,378,198,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867949.10/warc/CC-MAIN-20180526225551-20180527005551-00495.warc.gz | 556,385,370 | 13,804 | # Electricity
HideShow resource information
Static Electricity
When 2 insulators rub together electrons are scraped off one and is dumped on the other, Electrons = Negatively Charged leaving positive charge on one and negative on the other
1 of 13
Like Charge
Repel
2 of 13
Opposite Charges
Attract
3 of 13
Current
Rate of flow of charge around circuit, Measured in Amps
4 of 13
Equation for Current
Charge = Current x Time
5 of 13
Resistor in I-V Graph
Directly Proportional to current and voltage
6 of 13
Bulb in I-V Graph
Obeys Ohm's Law fro the middle bit but then current increase = resistance increases so no longer obeys Ohm's Law
7 of 13
Diode in I-V Graph
Obeys Ohm's Law for 1/2
8 of 13
Resistance
Increase with Temperature, Heat energy causes ions in conductor to vibrate more so harder for current to flow through
9 of 13
LDR's
As light increases, Resistance decreases
10 of 13
Thermistor's
As temperature increases, Resistance decreases
11 of 13
PIV Equation
Power= Current x Voltage
12 of 13
PET Equation
Energy= Power x Time
13 of 13
## Other cards in this set
Repel
Like Charge
Attract
### Card 4
#### Front
Rate of flow of charge around circuit, Measured in Amps
### Card 5
#### Front
Charge = Current x Time | 344 | 1,236 | {"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} | 3.296875 | 3 | CC-MAIN-2018-22 | latest | en | 0.864992 |
https://wiki-helper.com/2-a-y-b-2-a-y-b-4-solve-using-substution-method-please-answer-correctly-answer-is-2a-and-y-39737324-39/ | 1,716,561,651,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00590.warc.gz | 528,111,988 | 30,356 | 2x/a+y/b=2 , x/a-y/b=4 solve using substution method
answer is x=2a and y= -2b
### 1 thought on “2x/a+y/b=2 , x/a-y/b=4 solve using substution method<br />please answer correctly<br />answer is x=2a and y= -2b”
x=2
put x in eq 1
y=2-2x
y=2-2(2)
y=2-4
y=-2 | 115 | 264 | {"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} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.683104 |
http://fixitmaths.com/howto-dividing-by-10.html | 1,721,445,730,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00041.warc.gz | 11,420,459 | 6,986 | ## Dividing by 10
When a number is divided by 10, each part of that number becomes 10 times smaller. So a hundred becomes a ten, and a ten becomes a one. This can easily be seen using Dienes, reversing the process used when multiplying by 10, with a hundred flat becoming a ten stick and a ten stick becoming a one cube. This means that any placeholder in the ones position is no longer needed. If there was a digit in the ones place, this will move into the tenths place to give a decimal answer.
Using a place value grid, we can see that the digits move one place to the right as they are divided by 10, representing how the value of each has become 10 times smaller. The digits move the opposite way to when we multiply by 10, because division is the inverse, or opposite, of multiplication.
Download an activity pack of games to reinforce multiplying or dividing by 10, 100 or 1000
Click here to generate worksheet to practise multiplying and dividing by 10, 100 & 1000. | 224 | 977 | {"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-2024-30 | latest | en | 0.94366 |
https://www.scribd.com/document/58898790/Fibonacci-Surprises | 1,563,846,482,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528635.94/warc/CC-MAIN-20190723002417-20190723024417-00534.warc.gz | 810,466,667 | 67,628 | You are on page 1of 23
# 1
## Some FIBONACCI SURPRISES The Power of a Picture
www.jamestanton.com
******
## 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...
and many of their delightful properties and unexpected appearances in mathematics. TOUGH QUESTION : Are 1 and 144 the only square Fibonacci numbers?
Each number beyond the initial pair of 1s is the sum of the two entries preceding it. QUESTION: If this pattern persists, what number should precede the pair of 1 s? Extend the Fibonacci sequence towards the left. What pattern of numbers appears? In this essay we present a few more appearances of the Fibonacci numbers (some of which are possibly new to the world). We do this first as a series of puzzlers to mull upon, which is Part I of this essay. In Part II we present one final puzzler also intimately connected to Fibonacci numbers, a classic, about path counting, and then show in part III how this visual interpretation of these numbers unites the first twelve puzzles and explains their Fibonacci connection in one fabulous and spectacular fell swoop. Part IV ends the essay with some ideas for research. Just to be clear with notation we set:
F1 = 1, F2 = 1 with Fn = Fn 1 + Fn 2 for n 3 .
When it is needed well also declare F0 = 0 .
## PART I: THE PUZZLERS
1. PLACING PARENTHESES There are 13 ways to place parentheses within a string of three symbols avoiding nested of parentheses:
There are 2 ways to accomplish this feat with a string one symbol long: * and (*); and 5 ways with a string of two symbols: ** (*)* *(*) (**) and (*)(*). Verify that there are 34 ways to place parentheses in a string of four symbols. We have so far:
Every second Fibonacci number? 2. DOUBLING ONES IN PARTITIONS There are 4 ways to write the number three as a sum of positive integers, with order of terms considered important. (That is, there are 4 ordered partitions of three.)
3 = 2 +1 = 1+ 2 = 1+1+1
(ASIDE CHALLENGE: Show, in general, there are 2 N 1 ways to write a positive integer N into a sum of positive integers with order relevant.) Suzy is fond of the number 1 and will write it either using her red pen or her blue pen. She writes all other numbers with black pen. Given that there are now two different types of number 1 we find that there are 13 ways to partition the number three:
3 = 2 +1 = 2 +1 = 1+ 2 = 1+ 2 = 1 + 1 + 1 = 1 +1 + 1 = 1 + 1 +1 = 1 +1 +1 = 1 + 1 + 1 = 1 +1 + 1 = 1 + 1 +1 = 1 + 1 + 1
Thirteen is a Fibonacci number. Coincidence?
## James Tanton 2011
3. ABEEBA The language of ABEEBA contains only three letters, A, B and E, and most any combination of letters is a word in this language. The only stipulation is that the letter A can never be immediately followed by the letter E. For example, AABBBEEEBBBE and AABEA are words in ABEEBA, as is the name of the language itself, but AAABBAE and BEEEAEAAA are not. There are 3 one-letter word (A, B, and E), 8 two-letter words (AA, AB, BA, BB, BE, EA, EB, EE) and there are 21 three-letter words (all 27 strings of three letters but with words of the form AE* and *AE deleted). Prove that there are F2 N + 2 N -letter words in ABEEBA.
4. ABOOBA The language of ABOOBA contains only three letters, A, B and O. Any word in this language begins and ends with A and between any pair of As there are is even number of Os. For example, there is 1 two-letter word (AA), 2 three-letter words (AAA, ABA) and 5 four-letter words (AAAA, ABAA, AABA, ABBA, AOOA). a) Verify that there are 13 five-letter words. b) Prove that there are F2 N 3 N -letter words for N 2 . c) How many N -letter words DONT begin with AA?
5. ORDERED PARTITION PRODUCTS This puzzler was inspired by Sam Vandervelde. If we list all the ordered partitions of the number 3, multiply the terms of each partition together and sum the products, we obtain the answer 8. If we do the same for the number 4, we obtain the answer 21.
## James Tanton 2011
6. PRODUCTS OF DIFFERENCES Write the numbers 1 through N in a list and circle two or more of the numbers. Compute the differences between consecutive circled numbers and multiply them together. For example, with N = 3 , we obtain the following possibilities:
The sum of all products is 5. a) Verify that the sum of all possible products of differences for N = 4 is 17. Let D ( N ) denote the sum of all possible products of differences starting with the numbers 1 through
## N . (So D ( 3) = 5 and D ( 4 ) = 17 .) In general, D ( N ) is not a Fibonacci number.
b) Prove that D ( N ) = F2 N N 7. ODD PARTITIONS There are 8 ordered partitions of the number six into odd parts:
## 5 +1 = 1+ 5 = 3 + 3 = 3 +1+1+1 = 1+ 3 +1+1 = 1+1+ 3 +1 = 1+1+1+ 3 = 1+1+1+1+1+1
If Podd ( N ) denotes the count of odd ordered partitions of N , prove: Podd ( N ) = FN . 8. ONE-LESS PARTITIONS There are 5 ordered partitions of the number six that avoid use of the number 1:
6 = 4+ 2 = 2+ 4 = 3+3 = 2+ 2+ 2
If P ( N ) denotes the count of one-less ordered partitions of N , prove: P ( N ) = FN 1 1 1 While were at it, we should include 9. Let P 2 ( N ) denote the number of ordered partitions of N using only the 1, numbers 1 and 2 . Show that P 2 ( N ) = FN +1 1,
## James Tanton 2011
10. TRICKY FORMULAS Consider every second Fibonacci number (the ones with odd index): 1,2,5,13 34, .. Create triangular arrays with the first few terms of this sequence and add the entries in the array:
Prove:
N F1 + ( N 1) F3 + ( N 2 ) F5 + + 1 F2 N 1 = F2 N +1 1
## For the even-indexed terms: 1, 3, 8, 21, 55, prove:
N + ( N 1) F2 + ( N 2 ) F4 + + 1 F2 N 2 = F2 N
While were at it, we should include 11. Prove: Fa Fb + Fa +1 Fb +1 = Fa +b +1 and 12: Prove that if N is a multiple of a , then FN is a multiple of Fa . CHALLENGE: Find formulas for the quotient and remainder upon dividing FN by Fa even if N is not a multiple of a .
## PART II: THE UNITING PUZZLER
The Fibonacci numbers make an appearance in a classic path counting puzzle: 13. HONEYCOMB WALKS A bee, starting at cell marked S wants to move to the cell marked E in a honeycomb design consisting of two rows of hexagonal cells.
Each step must move to the right and to a neighbouring cell. This gives four possible types of steps: D = downward diagonal U = upward diagonal H = high horizontal step L = lower horizontal step
How many different routes are there from S to E using these four types of steps?
## James Tanton 2011
As with any complex problem it is often easier to start by examining smaller cases. Clearly there is only one route from S to the cell just below it: Take a single D step.
And there are two routes from S to the cell directly to its right: DU or H.
One can check that there are three routes to the next cell along the honeycomb: DL, DUD. HD.
To reach the fifth cell of the honeycomb (the third cell along the top row) one has two options: Either Reach the cell directly to its left and then take a step H to the right. (There are two ways to accomplish this,) or Reach the cell diagonally to its left and then take a diagonal step U. (There are three ways to accomplish this.)
## James Tanton 2011
In general, the number ways to reach any particular cell of the honeycomb must be the sum of the two counts of paths to each of the two cells to its left.
This matches the construction of the Fibonacci numbers: each value at one level is the sum of the values at the two previous levels. (And since our path counting pattern starts with the numbers 1, 2, 3 ,5 , the same as the Fibonacci numbers, we will indeed continue to match the Fibonacci numbers forever more.) It is compelling to say that there is 1 way to walk to cell S starting at cell S. (This matches the Fibonacci numbers precisely.) If we number the cells 1, 2, 3, 4, in a zig-zag pattern from left to right, then we can say:
## The number of paths from the leftmost cell to the N th cell is FN .
The count of paths that end on an upper cell is a Fibonacci number with an even index. The count of paths that end on a lower cell is a Fibonacci number with an odd index.
## PART III: INTERPRETING PATHS
For convenience, lets draw dots instead of hexagonal cells and denote the zigzag of cells with dotted line segments,
If we are walking to the N th dot, there are N 1 line segments along this zigzag line. There are FN honeycomb paths to that N th dot.
ASIDE: Consider the ( a + 1) th dot along the zigzag line and assume that N = a + b + 1 so that there are
## a dots to its left and b dots to its right.
Any one of the FN honeycomb paths in this diagram must either got through the ( a + 1) th dot or avoid it. To walk a path that goes through this ( a + 1) th dot we must first walk to it in one of Fa +1 ways, and then walk the a path from it through the remaining b dots, which can be done in one of Fb +1 ways. To walk a path that avoids this ( a + 1) th dot, we must take a horizontal step opposite it. This means first walking one of Fa ways to the a th dot, taking a horizontal step, and walking a path along the remaining
## James Tanton 2011
10
Lets now interpret honeycomb paths along the zigzag line in different ways.
PATH INTERPRETATION 1 Think of the zigzag line on N dots as a string of N 1 line segments. Each U or D step in a path to the N th dot traces a single line segment and each L or H step skips over two line segments. We thus see that each path corresponds to a way of writing N 1 as a sum of 1s and 2s. (And conversely, each partition of N 1 into 1s and 2s corresponds to a path.)
As there are FN paths, there are this many partitions of N 1 . If we adjust the index by one, we have explained puzzle 9: 9. There are FN +1 ways to partition N using the numbers 1 and 2 with order considered relevant.
QUESTION: How many ways are there to partition N into 1s and 2s if order is NOT considered relevant?
QUESTION: In how many ways can one tile a 2 N strip of square cells with 1 2 dominos?
## James Tanton 2011
11
PATH INTERPRETATION 2 Think of the zigzag line on N dots as a string of N dots with the N 1 line segments as the spaces between the dots. For any path to the N th dot interpret a U or a D step as a separator.
Since U and D steps move the path from the top to bottom row, or vice versa, there are an odd number of dots between each separator. Thus we see that honeycomb paths to the N th dot correspond to partitions of N into odd parts. (And, conversely, any ordered partition of N into odd parts can be interpreted as a path to the N th dot.) As there are FN paths in all, we have established puzzler 7: 7. Podd ( N ) = FN
## James Tanton 2011
12
PATH INTERPRETATION 3 Place false line segments at the beginning and end of a zigzag line of N 1 dots and think of each dot as a space between the ( N 2 ) + 2 = N line segments.
For any honeycomb path from the first to the ( N 1) th dot circle the dots it misses.
The first dot will not be circled, the final dot will not be circled, and no two dots that are consecutive along the zigzag line will both be circled. The circled dots thus partition the N line segments into parts containing 2 or more segments each. This represents a partition of the number N that avoids use of 1.
And, conversely, any ordered partition of the number N + 1 into parts of size greater than 1 corresponds to a honeycomb path to the ( N 1) th dot. This correspondence between partitions and paths solves puzzler 8: 8. P ( N ) = FN 1 1
## James Tanton 2011
13
PATH INTERPRETATION 4 Consider the set of honeycomb paths that end on a given odd-numbered dot 2 N + 1 .
There are N dots along the bottom row. If we regard a D step as a left parenthesis and a U step as a right parenthesis, then any honeycomb path to the (2 N + 1) th dot corresponds to a set of non-nested parentheses about the N symbols along the bottom row:
This solves puzzle 1: 1. The number of ways to place non-nested parentheses about some or all of the elements of a string of N symbols is F2 N +1 , the number of paths to the ( 2 N + 1) th dot in the honeycomb.
QUESTION: How many ways are there to place non-nested parentheses among N dots if the last dot must be within a parenthesis pair?
## James Tanton 2011
14
Lets go further and examine honeycomb paths touch certain dots on the bottom row. There is one honeycomb path that touches no dots on the bottom row.
## There are N honeycomb paths that touch just one dot:
Suppose a path touches just two consecutive dots on the bottom row:
There are ( N 1) choices for which pair of points this could be, and between those pairs of points there are F3 ways to connect them with a honeycomb path. (All other steps must be Hs.) Thus there are
( N 1) F3 honeycomb paths that touch two consecutive dots on the bottom row.
Suppose a path touches dots on the bottom row within a span of three dots:
There are ( N 2 ) possible locations for this span of dots and there are F5 honeycomb paths connecting the leftmost and rightmost bottom dots in this range. Thus there are ( N 2) F5 honeycomb paths in all with this property. Similarly, there are ( N 3) F7 honeycomb paths that touch a range of dots on the bottom row 3 places apart, and ( N 4 ) F9 paths that touch a range of dots on the bottom row 4 places apart, and so on. We have now categorized all F2 N +1 honeycomb paths in this diagram, and so we conclude:
1 + N + ( N 1) F3 + ( N 2 ) F5 + + 1 F2 N 1 = F2 N +1
This establishes the first part of puzzle 10: 10 a) N F1 + ( N 1) F3 + ( N 2 ) F5 + + 1 F2 N 1 = F2 N +1 1
## James Tanton 2011
15
PATH INTERPRETATION 5 Again consider honeycomb paths to a given odd numbered dot 2 N + 1 . There are N spaces between the N + 1 dots of the top row. Any honeycomb path to the ( 2 N + 1) th dot matches a partition of N with two types of one:
## James Tanton 2011
16
PATH INTERPRETATION 6 Consider the set of all honeycomb paths that end on a given even numbered dot 2 N + 2 .
Any honeycomb path is certainly described by a sequence of moves given by the letters D, U, L or H. In fact, one need not mention the letter D when listing the letters of a path! For example, from UUHLU and knowing that we are ending on a dot on the bottom row, we deduce that the path must contain the segments as shown below:
[The segment for the very first U must be at a position as to the far left as possible, otherwise a letter L or H must have appeared before it. In fact each line segment for each letter mentioned must appear as far to the left as it can.] We see that UUHLU must be the path DUDUHDLUD. [No double Ds will appear as they would contain a U between them, but such a U would have been mentioned.] Thus every path can be described in as a word with the three letters: U, L, and H. One checks that all pairs UU, UH, UL, HU, HH, HL and LU, LL can appear in a word, but the pair LH cannot. [We need a U between them.] Thus paths can be described as words in the language of ABEEBA with A = L, E = H and B = U!
ONE COMPLICATION: Do all N -letter words in ABEEBA correspond to paths of the same length? Suppose we have an N -letter word with a Us and a total of N a Ls and Hs. Since our honeycomb paths are ending on the bottom row, each path does wander between top and bottom rows. Each U that appears will be counteracted with one D, plus there will be an additional D to end us on the bottom row. Thus the complete word for the path will contain a + 1 Ds. Since each L and H is worth two steps along the zigzag line, the word with a Us, a + 1 Ds and N a Ls and Hs will be:
## James Tanton 2011
17
2 ( N a ) + a + a + 1 = 2N + 1
steps long. Thus all N -letter words correspond to paths ending on cell 2 N + 1 . As there are F2 N +1 honeycomb paths to this cell we have: 3. There are F2 N +1 N -letter words in the language of ABEEBA.
QUESTION: Quentin has a collection of wooden blocks painted red, blue and orange. Each red and each blue block is a 1 1 1 cube but the orange blocks come in all lengths: 1 1 k for k = 1, 2, 3, 4, ... . Quentin likes to line up his blocks in long rows to make trains. How many trains can he make of length N ? How many trains of length N can he make if a red block is never to be immediately followed by a blue block? [Assume Quentin has an infinite supply of each coloured black.]
## James Tanton 2011
18
PATH INTERPRETATION 7 The language of ABOOBA is easier to analyse. Consider honeycomb paths that end of a cell in the top row and suppose there are N + 1 dots along that top row (with N spaces between them). Label a path with the letters A, B and O above the spaces between the dots of the top row as suggested by the diagram:
Precisely: Each space covered by an H step is labeled A. Each space with one D or one U step underneath it is labeled O. Each space with two or zero diagonal steps below it is labeled B. OR going back to ordered partitions with two types of one set
## 1= A 1= B 2 = OO 3 = OBO 4 = OBBO 5 = OBBBO
There must be an even number of Os between any two As, as well as to the left of the leftmost A and to the right of the rightmost A. Thus the number of N -letter words with the letters A, B and O, with the Os satisfying this condition is the number of honeycomb paths to this ( N + 1) th dot on the top row. There are F2 N +1 such paths. This almost solves the ABOOBA problem. An N -letter word in ABOOBA contains an irrelevant beginning A and an irrelevant ending A: only the middle N 2 letters are of interest to us for paths. Thus we must adjust our index by 2. 4. There are F2( N 2 )+1 = F2 N 3 N -letter words in the language of ABOOBA. CHALLENGE: Answer part b) of the puzzler by looking at paths that end on the bottom row. (Place an imaginary additional dot to the left of the bottom row.)
## James Tanton 2011
19
PATH INTERPRATION 8 Consider honeycomb paths that end on a lower cell, cell number 2N , and the possible placement of the U steps in these paths.
Between any two Us, and to the left of the leftmost U and to the right of the rightmost U, there is a single D. In the above diagram with N = 12 (a zigzag line of 24 steps) there are 2 possible of choices for the placement of a D in the leftmost section, 4 choices for the placement of D in the second section, 1 in the third, 2 in the fourth and 3 in the final section. Thus there are
2 4 1 2 3
possible honeycomb paths with Us in these given positions. For a honeycomb path possessing every possible U step (with N = 12 still) there is just 1 1 1 1 1 1 1 1 1 1 1 1 option for what that path could be and for a honeycomb path containing no U steps there are 12 options. (There still must be a single D at some location.) Every partition of the number 12 corresponds to a count of honeycomb paths with Us in specific positions. In general, as there are a total number of honeycomb paths to cell 2N , the sum of all possible partitions of N with terms multiplied together must give F2 N . This solves puzzle 5. 5. Multiplying the terms of each partition of N and summing all the products yields F2 N .
## James Tanton 2011
20
PATH INTERPRETATION 9 Consider honeycomb paths that end on a lower cell, cell number 2N , and the possible placement of the D steps in these paths. Each path has at least one D.
## There are N paths with exactly one D.
If a path has two or more Ds, then there needs to be exactly one U between each pair of neighbouring Ds. Now, if we number of the dots on the top row 1 through N , choosing a D corresponds to circling one of these numbers. For two neighbouring Ds, the possible number of placements of a U between them is the difference of the two circled numbers for those Ds.
All other steps in the path once the locations of Ds and Us has been set must be Ls and Hs. Thus the number of honeycomb paths with two or more Ds fixed in position is given by the product of the differences between neighbouring circled numbers. This accounts for all F2 N possible paths and we have established puzzle 6: 6. F2 N = N + sum of all possible products of differences from circling two or more terms in the list of numbers 1 through N. QUESTION: Suppose we list all the numbers 1 through N, circle two or more numbers, and sum all possible products of differences of consecutive circled terms. Suppose we insist that the number 1 always be circled. Show that this sum of products of differences is sure to be F2 N 1 1 .
## James Tanton 2011
21
Going further Consider the span of possible circled numbers on the top row the number of dots from the leftmost circled dot to the rightmost circled dot. As we have seen, there are N honeycomb paths containing exactly one D. That is, there are N paths with a span of one.
Circling just two consecutive terms creates a picture with a span of two.
There are ( N 1) paths with just two Ds positioned this way. A picture with a span of three might or might not have the middle dot in the spread circled.
There are F4 paths between the two outer circled terms and there are ( N 2 ) possible positions for this span of three. Thus there are ( N 2 ) F4 honeycomb paths with a span of three. Similarly there are ( N 3) F6 honeycomb paths with a span of four.
## N + ( N 1) F2 + ( N 2 ) F4 + + 1 F2 N 2 = F2 N proving the second part of puzzler 10.
10 b) N + ( N 1) F2 + ( N 2 ) F4 + + 1 F2 N 2 = F2 N
## James Tanton 2011
22
ASIDE: Suppose N = qa + r for some positive integer a and for integer r with 0 r < a . (Thus r is the remainder that appears with N is divided by a and q is the quotient.) Lets find the remainder and quotient in dividing FN by Fa . Consider the FN honeycomb paths along N dots. How many paths go through the a th dot? Answer: Fa FN +1 a
How many path dont go through the a th dot but do go through dot 2a ? Answer: Fa 1 Fa FN +1 2 a
How many paths dont go through dots a and 2a but do go trhough dot 3a ? Answer: ( Fa 1 ) Fa FN +13a
2
And so on, all the way to How many paths dont go through dots a , 2a , 3a , , ( q 1) a but do go trhough dot qa ? Answer: ( Fa 1 ) and How many paths dont go through dots a , 2a , 3a , , ( q 1) a , and qa ? Anwser: ( Fa 1 ) Fr
q q 1
Fa FN +1 qa = ( Fa 1 )
q 1
Fa Fr +1 .
## This accounts for all FN paths and so:
FN = Fa FN +1 a + Fa 1 FN +1 2 a + ( Fa 1 ) FN +13a + + ( Fa 1 )
q 1
Fr +1 + ( Fa 1 ) Fr
23
q 2 q 1
F1 .)
## More generally, FN leaves a remainder of ( Fa 1 ) Fr upon division by Fa . (Here N = qa + r .) That is:
FN ( Fa 1 ) Fr mod Fa .
q
## PART IV: OTHER INTERPRETATIONS? OTHER IDENTITIES?
Might you be able to interpret honeycomb paths in other ways to establish other curious Fibonacci appearances? Might you be able to establish new Fibonacci identities a la the identities of puzzler 10 and 12? Here are some classic Fibonacci identities. Might any of these be proved via path walking?
## F1 + F2 + F3 + + FN = FN + 2 1 F1 + F3 + F5 + + F2 N 1 = F2 N F0 + F2 + F4 + + F2 N = F2 N +1 1 F1 + 2 F2 + 3F3 + + NFN = NFN + 2 FN +3 + 2 FN 1 FN +1 = ( FN ) + ( 1)
2 N
F12 + F2 2 + + FN 2 = FN FN +1
gcd ( Fa , Fb ) = Fgcd ( a , b ) | 6,362 | 22,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-30 | latest | en | 0.894891 |
https://www.jiskha.com/display.cgi?id=1233517407 | 1,516,497,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00504.warc.gz | 904,320,775 | 4,274 | # Literature;Crucible.
posted by .
i have to pick one character from the Crucible who decides that they have chosen the wrong action and change their behavior, this switch is often motivated by a number of factors...i have to tell their original course of action and how they change, their motivation [originally and after the change]...can you tell me who would be the best character to pick. someone who changes alot during the play..thanks
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