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# homework-08 - H OMEWORK 8 120202 ESM4A N UMERICAL M ETHODS...
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H OMEWORK 8 120202: ESM4A - N UMERICAL M ETHODS Spring 2009 Prof. Dr. Lars Linsen Zymantas Darbenas School of Engineering and Science Jacobs University Due: Friday, April 24, 2009 at noon (in the mailbox labeled “Linsen” in the entrance hall of Research I). Problem 22: Spline interpolation. (10 points) (a) Derive the collocation matrix for periodic quadratic spline interpolation using B-spline represen- tations over the knot sequence 3 2 , 5 2 , 7 2 , 9 2 , 11 2 . (b) Given knots u i = i + 2 for i = 1 , . . . , 4 , points ( p 1 , . . . , p 4 ) = (1 , 4 , 1 , 3) , and endpoint derivatives d 1 = d 4 = 1 . Use clamped cubic spline interpolation to compute an interpolating spline in B-spline representation. Sketch the graph of the spline curve and the control polygon. Problem 23: Uniform subdivision. (10 points) (a) Given uniform B-splines N n i ( u ) over knot sequence Z and uniform B-splines M n i ( u ) over knot sequence 1 2 Z with summationdisplay i c i N n i ( u ) = summationdisplay i b n i M n i ( u ) . Show that b n i = 1 2 ( b n - 1 i + b n - 1 i +1 ) with b 0 2 i = b 0 2 i +1 = c i for
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Ask a homework question - tutors are online | 458 | 1,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-13 | latest | en | 0.724658 |
https://www.reference.com/science/kind-motion-should-impart-stretched-coil-spring-provide-transverse-longitudinal-wave-9f438a1cd850b987 | 1,484,966,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280899.42/warc/CC-MAIN-20170116095120-00161-ip-10-171-10-70.ec2.internal.warc.gz | 973,937,736 | 21,045 | Q:
# What kind of motion should you impart to a stretched coil spring to provide a transverse and a longitudinal wave?
A:
In a stretched coil spring, longitudinal waves occur when the spring is pulled or pushed parallel to the line created by the stretched spring, and transverse waves are created when the spring is moved perpendicular to that line. If the coils bunch and then separate, the wave is longitudinal; if the spring curves to form moving crests and troughs, the wave is transverse.
## Keep Learning
In longitudinal waves, also called compression waves or primary waves, particles experience compression and rarefaction in the direction of the traveling wave. This phenomenon can be easily observed in a stretched spring or objects such as a Slinky, and it can illustrate how waves, such as sound waves, travel.
Transverse waves, sometimes called secondary waves, move across the direction of the wave. The particles move perpendicular to the direction of the traveling wave.
According to Daniel Russell of Pennsylvania State University, the motion of a water surface has examples of both longitudinal and transverse waves. When observed from above, a ripple looks like a longitudinal wave because the waves are pushing away from the center of the ripple, and the particles are moving in the direction of the wave. However, when viewed from the side, the wave looks like it is moving up and down as the wave travels across the surface of the water. Because some of the motion of the water is perpendicular to the direction of the wave, a ripple is also a transverse wave.
Sources:
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## #1 2016-09-30 06:16:04
بِسمِ اللَّهِ الرَّحمٰنِ الرَّحيمِ
From: Sahiwal Division
Registered: 2012-03-22
Posts: 22,147
Website
# General Knowledge QuizzesSequence Series
Directions:
Look carefully for the pattern, and then choose which pair of numbers comes next.
Question:
28 25 5 21 18 5 14
Option A):
11 8
Option B):
10 7
Option C):
5 10
Option D):
10 5
Option E):
11 5
11 5
Explanation:
This is an alternating subtraction series with the interpolation of a random number, 5, as every third number. In the subtraction series, 3 is subtracted, then 4, then 3, and so on.
Online Web Tutorials And Interview Questions With Answers Forum:
https://www.globalguideline.com/forum/
Offline | 220 | 719 | {"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 | 0.71378 |
https://www.physicsforums.com/threads/how-do-two-light-beams-combine-at-a-shear-angle.816186/ | 1,721,095,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00428.warc.gz | 822,818,306 | 17,451 | # How do two light beams combine at a shear angle?
• stedwards
In summary, the conversation discusses how two light beams combine at a shear angle and the resulting increase in energy density. The participants also bring up the principles of intensity, energy conservation, and interference. They also mention how this concept applies to other types of waves such as sound and water waves.
stedwards
How do two light beams combine at a shear angle?
Two electromagnetic beams cross at a shear angle. They have equal phase, intensity and polarization.
The angle is shear enough, so that in region in which they intersect, there is less than a quarter wave difference in phase over the cross-sectional area.
You can just take the superposition of each individual wave.
The combined energy [if that were so] would nearly double, wouldn't it?
Specifically, how do the electric and magnetic field components combine.
stedwards said:
The combined energy [if that were so] would nearly double, wouldn't it?
Specifically, how do the electric and magnetic field components combine.
The combined energy would be exactly double. See http://en.wikipedia.org/wiki/Poynting's_theorem
This is actually easiest to think about for a finite duration square wave pulse, IMO. The fields add together, and the energy density increases, but the volume decreases, so the end result is twice the total energy of a single pulse.
nasu said:
What is this "shear angle"?
It does not looks like you mean the usual meaning of it.
http://encyclopedia2.thefreedictionary.com/shear+angle
Really? It's thefreedictionary. Place two lasers closely together. Adjust the beams so they intersect at the other end of the optical bench or much further.
stedwards said:
The combined energy [if that were so] would nearly double, wouldn't it?
Specifically, how do the electric and magnetic field components combine.
DaleSpam said:
The combined energy would be exactly double. See http://en.wikipedia.org/wiki/Poynting's_theorem
This is actually easiest to think about for a finite duration square wave pulse, IMO. The fields add together, and the energy density increases, but the volume decreases, so the end result is twice the total energy of a single pulse.
1. No, I’m asking about combing the E and B fields at the beam intersection, not the intensities. The beam intensity at the intersection is not double the two contributing intensities, creating energy out of nothing, as we both know. So what went wrong?
If we naively add the two field amplitudes, the intensity nearly quadruples--shy of quadrupling due to phase variation across the intersection. I believe the error in this idealized set-up is from failure to consider the source apertures, but its just a guess.2. I post this thread is in sequel to https://www.physicsforums.com/threads/splitting-and-combining-em-waves-amplitude-intensity.815517. I appreciated BvU responses and references, but these did not satisfy the original poster nor I. We seem to have three mutually inconsistant principles.
1) $I_m = {A_m}^2$, $I_\Sigma = {A_\Sigma}^2$ –intensity (energy) is equal to the square of the amplitude
2)$I_{\Sigma} = I_{1} + I_{2}$ –conservation of energy
3)$A_{Sigma} = A_{1} + A^{2}$ –interference or superposition principleCombining, $2A_1 A_2 = 0$, which says that either $A_1$, $A_2$ or both must be zero. What happened?
stedwards said:
Really? It's thefreedictionary. Place two lasers closely together. Adjust the beams so they intersect at the other end of the optical bench or much further.
I did not give the reference to dictionary as an "authority" about the meaning but just to show the meaning that I was familiar with.
I understand now that you mean "a very small angle" when you mean a shear angle. I suppose it is a common use in your field.
sorry
Anyone?
Taken, in general, this is a broad concern beyond electromagnetic radiation to include: Sound Waves in air, solid material Transverse Waves and Electrical Power, Water Waves... and eventually quantum mechanics. Anything else?
There seems to be some general principle that never came up in school.
Last edited:
## Question 1: What is a shear angle?
A shear angle is the angle formed between two light beams when they intersect and combine.
## Question 2: How do two light beams combine at a shear angle?
When two light beams intersect at a shear angle, they combine by overlapping and interacting with each other's electric and magnetic fields, resulting in a new combined light beam with a different intensity and polarization.
## Question 3: What factors affect the combination of light beams at a shear angle?
The combination of light beams at a shear angle can be affected by the intensity, wavelength, and polarization of the individual light beams, as well as the angle at which they intersect and the properties of the medium they are passing through.
## Question 4: What is the difference between coherent and incoherent combination of light beams at a shear angle?
In coherent combination, the light beams have the same wavelength, phase, and polarization, resulting in a constructive interference and a stronger combined beam. In incoherent combination, the light beams have different wavelengths, phases, and polarizations, resulting in a less intense and more random combination.
## Question 5: What are the practical applications of combining light beams at a shear angle?
Combining light beams at a shear angle can be used in various applications such as optical communications, holography, and interferometry. It can also be used to manipulate and control the properties of light, leading to advancements in technologies such as lasers and optical sensors.
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1K | 1,383 | 6,090 | {"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} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.955178 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=56988 | 1,611,552,065,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00470.warc.gz | 426,746,498 | 11,696 | ## Homework 6D.7a
Moderators: Chem_Mod, Chem_Admin
Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am
### Homework 6D.7a
Find the initial concentration of the weak acid or base in the following aqueous solutions. a) solution of HClO with pH = 4.60. I know how to find the concentration of H3O+ using the given pH but I don't know what to do after this. Am I supposed to look up the Ka and set up a Ka expression with the initial concentration of the weak acid as the variable?
JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am
### Re: Homework 6D.7a
I believe you have to search up the Ka value. I found it to be 3.0*10^-8. Use the hydronium concentration you found and plug it into the ice table. Then take the values from the ice table and set up an equation equal to Ka to solve for the initial concentration
Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am
### Re: Homework 6D.7a
Yes, I believe that you should look up the Ka. I think that questions such as these are supposed to provide the Ka in the problem unless previously stated in a table of Ka, but I can't seem to find the Ka for HClO in the textbook.
After finding Ka, set up the Ka expression using information deduced by an ICE table. HClO should be represented as x and its final component should be described as $x-2.5\cdot 10^-5$.
Solving for x will give you the initial concentration in question
Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am
### Re: Homework 6D.7a
Yes, you are on the right track. For HClO, the Ka = 3.0 x 10^-8, and the variable in the Ka expression would be the amount of molar concentration of HClO, which you can find since you are given the pH. So, it would be Ka = 3.0 x 10^-8 = [HClO]^2/x-[HClO], and you would just solve for x.
Return to “Equilibrium Constants & Calculating Concentrations” | 549 | 1,837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.94765 |
https://dsp.stackexchange.com/questions/64237/what-is-the-difference-between-the-convolution-and-differentiation-over-image-in | 1,718,755,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00549.warc.gz | 189,094,713 | 41,024 | # what is the difference between the convolution and differentiation over image in image processing?
In image processing the differentiation and convolution are the terms that are used interchangeably.
What is the difference between applying convolution and differentiation over the image?
How we perform the differentiation over the image(Since in order to differentiate we need to represent it as equation that is differentiable How we do it?) ?
• The differentiation that is referred is nomerical. Numerical differntiation can be implemented with different coefficients. Any chosen set of coefficients can be imemented as filter, used with convolution Commented Feb 29, 2020 at 20:14
• Where did you see someone saying they are the same? Finite Differences, which can be seen as a discrete approximation to the derivative can be applied using Convolution. But certainly they are not the same.
– Royi
Commented Mar 1, 2020 at 5:52
• Images are discrete, so discrete approximation is used to approximate the differentiation. The expression can then, after analysis, be expressed with a certain matrix, and the approximated differentiation process can be achieved by convolution with that matrix. Commented May 2, 2020 at 3:12
In this context, convolution refers to the operation of LTI filtering performed on the image by the filter impulse response. Filtering can have many purposes such as blurring, sharpening, noise reduction etc.
Certain applications requires that you compute an approximation to the derivative of the image data. This can be accomplished by filering the image with a specific filter kernel (impulse response) that's some sort of a high pass characteristics.
Therefore, computing the (approximate) derivative of an image can be accomplished by LTI filtering with a highpass impulse response, which refers to a convolution operation. This is the connection in between.
Generally, differentiation can be seen a a special case of convolution.
In sampled images, and discrete signals in general, differentiation is made in a discrete manner: it becomes a discrete approximation of what you'd get if the images were continuous. The classical sampling context is linear, convolution reflects linearity, and a lot of classical discrete derivatives are linear combinations of pixel values. So, those derivatives can be implemented as convolutions, and convolutions commute.
In other words: as derivation may emphases noise, some believe that one should filter or smooth data before applying a derivative kernel. But with linear filters and derivatives, you can use any order, and you can even combine then into a single "smoothed derivative" operator.
However, remember that image processing sometimes uses non-linear derivatives or gradient estimators. In those cases, they don't communte with convoluton anymore.
• In other words: convolution is a general operation that implements linear time-invariant filtering: any linear time-invariant filter can be reduced to a convolution. Differentiation is a specific linear time-invariant filtering operation, so it can, of course, be implemented using convolution. Commented Mar 1, 2020 at 15:30
• Sounds way clearer than that I wrote Commented Mar 1, 2020 at 16:24 | 644 | 3,239 | {"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-2024-26 | latest | en | 0.952704 |
http://www.physicsforums.com/showthread.php?t=240840 | 1,410,881,358,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657118605.64/warc/CC-MAIN-20140914011158-00348-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 748,733,966 | 6,761 | # Vector Proof Centroid Theorem
by pjallen58
Tags: centroid, proof, theorem, vector
P: 12 I am trying to figure out how to prove the 2:1 ratio of a triangle's medians at the centroid using vectors. Example if I had a triangle ABC with midpoints D of BC, E of AC and F of AB. I know G is where the medians intersect. I have seen many proofs and understand the process that proves the addition of the vectors from the centroid to the vertices are zero i.e. GA+GB+GC=0. Does this prove the 2:1 ratio? I cannot find anything explaining how to prove the actual 2:1 ratio. I am not sure if I am missing something or what. Any help or suggestions would be appreciated. Thanks.
HW Helper
Thanks
P: 26,148
Quote by pjallen58 I am trying to figure out how to prove the 2:1 ratio of a triangle's medians at the centroid using vectors. Example if I had a triangle ABC with midpoints D of BC, E of AC and F of AB. I know G is where the medians intersect. I have seen many proofs and understand the process that proves the addition of the vectors from the centroid to the vertices are zero i.e. GA+GB+GC=0. Attachment 14402 Does this prove the 2:1 ratio? I cannot find anything explaining how to prove the actual 2:1 ratio. I am not sure if I am missing something or what. Any help or suggestions would be appreciated. Thanks.
Hi pjallen58!
Yes, because, for example, a + 2(d) is 1/3 of the way,
and that's a + 2(1/2(b + c)), = a + b + c!
Related Discussions Calculus & Beyond Homework 3 Calculus & Beyond Homework 1 Calculus & Beyond Homework 9 Introductory Physics Homework 7 Introductory Physics Homework 10 | 417 | 1,603 | {"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-2014-41 | latest | en | 0.921598 |
https://serverfault.com/questions/403724/number-of-subnets-and-total-hosts | 1,566,511,586,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317516.88/warc/CC-MAIN-20190822215308-20190823001308-00005.warc.gz | 633,951,429 | 26,590 | # Number of Subnets and total hosts [duplicate]
Possible Duplicate:
How does Subnetting Work?
Can someone please tell me what does this term means "A Class C network with 10 subnets" What exactly is 10 subnets? I know a subnet is something like 255.255.255.0 and in case of class C network it can host 254 computers. Is my calculation true?
So if we have 10 subnets it means we can have 2540 computers in the network? | 106 | 420 | {"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-2019-35 | latest | en | 0.951199 |
https://cracku.in/33-who-amongst-the-following-sit-exactly-between-t-an-x-ibps-rrb-clerk-2016-shift-3 | 1,721,667,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00568.warc.gz | 163,727,328 | 23,804 | Instructions
Study 1 the following information careful’ ly and answer the questions given below :
Nine persons – G, H, I, J, K, R, S, T and U — are seated in a straight line facing North, with equal distance between each other but not necessarily in the same order.
Only two persons sit to the left of I. Only one person sits between I and U. H sits fourth to the right of R. R is not an immediate neighbour of U. Less than three persons sit between R and U. Number of persons sitting between I and U is half as that between H and J. Only three persons sit between K and T. K is not an immediate neighbour of J. Only two persons sit between T and G.
Question 33
# Who amongst the following sit exactly between T and G?
Solution
Only two persons sit to the left of I, => I sits at the third position from left end.
Only one person sits between I and U, => U sits at the extreme left end of the line.
Since, less than 3 people sit between R and U, and R is not an immediate neighbor of U, => R sits third to the right of U and also, H sits fourth to the right of R.
Number of persons sitting between I and U is half as that between H and J, => 2 persons are sitting between H and J.
Since, K is not an immediate neighbor of J, => K sits second to the right of U and T sits to the immediate right of J.
Only two persons sit between T and G, => G sits at the extreme right end of the line and the remaining place is occupied by S.
The arrangement :
Clearly, S and H are sitting between T and G.
=> Ans - (D) | 367 | 1,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.96751 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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• how to do slopes in algebra | 943 | 3,911 | {"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} | 2.53125 | 3 | CC-MAIN-2018-34 | longest | en | 0.881231 |
https://www.spec2000.net/water-resistivity/formation-water-resistivity-from-spontaneous-potential.htm | 1,701,248,965,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00804.warc.gz | 1,139,706,991 | 5,379 | Water Resistivity FROM SP LOG
Calculation from knowledge of the SP value in a clean zone has been a traditional method for finding RW. It works best in clean water bearing zones, but the Rwa or R0 method would be better in this case. Shale content and hydrocarbon content reduce the SP value and cause RW to be too high, giving pessimistic saturation results. However, if you have no other helpful RW data, start here and use good judgement.
RW from SP MODEL
This is the official 1980 vintage algorithm courtesy of the Bateman and Konen programmable calculator paper. It honours the chartbook solution quite closely:
1: FT = SUFT + (BHT - SUFT) / BHTDEP * DEPTH
2: IF LOGUNITS\$ = "METRIC"
3: THEN FT1 = 9 / 5 * FT + 32
4: OTHERWISE FT1 = FT
5: KSP = 60 + 0.122 * FT1
6: RSP = 10 ^ (-SSP / KSP)
7: IF RMF@FT > 0.1
8: THEN RMFE = 0.85 * RMF@FT
9: IF RMF@FT <= 0.1
10: THEN RMFE = (146 * RMF@FT - 5) / (337 * RMF@FT + 77)
11: RWE = RMFE / RSP
12: IF RWE > 0.12
13: THEN RW@FT = - (0.58 - 10 ^ (0.69 * RWE - 0.24))
14: IF RWE <= 0.12
15: THEN RW@FT = (77 * RWE + 5) / (146 - 337 * RWE)
Where:
BHT = bottom hole temperature (degrees Fahrenheit or Celsius)
BHTDEP = depth at which BHT was measured (feet or meters)
DEPTH = mid-point depth of reservoir (feet or meters)
FT1 = formation temperature (degrees Fahrenheit)
KSP = temperature factor (degrees Fahrenheit)
RSP = RWE / RMFE (fractional)
RMFE = equivalent mud filtrate resistivity (ohm-m)
RMF@FT = mud filtrate resistivity at formation temperature (ohm-m)
RWE = equivalent water resistivity (ohm-m)
RW@FT = water resistivity at formation temperatures (ohm-m)
SSP = static SP reading in clean zone (ohm-m)
SUFT = surface temperature (degrees Fahrenheit or Celsius)
This algorithm should only be used IF the SP log has sufficient character, the zone of interest is a clean water bearing sandstone, and the result is reasonable for the area. Use caution since many SP logs are not calibrated, and RMF or RW can be measured carelessly.
Solution of these formulae can be done on the two charts below. An Excel macro for RW from SP, published by SPWLA, is located HERE.
Nomographs for RW from SP
NUMERICAL EXAMPLE:
1. Data for Sand C:
SUFT = 25 degrees C
BHT = 65 degrees C
BHTDEP = 2225 m
DEPTH = 1000 m
RMF = 0.75 @ 25 degrees C
SP = -90 mv
FT = 25 + (65 - 25) / 2225 * 1000 = 43 degrees C
FT1 = 9 / 5 * 43 + 32 = 109 degrees F
RMF@FT = 0.75 * (25 + 21.5) / (43 + 21.5) = 0.54
KSP = 60 + 0.122 * 109 = 73.3
RSP = 10 ^ (90 / 73.3) = 16.9
RMFE = 0.85 * 0.54 = 0.46
RWE = 0.46 / 16.9 = 0.027
RW@FT = (77 * 0.027 + 5) / (146 - 337 * 0.027) = 0.051
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posted by .
The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find the lengths of the sides of the parallelogram. (Hint: The diagonals bisect each other.)
• Trig -
if the sides have length a and b,
a^2 = 6^2 + 3.5^2 - 2(6)(3.5)cos42
b^2 = 6^2 + 3.5^2 + 2(6)(3.5)cos42
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posted by .
If it takes painter # 1 three hours to paint a house, and painter # 2 eight hours to paint a house, how long would it take them if they work together?
• Math -
#1's rate = house/3
#2's rate = house/8
combined rate = house/3 + house/8
= 11house/24
so time for combined effort = house/(11house/24)
= 24/11 hours
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https://brilliant.org/problems/x-y-and-x-ponents/ | 1,500,591,172,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423512.93/warc/CC-MAIN-20170720222017-20170721002017-00149.warc.gz | 630,382,234 | 18,182 | # $$x$$, $$y$$, and $$x$$-ponents
Algebra Level 1
If $$\frac{4^{x}}{2^{x + y}}$$ $$= 8$$ and $$\frac{9^{x + y}}{3^{5y}}$$ $$= 243$$, where $$x$$ and $$y$$ are real numbers, then what is the sum $$x + y$$?
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# Reverse the values of an Object in R Programming – rev() Function
• Last Updated : 17 Jun, 2020
rev() function in R Language is used to return the reverse version of data objects. The data objects can be defined as Vectors, Data Frames by Columns & by Rows, etc.
Syntax: rev(x)
Parameter:
x: Data object
Returns: Reverse of the data object passed
Example 1:
# R program to reverse a vector # Create a vectorvec <- 1:5 vec # Apply rev() function to vector vec_rev <- rev(vec) vec_rev
Output:
[1] 1 2 3 4 5
[1] 5 4 3 2 1
Example 2: Reverse Columns of a Data Frame
# R program to reverse # columns of a Data Frame # Creating a data.framedata <- data.frame(x1 = 1:5, x2 = 6:10, x3 = 11:15)data # Print reversed example data framedata_rev <- rev(data) data_rev
Output:
x1 x2 x3
1 1 6 11
2 2 7 12
3 3 8 13
4 4 9 14
5 5 10 15
x3 x2 x1
1 11 6 1
2 12 7 2
3 13 8 3
4 14 9 4
5 15 10 5
Example 3: Reverse Rows of a Data Frame
# R program to reverse# rows of a data frame # Creating a data framedata <- data.frame(x1 = 1:5, x2 = 6:10, x3 = 11:15)data # Calling rev() & apply() functions combineddata_rev_row_a <- apply(data, 2, rev) data_rev_row_a # Alternative without rev() data_rev_row_b <- data[nrow(data):1, ] data_rev_row_b
Output:
x1 x2 x3
1 1 6 11
2 2 7 12
3 3 8 13
4 4 9 14
5 5 10 15
x1 x2 x3
[1,] 5 10 15
[2,] 4 9 14
[3,] 3 8 13
[4,] 2 7 12
[5,] 1 6 11
x1 x2 x3
5 5 10 15
4 4 9 14
3 3 8 13
2 2 7 12
1 1 6 11
My Personal Notes arrow_drop_up | 684 | 1,708 | {"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} | 2.59375 | 3 | CC-MAIN-2021-39 | latest | en | 0.467381 |
https://philoid.com/question/108650-mark-the-tick-against-the-correct-answer-in-the-following-the-value-of-is | 1,720,980,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00667.warc.gz | 415,457,636 | 8,576 | ##### Mark the tick against the correct answer in the following:The value of is
To Find: The value of
Now, let x =
cos x =cos ()
Here ,range of principle value of cos is [0,]
x = [0,]
Hence for all values of x in range [0,] ,the value of
is
cos x =cos (2) (cos ()= cos () )
cos x =cos () (cos ()= cos )
x =
1 | 101 | 319 | {"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} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.828634 |
https://www.infoplease.com/math-science/mathematics/algebra/equations-with-multiple-variables | 1,718,711,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00516.warc.gz | 750,758,494 | 20,942 | # Equations with Multiple Variables
## Equations with Multiple Variables
So far, when I've asked you to solve an equation for a variable, it was pretty obvious which one I was talking about. For example, to solve the equation 3x + 2 = 23 , you'd solve for (isolate) the x variable. Why? Because it's the only variable in there! That x is enjoying all the attention, like the only girl in an all-boys school.
I need to add another skill to your equation-solving repertoire that will be extremely important in Graphing Linear Equations: how to solve for a variable when there's more than one variable in the equation. Don't worryyou don't need to learn a whole new set of steps to follow or anything; it's basically done the same way you're used to. Here's the only difference: When you're finished, you'll have variables on both sides of the equation, rather than just one side.
Example 4: Solve the equation -2(x - 1) + 4y = 5 for y.
Solution: Start by simplifying the left side of the equation.
• -2x + 2 + 4y = 5
Now it's time to separate the variable term. Since you're trying to isolate the y, eliminate every term not containing a y from the left side of the equation. In other words, add 2x to, and subtract 2 from, both sides of the equation.
• 4y = 2x + 3
It might have felt weird to move that -2x term, but it had to go, since you're solving for y, not x. All that's left to do now is eliminate y's coefficient by dividing both sides by 4.
• y = 2x + 34
##### You've Got Problems
Problem 5: Solve the equation 9x + 3y = 5 for y.
Even though that answer is correct, it will make more sense in Graphing Linear Equations if you rewrite it slightly. Any time two or more things are added or subtracted in the numerator of a fraction, you can break that fraction into smaller fractions, each of which will contain one term of the original numerator and a copy of the denominator. (If the terms contain variables, just stick them to the right of the fraction.)
• y = 24x + 34
• y = 12x + 34
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc. | 585 | 2,265 | {"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} | 4.75 | 5 | CC-MAIN-2024-26 | latest | en | 0.950644 |
https://forums.civfanatics.com/threads/tech-tree-statistics-and-ingame-tech-tree.625036/ | 1,723,366,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00539.warc.gz | 211,863,431 | 36,122 | # Tech tree statistics and ingame tech tree.
#### raxo2222
##### Time Traveller
This is thread dedicated to worship of our Tech Tree ;^)
No other games even modded ones have that many techs and that big tech tree.
13 eras.
942 techs spread over 160 columns.
And this game isn't even fully developed! We could have end up with nice 1000 techs
Era - Era Column Start - Amount of columns and techs - Techs/Column
Prehistoric - x = 1, 18 columns and 98 techs. That is 5.44 techs per column.
Ancient - x = 19, 13 columns and 93 techs. That is 7.15 techs per column.
Classical - x = 32, 9 columns and 60 techs. That is 6.67 techs per column.
Medieval - x = 41, 9 columns and 54 techs. That is 6.00 techs per column.
Renaissance - x = 50, 11 columns and 59 techs. That is 5.36 techs per column.
Industrial - x = 61, 10 columns and 67 techs. That is 6.70 techs per column.
Atomic - x = 71, 15 columns and 102 techs. That is 6.80 techs per column.
Information - x = 86, 13 columns and 79 techs. That is 6.08 techs per column.
Nanotech - x = 99, 11 columns and 85 techs. That is 7.72 techs per column.
Transhuman - x = 110, 13 columns and 76 techs. That is 5.85 techs per column.
Galactic - x = 123, 14 columns and 65 techs. That is 4.64 techs per column.
Cosmic - x = 137, 11 columns and 51 techs. That is 4.64 techs per column.
Transcendent - x = 148, 12 columns and 52 techs. That is 4.33 techs per column.
Future - x = 160, 1 columns and 1 tech. That is 1 tech per column.
You can have up to 10 techs per column.
Average for whole tech tree not counting Future Era is 5.89 techs per column.
Spoiler Tech tree as of SVN 11094 :
Prehistoric
200 000 BC - 12 000 BC
Sticks, Stones and Words
Ancient
12 000 BC - 1200 BC
Wisdom of the Ancients
Classical
Veni, Vidi, Vici
Medieval
Kings and Castles
Renaissance
Legacy of Da Vinci
Industrial
Viva la Industrial Revolution
Atomic
Nuclear Option
Information
The Dankest Timeline
Nanotech
Pepper Space Program
Transhuman
Faster, Stronger, Better
Galactic
Interstellar Voyage
Cosmic
Final Frontier
Transcendent
Tales of Infinity and Eternity
941 tech buttons - some religious techs are in modules, and there are 5 special purpose techs.
There is 11 Punks, 29 Religions and 2 Special techs scattered across tech tree.
Spoiler :
Prehistoric - 3 R and 1 P -> 4 absolutely non mandatory.
Megafauna Domestication
Ancient - 10 R -> 10 absolutely non mandatory.
Mesopotamism, Yoruba, Shinto, Ngaiism, Andeaism, Zoroastrianism, Judaism, Hinduism, Canaaism, Kementism
Classical - 8 R, 1 S -> 9 absolutely non mandatory.
Rodnovera, Asatru, Buddhism, Hellenism, Naghualism, Confucianism, Taoism, Jainism
Waterproof Concrete - requires building.
Medieval - 3 R, 1 P -> 4 absolutely non mandatory.
Christianity, Islam, Voodoo
Clockpunk
Renaissance - 1 R, 1 S -> 2 absolutely non mandatory.
Sikhism
Industrial - 2 R and 1 P -> 3 absolutely non mandatory.
Baha'i, Mormon
Steampunk
Atomic - 2 R and 2 P -> 4 absolutely non mandatory.
CaoDai, Scientology
Dieselpunk, Atompunk
Information - 0 P -> 0 absolutely non mandatory.
Memes aside this era is bit boring
Nanotech - 2 P -> 2 absolutely non mandatory.
Biopunk, Nanopunk
Transhuman - 2 P -> 2 absolutely non mandatory.
Crystalpunk, Cyberpunk
Galactic - 1 P -> 1 absolutely non mandatory.
Spacepunk
Cosmic - 1 P -> 1 absolutely non mandatory.
Starpunk
Transcendent - Nothing. Or everythingpunk?
Atomic - 102 (98) techs.
Prehistoric - 98 (94) techs.
Ancient - 93 (83) techs.
Nanotech - 85 (83) techs.
Information - 79 (79) techs.
Transhuman - 76 (74) techs.
Industrial - 67 (64) techs.
Galactic - 65 (64) techs.
Classical - 60 (51) techs.
Renaissance - 59 (57) techs.
Medieval - 54 (50) techs.
Transcendent - 52 (52) techs.
Cosmic - 51 (50) techs.
Fun fact:
Vast majority of techs will be researched two eras later.
There is plenty of techs, that can wait one era though.
Excluding religions, punks, and few other techs (Megafauna Domestication, Waterproof Concrete and Lead Glass) all techs will be eventually researched.
Techs, that won't be researched (if you don't choose them), if you reach X lifestyle (only listing techs with at least two eras of space):
Spoiler :
Classical Lifestyle:
Prehistoric - Feline/Elephant/Camelid Domestication, Animal Riding.
Medieval Lifestyle:
Prehistoric - Feline/Elephant/Camelid Domestication, Animal Riding.
Ancient - Mummification, Soap/Candle Making, Veterinary Medicine.
Renaissance Lifestyle:
Prehistoric - Camelid Domestication.
Ancient - Mummification.
Classical - Mounted Archery.
Industrial Lifestyle:
Ancient - Mummification.
Medieval - Fire Brigades, Crop Rotation, Commercial Whaling.
Atomic Lifestyle:
Renaissance - Oil Painting, Free Artistry, Mine Warfare, Hot Air Balloon.
Information Lifestyle
Industrial - Meteorology
Those lifestyle techs below have OR requirements.
None to all of those techs might be researched by selecting one OR prereq (in bracket OR prereq, that doesn't require this tech down in tech tree).
Nanotech Lifestyle <Genesis Biology OR Graphene Alloys OR Disaster Robots>:
Atomic - Radioastronomy (1, 3), Lunar Exploration (1, 3), Astrobiology (1, 3), Astrogeology (1, 3).
Transhuman Lifestyle <Cloaking OR Personal Robots OR Lunar Trade OR Supercharged Crystallography OR Designer Microbiology OR Hypothetical Biochemistry>:
No tech left behind for two eras.
Galactic Lifestyle:<Nuclear Pulse Propulsion OR Nanotroids>:
No tech left behind for two eras.
Cosmic Lifestyle: <Neutrino Communications OR Folding Space>:
No tech left behind for two eras.
Transcendent Lifestyle: <Mensiokinesis OR Fantasy Materials OR Grand Unification Biology OR Stable Time Loops OR Cosmic Engineering or Tachyon Communication>:
No tech left behind for two eras.
Tech tree in video
Last edited:
Information - x = 86, 13 columns and 81 techs. That is 6.23 techs per column.
Nanotech - x = 99, 11 columns and 83 techs. That is 7.55 techs per column.
Transhuman - x = 110, 13 columns and 75 techs. That is 5.77 techs per column.
Galactic - x = 123, 14 columns and 64 techs. That is 4.57 techs per column.
Cosmic - x = 137, 11 columns and 51 techs. That is 4.64 techs per column.
Transcendent - x = 148, 12 columns and 52 techs. That is 4.33 techs per column.
Future - x = 160, 1 columns and 1 tech. That is 1 tech per column.
Is this a recent techtree count? pepper2000 just gave me a list for the number of techs he has added to each era and these last eras number of techs don't agree with his count. We should have almost 1000 tech now. A Recount is in progress.
Is this a recent techtree count? pepper2000 just gave me a list for the number of techs he has added to each era and these last eras number of techs don't agree with his count. We should have almost 1000 tech now. A Recount is in progress.
@Toffer90 civilopedia doesn't lie - it shows how many entries are in each subcategory.
Basically this is count from latest SVN + Toffers modmod.
v0.5.9.0.6
• Pedia now respects the hard limit on tech "tech prereqs".
• Added a list entry count to all the category lists.
Spoiler Civilopedia screenshots :
Last edited:
I think those numbers are right. I only reported how many I added, not how may are there total.
That is super cool to have each era separated out in the pedia! Each era is really like its own game!
I agree that is cool! On buildings and such, it might be nice to also show what hasn't obsoleted or been replaced yet as an era starts as well, not just what is unlocked in that era - though you'd want to be able to see it either way so offering an option switch somewhere might work.
I agree that is cool! On buildings and such, it might be nice to also show what hasn't obsoleted or been replaced yet as an era starts as well, not just what is unlocked in that era - though you'd want to be able to see it either way so offering an option switch somewhere might work.
We can always ask @Toffer90 if this is possible in pedia.
We can always ask @Toffer90 if this is possible in pedia.
I don't know... probably.
It would be a pretty ambitious project to extend the pedia in such a way.
I don't know... probably.
It would be a pretty ambitious project to extend the pedia in such a way.
I figured, which is why I threw the idea out there rather than made an official request of it. It could help with some design visibility factors is all.
I'll post screenshots of tech tree (latest SVN)
Edit: pics look funny in spoiler, as they are scaled to have constant width, while they have constant height
Screenshots are in first post
Last edited:
That look soooo cool ! I Really can't await play v38. I hoped it would be appear the 23. december as a christmas gift. I doesn't was really motivated to finish my earth large game cause I tought soon will be finnally v38 released so I played more Assasins Creed. Now I continue my large earth map game. How long must I wait for v38 ? For only one turn I play 1 hour, sometimes more. I have even bought a new PC only for this game. I haven't overclocked my CPU but it works very well with 4,2 ghz by my Intel 7700k CPU. Not complete without a lag, but really really fast and it is playable. When will v38 release ?
That look soooo cool ! I Really can't await play v38. I hoped it would be appear the 23. december as a christmas gift. I doesn't was really motivated to finish my earth large game cause I tought soon will be finnally v38 released so I played more Assasins Creed. Now I continue my large earth map game. How long must I wait for v38 ? For only one turn I play 1 hour, sometimes more. I have even bought a new PC only for this game. I haven't overclocked my CPU but it works very well with 4,2 ghz by my Intel 7700k CPU. Not complete without a lag, but really really fast and it is playable. When will v38 release ?
We were hoping NewYear, but T-brd has the flu and a Bug to fix.
If you waiting so eagerly, just get the SVN.
That look soooo cool ! I Really can't await play v38. I hoped it would be appear the 23. december as a christmas gift. I doesn't was really motivated to finish my earth large game cause I tought soon will be finnally v38 released so I played more Assasins Creed. Now I continue my large earth map game. How long must I wait for v38 ? For only one turn I play 1 hour, sometimes more. I have even bought a new PC only for this game. I haven't overclocked my CPU but it works very well with 4,2 ghz by my Intel 7700k CPU. Not complete without a lag, but really really fast and it is playable. When will v38 release ?
I'd like to get that to you quickly here but I keep finding big reasons to delay for a quality release. When we put our stamp on it we're basically saying this is the best version it can be at the moment based on current implementations. So we have to take our time and do it right here. There's been a few major MAJOR bugs that have come up and they aren't going to be the quickest to solve. Very hard to guess how long now. Should be fairly soon.
I did get a breakthrough in the OOS error dept. It looks like if you cut off events you'll be able to go quite a ways without an OOS (Though there's a mean little one that can often happen in turn one that has something to do with founding a city. Strange it doesn't always happen.) We can allow this to be the state of being for v38 so long as we don't keep running into more repeating ones.
There is more techs during prehistoric age of C2C than in whole game civ 5 or civ 6. x-D
There is more techs during prehistoric age of C2C than in whole game civ 5 or civ 6. x-D
And vanilla Civ IV.
tmv
Even Civ4 with BTS, that adds 6 techs, has only 92 techs in entire tech tree.
Tech tree got reorganized, its time to remake screenshots of it
Were some X tech positions swapped?
If yes, then some buildings would need recalibrating.
Tech tree got reorganized, its time to remake screenshots of it
Were some X tech positions swapped?
If yes, then some buildings would need recalibrating.
No. That was the big no no barrier in the project. Only y positioning and prerequisites were adjusted. That reminds me... I updated my excel document but not the online one and I need to also update the building with techs document.
No. That was the big no no barrier in the project. Only y positioning and prerequisites were adjusted. That reminds me... I updated my excel document but not the online one and I need to also update the building with techs document.
While tech tree is neat, there are redundancies.
Also sometimes techs and their OR/AND prereqs are in same column.
Will be these addressed later after V38 release?
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1K | 3,432 | 12,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.748801 |
https://developer.unigine.com/en/docs/2.16/api/library/math/cs/quat?rlang=cpp | 1,725,730,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00104.warc.gz | 198,974,789 | 57,122 | Warning! This version of documentation is OUTDATED, as it describes an older SDK version! Please switch to the documentation for the latest SDK version.
Warning! This version of documentation describes an old SDK version which is no longer supported! Please upgrade to the latest SDK version.
# Unigine.quat Struct
Notice
The functions listed below are the members of the Unigine.MathLib namespace.
## floatx#
The X component of the vector.
## floaty#
The Y component of the vector.
## floatz#
The Z component of the vector.
## floatw#
The W component of the vector.
## floatLength#
The Length of the vector.
## floatMinimum#
The Minimum value among all components.
## floatMaximum#
The Maximum value among all components.
## floatLength2#
The Squared length of the vector. This method is much faster than length() — the calculation is basically the same only without the slow Sqrt call. If you are using lengths simply to compare distances, then it is faster to compare squared lengths against the squares of distances as the comparison gives the same result.
## floatILength#
The Inverted length of the vector.
## floatSum#
The Sum of vector components.
## quatNormalized#
The Returns a vector with the same direction as the specified vector, but with a length of one.
## mat3Mat3#
The Mat3 that this quaternion represents.
## mat4Mat4#
The Mat4 that this quaternion represents.
## dmat4DMat4#
The DMat4 that this quaternion represents.
## vec3Euler#
The Euler angles as a vec3 that this quaternion represents.
## floatEulerX#
The X Euler angle (yaw) of this quaternion.
## floatEulerY#
The Y Euler angle (roll) of this quaternion.
## floatEulerZ#
The Z Euler angle (pitch) of this quaternion.
## vec3Normal#
The Quaternion normal vector.
## vec3Tangent#
The Quaternion tangent vector as a vec3.
## vec4Tangent4#
The Quaternion tangent vector as a vec4.
## vec3Binormal#
The Quaternion binormal vector with respect to orientation as a vec3.
## quatZERO#
The Quaternion, filled with zeros (0).
## quatIDENTITY#
The Identity matrix.
## byteNUM_ELEMENTS#
The Number of elements in the vector.
## quatoperator* ( quat q, float v ) #
Multiplication.
### Arguments
• quat q - Quaternion.
• float v - Value.
Multiplication.
Multiplication.
Multiplication.
Multiplication.
## quatoperator* ( quat q0, quat q1 ) #
Multiplication.
### Arguments
• quat q0 - First value.
• quat q1 - Second value.
## quatoperator/ ( quat q, float v ) #
Division.
### Arguments
• quat q - Quaternion.
• float v - Value.
## quatoperator/ ( quat q0, quat q1 ) #
Division.
### Arguments
• quat q0 - First value.
• quat q1 - Second value.
## quatoperator+ ( quat q0, quat q1 ) #
### Arguments
• quat q0 - First value.
• quat q1 - Second value.
## quatoperator- ( quat q0, quat q1 ) #
Subtraction.
### Arguments
• quat q0 - First value.
• quat q1 - Second value.
## booloperator== ( quat v0, quat v1 ) #
Performs equal comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperator!= ( quat v0, quat v1 ) #
Not equal comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperator> ( quat v0, quat v1 ) #
Greater comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperator< ( quat v0, quat v1 ) #
Greater comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperator>= ( quat v0, quat v1 ) #
Greater than or equal to comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperator<= ( quat v0, quat v1 ) #
Less than or equal to comparison.
### Arguments
• quat v0 - First value.
• quat v1 - Second value.
## booloperatortrue ( quat v ) #
Returns true if the operand is both, not null and not NaN.
## booloperatorfalse ( quat v ) #
Returns true if the operand is both, null and NaN.
Subtraction.
## voidSet ( quat v ) #
Sets the value using the specified argument(s).
### Arguments
• quat v - Source vector.
## voidSet ( vec4 v ) #
Sets the value using the specified argument(s).
### Arguments
• vec4 v - Source vector.
## voidSet ( vec3 axis, float angle ) #
Sets the value using the specified argument(s).
### Arguments
• vec3 axis - Axis of rotation.
• float angle - Angle, in degrees.
## voidSet ( float[] q ) #
Sets the value using the specified argument(s).
### Arguments
• float[] q - Source quaternion.
## voidSet ( vec3 t, vec3 b, vec3 n ) #
Sets the quaternion using the specified argument(s).
### Arguments
• vec3 t - Tangent vector.
• vec3 b - Binormal vector.
• vec3 n - Normal vector.
## voidGet ( vec3 axis, float angle ) #
Gets the axis of rotation and the angle from the quaternion and stores it to the provided arguments.
### Arguments
• vec3 axis - Returned axis of rotation.
• float angle - Returned angle, in degrees.
## voidClear ( ) #
Clears the value by setting all components/elements to 0.
## voidAdd ( vec4 v ) #
Performs addition of the specified argument.
## voidAdd ( float v ) #
Performs addition of the specified argument.
### Arguments
• float v - Value.
## voidSub ( vec4 v ) #
Subtracts each element of the specified argument from ther corresponding element.
## voidSub ( float v ) #
Subtracts each element of the specified argument from ther corresponding element.
### Arguments
• float v - Value.
## voidMul ( vec4 v ) #
Multiplies the vector by the value of the specified argument.
### Arguments
• vec4 v - Vector multiplier.
## voidMul ( float v ) #
Multiplies the vector by the value of the specified argument.
### Arguments
• float v - A float multiplier.
## voidDiv ( vec4 v ) #
Returns the result of division of the vector by the value of the specified arguments.
### Arguments
• vec4 v - A vec4 divisor value.
## voidDiv ( float v ) #
Returns the result of division of the vector by the value of the specified arguments.
### Arguments
• float v - A float divisor value.
## voidNormalize ( ) #
Returns a vector with the same direction, but with a length of 1.
## floatGetAngle ( vec3 axis ) #
Returns the angle of rotation using the provided axis.
### Arguments
• vec3 axis - Axis of rotation.
### Return value
Resulting float value.
## boolEquals ( quat other ) #
Checks if the vector and the specified argument are equal (epsilon).
### Arguments
• quat other - Value to be checked for equality.
Return value.
## boolEqualsNearly ( quat other, float epsilon ) #
Checks if the argument represents the same value with regard to the specified accuracy (epsilon).
### Arguments
• quat other - Value to be checked for equality.
• float epsilon - Epsilon value, that determines accuracy of comparison.
Return value.
## boolEquals ( object obj ) #
Checks if the vector and the specified argument are equal (epsilon).
Return value.
## intGetHashCode ( ) #
Returns a hash code for the current object. Serves as the default hash function.
### Return value
Resulting int value.
## stringToString ( ) #
Converts the current value to a string value.
### Return value
Resulting string value.
## stringToString ( string format ) #
Converts the current value to a string value.
### Arguments
• string format - String formatting to be used. A format string is composed of zero or more ordinary characters (excluding %) that are copied directly to the result string and control sequences, each of which results in fetching its own parameter. Each control sequence consists of a percent sign (%) followed by one or more of these elements, in order:
• An optional number, a width specifier, that says how many characters (minimum) this conversion should result in.
• An optional precision specifier that says how many decimal digits should be displayed for floating-point numbers.
• A type specifier that says what type the argument data should be treated as. Possible types:
• c: the argument is treated as an integer and presented as a character with that ASCII value.
• d or i: the argument is treated as an integer and presented as a (signed) decimal number.
• o: the argument is treated as an integer and presented as an octal number.
• u: the argument is treated as an integer and presented as an unsigned decimal number.
• x: the argument is treated as an integer and presented as a hexadecimal number (with lower-case letters).
• X: the argument is treated as an integer and presented as a hexadecimal number (with upper-case letters).
• f: the argument is treated as a float and presented as a floating-point number.
• g: the same as e or f, the shortest one is selected.
• G: the same as E or F, the shortest one is selected.
• e: the argument is treated as using the scientific notation with lower-case 'e' (e.g. 1.2e+2).
• E: the argument is treated as using the scientific notation with upper-case 'E' (e.g. 1.2E+2).
• s: the argument is treated as and presented as a string.
• p: the argument is treated as and presented as a pointer address.
• %: a literal percent character. No argument is required.
### Return value
Resulting string value.
## stringToString ( string format, IFormatProvider formatProvider ) #
Converts the current value to a string value.
### Arguments
• string format - String formatting to be used. A format string is composed of zero or more ordinary characters (excluding %) that are copied directly to the result string and control sequences, each of which results in fetching its own parameter. Each control sequence consists of a percent sign (%) followed by one or more of these elements, in order:
• An optional number, a width specifier, that says how many characters (minimum) this conversion should result in.
• An optional precision specifier that says how many decimal digits should be displayed for floating-point numbers.
• A type specifier that says what type the argument data should be treated as. Possible types:
• c: the argument is treated as an integer and presented as a character with that ASCII value.
• d or i: the argument is treated as an integer and presented as a (signed) decimal number.
• o: the argument is treated as an integer and presented as an octal number.
• u: the argument is treated as an integer and presented as an unsigned decimal number.
• x: the argument is treated as an integer and presented as a hexadecimal number (with lower-case letters).
• X: the argument is treated as an integer and presented as a hexadecimal number (with upper-case letters).
• f: the argument is treated as a float and presented as a floating-point number.
• g: the same as e or f, the shortest one is selected.
• G: the same as E or F, the shortest one is selected.
• e: the argument is treated as using the scientific notation with lower-case 'e' (e.g. 1.2e+2).
• E: the argument is treated as using the scientific notation with upper-case 'E' (e.g. 1.2E+2).
• s: the argument is treated as and presented as a string.
• p: the argument is treated as and presented as a pointer address.
• %: a literal percent character. No argument is required.
• IFormatProvider formatProvider - Provider to be used to format the value. Pass a null reference to obtain the numeric format information from the current locale setting of the operating system.
### Return value
Resulting string value.
## IEnumerator<float>GetEnumerator ( ) #
Returns an IEnumerator for the object.
Return value.
## IEnumeratorGetEnumerator ( ) #
Returns an IEnumerator for the object.
### Return value
Return value.
Last update: 2022-10-10 | 2,656 | 11,518 | {"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-2024-38 | latest | en | 0.786402 |
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# Articulate started the e-learning blog, where the theory was simple
Author Message
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Joined: 18 Sep 2016
Posts: 26
Articulate started the e-learning blog, where the theory was simple [#permalink]
### Show Tags
10 Nov 2017, 11:47
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Question Stats:
44% (01:04) correct 56% (01:03) wrong based on 81 sessions
### HideShow timer Statistics
Articulate started the e-learning blog, where the theory was simple – consumers showing a greater willingness to buy its tools if they find multiple uses for them.
A. where the theory was simple – consumers showing
B. which had the simple theory of consumers showing
C. with a simple theory of consumers showing
D. whose theory was simple – consumers show
E. with a theory that is a simple one – consumers show
Can someone explain what is wrong with B?
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Re: Articulate started the e-learning blog, where the theory was simple [#permalink]
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10 Nov 2017, 21:14
1
tejasridarsi wrote:
Articulate started the e-learning blog, where the theory was simple – consumers showing a greater willingness to buy its tools if they find multiple uses for them.
A. where the theory was simple – consumers showing
"where" here wrongly refers to "the e-learning blog". Should use which instead.
B. which had the simple theory of consumers showing
"its" here should refer back to "the e-learning blog". However, this choice makes "its" refer to "a simple theory"
C. with a simple theory of consumers showing
"its tool", then "its" here should refer back to "the e-learning blog". However, this choice makes "its" refer to "a simple theory"
D. whose theory was simple – consumers show
Correct. "its" here refers to "whose" that refers to "the e-learning blog"
E. with a theory that is a simple one – consumers show
"its" here refers to "a theory"
--== Message from GMAT Club Team ==--
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# Grinding Ball Selection Method For Cement Ball Mill
• ### Cement ball millmachine
The cement mill is the key equipment for re grinding cement clinker after pre grinding by system ingredients Cement grinding is one of the important cement equipment It plays a vital role in the cement production line and cement production process Cement mills are mainly used in cement silicate products new building materials refractory
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• ### Operations and maintenance training for ball mills
Ball mills operations and maintenance seminar Learn how to optimise your ball mill systems in this 5 day training seminar focused on best practices for operations and maintenance preventive and reactive to achieve energy savings reduced maintenance costs and overall improved productivity of the ball mill
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• ### Calculate Top Ball Size of Grinding MediaEquation
Calculate Top Ball Size of Grinding Media FRED C BOND Equation Method Although it was developed nearly 50 years ago Bond s method is still useful for calculating necessary mill sizes and power consumption for ball and rod mills This paper discusses the basic development of the Bond method the determination of the efficiency
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• ### Ball Mill for Sale Mining and Cement Milling Equipment
We provide ball mill machine for cement plant power plant mining industry metallurgy industry etc Ball mill machine can grind a wide range of materials with enough continuous production capacity simple maintenance Capacity range from 5t/h to 210t/h The feeding size is
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• ### Modeling and control of cement grinding processes
In this study a nonlinear dynamic model of a cement grinding process including a ball mill and an air separator in closed loop is developed This gray box model consists of a set of algebraic and partial differential equations containing a set of unknown parameters The selection of a model parametrization the design of experiments the estimation of unknown parameters from experimental
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• ### Manufacturing process of grinding media balls
Method of balls rolling on ball rolling mills is one of the most modern production methods Transversal rolling of balls from the round billet is economical It differs from other methods due to the possibility to produce grinding balls on the uninterrupted automated aggregates which allows 2 to 8 times to increase the output and 10 15 to
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• ### Grinding Mill For Cementriedel zeller
Long Working Life Cement Grinding Ball Mill Mining Cement Industry Use Introduction of Cement Ball Mill Cement ball mill is mainly used to grind the clinker and raw materials in cement industry It is also applied in metallurgy chemical electric power and other industries to grind all kinds of ores.
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• ### Jinan Jinchi Steel Ball Co Ltd.Grinding Ball Steel
Best Products about JC GRINDING CYLPEBS JC GRINDING RODS/BARS and JCC CASTING STEEL BALL SGS service casting grinding ball for ball mill in power stations FOB Price 700 Metric Ton Min Order 20 Metric Tons chrome casting balls for cement plants FOB Price 700
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• ### Ball Mill Manufacturer Ball Mill Grinder Ball Mill Price
Ball Mills Manufacturer Ball Mill Working Principle A ball mill is a type of grinding mill it is an aggregate for grinding and crushing grinder of hard materials that has the same goal as other grinding machinery and crushing machinery.Ball mills are used for crushing and mixing of raw materials While rotating the grinding media balls beads pulps etc and the raw material rotate
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• ### Cement mill notebookSlideShare
Raw mills usually operate at 72 74 critical speed and cement mills at 74 76 3.2 Calculation of the Critical Mill Speed G weight of a grinding ball in kg w Angular velocity of the mill tube in radial/second w = 2 3.14 n/60 Di inside mill diameter in meter effective mill diameter n Revolution per minute in rpm 7.
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• ### The specific selection function effect on clinker grinding
Dry grinding experiments on cement clinker were carried out using a laboratory batch ball mill equipped with a torque measurement The influence of the ball size distribution on the specific selection function can be approached by laboratory runs using mono size balls The
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• ### A generic wear prediction procedure based on the discrete
Ball mills i.e rotating cylindrical drums filled with a feed material and several hundred thousand metal balls also known as the charge are a major category of grinding devices in mineral processing and cement production .Grinding is the final stage of particle size reduction also known as comminution which consists in breaking already small ore or clinker particles into much smaller
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• ### Tangshan Fengrun Shougang Metallurgy And Building
Tangshan Fengrun Shougang Metallurgy And Building Materials Co Ltd is located in Tangshan city Hebei Province shortened form Shougang Steel Ball is a professional grinding media manufacturer in research serving Mainly product grinding steel balls grinding steel cylpebs mill liners crusher hammers etc.
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• ### Method for grinding of material especially raw material
22 23.10.2019 43 28.04.2021 57 A method of grinding material in particular a raw material mixture for firing clinker in mills in particular ball mills in a tubular ball mill with or without discharge and/or in a vertical mill in circulation and/or in a grinding circuit by separating fine gravel so called semolina remaining after grinding the feed material with subsequent
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• ### Chapter 10 Selection of Fine Grinding MillsScienceDirect
The fine grinding mills are classified often into five major groups 1 impact mills 2 ball media mills 3 air jet mills 4 roller mills and 5 shearing attrition mills from the viewpoints of grinding machines For the purpose of size reduction of solid particles a number of different types of grinding mills are used in various
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• ### Ball Mill for Cement Grinding Process
Cement Ball Mill Structure When Ball Mill is working raw material enters the mill cylinder through the hollow shaft of the feed The inside of the cylinder is filled with grinding media of various diameters steel balls steel segments etc when the cylinder rotates around the horizontal axis at a certain speed Under the action of centrifugal force and friction force the medium and the
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• ### Durable Mining Grinding Ball Mill for Ore Cement Final
Durable Mining Grinding Ball Mill for Ore Cement Final Product Size 100325 Meshes Introduction Ball mill is used to grind gold calcite potash feldspar talcum marble limestone cream and slag below scale 7 of Moths hardness and 280 odd sorts of non inflammable and non explosive stuff in the industries of mining building material chemical engineering and metallurgy for power
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• ### The difference between forged steel balls and cast steel balls
In addition the advanced heat treatment process makes the grinding ball more resistant to corrosion Under normal conditions the mill is dry Both grinding temperature not greater than 100 degrees and wet grinding are applicable cast balls are more suitable for dry grinding For example cast steel balls must be used in the cement industry.
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• ### THE OPTIMAL BALL DIAMETER IN A MILL
The basic condition which must be met while grinding the material in a mill is that the ball while breaking the material grain causes in it stress which is higher than the grain hardness Bond 1962 Razumov 1947 Supov 1962 Therefore for the biggest grain size it is necessary to have a definite number of the biggest balls in the charge
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• ### Grinding process within vertical roller mills
Based on screening analysis laser size analysis grindability and rigidity tests of samples collected on line from a cement and a power plant a simulation of the grinding process in vertical roller mills was carried out The simulation calculation used a breakage function B The results indicate that the breakage function B and the selection function S in the form of a matrix can be
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• ### Ball charges calculatorsthecementgrindingoffice
Ball top size bond formula calculation of the top size grinding media balls or cylpebs Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency
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• ### Calculate and Select Ball Mill Ball Size for Optimum Grinding
In Grinding selecting calculate the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do Often the ball used in ball mills is oversize just in case .
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• ### Heat treatment process for high chromium grinding ball
Heat treatment of high chromium grinding ball is the main method to obtain good wear resistance The research on high chromium cast iron ball mainly focuses on chemical composition selection heat treatment process determination modifier selection
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• ### Ball charges calculatorsthecementgrindingoffice
Ball top size bond formula calculation of the top size grinding media balls or cylpebs Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency
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• ### Ball Millsan overview ScienceDirect Topics
8.3.2.2 Ball mills The ball mill is a tumbling mill that uses steel balls as the grinding media The length of the cylindrical shell is usually 1–1.5 times the shell diameter Figure 8.11 The feed can be dry with less than 3 moisture to minimize ball coating or slurry containing 20–40 water by weight.
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• ### The specific selection function effect on clinker grinding
Batch dry grinding tests of cement clinker were performed in a ball mill measuring the power input The effect of ball size distribution on specific selection functions was investigated At the initial size reduction stage the experimental results show that the breakage process is more efficient with a maximal specific selection function.
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• ### Grinding MillZENITH Crusher
Raymond Mill Raymond mill has a history of over 100 years So it s classic undoubtedly Recent years with the growth of non metallic mineral grinding industry ZENITH upgraded Raymond mill to make it have more application areas and meanwhile own high degree of reliability and automation.
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• ### Mill sizing method
Please find below two calculators for sizing mills using the Bond and Rowland methods Ball mill sizing Calculator for ball mill s in a single stage circuit Rod ball mill sizing Calculator for rod mill s as first stage of the circuit and ball mill s as second stage of the circuit.
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• ### A generic wear prediction procedure based on the discrete
Ball mills i.e rotating cylindrical drums filled with a feed material and several hundred thousand metal balls also known as the charge are a major category of grinding devices in mineral processing and cement production .Grinding is the final stage of particle size reduction also known as comminution which consists in breaking already small ore or clinker particles into much smaller
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• ### The operating principle of the ball millenergosteel
The operating principle of the ball mill consists of following steps In a continuously operating ball mill feed material fed through the central hole one of the caps into the drum and moves therealong being exposed by grinding media The material grinding occurs during impact falling grinding balls and abrasion the particles between the balls Then discharge of ground material performed
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• ### grinding ball selection method for cement ball mill
Ball mill 16 electrical consumption ASCERIWintergames a new approach for the calculation of the power draw of cement grinding ball mills is proposed For this purpose cement grinding Read More Ball Mill DesignPower Calculation Mineral Processing Extractive Mar 31 The basic parameters used in ball mill design PDF A Parison Of Wear.
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• ### Cement MillGrinding MachineryCement Ball MillCement
This kind cement grinding machinery is stable to operate and reliable to work As the cement mill has the similar working principle with the ball mill it is also called the cement ball mill With other names of cement grinder and cement grinding machine the cement mill owns multiple cabin separating boards which do better in grinding
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• ### Use Cement Ball Mill Grinding Cement Clinker Cement
Cement ball mill is a kind of cement grinding equipment commonly used in cement plants It is mainly used for grinding materials in cement clinker section Cement ball mill is widely used in cement production silicate products new building materials refractories fertilizers black and non ferrous metal mineral processing glass ceramics and other industries.
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• ### TongLi mineral processing operational setting for roll
TongLi is a large scale crusher manufacturer and supplier of sand making machines It specializes in the production of various types of crushers mills sand making machines and other equipment with various models and complete functions which can meet the professional needs of mines quarries and coal mines Broken processing needs.
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• ### EFFECT OF BALL SIZE DISTRIBUTION ON MILLING
various mill load conditions D=195 mm d=25.4 mm c=0.7 41 3.1 Snapshot of the laboratory mill 44 4.1 Selection functions as obtained for three media diameters grinding mono sized coal materials In this case 2360 1700 microns 54 4.2 Effect of ball diameter on the selection function 55 4.3 Reduced selection function graph 56
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• ### IMPROVED CEMENT QUALITY AND GRINDING
iii ABSTRACT Improved Cement Quality and Grinding Efficiency by Means of Closed Mill Circuit Modeling December 2007 Gleb Gennadievich Mejeoumov B.S Ivanovo State Power University Russia
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• ### Chapter 10 Selection of Fine Grinding MillsScienceDirect
An example of dry grinding system of ball agitation mill The grinding rate of the ball media mills depends upon the intensity and frequency of the collisions of the ball media against the balls the casing or the agitator The intensity is controlled by the mechanical conditions of the mill as well as the properties of balls.
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• ### Ball Mill Designfreeshell
involve grinding With Lloyd s ball milling book having sold over 2000 copies there are probably over 1000 home built ball mills operating in just America alone This article borrows from Lloyd s research which was obtained from the commercial ball milling industry and explains some of the key design criteria for making your own ball mill.
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• ### Cement grinding optimisation Request PDF
A total of 40 of the total energy consumption of a cement plant is used in clinker grinding in a ball mill to produce the final cement product 3 Figure 1 shows the percentage of energy usage
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• ### Cement Ball MillJXSC Machine
Ball mill cement grinding circuit The application of ball mill in cement industry dates back more than 100 years The ball mill for cement grinding plant is mainly of high fineness dry grinding method and the process is mainly of open circuit process and closed circuit process.
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• ### Ball Mill Parameter Selection CalculationPower
1 Calculation of ball mill capacity The production capacity of the ball mill is determined by the amount of material required to be ground and it must have a certain margin when designing and selecting There are many factors affecting the production capacity of the ball mill in addition to the nature of the material grain size hardness density temperature and humidity the degree of
Get Price | 3,123 | 16,001 | {"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-2022-33 | latest | en | 0.87441 |
https://www.taylorfrancis.com/books/e/9781315885551/chapters/10.4324/9781315885551-20 | 1,581,924,230,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00307.warc.gz | 938,219,890 | 38,152 | chapter
24 Pages
## Quantity and Number
WithJAMES FRANKLIN
Quantity is the fi rst category that Aristotle lists after substance. More than any other category, it has an extraordinary epistemological clarity. “2 + 2 = 4” is the paradigm of objective and irrefutable knowledge, and “2 million + 2 million = 4 million” is not far behind in certainty, despite its distance from immediate perception. Indeed, certainties about quantity extend to the infi nite-for example, we know that the counting numbers do not run out. Nor does this certainty come at the expense of application to reality. If we put two rabbits and two rabbits in a box and later fi nd fi ve rabbits in there, it is our absolute certainty that 2 + 2 = 4 that allows us to infer that the rabbits must have bred. Continuous quantities are no less open to perfection of knowledge: The quantity π, the ratio of the circumference of any circle to its diameter, is calculable to any degree of precision that computers can cope with (currently claimed to be ten trillion decimal places). 1 The mathematics of quantity delivers certainty about reality, to the envy of other disciplines, including philosophy. | 250 | 1,169 | {"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-2020-10 | latest | en | 0.923054 |
http://priorart.ip.com/IPCOM/000025833 | 1,519,333,346,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814290.11/warc/CC-MAIN-20180222200259-20180222220259-00271.warc.gz | 276,863,179 | 8,200 | Browse Prior Art Database
# GEOMETRY INDEPENDENT - WIDE RANGE END EFFECTOR
IP.com Disclosure Number: IPCOM000025833D
Original Publication Date: 1988-Jun-30
Included in the Prior Art Database: 2004-Apr-04
Document File: 2 page(s) / 99K
## Publishing Venue
Xerox Disclosure Journal
## Abstract
Frequently, it is required to unload parts from three turning centers and then place these parts in a specific position for transportation to the next successive operation. It is desirable to utilize a low cost robots to unload all three machines, pick up the parts in the same relative position to the machine chuck, use one end effector, and place all the parts in the same relative conveyance position. To achieve this, the end effector must be able to handle various materials without surface damage, a wide range of diameters and lengths must be compliant and accurate, and must be capable of overcoming forces and moments exerted by the parts and the dynamics of motion. Thus, the robotic end effector must be capable of handling cylindrical stock of a wide range of diameters and lengths. To achieve this, the end effector utilizes multiple grippers simultaneously activated but acting independently to accommodate a variety of shapes. The levers are capable of rotation and maintained within the same plane. They are pivoted such that rotation of these levers will cause the levers to converge about a common centerline and provide the greatest latitude in grasping various sized objects in the smallest amount of space. As shown in the drawing, levers 10 are controlled by return springs 12, ropes 14 and braking mechanism 16. Tension and compliance springs 18 are employed as well as movable plate 20 A handle 22 is utilized to control the foregoing. Normal force attainment, compliance and stability is achieved by driving these levers by the wire rope attached to the spring connected to the movable plate. When the plate is moved, such that the levers are about an object, the plate is moved further so that tension is applied to the rope creating a normal force on the object being grasped. The plate is then moved further so that the brake/pinch mechanism is engaged in position on the wire rope, between the spring and lever holding the wire rope from movement and maintaining the tension on the wire rope by the spring force.
This text was extracted from a PDF file.
At least one non-text object (such as an image or picture) has been suppressed.
This is the abbreviated version, containing approximately 56% of the total text.
Page 1 of 2
CEROX DISCLOSURE JOURNAL
GEOMETRY INDEPENDENT - WIDE RANGE END EFFECTOR U.S. c1.355/3 Edward J. Wrobbel
Proposed Classification Int. C1. G03g 15/00
0
Volume 13 Number 2 May/June 1988 143 | 605 | 2,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-09 | latest | en | 0.907152 |
http://kb.compute.info/foundations/algorithms/sorting/bubble-sort | 1,627,989,584,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00460.warc.gz | 25,261,613 | 6,999 | Foundations > Algorithms > Sorting >
### Bubble sort
- Time Complexity - `O``(``n``2````) ``` - Pseudocode -``````BUBBLE-SORT(A) 1 ````do```` 2 swapped ← false 3 ````for```` i ← 0 to length[A] - 1 4 ````if`` A[i] > A[i + 1] ``then```` 5 swap(A[i], A[i + 1]) 6 swapped ← true 7 ````while`` swapped```` | 127 | 310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | longest | en | 0.483548 |
https://www.physicsforums.com/threads/vector-valued-function-tangent.428714/ | 1,544,965,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00241.warc.gz | 996,044,936 | 13,454 | # Homework Help: Vector-valued function tangent
1. Sep 13, 2010
### Jonnyb42
1. The problem statement, all variables and given/known data
If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on the sphere with center at the origin.
2. Relevant equations
I know dot product might help:
r(t) . r'(t) = 0
and the equation of a sphere in 3-space:
r2 = x2 + y2 + z2
3. The attempt at a solution
if I write out the components of the dot product...
r(t) . r'(t) = fx(t)*fx'(t) + fy(t)*fy'(t) + fz(t)*fz'(t) = 0
From there, I am not sure what to do, if that even is the right way to start.
2. Sep 13, 2010
### Staff: Mentor
What if you integrate both sides of your last equation?
3. Sep 13, 2010
### Jonnyb42
Wow, how did you think of that?
It seems to work. The one thing I need help with is integrating the right side of 0, I think it's my lack of calculus knowledge. Does it become a constant?
4. Sep 13, 2010
### Staff: Mentor
I don't know - it just occurred to me because of those terms fx fx'.
Yes.
5. Sep 13, 2010
### Jonnyb42
Thank you very much, just for completion's sake, I'll show the rest of the work:
It is easier to write functions with different letters, so from before, fx(t) will now be f(t), fy(t) will now be g(t), and fz(t) is now h(t).
$$\int f(t)df(t)$$ + $$\int g(t)dg(t)$$ + $$\int h(t)dh(t)$$ = $$\int 0dt$$
$$\stackrel{1}{2}$$ f2(t) + $$\stackrel{1}{2}$$ g2(t) + $$\stackrel{1}{2}$$ h2(t) = C
f2(t) + g2(t) + h2(t) = r2 <-- form of a sphere.
Last edited: Sep 13, 2010 | 523 | 1,600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-51 | latest | en | 0.885553 |
https://myassignmenthelp.com/free-samples/bmat230-business-mathematics/annual-rate.html | 1,722,717,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00220.warc.gz | 342,325,171 | 23,278 | Get Instant Help From 5000+ Experts For
Writing: Get your essay and assignment written from scratch by PhD expert
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1. Apply compound interest formulas to solve basic financial problems.
2. Prepare a Word document to present financial findings.
The loan amount provided by the angel investor could be calculated as the present value of the amount paid after two years of time. The interest rate applicable for the transaction was around 6% payable semi-annually over the two-year time. The savings of \$80,654 will be pulled back two years back at the semiannual rate of 6% payable semiannually (Eisenberg & Mishura, 2018).
Future Value = 80,654
Interest Rate = 6% payable semiannually.
Time= 2years
Calculated Present Value = 71,660.34
Int Rate Compounded Monthly Future Value 80654 Time (In Years) 12 Interest Rate 5.93% Int Rate Comp. Semi-Annually 0.004938 Factor 1.004938 Semi Annual Rate 1.060886 Present Value 71661.95
It would take around 71 month of time for the repayment and for reaching the debt of \$100,000 assuming the interest rate of around 6% payable semiannually. The Present Value considered for the calculation was around 71,660 and the N (Time period was calculated using the back calculation for solving the same).
Future Value 80654 Time (In Years) 1.5 Interest Rate 8% Int Rate Comp. Semi-Annually 0.0410067 Factor 1.0410067 Semi Annual Rate 1.062 Present Value (Value of the Loan) 71660.03
Future Value 80654 Time (In Years) 2 Interest Rate 6% Int Rate Comp. Annually 6% Factor 1.06 Semi Annual Rate 1.1236 Present Value (Value of the Loan) 63885.522
If the contract was for two years of time and the loan had to be settled in the same amount of time then the following amount 25,000 and 40,000. It would be repaid at the respective year end would be discounted at the todays rate at the all in interest rate of 6% to determine the fair value of the loan amount that can be granted. The final amount calculated for the loan amount was around \$59,184.76 as the total amount (Xiong et al. 2016).
The final amount of \$80,654 will be pulled back at the respective mentioned period of two years with the rate of 6% compounded semiannually and 8% compounded quarterly. The rate calculated for the first year was 6.09% and the return for the second year was calculated to be around 8.24% that was the return calculated on a quarterly compounded basis.
The amount for the second year will be calculated as = (80,654/ (1.0824)) = 74,514.
The amount derived for the second year will then be pulled back for the first year as = (74514/1.0609) = 70,236.63
The size of the loan will be around 70,236.63 for the same to be repaid within the two year of frame time.
References
Eisenberg, J., & Mishura, Y. (2018). An Exponential Cox-Ingersoll-Ross Process as Discounting Factor. arXiv preprint arXiv:1808.10355.
Xiong, J., Zhang, Q., Sun, G., Zhu, X., Liu, M., & Li, Z. (2016). An information fusion fault diagnosis method based on dimensionless indicators with static discounting factor and KNN. IEEE Sensors Journal, 16(7), 2060-2069.
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Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost | 1,080 | 4,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-33 | latest | en | 0.938379 |
https://www.coursehero.com/file/p362g3q/Lizhi-Wang-lzwangiastateedu-IE-534-Linear-Programming-October-22-2012-29-30/ | 1,603,895,591,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00574.warc.gz | 670,462,192 | 58,103 | Lizhi Wang lzwangiastateedu IE 534 Linear Programming 29 30
# Lizhi wang lzwangiastateedu ie 534 linear programming
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Lizhi Wang ([email protected]) IE 534 Linear Programming October 22, 2012 29 / 30
Example 4 Suppose we add a new variable x 3 : x 0 = x 1 x 2 x 3 , c 0 = - 1 - 1 1 , A 0 = - 2 - 1 3 - 2 4 - 2 - 1 3 1 . The old optimal solution: ( x * 1 = 7 , x * 2 = 0) and ( y * 1 = 0 , y * 2 = 0 , y * 3 = 1) . y * satisfies the constraint 3 - 2 1 y * 1 . The new optimal solution is ( x 0 1 = 7 , x 0 2 = 0 , x 0 3 = 0) . Lizhi Wang ([email protected]) IE 534 Linear Programming October 22, 2012 30 / 30
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Want to read all 30 pages? | 292 | 736 | {"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} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.644717 |
https://answerprime.com/solve-for-z-cz6zto83-3/ | 1,675,231,748,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00792.warc.gz | 116,561,275 | 18,410 | # Solve for z -cz+6z=to+83
Solve for z -cz+6z=to+83
answer: o Step-by-step explanation: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
-c*z+6*z-(t*o+83)=0 Step by step solution :
Step 1 :
Pulling out like terms :
1.1 Pull out like factors :
-cz + 6z – to – 83 = -1 • (cz – 6z + to + 83) Equation at the end of step 1 :
Step 2 :
Solving a Single Variable Equation :
2.1 Solve -cz+6z-to-83 = 0 hope you understand brainliest
z = 83/( -c+6-t) Step-by-step explanation: -cz + 6z = tz + 83 Subtract tz from each side -cz + 6z -tz= tz-tz + 83 -cz + 6z – tz = 83 Factor out z z( -c+6-t) = 83 Divide each side by ( -c+6-t) z( -c+6-t)/( -c+6-t) = 83/( -c+6-t) z = 83/( -c+6-t) | 304 | 760 | {"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} | 4.25 | 4 | CC-MAIN-2023-06 | latest | en | 0.502455 |
https://eigenfoo.xyz/_posts/2018-04-13-bias-variance-cookbook/ | 1,534,729,544,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215487.79/warc/CC-MAIN-20180820003554-20180820023554-00621.warc.gz | 661,463,004 | 5,729 | The bias-variance tradeoff is a unique result in machine learning: it sits on extremely solid theoretical foundations, and has a ludicrously far-reaching scope of applicability. Which might explain why it’s treated so… interestingly… in introductory machine learning courses (or at least, my introductory course). Quickly go over the proof (it’s easy enough that even undergraduates can digest it), build some intuition about what it means for real-life predictors, and move on.
Enough has been said about what the bias-variance tradeoff (a.k.a. the bias-variance decomposition) is (read this excellent essay if you’re shaky). However, not that much has been said about how to manage the bias or variance of your model, which seems to me to be infinitely more important to know. There’s neat advice scattered thinly throughout the interwebs, but I figure I’d consolidate them in one place. I’d also highly recommend these slides from Andrew Ng at Stanford, in which he outlines some practical advice for machine learning models. I basically recreated some of his graphs for two of the three images below.
Of course, take this with a pinch of salt: the most important thing in statistical modelling is not overbearing guidelines or rigid dogma, but the situation you have at hand. Clear thinking is always good!
## High Variance
### Diagnostics
• The testing error looks like it continues to decrease with the training set size, suggesting that more data (specifically, more examples) will help.
• There is a significant difference between the training and testing errors.
### Remedies
• Try to obtain more examples.
• Perform some feature selection or feature engineering to get a high-quality set of features.
• Considering ensembling (e.g. bagging or boosting your predictors).
• Consider regularizing your model (or pruning it, or otherwise enforcing model parsimony).
• Consider using a less flexible model: perhaps a parametric (or even linear!) model will suffice for your purposes?
## High Bias
### Diagnostics
• Even the training error is unacceptably high.
• The training and testing errors quickly converge to a common value.
### Remedies
• Try to obtain/engineer more features (consider polynomial or interaction terms).
• Consider using a more flexible model: perhaps nonlinear or even nonparametric models would work better for your purposes? | 476 | 2,374 | {"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-2018-34 | latest | en | 0.936071 |
https://www.icicidirect.com/futures-and-options/articles/understanding-the-power-of-moving-average-convergence-divergence | 1,716,568,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00427.warc.gz | 710,526,891 | 26,280 | # Understanding the Power of Moving Average Convergence Divergence (MACD) in the Stock Market
In the ever-evolving landscape of the stock market, traders rely on a range of technical indicators to gain insights and make informed decisions. One such indispensable tool is the Moving Average Convergence Divergence (MACD) indicator. The MACD is a versatile and widely used indicator that helps traders identify potential trend reversals, gauge market momentum, and generate timely buy and sell signals. In this blog post, we will explore the significance of the Moving Average Convergence Divergence as a technical indicator in the stock market.
## Understanding Moving Average Convergence Divergence (MACD)
The Moving Average Convergence Divergence (MACD) is a trend-following momentum oscillator that consists of three components: the MACD line, the signal line, and the histogram. The MACD line is calculated by subtracting a longer-term Exponential Moving Average (EMA) from a shorter-term EMA. The signal line, often a 9-period EMA of the MACD line, acts as a trigger line for buy and sell signals. The histogram represents the difference between the MACD line and the signal line, providing visual cues about the strength and direction of the trend.
## How to apply MACD indicator in real life
Let's take an example of a real-life Indian stock, such as Tata Consultancy Services (TCS), and use the Moving Average Convergence Divergence (MACD) indicator to explain changes in position.
The MACD is a popular trend-following momentum indicator that helps traders identify the strength and direction of a stock's trend. It consists of two lines: the MACD line and the signal line.
### Here's how the MACD is calculated:
• Calculate the 12-day Exponential Moving Average (EMA) of the stock's closing prices.
• Calculate the 26-day Exponential Moving Average (EMA) of the stock's closing prices.
• Subtract the 26-day EMA from the 12-day EMA to get the MACD line.
• Calculate a 9-day Exponential Moving Average (EMA) of the MACD line to get the signal line.
Let’s look at an example of Tata Consultancy Services (TCS) using the Moving Average Convergence Divergence (MACD) indicator presented in a tabular format:
In the table above:
- "Closing Price" represents the daily closing price of TCS stock.
- "12-day EMA" and "26-day EMA" are the calculated 12-day and 26-day Exponential Moving Averages, respectively.
- "MACD Line" is the difference between the 12-day EMA and the 26-day EMA.
- "Signal Line" is the 9-day Exponential Moving Average of the MACD Line.
### Now, let's understand the changes in position using the MACD indicator:
When the MACD line crosses above the signal line, it suggests a potential bullish signal, indicating that the stock's upward momentum might be strengthening. This could be a signal to consider buying the stock.
When the MACD line crosses below the signal line, it suggests a potential bearish signal, indicating that the stock's upward momentum might be weakening, and it could be heading for a downward move. This could be a signal to consider selling the stock.
Traders often look for crossovers between the MACD line and the signal line as potential entry or exit points for their positions. For example, a bullish crossover (MACD line crossing above the signal line) might be seen as a buy signal, while a bearish crossover (MACD line crossing below the signal line) might be seen as a sell signal.
## How traders can benefit from MACD indicator?
1. Identifying Trend Reversals: The MACD is particularly effective in identifying potential trend reversals in the stock market. When the MACD line crosses above the signal line, it generates a bullish signal, indicating a potential shift from a downtrend to an uptrend. Conversely, when the MACD line crosses below the signal line, it generates a bearish signal, suggesting a potential reversal from an uptrend to a downtrend. These crossover points serve as crucial indicators for traders to enter or exit positions, capturing profit potential in trend reversals.
2. Assessing Market Momentum: The MACD provides valuable insights into market momentum. When the MACD line and the signal line are both above the zero line, it indicates a bullish momentum, suggesting a higher probability of price increases. Conversely, when both lines are below the zero line, it signals a bearish momentum, implying a higher likelihood of price declines. Traders use this information to gauge the overall strength of the market and adjust their trading strategies accordingly.
3. Generating Buy and Sell Signals: The MACD generates buy and sell signals based on the crossovers between the MACD line and the signal line. A bullish signal is generated when the MACD line crosses above the signal line, indicating a potential buying opportunity. Conversely, a bearish signal is generated when the MACD line crosses below the signal line, suggesting a potential selling opportunity. These signals help traders time their trades more effectively and capture profitable opportunities in the market.
4. Divergence Analysis: Another significant aspect of the MACD is divergence analysis. Divergence occurs when the direction of the MACD and the price chart diverge. Bullish divergence happens when the price makes lower lows, but the MACD makes higher lows. This suggests a potential trend reversal to the upside. Conversely, bearish divergence occurs when the price makes higher highs, but the MACD makes lower highs. This indicates a potential trend reversal to the downside. Traders use divergence signals in conjunction with other analysis tools to enhance their decision-making process.
## Conclusion
The Moving Average Convergence Divergence (MACD) indicator is a vital tool for stock market traders seeking to identify trends and make informed decisions. Its ability to generate clear buy and sell signals makes it a favorite among traders of all levels. However, like any indicator, the MACD should be used in conjunction with other tools to validate signals and minimize risks.
By mastering the MACD indicator and understanding its nuances, traders can gain a competitive edge in the dynamic and ever-evolving world of stock market trading. Remember, prudent risk management and a well-defined trading strategy are essential to successful trading in the stock market.
Disclaimer: ICICI Securities Ltd. ( I-Sec). Registered office of I-Sec is at ICICI Securities Ltd. - ICICI Venture House, Appasaheb Marathe Marg, Prabhadevi, Mumbai - 400 025, India, Tel No : 022 - 6807 7100. I-Sec is a Member of National Stock Exchange of India Ltd (Member Code :07730), BSE Ltd (Member Code :103) and Member of Multi Commodity Exchange of India Ltd. ( Member Code : 56250) and having SEBI registration no. INZ000183631. Name of the Compliance officer (broking): Ms. Mamta Shetty, Contact number: 022-40701000, E-mail address: complianceofficer@icicisecurities.com. Investments in securities market are subject to market risks, read all the related documents carefully before investing. The contents herein above shall not be considered as an invitation or persuasion to trade or invest. I-Sec and affiliates accept no liabilities for any loss or damage of any kind arising out of any actions taken in reliance thereon. Investors should consult their financial advisers whether the product is suitable for them before taking any decision. The client shall not have any claim against I-Sec and/or its employees on account of any suspension, interruption, non-availability or malfunctioning of I-Sec system or service or non-execution of algo orders due to any link/system failure for any reason beyond I-Sec control. I-Sec reserves the right to pause, stop or call back any of the execution algos in case of any technical or mechanical exigency. | 1,654 | 7,831 | {"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} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.870236 |
https://freezingblue.com/flashcards/201099/preview/ph-204-chap-22-current-and-resistance | 1,716,603,243,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00192.warc.gz | 225,827,394 | 4,302 | # Ph-204 Chap 22 Current and Resistance
Current (Ι)is defined to be: the motion of positive charges.I=Δq/Δt emf: is the potential difference of the battery. The battery does work to raise the potential of the charges. Conservation of current: Dictates that the current is the same at all points in the circuit. A battery is: a source of potential difference. Chemical processes in the battery separate charges. We use a charge escalator model to show the lifting of charges to higher potential. The actual charge carriers are: electrons. Their random collisions with atoms impede the flow of charge and are the source of resistance. The collisions increase the thermal energy of the resistor. We use the ideal wire model in which: we assume that there is no resistance in the wires. The resistivity p is a property of a material: a measure of how good a conductor the material is.Good conductors have low resistivity.Poor conductors have high resistivity. Resistance is: The property of a particular wire or conductor. It depends on its resitivity and dimensions. R=ρL/A Ohm's Law: describes the relationship between potential difference and current in a resistor. I=ΔV/R A battery supplies power at the rate: Pemf=Iε The resistor dissipates power at the rate: PR=IΔVR=I2R=(ΔVR)2/R Define Ohmic: When a potential difference is applied to a wire, if the relationship between potential difference and current is linear, the material is Ohmic. Resistors are: made of ohmic materials and have a well-defined value of resistance: R=ΔV/I If the variation is not linear then the material is: Nonohmic. AuthorAllistermark ID201099 Card SetPh-204 Chap 22 Current and Resistance DescriptionCurrent and Resistance Updated2013-02-20T18:21:10Z Show Answers | 412 | 1,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.853222 |
http://www.wjhsh.net/pk28-p-7207875.html | 1,638,480,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00240.warc.gz | 129,335,239 | 4,904 | 38. Count and Say
### 38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
```1. 1
2. 11
3. 21
4. 1211
5. 111221
```
`1` is read off as `"one 1"` or `11`.
`11` is read off as `"two 1s"` or `21`.
`21` is read off as `"one 2`, then `one 1"` or `1211`.
Given an integer n, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
```Input: 1
Output: "1"
```
Example 2:
```Input: 4
Output: "1211"
```
递归读取前面的数。注意读取时边界的问题
```class Solution {
public:
string cv(int num) {
string t = "";
while(num > 0) {
t += (char)(num % 10 + '0');
num /= 10;
}
reverse(t.begin(), t.end());
return t;
}
string countAndSay(int n) {
if (n == 1) return "1";
else {
string k = countAndSay(n - 1);
string tmp("");
int len = k.length();
if (len == 1) return "11";
int num = 1;
for (int i = 1; i < k.length(); ++i) {
if(k[i] == k[i - 1]) {
num++;
if (i == k.length() - 1) {
tmp += cv(num);
tmp += k[i - 1];
}
}
else {
tmp += cv(num);
tmp += k[i - 1];
num = 1;
}
}
if (num == 1) {
tmp += cv(num);
tmp += k[len - 1];
}
//cout<<tmp<<"-"<<k<<endl;
return tmp;
}
}
};```
数字转字符串还有 std::to_string（int n）这东西 | 475 | 1,249 | {"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} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.333749 |
https://physics.stackexchange.com/questions/372634/the-hamiltons-equations-of-a-charged-particle-in-electromagnetic-field | 1,713,040,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00009.warc.gz | 443,277,669 | 45,904 | # The Hamilton's equations of a charged particle in electromagnetic field
For the relativistic charged particle in EM field we have the following equation for the hamiltonian $$H\left( {\vec r,\vec P,t} \right) = c\sqrt {{m^2}{c^2} + {p^2}} + e\varphi = c\sqrt {{m^2}{c^2} + {{\left( {\vec P - e\vec A\left( {\vec r,t} \right)} \right)}^2}} + e\varphi.$$
Then the hamiltonian's equations of motion can be written as
$$\frac{{d\vec r}}{{dt}} = \frac{{\partial H}}{{\partial \vec P}} = \frac{{c\vec p}}{{\sqrt {{m^2}{c^2} + {p^2}} }}$$
and
$$\frac{{d\vec P}}{{dt}} = - \frac{{\partial H}}{{\partial \vec r}} = \vec ve\frac{{\partial \vec A}}{{\partial \vec r}} - e\frac{{\partial \phi }}{{\partial \vec r}}$$
Where ${\vec P}$ is the generalised momentum. I don't understand why the generalized momentum is used in the last equation instead of the ordinary momentum.
Futhermore, the expression for the ordinary momentum ${\vec p}$ is obtained and has such a form
$$\frac{{d\vec p}}{{dt}} = - e\frac{{\partial \vec A}}{{\partial t}} - e\frac{{\partial \phi }}{{\partial \vec r}} + e\left( {\vec v\frac{{\partial \vec A}}{{\partial \vec r}} - \frac{{\partial \vec A}}{{\partial \vec r}}\vec v} \right)$$
And it is not clear for me why the last term in this equation $e\left( {\vec v\frac{{\partial \vec A}}{{\partial \vec r}} - \frac{{\partial \vec A}}{{\partial \vec r}}\vec v} \right)$ is not zero and what ${\frac{{\partial \vec A}}{{\partial \vec r}}}$ means? If ${\vec A}$ was a scalar it would be just a gradient value but the vector quantity confuses me
I don't understand why the generalized momentum is used in the last equation instead of the ordinary momentum.
The generalized momentum $\vec P$ is used because Hamilton's equations of motion relate the time-derivative of the generalized momentum $d \vec P/dt$ -- not the time-derivative of the kinetic momentum $d \vec p/dt$ -- to the negative partial derivatives of the Hamiltonian with respect to the generalized position $-\partial H/\partial \vec r$.
And it is not clear for me why the last term in this equation $e\left( {\vec v\frac{{\partial \vec A}}{{\partial \vec r}} - \frac{{\partial \vec A}}{{\partial \vec r}}\vec v} \right)$ is not zero and what ${\frac{{\partial \vec A}}{{\partial \vec r}}}$ means? If ${\vec A}$ was a scalar it would be just a gradient value but the vector quantity confuses me
Correct. If $\vec A$ were instead a scalar field, that term would denote the gradient of a scalar field. It turns out that we can apply the concept of a gradient not only to scalars but also to vectors and, more generally, tensors, a class of geometric objects to which scalars, or zeroth-order tensors, and vectors, or first-order tensors, belong. As the gradient of a zeroth-order tensor yields a first-order tensor, you might guess that the gradient of a first-order tensor yields a second-order tensor, a geometric object which can be represented using an $n \times n$ matrix; you'd be correct, and while it is not as clear in the notation you have chosen, the fact that $\partial \vec A/\partial \vec r$ -- the gradient of the vector field $\vec A$ -- is a second-order tensor is exactly the reason why $\vec v (\partial \vec A/\partial \vec r) - (\partial \vec A/\partial \vec r)\vec v \neq 0$.
In fact, just by inspection, you should be able to at least convince yourself that since
$$- e\frac{\partial \vec A}{\partial t} - e\frac{\partial \phi}{\partial \vec r} = e \left(-\frac{\partial \vec A}{\partial t} - \frac{\partial \phi}{\partial \vec r}\right) = e \vec E$$
it must be that $\vec v (\partial \vec A/\partial \vec r) - (\partial \vec A/\partial \vec r)\vec v = \vec v \times \vec B$, since the right-hand side of the equation for $d \vec p/dt$ should yield the correct expression for the Lorentz force.
Now let's prove our conviction.
# Tensors in $\mathbb R^3$
Consider a vector basis $\{\mathbf e_i\}$, where the index $i$ ranges from 1 to 3, and for simplicity, assume that this vector basis is Euclidean; in other words, the inner product of basis vectors $\mathbf e_i$ and $\mathbf e_j$ gives,
$$\mathbf e_i \cdot \mathbf e_j = \delta_{ij} \tag{1}$$
where
$$\left(\delta_{ij}\right) = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \tag{2}$$
is the identity matrix.
A vector $\mathbf f$ may be expressed in this coordinate basis as,
$$\mathbf f = \sum_{i=1}^3 f_i\, \mathbf e_i = \left(\begin{matrix} f_1 & f_2 & f_3 \end{matrix}\right)_{\{\mathbf e_i\}} \tag{3}$$
while a second-order tensor $\mathbf F$ may be expressed as,
$$\mathbf F = \sum_{i=1}^3 \sum_{j=1}^3 F_{ij}\, \mathbf e_i \otimes \mathbf e_j = \left(\begin{matrix} F_{11} & F_{12} & F_{13} \\ F_{21} & F_{22} & F_{23} \\ F_{31} & F_{32} & F_{33} \end{matrix}\right)_{\{\mathbf e_i \otimes \mathbf e_j\}} \tag{4}$$
where $\mathbf e_i \otimes \mathbf e_j$ is called the outer product of $\mathbf e_i$ and $\mathbf e_j$. The outer product is defined such that, given vectors $\mathbf a$, $\mathbf b$, $\mathbf c$, and $\mathbf d$,
$$(\mathbf a \otimes \mathbf b) \cdot (\mathbf c \otimes \mathbf d) = (\mathbf b \cdot \mathbf c)(\mathbf a \otimes \mathbf d) \tag{5}$$
or, equivalently,
$$(\mathbf a \otimes \mathbf b) \cdot \mathbf c = (\mathbf b \cdot \mathbf c)\mathbf a$$ $$\mathbf c \cdot (\mathbf a \otimes \mathbf b) = (\mathbf c \cdot \mathbf a)\mathbf b \tag{6}$$
The transpose of $\mathbf a \otimes \mathbf b$, denoted as $(\mathbf a \otimes \mathbf b)^T$, is defined as,
$$(\mathbf a \otimes \mathbf b)^T = \mathbf b \otimes \mathbf a \tag{7}$$
and so (6) may be rewritten as,
$$(\mathbf a \otimes \mathbf b) \cdot \mathbf c = \mathbf c \cdot (\mathbf a \otimes \mathbf b)^T = (\mathbf b \cdot \mathbf c)\mathbf a$$ $$\mathbf c \cdot (\mathbf a \otimes \mathbf b) = (\mathbf a \otimes \mathbf b)^T \cdot \mathbf c = (\mathbf c \cdot \mathbf a)\mathbf b \tag{8}$$
Additionally, applying (7) to (4), the transpose of $\mathbf F$, denoted $\mathbf F^T$, is,
$$\mathbf F^T = \sum_{i=1}^3 \sum_{j=1}^3 F_{ij}\, \mathbf e_j \otimes \mathbf e_i = \sum_{i=1}^3 \sum_{j=1}^3 F_{ji}\, \mathbf e_i \otimes \mathbf e_j = \left(\begin{matrix} F_{11} & F_{21} & F_{31} \\ F_{12} & F_{22} & F_{32} \\ F_{13} & F_{23} & F_{33} \end{matrix}\right)_{\{\mathbf e_i \otimes \mathbf e_j\}} \tag{9}$$
If the second-order tensor $\mathbf F^T = \mathbf F$, then $\mathbf F$ is said to be symmetric; if $\mathbf F^T = -\mathbf F$, then $\mathbf F$ is said to be antisymmetric.
A third-order tensor $\mathbf \Phi$ may be expressed as,
$$\mathbf \Phi = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \Phi_{ijk}\, \mathbf e_i \otimes \mathbf e_j \otimes \mathbf e_k \tag{10}$$
One commonly-encountered third-order tensor known as the alternating tensor, denoted as $\mathbf \epsilon$, is defined as,
$$\mathbf \epsilon = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\, \mathbf e_i \otimes \mathbf e_j \otimes \mathbf e_k \tag{11}$$
such that $\mathbf \epsilon$ is antisymmetric under an exchange of any two indices (e.g $\epsilon_{jik} = -\epsilon_{ijk}$) and $\epsilon_{123} = 1$. Note that this implies that any component $\epsilon_{ijk}$ that has two or more indices set to the same value is equal to zero (e.g $\epsilon_{112} = 0$).
A common practice in many publications is to omit the summation symbols found in the expressions for the tensors given above; this is known as the Einstein summation convention, and this convention will be used from this point onward. The rules of the Einstein summation convention are as follows:
1. In a term which involves only the product of tensor components, if an index $i$ only appears once in the term, then $i$ is referred to as a free index, and a summation on that index is not implied.
2. In a term which involves only the product of tensor components, if an index $i$ appears twice in the term, then $i$ is referred to as a summation index, and a summation on that index is implied.
3. For any index $i$, you may denote it using any other letter that is not already being used as an index in the same term.
4. In order to avoid ambiguity, no index can appear more than twice in the same term.
## Relevant Tensor Operations
The scalar product of two vectors $\mathbf a$ and $\mathbf b$, denoted $\mathbf a \cdot \mathbf b$, is given by,
$$\mathbf a \cdot \mathbf b = a_i b_j\, \mathbf e_i \cdot \mathbf e_j = a_i b_j \delta_{ij} = a_i b_i \tag{12}$$
Similarly, the inner product for a vector $\mathbf a$ and a second-order tensor $\mathbf B$ is given by,
$$(\mathbf B \cdot \mathbf a)_i = B_{ij} \delta_{jk} a_k = B_{ij}a_j$$ $$(\mathbf a \cdot \mathbf B)_i = a_k \delta_{kj} B_{ji} = B_{ji}a_j \tag{13}$$
where $(\mathbf B \cdot \mathbf a)_i$ and $(\mathbf a \cdot \mathbf B)_i$ are, respectively, the $i$-th components of the vectors $\mathbf B \cdot \mathbf a$ and $\mathbf a \cdot \mathbf B$.
The vector product of two vectors $\mathbf a$ and $\mathbf b$, denoted $\mathbf a \times \mathbf b$, is given by,
$$(\mathbf a \times \mathbf b)_i = \epsilon_{ijk} a_j b_k \tag{14}$$
where $(\mathbf a \times \mathbf b)_i$ is the $i$-th component of $\mathbf a \times \mathbf b$. Similarly, the curl of a vector $\mathbf a$ is given by,
$$(\nabla \times \mathbf a)_i = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \tag{15}$$
where $x_j$ is the $j$-th component of the position vector $\mathbf r$ relative to the origin of our coordinate basis.
The gradient of a vector $\mathbf a$, a second-order tensor denoted $\nabla \mathbf a$, is defined as,
$$(\nabla \mathbf a)_{ij} = \frac{\partial a_i}{\partial x_j} \tag{16}$$
where $(\nabla \mathbf a)_{ij}$ is the component of $\nabla \mathbf a$ associated with the outer product $\mathbf e_i \otimes \mathbf e_j$.
## A Note About Antisymmetric Second-Order Tensors
If a second-order tensor $\mathbf F$ is antisymmetric (i.e. $F_{ji} = -F_{ij}$), then there exists some scalars $f_k$ such that,
$$F_{ij} = \epsilon_{ijk} f_k \tag{17}$$
or, equivalently,
$$f_k = \frac{1}{2}\epsilon_{ijk}F_{ij} \tag{18}$$
and the vector $\mathbf f$ for which $f_k$ is the $k$-th component is called the axial vector of $\mathbf F$. Note that the definition for the vector product in (14) also involves the components of the alternating tensor. This is no accident, as the vector product in $\mathbb R^3$ between two vectors $\mathbf a$ and $\mathbf f$ is actually the result of an inner product between an antisymmetric tensor $\mathbf F$ and a vector $\mathbf a$,
$$F_{ij}a_j = \epsilon_{ijk} a_j f_k \tag{19}$$
## The Lorentz Force
Now we are in a position to consider the equation you have written for $d \vec p/dt$ using clearer notation,
\begin{align} \frac{d\mathbf p}{dt} & = e\left[\left(-\frac{\partial \mathbf A}{\partial t} - \nabla \phi \right) + \mathbf v \cdot \nabla \mathbf A - \nabla \mathbf A \cdot \mathbf v\right] \\ & = e\left(\mathbf E + \mathbf v \cdot \nabla \mathbf A - \nabla \mathbf A \cdot \mathbf v\right) \end{align} \tag{20}
Let's rewrite this in summation notation:
\begin{align} \frac{dp_i}{dt} & = e\left(E_i + v_l \delta_{lj}\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \delta_{jl} v_l\right) \\ & = e\left(E_i + \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) v_j\right) \end{align} \tag{21}
The term $(\partial A_j/\partial x_i) - (\partial A_i/\partial x_j)$ is clearly the component $F_{ij}$ of an antisymmetric second-order tensor $\mathbf F$, and so the components of its corresponding axial vector $f_k$ are,
\begin{align} f_k & = \frac{1}{2} \epsilon_{ijk} \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left(\epsilon_{ijk}\frac{\partial A_j}{\partial x_i} - \epsilon_{ijk}\frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left(\epsilon_{kij}\frac{\partial A_j}{\partial x_i} + \epsilon_{kji}\frac{\partial A_i}{\partial x_j}\right) \\ & = \frac{1}{2} \left[\left(\nabla \times \mathbf A\right)_k + \left(\nabla \times \mathbf A\right)_k\right] \\ & = \left(\nabla \times \mathbf A\right)_k \\ & = B_k \end{align} \tag{22}
Thus,
\begin{align} \frac{dp_i}{dt} & = e\left(E_i + \left(\frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}\right) v_j\right) \\ & = e\left(E_i + \epsilon_{ijk} B_k v_j\right) \\ & = e\left(E_i + \left(\mathbf v \times \mathbf B\right)_i \right) \\ \end{align} \tag{23}
## A Final Note on the Tensor Formulation of Electromagnetism
When studying electromagnetic phenomena in the context of relativity, due to the non-Euclidean nature of the geometry of spacetime, we cannot simplify our mathematical analysis by working in a 3-dimensional Euclidean basis; however, since tensors are geometric objects that exist independently of any coordinate basis used to describe them, a similar analysis can be performed to yield the correct result, and in Minkowski spacetime you will end up constructing a $4 \times 4$ antisymmetric tensor of the form,
$$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} = \left(\begin{matrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & B_z & -B_y \\ E_y/c & -B_z & 0 & B_x \\ E_z/c & B_y & -B_x & 0 \\\end{matrix}\right) \tag{24}$$
more commonly know as the electromagnetic field tensor, and the Lorentz Force Law for a charge $q$ will take the form of
$$\frac{dp_\mu}{d \tau} = qF_{\mu\nu}U^{\nu} \tag{25}$$
where $p_\mu$ are the covariant components of the charge's four-momentum, $\tau$ is the proper time experienced by the charge, and $U^{\nu}$ are the contravariant components of the charge's four-velocity.
• Thank you vary much. It is an awesome answer. I'm not that person who I was before your post, my mind is different now) p.s. There is shoud be the "a" insted of "f" in the 16th expression. Dec 5, 2017 at 14:55
• You're welcome, and thank you for catching that typo; it has been fixed.
– JM1
Dec 5, 2017 at 15:10
• Just a little question: from where did epsilon emerge in the 23 equation? Dec 5, 2017 at 15:11
• I used the relationship from (17) to rewrite $(\partial A_j/\partial x_i) - (\partial A_i/\partial x_j)$ in terms of $B_k$
– JM1
Dec 5, 2017 at 15:15
Here: let's do it a little bit differently that will help make things more clear. For a charged body, with charge $$e$$, the potential energy $$eφ$$ has the same relation to the body's energy $$E$$, as $$e𝐀$$ does to the body's momentum $$𝐩$$. So, you can think of it as "potential momentum". The field adds to the body's momentum in the same way that it adds to the body's energy. Hence, the relations: $$H = E + eφ, \hspace 1em 𝐏 = 𝐩 + e𝐀.$$ Correspondingly, you may refer to $$E$$ and $$𝐩$$, respectively, as the "kinetic" parts ... or more accurately, after subtracting out $$mc^2$$, to refer to $$E - mc^2$$ is the kinetic part of the energy.
The kinetic momentum arises entirely from the motion of the body and is proportional to its velocity $$𝐯$$, as $$𝐩 = M𝐯$$, where $$M$$ is the "moving" mass. The total energy, itself, can be expressed by That Equation $$E = Mc^2$$, with the kinetic part being $$(M - m)c^2 = Mw$$, where $$w$$ is that all-important ubiquitous factor that nobody ever talks about in Relativity that is given as the solution to: $$v^2 - 2w + \frac{w^2}{c^2} = 0.$$ Specifically: $$w = c^2\left(1 - \sqrt{1 - (v/c)^2}\right) = \frac{v^2}{1 + \sqrt{1 - (v/c)^2}}.$$ And it just so happens that we may also write: $$c^2\left(\frac{1}{\sqrt{1 - (v/c)^2}} - 1\right) = \frac{w}{\sqrt{1 - (v/c)^2}}.$$
So, we can write the respective quantities as: $$H = Mc^2 + eφ = mc^2 + Mw + eφ, \hspace 1em 𝐏 = M𝐯 + e𝐀.$$ The relation between the moving mass and (kinetic) momentum comes out of the mass-energy-momentum relation $$E^2 - p^2c^2 = m^2c^4$$ (with $$\text{sgn}(E) = \text{sgn}(M) = \text{sgn}(m)$$ used to select out the root). This works out to be: $$M = m\sqrt{1 + \frac{p^2}{m^2c^2}}.$$
Substituting for $$𝐯$$ in terms of $$M$$ and $$𝐩$$, we can also write $$w$$ in momentum space form as: $$w = \frac{p^2}{M(m + M)}.$$
So, after substituting and reverting to the total momentum, you get: $$H = mc^2 + \frac{|𝐏 - e𝐀|^2}{m + M} + eφ, \hspace 1em M = m\sqrt{1 + \frac{|𝐏 - e𝐀|^2}{m^2c^2}},$$ or after putting back in the $$mc^2$$: $$H = Mc^2 + eφ = mc^2\sqrt{1 + \frac{|𝐏 - e𝐀|^2}{m^2c^2}} + eφ.$$
The alternative forms are preferable if you want to use this in regimes where Newtonian theory is good as an approximation, since - in that setting - $$w = v^2/2 = p^2/(2m^2)$$ and $$M = m$$; which is why I confused the issue by bringing them into play. It's always good to confuse the issue. It keeps your mind sharp!
Now ... the second part of your question: differentiation. The best way to address that is to just write everything out as differential relations and (true to the name "differential coefficient") extract out the coefficients. For Hamiltonians, the total differential, combined with the equations involving their coefficients, is: $$dH = 𝐯·d𝐏 - 𝐅·d𝐫 + \frac{∂H}{∂t} dt, \hspace 1em 𝐯 = \frac{d𝐫}{dt}, \hspace 1em 𝐅 = \frac{d𝐏}{dt}.$$
Using the relation $$M^2 = m^2 + p^2/c^2$$, we may write $$M dM = \frac{𝐩·d𝐩}{c^2} = \frac{𝐩·d𝐏 - e𝐩·d𝐀}{c^2}.$$ Substituting into the expression $$H = Mc^2 + eφ$$, we get: $$dH = c^2 dM + edφ = \frac{𝐩·d𝐏 - e𝐩·d𝐀}{M} + edφ.$$ Use the chain rule on the fields: $$dφ = \frac{∂φ}{∂t}dt + d𝐫·∇φ, \hspace 1em d𝐀 = \frac{∂𝐀}{∂t}dt + d𝐫·∇𝐀.$$ Thus, $$dH = \frac{𝐩}{M}·d𝐏 - e\frac{𝐩}{M}·\frac{∂𝐀}{∂t}dt - e\frac{𝐩}{M}(d𝐫·∇)𝐀 + e\frac{∂φ}{∂t}dt + ed𝐫·∇φ.$$
The math that's going on in the extra term $$(e/M) 𝐩·(d𝐫·∇)𝐀$$ is really what you were really asking about. To handle stuff like that, when Heaviside originally developed vector notation, he also developed notation for dyads and higher order tensors and they are still in common use (except in the theoretical literature). Tensor products are denoted by direct juxtaposition and are associative with respect to itself and the dot product, e.g. $$𝐚·(𝐛𝐜) = (𝐚·𝐛)𝐜$$, and $$(𝐚𝐛)·𝐜 = 𝐚(𝐛·𝐜)$$. So, the total differential can just as well be be written as $$dH = \frac{𝐩}{M}·d𝐏 - e\frac{𝐩}{M}·\frac{∂𝐀}{∂t}dt - ed𝐫·∇𝐀·\frac{𝐩}{M} + e\frac{∂φ}{∂t}dt + ed𝐫·∇φ.$$
From this, we can directly read off the differential coefficients: $$𝐯 = \frac{𝐩}{M}, \hspace 1em 𝐅 = e\left(∇𝐀·\frac{𝐩}{M} - ∇φ\right), \hspace 1em \frac{∂H}{∂t} = e\left(\frac{∂φ}{∂t} - \frac{𝐩}{M}·\frac{∂𝐀}{∂t}\right),$$ and after substituting in $$𝐯$$: $$𝐅 = e\left(∇𝐀·𝐯 - ∇φ\right), \hspace 1em \frac{∂H}{∂t} = e\left(\frac{∂φ}{∂t} - 𝐯·\frac{∂𝐀}{∂t}\right).$$
Using dyads, the double cross product can be more succinctly written as $$𝐚×(𝐛×𝐜) = 𝐚·(𝐜𝐛 - 𝐛𝐜) = (𝐛𝐜 - 𝐜𝐛)·𝐚.$$ You can also write it as $$𝐚×(𝐛×𝐜) = 𝐚·𝐜𝐛 - 𝐜𝐛·𝐚 = 𝐛𝐜·𝐚 - 𝐚·𝐛𝐜.$$ In particular $$𝐯×(∇×𝐀) = ∇𝐀·𝐯 - 𝐯·∇𝐀.$$
So, applying this to the equation of motion for the position $$𝐫$$ and total momentum $$𝐏$$, you get: $$\frac{d𝐫}{dt} = 𝐯 = \frac{𝐩}{M}, \hspace 1em \frac{d𝐏}{dt} = 𝐅 = e\left(∇𝐀·𝐯 - ∇φ\right).$$ Break this down to get the equation for the kinetic momentum $$\frac{d𝐩}{dt} + e\frac{d𝐀}{dt} = e\left(∇𝐀·𝐯 - ∇φ\right).$$ Use the chain rule on $$𝐀$$: $$\frac{d𝐀}{dt} = \frac{∂𝐀}{∂t} + \frac{d𝐫}{dt}·∇𝐀 = \frac{∂𝐀}{∂t} + 𝐯·∇𝐀$$ to get: $$\frac{d𝐩}{dt} + e\left(\frac{∂𝐀}{∂t} + 𝐯·∇𝐀\right) = e\left(∇𝐀·𝐯 - ∇φ\right).$$ Collect all the $$e$$ terms on the right side: $$\frac{d𝐩}{dt} = e\left(-∇φ - \frac{∂𝐀}{∂t} + ∇𝐀·𝐯 - 𝐯·∇𝐀\right) = e\left(-∇φ - \frac{∂𝐀}{∂t} + 𝐯×(∇×𝐀)\right).$$ Finally, use the Maxwell equations for potentials versus forces: $$𝐁 = ∇×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t},$$ to reduce this to more familiar form: $$\frac{d𝐩}{dt} = e(𝐄 + 𝐯×𝐁).$$
This leads, also, to equations for the moving mass $$M$$ and total energy $$E$$: $$Mc^2 \frac{dM}{dt} = 𝐩·\frac{d𝐩}{dt} \\ ⇒ \\ \frac{dE}{dt} = c^2 \frac{dM}{dt} = \frac{𝐩}{M}·\frac{d𝐩}{dt} = 𝐯·e(𝐄 + 𝐯×𝐁) = e𝐯·𝐄,$$ since $$𝐯·𝐯×𝐁 = 0$$. | 7,199 | 19,826 | {"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": 65, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.72126 |
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This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
$z$ test for the difference between two proportions
Paired sample $t$ test
Sign test
Two sample $z$ test
Paired sample $t$ test
Regression (OLS)
Kruskal-Wallis test
Independent variablesIndependent/grouping variableIndependent variableIndependent variableIndependent/grouping variableIndependent variableIndependent variablesIndependent/grouping variable
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne categorical with 2 independent groups2 paired groups2 paired groupsOne categorical with 2 independent groups2 paired groupsOne or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent variableDependent variableDependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne of ordinal levelOne quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesis
$F$ test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
or equivalenty
• H0: the variance explained by all the independent variables together (the complete model) is 0 in the population, i.e. $\rho^2 = 0$
$t$ test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
in the regression equation $\mu_y = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\mu_y$ represents the population mean of the dependent variable $y$ given the scores on the independent variables.
H0: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.
H0: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
• H0: P(first score of a pair exceeds second score of a pair) = P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
• H0: the population median of the difference scores is equal to zero
A difference score is the difference between the first score of a pair and the second score of a pair.
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
H0: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
$F$ test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
or equivalenty
• H0: the variance explained by all the independent variables together (the complete model) is 0 in the population, i.e. $\rho^2 = 0$
$t$ test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
in the regression equation $\mu_y = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\mu_y$ represents the population mean of the dependent variable $y$ given the scores on the independent variables.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
$F$ test for the complete regression model:
• H1: not all population regression coefficients are 0
or equivalenty
• H1: the variance explained by all the independent variables together (the complete model) is larger than 0 in the population, i.e. $\rho^2 > 0$
$t$ test for individual regression coefficient $\beta_k$:
• H1 two sided: $\beta_k \neq 0$
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
• H1 two sided: P(first score of a pair exceeds second score of a pair) $\neq$ P(second score of a pair exceeds first score of a pair)
• H1 right sided: P(first score of a pair exceeds second score of a pair) > P(second score of a pair exceeds first score of a pair)
• H1 left sided: P(first score of a pair exceeds second score of a pair) < P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
• H1 two sided: the population median of the difference scores is different from zero
• H1 right sided: the population median of the difference scores is larger than zero
• H1 left sided: the population median of the difference scores is smaller than zero
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
$F$ test for the complete regression model:
• H1: not all population regression coefficients are 0
or equivalenty
• H1: the variance explained by all the independent variables together (the complete model) is larger than 0 in the population, i.e. $\rho^2 > 0$
$t$ test for individual regression coefficient $\beta_k$:
• H1 two sided: $\beta_k \neq 0$
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
• In the population, the residuals are normally distributed at each combination of values of the independent variables
• In the population, the standard deviation $\sigma$ of the residuals is the same for each combination of values of the independent variables (homoscedasticity)
• In the population, the relationship between the independent variables and the mean of the dependent variable $\mu_y$ is linear. If this linearity assumption holds, the mean of the residuals is 0 for each combination of values of the independent variables
• The residuals are independent of one another
Often ignored additional assumption:
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
• Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
• Significance test: number of successes and number of failures are each 5 or more in both sample groups
• Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
• Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Difference scores are normally distributed in the population
• Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• Population standard deviations $\sigma_1$ and $\sigma_2$ are known
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Difference scores are normally distributed in the population
• Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
• In the population, the residuals are normally distributed at each combination of values of the independent variables
• In the population, the standard deviation $\sigma$ of the residuals is the same for each combination of values of the independent variables (homoscedasticity)
• In the population, the relationship between the independent variables and the mean of the dependent variable $\mu_y$ is linear. If this linearity assumption holds, the mean of the residuals is 0 for each combination of values of the independent variables
• The residuals are independent of one another
Often ignored additional assumption:
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
Test statisticTest statisticTest statisticTest statisticTest statisticTest statisticTest statisticTest statistic
$F$ test for the complete regression model:
• \begin{aligned}[t] F &= \dfrac{\sum (\hat{y}_j - \bar{y})^2 / K}{\sum (y_j - \hat{y}_j)^2 / (N - K - 1)}\\ &= \dfrac{\mbox{sum of squares model} / \mbox{degrees of freedom model}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square model}}{\mbox{mean square error}} \end{aligned}
where $\hat{y}_j$ is the predicted score on the dependent variable $y$ of subject $j$, $\bar{y}$ is the mean of $y$, $y_j$ is the score on $y$ of subject $j$, $N$ is the total sample size, and $K$ is the number of independent variables.
$t$ test for individual $\beta_k$:
• $t = \dfrac{b_k}{SE_{b_k}}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$
with $s$ the sample standard deviation of the residuals, $x_j$ the score of subject $j$ on the independent variable $x$, and $\bar{x}$ the mean of $x$. For models with more than one independent variable, computing $SE_{b_k}$ is more complicated.
Note 1: mean square model is also known as mean square regression, and mean square error is also known as mean square residual.
Note 2: if there is only one independent variable in the model ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1.$
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2.
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
$W =$ number of difference scores that is larger than 0$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
$F$ test for the complete regression model:
• \begin{aligned}[t] F &= \dfrac{\sum (\hat{y}_j - \bar{y})^2 / K}{\sum (y_j - \hat{y}_j)^2 / (N - K - 1)}\\ &= \dfrac{\mbox{sum of squares model} / \mbox{degrees of freedom model}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square model}}{\mbox{mean square error}} \end{aligned}
where $\hat{y}_j$ is the predicted score on the dependent variable $y$ of subject $j$, $\bar{y}$ is the mean of $y$, $y_j$ is the score on $y$ of subject $j$, $N$ is the total sample size, and $K$ is the number of independent variables.
$t$ test for individual $\beta_k$:
• $t = \dfrac{b_k}{SE_{b_k}}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$
with $s$ the sample standard deviation of the residuals, $x_j$ the score of subject $j$ on the independent variable $x$, and $\bar{x}$ the mean of $x$. For models with more than one independent variable, computing $SE_{b_k}$ is more complicated.
Note 1: mean square model is also known as mean square regression, and mean square error is also known as mean square residual.
Note 2: if there is only one independent variable in the model ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1.$
$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$
Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.
Note: if ties are present in the data, the formula for $H$ is more complicated.
Sample standard deviation of the residuals $s$n.a.n.a.n.a.n.a.n.a.Sample standard deviation of the residuals $s$n.a.
\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}-----\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}-
Sampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $W$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $H$ if H0 were true
Sampling distribution of $F$:
• $F$ distribution with $K$ (df model, numerator) and $N - K - 1$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - K - 1$ (df error) degrees of freedom
Approximately the standard normal distribution$t$ distribution with $N - 1$ degrees of freedomThe exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
Standard normal distribution$t$ distribution with $N - 1$ degrees of freedomSampling distribution of $F$:
• $F$ distribution with $K$ (df model, numerator) and $N - K - 1$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - K - 1$ (df error) degrees of freedom
For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.
For small samples, the exact distribution of $H$ should be used.
Significant?Significant?Significant?Significant?Significant?Significant?Significant?Significant?
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
$t$ Test two sided:
$t$ Test right sided:
$t$ Test left sided:
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
If $n$ is small, the table for the binomial distribution should be used:
Two sided:
• Check if $W$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $W$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $W$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
$t$ Test two sided:
$t$ Test right sided:
$t$ Test left sided:
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$C\%$ confidence interval for $\beta_k$ and for $\mu_y$, $C\%$ prediction interval for $y_{new}$Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$$C\% confidence interval for \mun.a.C\% confidence interval for \mu_1 - \mu_2$$C\%$ confidence interval for \mu$$C\% confidence interval for \beta_k and for \mu_y, C\% prediction interval for y_{new}n.a. Confidence interval for \beta_k: • b_k \pm t^* \times SE_{b_k} • If only one independent variable: SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}} Confidence interval for \mu_y, the population mean of y given the values on the independent variables: • \hat{y} \pm t^* \times SE_{\hat{y}} • If only one independent variable: SE_{\hat{y}} = s \sqrt{\dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} Prediction interval for y_{new}, the score on y of a future respondent: • \hat{y} \pm t^* \times SE_{y_{new}} • If only one independent variable: SE_{y_{new}} = s \sqrt{1 + \dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} In all formulas, the critical value t^* is the value under the t_{N - K - 1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). Regular (large sample): • (p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) With plus four method: • (p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}} where p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}, p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}, and the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval) \bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}} where the critical value t^* is the value under the t_{N-1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu can also be used as significance test. -(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}} where the critical value z^* is the value under the normal curve with the area C / 100 between -z^* and z^* (e.g. z^* = 1.96 for a 95% confidence interval). The confidence interval for \mu_1 - \mu_2 can also be used as significance test. \bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}} where the critical value t^* is the value under the t_{N-1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu can also be used as significance test. Confidence interval for \beta_k: • b_k \pm t^* \times SE_{b_k} • If only one independent variable: SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}} Confidence interval for \mu_y, the population mean of y given the values on the independent variables: • \hat{y} \pm t^* \times SE_{\hat{y}} • If only one independent variable: SE_{\hat{y}} = s \sqrt{\dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} Prediction interval for y_{new}, the score on y of a future respondent: • \hat{y} \pm t^* \times SE_{y_{new}} • If only one independent variable: SE_{y_{new}} = s \sqrt{1 + \dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} In all formulas, the critical value t^* is the value under the t_{N - K - 1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). - Effect sizen.a.Effect sizen.a.n.a.Effect sizeEffect sizen.a. Complete model: • Proportion variance explained R^2: Proportion variance of the dependent variable y explained by the sample regression equation (the independent variables):$$ \begin{align} R^2 &= \dfrac{\sum (\hat{y}_j - \bar{y})^2}{\sum (y_j - \bar{y})^2}\\ &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}\\ &= 1 - \dfrac{\mbox{sum of squares error}}{\mbox{sum of squares total}}\\ &= r(y, \hat{y})^2 \end{align} $$R^2 is the proportion variance explained in the sample by the sample regression equation. It is a positively biased estimate of the proportion variance explained in the population by the population regression equation, \rho^2. If there is only one independent variable, R^2 = r^2: the correlation between the independent variable x and dependent variable y squared. • Wherry's R^2 / shrunken R^2: Corrects for the positive bias in R^2 and is equal to$$R^2_W = 1 - \frac{N - 1}{N - K - 1}(1 - R^2)$$R^2_W is a less biased estimate than R^2 of the proportion variance explained in the population by the population regression equation, \rho^2. • Stein's R^2: Estimates the proportion of variance in y that we expect the current sample regression equation to explain in a different sample drawn from the same population. It is equal to$$R^2_S = 1 - \frac{(N - 1)(N - 2)(N + 1)}{(N - K - 1)(N - K - 2)(N)}(1 - R^2)$$Per independent variable: • Correlation squared r^2_k: the proportion of the total variance in the dependent variable y that is explained by the independent variable x_k, not corrected for the other independent variables in the model • Semi-partial correlation squared sr^2_k: the proportion of the total variance in the dependent variable y that is uniquely explained by the independent variable x_k, beyond the part that is already explained by the other independent variables in the model • Partial correlation squared pr^2_k: the proportion of the variance in the dependent variable y not explained by the other independent variables, that is uniquely explained by the independent variable x_k -Cohen's d: Standardized difference between the sample mean of the difference scores and \mu_0:$$d = \frac{\bar{y} - \mu_0}{s}$$Cohen's d indicates how many standard deviations s the sample mean of the difference scores \bar{y} is removed from \mu_0. --Cohen's d: Standardized difference between the sample mean of the difference scores and \mu_0:$$d = \frac{\bar{y} - \mu_0}{s}$$Cohen's d indicates how many standard deviations s the sample mean of the difference scores \bar{y} is removed from \mu_0. Complete model: • Proportion variance explained R^2: Proportion variance of the dependent variable y explained by the sample regression equation (the independent variables):$$ \begin{align} R^2 &= \dfrac{\sum (\hat{y}_j - \bar{y})^2}{\sum (y_j - \bar{y})^2}\\ &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}\\ &= 1 - \dfrac{\mbox{sum of squares error}}{\mbox{sum of squares total}}\\ &= r(y, \hat{y})^2 \end{align} $$R^2 is the proportion variance explained in the sample by the sample regression equation. It is a positively biased estimate of the proportion variance explained in the population by the population regression equation, \rho^2. If there is only one independent variable, R^2 = r^2: the correlation between the independent variable x and dependent variable y squared. • Wherry's R^2 / shrunken R^2: Corrects for the positive bias in R^2 and is equal to$$R^2_W = 1 - \frac{N - 1}{N - K - 1}(1 - R^2)$$R^2_W is a less biased estimate than R^2 of the proportion variance explained in the population by the population regression equation, \rho^2. • Stein's R^2: Estimates the proportion of variance in y that we expect the current sample regression equation to explain in a different sample drawn from the same population. It is equal to$$R^2_S = 1 - \frac{(N - 1)(N - 2)(N + 1)}{(N - K - 1)(N - K - 2)(N)}(1 - R^2)$Per independent variable: • Correlation squared$r^2_k$: the proportion of the total variance in the dependent variable$y$that is explained by the independent variable$x_k$, not corrected for the other independent variables in the model • Semi-partial correlation squared$sr^2_k$: the proportion of the total variance in the dependent variable$y$that is uniquely explained by the independent variable$x_k$, beyond the part that is already explained by the other independent variables in the model • Partial correlation squared$pr^2_k$: the proportion of the variance in the dependent variable$y$not explained by the other independent variables, that is uniquely explained by the independent variable$x_k$- Visual representationn.a.Visual representationn.a.Visual representationVisual representationVisual representationn.a. Regression equations with: --Regression equations with: - ANOVA tablen.a.n.a.n.a.n.a.n.a.ANOVA tablen.a. ----- - n.a.Equivalent toEquivalent toEquivalent ton.a.Equivalent ton.a.n.a. -When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. • One sample$t$test on the difference scores. • Repeated measures ANOVA with one dichotomous within subjects factor. Two sided sign test is equivalent to - • One sample$t$test on the difference scores. • Repeated measures ANOVA with one dichotomous within subjects factor. -- Example contextExample contextExample contextExample contextExample contextExample contextExample contextExample context Can mental health be predicted from fysical health, economic class, and gender?Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.Is the average difference between the mental health scores before and after an intervention different from$\mu_0 = 0$?Do people tend to score higher on mental health after a mindfulness course?Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is$\sigma_1 = 2$amongst men and$\sigma_2 = 2.5$amongst women.Is the average difference between the mental health scores before and after an intervention different from$\mu_0 = 0$?Can mental health be predicted from fysical health, economic class, and gender?Do people from different religions tend to score differently on social economic status? SPSSSPSSSPSSSPSSn.a.SPSSSPSSSPSS Analyze > Regression > Linear... • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Independent(s) SPSS does not have a specific option for the$z$test for the difference between two proportions. However, you can do the chi-squared test instead. The$p$value resulting from this chi-squared test is equivalent to the two sided$p$value that would have resulted from the$z$test. Go to: Analyze > Descriptive Statistics > Crosstabs... • Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s) • Click the Statistics... button, and click on the square in front of Chi-square • Continue and click OK Analyze > Compare Means > Paired-Samples T Test... • Put the two paired variables in the boxes below Variable 1 and Variable 2 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... • Put the two paired variables in the boxes below Variable 1 and Variable 2 • Under Test Type, select the Sign test -Analyze > Compare Means > Paired-Samples T Test... • Put the two paired variables in the boxes below Variable 1 and Variable 2 Analyze > Regression > Linear... • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Independent(s) Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples... • Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable • Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow • Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum • Continue and click OK JamoviJamoviJamoviJamovin.a.JamoviJamoviJamovi Regression > Linear Regression • Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates • If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well • Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes' Jamovi does not have a specific option for the$z$test for the difference between two proportions. However, you can do the chi-squared test instead. The$p$value resulting from this chi-squared test is equivalent to the two sided$p$value that would have resulted from the$z$test. Go to: Frequencies > Independent Samples -$\chi^2$test of association • Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns T-Tests > Paired Samples T-Test • Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line • Under Hypothesis, select your alternative hypothesis Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The$p$value resulting from this Friedman test is equivalent to the two sided$p\$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
• Put the two paired variables in the box below Measures
-T-Tests > Paired Samples T-Test
• Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
• Under Hypothesis, select your alternative hypothesis
Regression > Linear Regression
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
ANOVA > One Way ANOVA - Kruskal-Wallis
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
Practice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questionsPractice questions | 9,713 | 36,135 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-45 | latest | en | 0.666526 |
https://www.hackmath.net/en/example/517 | 1,556,013,541,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578596571.63/warc/CC-MAIN-20190423094921-20190423120921-00368.warc.gz | 710,356,599 | 6,743 | # Four pupils
Four pupils divided \$ 1485 so that the second received 50% less than the first, the third 1/2 less than a fourth and fourth \$ 154 less than the first. How much money had each of them?
Result
a = 572 \$
b = 286 \$
c = 209 \$
d = 418 \$
#### Solution:
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How many seeds germinated from 1000 pcs, when 23% no emergence? | 725 | 2,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-18 | latest | en | 0.938122 |
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# Physics 2 - PowerPoint PPT Presentation
Physics 2. Rotational Motion. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. First some quick geometry review:. We need this formula for arc length to see the connection between rotational motion and linear motion. x = r θ. θ.
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### Physics 2
Rotational Motion
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We need this formula for arc length to see the connection between rotational motion and linear motion.
x = rθ
θ
We will also need to be able to convert from revolutions to radians.
There are 2π radians in one complete revolution.
r
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion.
x = rθ
This is the Greek letter omega (not w)
θ
Angular Velocity = ω =
r
Angular Acceleration = α =
This is the Greek letter alpha (looks kinda like a fish)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion.
x = rθ
This is the Greek letter omega (not w)
θ
Angular Velocity = ω =
r
Angular Acceleration = α =
This is the Greek letter alpha (looks kinda like a fish)
Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.
Find the final angular velocity and the angular acceleration (assume constant).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion.
x = rθ
This is the Greek letter omega (not w)
θ
Angular Velocity = ω =
r
Angular Acceleration = α =
This is the Greek letter alpha (looks kinda like a fish)
Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.
Find the final angular velocity and the angular acceleration (assume constant).
rpm stands for “revolutions per minute” – we can just do a unit conversion:
Standard units for angular velocity are radians per second
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion.
x = rθ
This is the Greek letter omega (not w)
θ
Angular Velocity = ω =
r
Angular Acceleration = α =
This is the Greek letter alpha (looks kinda like a fish)
Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.
Find the final angular velocity and the angular acceleration (assume constant).
rpm stands for “revolutions per minute” – we can just do a unit conversion:
Standard units for angular velocity are radians per second
Now we can use the definition of angular acceleration:
Standard units for angular acceleration are radians per second2.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We already know how to deal with linear motion. analogous to what we had for linear motion.
We have formulas for kinematics, forces, energy and momentum.
We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones!
All you have to do is translate the variables.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We already know how to deal with linear motion. analogous to what we had for linear motion.
We have formulas for kinematics, forces, energy and momentum.
We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones!
All you have to do is translate the variables.
We have already seen one case:
x = rθ This translates between distance (linear) and angle (rotational)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We already know how to deal with linear motion. analogous to what we had for linear motion.
We have formulas for kinematics, forces, energy and momentum.
We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones!
All you have to do is translate the variables.
We have already seen one case:
x = rθ This translates between distance (linear) and angle (rotational)
Here are the other variables:
v = rω linear velocity relates to angular velocity
atan = rα linear acceleration relates to angular acceleration
Notice a pattern here?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We already know how to deal with linear motion. analogous to what we had for linear motion.
We have formulas for kinematics, forces, energy and momentum.
We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones!
All you have to do is translate the variables.
We have already seen one case:
x = rθ This translates between distance (linear) and angle (rotational)
Here are the other variables:
v = rω linear velocity relates to angular velocity
atan = rα linear acceleration relates to angular acceleration
Multiply the angular quantity by the radius to get the linear quantity.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
We basically have two options on how to proceed. We can switch to angular variables right away, or we can do the corresponding problem in linear variables and translate at the end.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
Switching to angular variables right away:
Convert to angular velocity:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
Switching to angular variables right away:
Convert to angular velocity:
Find angular acceleration:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
Switching to angular variables right away:
Convert to angular velocity:
Use a kinematics equation:
Find angular acceleration:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
Switching to angular variables right away:
Convert to angular velocity:
Use a kinematics equation:
Find angular acceleration:
Convert to revolutions:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book)
Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.
Find the total number of revolutions that the wheels make during the 25 second interval.
This time do the linear problem first:
Find linear acceleration:
Convert to revolutions:
Use a kinematics equation:
(we did some rounding off)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
All of these have analogues for rotational motion as well. Forces and Momentum will be covered in chapter 10. That leaves us with Energy.
Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula:
We know that ω is angular velocity. Comparing with the formula for linear kinetic energy, what do you think I is?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
All of these have analogues for rotational motion as well. Forces and Momentum will be covered in chapter 10. That leaves us with Energy.
Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula:
The I in our formula takes the place of m (mass) in the linear formula. We call it Moment of Inertia (or rotational inertia). It plays the same role in rotational motion that mass plays in linear motion (I quantifies how difficult it is to produce an angular acceleration.
The value for I will depend on the shape of your object, but the basic rule of thumb is that the farther the mass is from the axis of rotation, the larger the inertia.
Page 299 in your book has a table of formulas for different shapes.
These are all based on the formula for a point particle.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
Hollow Sphere
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Problem 9.48 Momentum.
A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.
a) How fast is each sphere moving when it reaches the bottom of the hill?
b) Which sphere will reach the bottom first, the hollow one or the solid one?
We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.
h
θ
we can replace ω with v/r so everything is in terms of the desired unknown
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
Hollow Sphere
The solid sphere is faster because its moment of inertia is smaller.
It reaches the bottom first.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB | 4,125 | 19,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-39 | latest | en | 0.900569 |
https://physics.stackexchange.com/questions/355751/charge-distribution-that-generates-electrostatic-field-with-constant-magnitude | 1,653,793,558,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00360.warc.gz | 514,167,982 | 65,090 | # Charge distribution that generates Electrostatic Field with constant magnitude
I have founde the following problem. I am given a sphere of radius $R$, with charge density $\rho$ and the fact that the electrostatic field inside the sphere has constant magnitude $E_0$. The question is to find the charge distribution.
Gauss' Law states that $\nabla\cdot \vec{E} = \frac{\rho}{\epsilon_0}$. Thus, if I know a mathematical expression for the electrostatic field, I can calculate the charge density.
Obviously, the uniform field $\vec{E}=E_0 \cdot \hat{n}$ satisfies the constant magnitude constraint, but can there be another solution other than that trivial case? How can I prove that this is the only solution?
From Gauss' law, using a Gaussian surface of radius $R$ with $R$ inside your charged sphere: $$\epsilon\int \vec E\cdot d\vec S=\oint \rho dV$$ Assuming $\rho$ is spherically symmetric we obtain $$\epsilon \vert \vec E\vert\, 4\pi R^2 = 4\pi \int_0^R r^2 \rho dr$$ Thus, if $R_0$ is the radius of your charged sphere, any charge density $\rho$ of the form $$\rho=\rho_0 R_0/r$$ for $\rho_0$ constant will produce $$4\pi \int_0^R r^2 \rho dr = 4\pi \int_0^R r \rho_0 R_0 = 2\pi \rho_0 R_0 R^2$$ and thus $$\vert \vec E \vert = \frac{2\pi \rho_0 R_0 R^2 }{4\pi \epsilon R^2}= \frac{\rho_0 R_0}{2\epsilon}$$ will be constant.
• For the spherical geometry this is the only charge distribution that works. Note that the distribution is at $r=0$ so you might want to re-examine the situation if your sphere is hollow. Sep 6, 2017 at 12:34 | 462 | 1,548 | {"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} | 3.734375 | 4 | CC-MAIN-2022-21 | longest | en | 0.827315 |
https://www.r-bloggers.com/2016/11/repeated-measures-anova-in-r-exercises/ | 1,718,722,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00314.warc.gz | 830,153,447 | 22,728 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
One way, two way and n way ANOVA are used to test difference in means when we have one, two and n factor variables. A key assumption when performing these ANOVAs is that the measurements are independent. When we have repeated measures this assumption is violated, so we have to use repeated measures ANOVA. Repeated measures designs occur often in longitudinal studies where we are interested in understanding change over time. For example a medical researcher would be interested in assessing the level of depression before and after a surgery procedure. Repeated measures designs are not limited to longitudinal studies, they can also be used when you have an important variable you would like to repeat measures. For example in a fitness experiment you can repeat your measures at different intensity levels. Repeated measures ANOVA can be considered an extension of the paired t test.
Before diving deeper into repeated measures ANOVA you need to understand terminology used. A subject is a member of the sample under consideration. In our medical study introduced earlier an individual patient is a subject. The within-subjects factor is the variable that identifies how the dependent variable has been repeatedly measured. In our medical study we would measure depression 4 weeks before surgery, 4 weeks after surgery and 8 weeks after surgery. The different conditions when repeated measurements are made are referred to as trials. A between-subjects factor identifies independent groups in the study. For example if we had two different procedures this would be the between subjects factor. These conditions are referred to as groups. Repeated measures analysis requires balance in between-subjects factor. For example subjects in each of surgery procedures need to be equal.
With a repeated measures design we are able to test the following hypotheses.
1. There is no within-subjects main effect
2. There is no between-subjects main effect
3. There is no between subjects interaction effect
4. There is no within subject by between subject interaction effect
There are two assumptions that need to be satisfied when using repeated measures.
1. The dependent variable is normally distributed in each level of the within-subjects factor. Repeated measures analysis is robust to violations of normality with a large sample size which is considered at least 30 subjects. However the accuracy of p values is questionable when the distribution is heavily skewed or thick tailed.
2. The variance across the within subject factor is equal. This is the sphericity assumption. Repeated measures analysis is not robust to this assumption so when there is a violation power decreases and a corresponding increase in probability of a type II error occurs. A Mauchly’s test assesses the null hypothesis variance is equal. The sphericity assumption is only relevant when there are more than 2 levels of the within subjects factor.
When the sphericity assumption is violated we make corrections by adjusting the degrees of freedom. Corrections available are Greenhouse-Geisser, Huynh-Feldt and Lower bound. To make a decision on appropriate correction we use a Greenhouse-Geisser estimate of sphericity (ξ). When ξ < 0.75 or we do not know anything about sphericity the Greenhouse-Geisser is the appropriate correction. When ξ > 0.75 Huynh-Feldt is the appropriate correction.
For this exercise we will use data on pulse rate exer. People were randomized to two diets, three exercise types and pulse was measured at three different time points. For this data time points is the within-subjects factor. The between-subjects factors are diet and exercise type
The solutions to the exercises below can be found here
Exercise 1
Load the data and inspect its structure
Exercise 2
Check for missing values
Exercise 3
Check for balance in between-subjects factor
Exercise 4
Generate descriptive statistics for the sex variable which is a between subjects factor
Exercise 5
Generate descriptive statistics for the treatment level variable which is a between subjects factor
Exercise 6
Generate descriptive statistics for the weeks variable which is the within subjects factor
Exercise 7
Use histograms to assess distribution across within subjects factor.
Exercise 8
Perform a repeated measures analysis with only the within subjects factor
Exercise 9
Perform a repeated measures analysis with the within subjects factor and one between subjects factor
Exercise 10
Perform a repeated measures analysis with the within subjects factor and two between subjects factors | 917 | 4,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.96073 |
https://forum.mustachianpost.com/t/buy-monthly-or-bi-monthly-vt-on-ib/4415 | 1,653,172,580,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00023.warc.gz | 327,286,131 | 5,709 | # Buy monthly or bi-monthly VT on IB?
Hi,
I currently buy CHF 10k worth of VT on IB every two months instead of 5k every month in order to save on the IB currency exchange fee of around CHF 1.80 (USD 2.00 if I remember correctly) for the exchange of my CHF to USD.
Now I was wondering if there is any advantage, even if small, in saving some money/fees if I chose another option such as:
a) buy every month 5k of VT
b) buy every 3 month (quarter) 15k of VT
So to resume my question in another way: would I earn more by immediately placing my money into VT every month (and not save on the currency exchange fee) or better save on the currency exchange fee and place my money bi-monthly or even quarterly?
Cheers!
Time in the market is what matters. IB fees are minimal.
1 Like
Pretty simple math.
For purchase of 60000 / year.
Monthly schedule costs you 24USD of fees.
Turns to 0.0004, i.e. 0.04% / year.
With a bi-monthly you then save 0.02%, with a quarterly 0.03%.
I would say irrelevant, because you (and nobody else) don’t know the opportunity costs (or profits) of the money waiting aside.
1 Like
True, that this difference is not much but I thought as I true mustachian one should also save on such currency exchange fees.
Basically I am trying to find out if the advantage of being longer in the market by investing monthly instead of bi-monthly weighs more in the balance than saving on these 12 CHF fees per year. But as you also say @dbu these opportunity costs are sort of impossible to predict so I will never really know…
You don’t know the opportunity costs, but you do invest in the hope of getting returns, otherwise, why would you do it? We say that historically you had an expected annual return of 7%. Let’s say that you can expect 0.5% return per month. Of course, it can be +10% or -20%, it’s a probability distribution, but we agree that there is a slightly higher chance of gain vs loss.
So if you have 1000 CHF ready to invest right now, you can say that 1 month in the market will be worth about 5 CHF. Already that is higher than the IB fee, so it’s preferable to invest now than wait 1 month.
I have no love for such penny-pinching. We’re all guilty of that, but in the end it’s not worth the time and the nerves. No point to discuss if it’s worth it to spend 2 CHF or not, really.
3 Likes
Thanks @Bojack for your clear example, explained like that it totally makes sense to invest on a monthly basis and ignore the additional currency exchange fees.
I enjoy buying ETFs too much to have to wait 2 months !
2 Likes | 655 | 2,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.945782 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-10-quadratic-equations-10-3-quadratic-formula-problem-set-10-3-page-453/30 | 1,534,354,837,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210243.28/warc/CC-MAIN-20180815161419-20180815181419-00455.warc.gz | 875,023,257 | 13,344 | ## Elementary Algebra
Published by Cengage Learning
# Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 30
#### Answer
{$\frac{-3 - \sqrt {33}}{4}\frac{-3 + \sqrt {33}}{4}$}
#### Work Step by Step
Step 1: Comparing $2x^{2}+3x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=2$, $b=3$ and $c=-3$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b, and c in the formula: $x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(-3)}}{2(2)}$ Step 4: $x=\frac{-3 \pm \sqrt {9+24}}{4}$ Step 5: $x=\frac{-3 \pm \sqrt {33}}{4}$ Step 6: $x=\frac{-3 - \sqrt {33}}{4}$ or $x=\frac{-3 + \sqrt {33}}{4}$ Step 7: Therefore, the solution set is {$\frac{-3 - \sqrt {33}}{4},\frac{-3 + \sqrt {33}}{4}$}.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 366 | 960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-34 | longest | en | 0.569206 |
https://developer.salesforce.com/blogs/developer-relations/2013/07/selecting-random-numbers-and-records-on-the-force-com-platform-part-1 | 1,709,302,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00617.warc.gz | 200,339,867 | 11,254 | Summertime air travel is in full swing here in the US and the Transportation Security Agency is out performing extra, “random” screening at various gates. I’ve watched a few of these recently and I’ve been left with the impression that random is, apparently, decided by the screeners and doesn’t appear random at all. Being a problem solver, I started thinking of how I could use the Force.com platform to help the TSA choose random travelers in a more efficient fashion.
Random Numbers
First, we need some randomness. There are a couple of ways to generate random numbers on the Force.com platform. The first is to call either of the static methods getRandomInteger() or getRandomLong() in the Crypto class. The second is to call either the random() or rint() methods of the Math class. Armed with this knowledge, let’s create a REST based web service that agents can use, at the jet bridge, from their mobile device that instructs them which passengers to choose for extra screening.
There’s a number of ways to do this. The first selects passengers by name, from a list, for random screening. The second, and I believe better, is to provide a list of numbers that show the line positions of people to pull without knowing their identity. For example {1,10,19,103} would tell me to pull the first, tenth, nineteenth and hundred and third passenger out for extra screening. We could combine the two as well.
Getting Started
Figure 1: Passenger Data Model
Figure 1 shows the simple data model that I am using for this example. I’ve also created a package with the objects and code you can install from this link
The Code
With the data model in place, let’s start writing some code. The first thing we’ll need is a method that generates a random number. We could easily write it like this:
A couple things are happening here. First, on line two we are calling Math.random() which will return a number greater than 0.0 and less than 1. I multiply by 1000 because I want a large range (to cover hundreds of passengers) and when I round, I get an integer with a range from 0 to 999. Since my flights will have different passenger totals and since it doesn’t make sense to select more passengers for screening than are on the flight, I put a limit on the return value. By moding the random number with the upper limit, I make sure I never return an integer bigger than what I want. For instance, if I have a flight with 139 people on it, I would pass that as the limit so I never got a result that said pull the two hundred and tenth passenger out of line. Since we don’t want to randomly “select” every passenger let’s put another limit on our method. Let’s specify how many passengers we want to select. We’ll change our code a bit as follows
Here’s the test:
Now we have a method that returns n number of random integers between 0 and upperLimit, which we want. The last thing left to do is make this code mobile friendly. We’ll do that by writing the following Apex REST service.
APEX Rest
Of course we’ll need to test our Apex Rest service before we can deploy it outside of a sandbox. Here’s the test class
All that’s left to do is build, or add this functionality, to a mobile app. Here I show testing the REST Service with Workbench. You could also use another client like curl or Postman but you would have to handle the OAuth authentication and URL escaping yourself.
Here I’ve logged into Workbench. Selected Utilities->Rest Explorer from the menu and inserted the following test URL: /services/apexrest/screeninglist/?flightNumber=99&date=7/10/2013&rands=5 (your data may be different)
In part 1 I’ve shown how to generate random numbers and use them as part of a selection process. In part 2, I’ll modify this slightly to show pulling random records from salesforce. This isn’t a perfect solution and there are some additional details that would need to be added to a production implementation. One thing is that I don’t check for duplicate randoms within my list. It’s possible my return value could look like this: {1,3,100,19,19}. I’m sure readers will find others. I’ll leave those modifications as an exercise to the reader. As always, if you think of a way to make this better please comment. | 940 | 4,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-10 | latest | en | 0.930801 |
https://realonomics.net/how-many-moles-of-carbon-atoms-are-there-in-0-500-mole-of-c2h6/ | 1,656,305,252,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00444.warc.gz | 528,047,433 | 55,102 | FAQ
# How Many Moles Of Carbon Atoms Are There In 0.500 Mole Of C2H6?
one mole
## How many moles of carbon are in c2h6?
(i) We know that in 1 mole of ethane there are 2 moles of carbon atom because 2 carbon atoms are present in one molecule of ethane. – So there are 6 moles of carbon atom present in 3 moles of ethane.
## How many atoms are 0.500 moles?
(0.500 mol Al)* (6.022 x 1023 atoms/mol)= 3.011 x 1023 atoms Al.
## How many grams of carbon are in 0.500 mole of carbon?
For every one mole of carbon the massive carbon is 12 g roughly according to the periodic table. So most of carbon cancel out. So then all I have now is 0.500 times 12 times 12 again gives me 72 g of carbon.
## How many atoms are in one mole of caoh2?
∴1mole ofCa(OH)2 contains 5N atoms.
## How many moles of carbon atoms and hydrogen atoms respectively are present in 3 moles of ethane?
so 3 moles of ethane contains 6 mole of carbon atoms. so 3 moles of ethane contains 18 mole of hydrogen atoms.
## How many atoms are in a mole of carbon?
6.022 × 1023
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro’s Number (6.0221421 x 1023).
## How many atoms are in 0.5 moles of argon?
so 0.500 of a mol x 6.02 x 10^23 = 3.01 x 10^23.
## How many atoms of argon are there in 0.500 moles AR?
Using the older standard as indicated in the explanation 11.2 L of Ar atoms contain 0.500 mol .
## How many moles of carbon atoms are there in 30.07 g of C2H6?
use molar mass to find moles: 77.28 g of ethane (C2H6) @ 1 mole / 30.07 grams C2H6 = 2.570 moles C2H6 2.570 moles C2H6 has twice as many carbon atoms …
## How many atoms does ethanol have?
There are nine atoms in a molecule of ethanol. The chemical formula can determine the number of atoms in a molecule.
## How many grams of Fe2O3 are there in 0.500 mol of Fe2O3?
79.9 grams
79.9 grams are in 0.500 moles of Fe2O3.
## How many oxygen atoms are there in 0.5 mol of CO2?
6.022 x 1023 x 0.50 = 3.011 x 1023.
## How many molecules are there in 0.5 mol of CO2?
since 1 mol represents 6.02 x 10^23 particles there are 0.5×6. 02×10^23 particles (molecules) of carbon dioxide.
## How many atoms are in an oxygen molecule o2 )?
two oxygen atoms
One mole of oxygen gas which has the formula O2 has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms because each molecule of oxygen contains two oxygen atoms.
## How many atoms are in caoh2?
Calcium hydroxide Ca(OH) 2 consists of one calcium atom (A r = 40) two oxygen atoms (A r = 16) and two hydrogen atoms (A r = 1).
## How many atoms are present in caoh2?
Answer: Calcium Hydroxide has three atoms 5 atoms in each molecule. molecules.
## How many atoms are present in a 52u of HE B 52 g of he?
Therefore in 52 u of Helium 13 atoms are present. Therefore in 52 g of Helium 7.83×1024 atoms are present.
## How do you find the number of atoms in 3 moles of carbon atoms?
Explanation:
1. 1 Mole is 6.02⋅1023 molecules.
2. Therefore 3 Moles of Carbon Dioxide CO2 =3(6.02⋅1023) or 18.06⋅1023 molecules of CO2.
3. There is one atom of Carbon and two atoms of Oxygen in each molecule of Carbon Dioxide CO2.
4. (3 atoms) 18.06⋅1023 molecules =54.18⋅1023 atoms.
5. 5.418⋅1024 atoms.
## How many molecules are present in 3 moles of a compound?
This is given by 3 moles of \${{H}_{2}}Stimes dfrac{6.022times {{10}^{23}}}{1}\$ . This gives the value = \$1.8times {{10}^{24}}\$ molecules of \${{H}_{2}}S\$ . Therefore the correct answer is 3 moles of \${{H}_{2}}S\$ contains \$1.8times {{10}^{24}}\$ number of molecules.
## How many atoms are there in carbon?
The value of the mole in precisely 12 grammes of pure carbon-12 is equal to the number of atoms. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 x 1023 atoms. Avogadro’s Number of Particles is called the number of particles in 1 mole (6.0221421 x 1023).
## How many moles of carbon atoms are in 1 mole of carbon dioxide?
1 mole carbon dioxide contains 6.02 x 1023 molecules. 1 mole of carbon atoms per mole of carbon dioxide = 6.02 x 1023 carbon atoms. 2 mole of oxygen atoms per mole of oxygen atoms = 2 x 6.02 x 1023 oxygen atoms.
## What is meant by 1 mole of carbon atoms?
A mole can be defined as the amount of a pure substance containing the same number of atoms molecules or ions as present in exactly 12g of carbon. … Therefore a mole of carbon atoms refers to a group of 6.022 × 1023 carbon atoms.
## What is the number of moles in 0.025 g NH4 2Cr2O7?
Description Match:
How many MOLES in 0.025 g (NH4)2Cr2O7? 9.91 X 10^-5 or 0.000099 mol
Mr. Adamek’s First Name What is Dustin?
The percent composition of Barium in BaCrO4 What is 54.21%?
The lowest whole-number ratio of the elements in a compound What is empirical formula?
## How many atoms are in a mole calculator?
Avogadro’s number is a very important relationship to remember: 1 mole = 6.022×1023 6.022 × 10 23 atoms molecules protons etc. To convert from moles to atoms multiply the molar amount by Avogadro’s number.
## How many atoms are in 1 mol of sulfur?
One mole of sulfur atoms is Avogadro’s number of atoms which is 6.02 x 1023 atoms.
## How many moles of carbon atoms and sulfur atoms does it take to make one mole of carbon disulfide?
Explanation: And this quotient gives…… 13.9⋅g76.14⋅g⋅mol−1=0.188⋅mol with respect to CS2 . Since each mole/molecule of CS2 contains 1 mole of carbon and 2 moles of sulfur there are 3×0.188⋅mol×6.022×1023⋅mol−1=3.39×1023 individual atoms…….
## What is the mole of oxygen?
15.998 grams
The mass of oxygen equal to one mole of oxygen is 15.998 grams and the mass of one mole of hydrogen is 1.008 g.
## How many atoms of oxygen are in 4 molecules of HNO3?
Note the amounts of atoms of all the component in HNO3 which are 1 atom of Hydrogen 1 atom of Nitrogen and 3 atoms of Oxygen.
## How many carbon atoms are in ethane?
2 atoms
C2H6 2 atoms of carbon combine with 6 atoms of hydrogen to form ethane.
## How many moles are in H2O?
one mole
The number of atoms is an exact number the number of mole is an exact number they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol. Notice that the molar mass and the formula mass are numerically the same.
## How many moles of carbon dioxide are in 211g of carbon dioxide?
Answer : The number of moles of carbon dioxide is 4.795 moles.
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A342262 Numbers divisible both by the product of their nonzero digits (A055471) and by the sum of their digits (A005349). 2
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 20, 24, 30, 36, 40, 50, 60, 70, 80, 90, 100, 102, 110, 111, 112, 120, 132, 135, 140, 144, 150, 200, 210, 216, 220, 224, 240, 300, 306, 312, 315, 360, 400, 432, 480, 500, 510, 540, 550, 600, 612, 624, 630, 700, 735, 800, 900, 1000, 1002, 1008 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Equivalently, Niven numbers that are divisible by the product of their nonzero digits. A Niven number (A005349) is a number that is divisible by the sum of its digits. Niven numbers without zero digit that are divisible by the product of their digits are in A038186. Differs from super Niven numbers, the first 16 terms are the same, then A328273(17) = 48 while a(17) = 50. This sequence is infinite since if m is a term, then 10*m is another term. LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 EXAMPLE The product of the nonzero digits of 306 = 3*6 = 18, and 306 divided by 18 = 17. The sum of the digits of 306 = 3 + 0 + 6 = 9, and 306 divided by 9 = 34. Thus 306 is a term. MATHEMATICA q[n_] := And @@ Divisible[n, {Times @@ (d = Select[IntegerDigits[n], # > 0 &]), Plus @@ d}]; Select[Range[1000], q] (* Amiram Eldar, Mar 27 2021 *) Select[Range[1200], Mod[#, Times@@(IntegerDigits[#]/.(0->1))]== Mod[#, Total[ IntegerDigits[#]]]==0&] (* Harvey P. Dale, Sep 26 2021 *) PROG (PARI) isok(m) = my(d=select(x->(x!=0), digits(m))); !(m % vecprod(d)) && !(m % vecsum(d)); \\ Michel Marcus, Mar 27 2021 CROSSREFS Intersection of A005349 and A055471. Supersequence of A038186. Cf. A051004, A328273, A342650. Sequence in context: A365420 A342650 A328273 * A255734 A357142 A033075 Adjacent sequences: A342259 A342260 A342261 * A342263 A342264 A342265 KEYWORD nonn,base AUTHOR Bernard Schott, Mar 27 2021 EXTENSIONS Example clarified by Harvey P. Dale, Sep 26 2021 STATUS approved
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Last modified September 26 17:49 EDT 2023. Contains 365666 sequences. (Running on oeis4.) | 842 | 2,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.691074 |
http://www.downloadplex.com/Scripts/Matlab/Development-Tools/shuffle-sripts_462777.html | 1,477,403,867,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720153.61/warc/CC-MAIN-20161020183840-00361-ip-10-171-6-4.ec2.internal.warc.gz | 412,145,777 | 12,771 | # Shuffle (Sripts) 1.0
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## Shuffle (Sripts) Publisher's description
### Shuffle - Random permutation of array elements
Shuffle - Random permutation of array elements
This function is equivalent to X(RANDPERM(LENGTH(X)), but 50% to 85% faster. It uses D.E. Knuth's shuffle algorithm (also called Fisher-Yates) and the cute KISS random number generator (G. Marsaglia). While RANDPERM needs 2*LENGTH(X)*8 bytes as temporary memory, SHUFFLE needs just a fixed small number of bytes.
1. Inplace shuffling: Y = Shuffle(X, Dim)
INPUT:
X: DOUBLE, SINGLE, CHAR, LOGICAL, (U)INT64/32/16/8 array.
Dim: Dimension to operate on. Optional, default: 1st non-singleton dimension.
OUTPUT:
Y: Array of same type and size as X with shuffled elements.
2. Create a shuffle index: Index = Shuffle(N, 'index', NOut)
This is equivalent to Matlab's RANDPERM, but much faster, if N is large and NOut is small.
INPUT:
N: Integer number.
NOut: The number of output elements. Optional, default: N.
OUTPUT:
Index: [1:NOut] elements of shuffled [1:N] vector in the smallest possible integer type.
3. Derangement index:
Index = Shuffle(N, 'derange', NOut)
Equivalent to the index method, but all Index[i] ~= i. A rejection method is used: Create an index vector until a derangement is gained.
EXAMPLES:
R = Shuffle(1:8) % [8, 1, 2, 6, 4, 3, 5, 7]
R = Shuffle([1:4; 5:8], 2) % [3, 2, 1, 4; 6, 8, 7, 5]
I = Shuffle(8, 'index'); % UINT8([1, 5, 7, 6, 2, 3, 4, 8])
Choose 10 different rows from a 1000 x 100 matrix:
X = rand(1000, 100); Y = X(Shuffle(1000, 'index', 10), :);
Operate on cells or complex arrays:
C = {9, 's', 1:5}; SC = C(Shuffle(numel(C), 'index'));
M = rand(3) + i * rand(3); SM = M(:, Shuffle(size(C, 2), 'index'))
NOTES: There are several other shuffle functions in the FEX. Some use Knuth's method also, some call RANDPERM. This implementation is faster due to calling a compiled MEX file and it has a smaller memory footprint. The KISS random numbers are much better than the RAND() of the C-standard libs.
Run the unit-test TestShuffle to test validity and speed (see screenshot).
Tested: Matlab 6.5, 7.7, 7.8, 32bit, WinXP,
Compiler: LCC 2.4/3.8, BCC 5.5, Open Watcom 1.8, MSVC 2008.
Compatibility to 64 bit, Linux and Mac is assumed.
Pre-compiled Mex: http://www.n-simon.de/mex
#### System Requirements:
MATLAB 7.8 (R2009a)
Program Release Status: New Release
Program Install Support: Install and Uninstall
#### Shuffle (Sripts) Tags:
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ASK, OOK, FSK, BPSK, QPSK, 8PSK modulation 1.1
ASK, OOK, FSK, BPSK, QPSK, 8PSK modulation contain several functions for digital modulation simulation | 851 | 2,725 | {"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} | 2.8125 | 3 | CC-MAIN-2016-44 | longest | en | 0.78818 |
https://couchsurfingcook.com/write-a-program-that-will-ask-the-user-to-enter-the-amount-of-a-purchase/ | 1,657,102,478,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00594.warc.gz | 237,501,016 | 4,610 | 2. Input,processing and also Output Checkpoint Multiple selection True Or False quick Answer Algorithm Workbench Programming exercises
You are watching: Write a program that will ask the user to enter the amount of a purchase
### 22 Feb" data-html="true" data-toggle="tooltip" data-placement="bottom"> Question
Write a regime that will ask the user to enter the quantity of a purchase. The program have to then compute the state and county sales tax. Assume the state sales taxation is 4 percent and also the county sales taxation is 2 percent. The regimen should display screen the quantity of the purchase, the state sales tax, the ar sales tax, the full sales tax, and the full of the sale (which is the amount of the lot of acquisition plus the total sales tax).
Hint: use the worth 0.02 to stand for 2 percent, and 0.04 to represent 4 percent.
### 1Helped junior by posting 31+ answers. " data-html="true" data-toggle="tooltip" data-placement="bottom"> Answer
Python Program:#Get the acquisition amountpurchaseAmount = float(input("Enter the amount of a purchase:"))#Assign the value to State sales taxation percentageStatesalestaxpercentage = 0.04#Assign the worth to county sales taxes percentagecountysalestaxpercentage =0.02#Calculate the state sales taxstatesalestax = purchaseAmount * Statesalestaxpercentage#Calculate the county sales taxcountysalestax = purchaseAmount * countysalestaxpercentage#Calculate the full sales taxTotalsalestax = statesalestax + countysalestax#Calculate the full of the saleTotalofthesale = purchaseAmount + Totalsalestax#Display the required outputprint("The amount of the purchase:",purchaseAmount)print("The state sales tax:",statesalestax)print("The county sales tax:",countysalestax)print("The full sales tax:",Totalsalestax)print("The total of the sale:",Totalofthesale)Sample output:Enter the amount of a purchase:2000The quantity of the purchase: 2000.0The state sales tax: 80.0The county sales tax: 40.0The full sales tax: 120.0The full of the sale: 2120.0
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### 0Helped small by posting 87+ answers. " data-html="true" data-toggle="tooltip" data-placement="bottom"> S
purchaseAmount = float(input("Enter the lot of a purchase:"))Statesalestaxpercentage = 0.04countysalestaxpercentage =0.02statesalestax = purchaseAmount * Statesalestaxpercentagecountysalestax = purchaseAmount * countysalestaxpercentageTotalsalestax = statesalestax + countysalestaxTotalofthesale = purchaseAmount + Totalsalestaxprint("The lot of the purchase:",purchaseAmount)print("The state sales tax:",statesalestax)print("The ar sales tax:",countysalestax)print("The complete sales tax:",Totalsalestax)print("The complete of the sale:",Totalofthesale) | 704 | 2,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-27 | latest | en | 0.823695 |
https://short-question.com/what-lines-run-parallel-to-the-prime-meridian/ | 1,721,331,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00160.warc.gz | 458,776,701 | 42,711 | Short-Question
# What lines run parallel to the prime meridian?
## What lines run parallel to the prime meridian?
Latitude is a measure of how far north or south somewhere is from the Equator; longitude is a measure of how far east or west it is from the Prime Meridian. Whilst lines (or parallels) of latitude all run parallel to the Equator, lines (or meridians) of longitude all converge at the Earth’s North and South Poles.
What are the lines called that run from east to west?
The lines run east-west are known as lines of latitude. The lines running north-south are known as lines of longitude.
Are lines of latitude also called meridians?
Merdians and Parallels You’ve seen lines running across maps your whole life and may not have noticed them. The lines running North to South are called “Meridians” or “lines of longitude” (Figure 2), while the lines running East to West are called “Parallels” or “lines of latitude” (Figure 3).
### What line runs from the North Pole to the South Pole?
The imaginary vertical lines that run from the North pole to the South pole on a map are called longitudinal lines. The Prime Meridian is the longitudinal line that has a value of 0 degrees. On a map, longitudinal lines are measured in increments of 15 degrees from the Prime Meridian.
What is the name for the lines that run North to south on a map or a globe and are not parallel?
meridians
Latitude is measured from 0 to 90 north and 0 to 90 south? 90 north is the North Pole and 90 south is the South Pole. Imaginary lines, also called meridians, running vertically around the globe. Unlike latitude lines, longitude lines are not parallel.
Where does the Meridian run between the north and South Poles?
Meridians run between the North and South poles. A (geographic) meridian (or line of longitude) is the half of an imaginary great circle on the Earth’s surface, terminated by the North Pole and the South Pole, connecting points of equal longitude, as measured in angular degrees east or west of the Prime Meridian.
#### What are the lines that run north and South?
They are used to measure distances north and south of the equator. The lines circling the globe in a north-south direction are called lines of longitude (or meridians).
Where do imaginary lines run around the globe?
Imaginary lines, also called meridians, running vertically around the globe. Unlike latitude lines, longitude lines are not parallel. Meridians meet at the poles and are widest apart at the equator.
Where are the coordinates of the Guide Meridian?
Based on the BLM manual’s 1973 publication date, and the reference to Clarke’s Spheroid of 1866 in section 2-82, coordinates appear to be in the NAD27 datum. Guide meridian created 12 miles east of Willamette Meridian, which is in Salish Sea at this latitude. | 639 | 2,815 | {"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} | 3.953125 | 4 | CC-MAIN-2024-30 | latest | en | 0.93255 |
https://library.curriki.org/oer/Covenlent-Bonding-Hints-and-Rules/?mrid=43186-43421-43415 | 1,638,690,906,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00371.warc.gz | 418,201,729 | 21,015 | Molecular Bonding Hints and Rules
General Hints for Making the “Skeleton” of Molecules
1) Carbon is usually central.
2) Hydrogen is NEVER central.
3) Halogens are seldom central.
4) Oxygen is seldom central, but may link carbons.
5) An atom that appears only once is usually central.
6) If you can’t decide which atom should be central, pick the one with the lowest electronegativity.
7) Balance out the atoms around the central atom (make it symmetrical).
Steps for Placing the Valence Electrons
1) First, calculate the total number of valence electrons for the entire molecule (i.e. H2O has 8 total valence electrons)
2) Remember if the substance is an ion it has gained or lost electrons. Add or subtract electrons as needed. DO NOT forget your brackets with the charge of the ion on the outside…
3) Place pairs of electrons between every 2 atoms in the skeleton (Bonding Pairs).
4) Use the remaining electrons to complete the octets of all outer atoms (Non-Bonding Pairs). CAUTION: H and He are complete with only 2 valence electrons.
5) Any left over electrons are added in pairs to the central atom.
6) When all electrons have been placed, the outer atoms will all have octets (8 electrons around it…unless it is H or He then they will only have 2). The central atom may or may not have an octet.
a) If the central atom has an octet it is complete.
b) It is OK if the central atom has fewer than 8 electrons if it is BORON. If it isn’t, make multiple bonds to get an octet.
c) A central atom may have more than an octet if the central atom is in Periods 3-7. If it is in Period 2, it cannot have more than 8.
7) Draw the line structure. To do so, replace any bonding pair with a single line. Represent all lone pairs or free radicals by leaving the dots. Do not include non-bonding pairs. Double check to make sure you have included your brackets and charge if you drawing an ion.
Non-profit Tax ID # 203478467 | 472 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.869571 |
https://www.mathsisfun.com/puzzles/weighing-pool-balls-solution.html | 1,519,049,498,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00504.warc.gz | 882,712,040 | 3,843 | # Weighing Pool Balls Puzzle - Solution
### The Puzzle:
One of twelve pool balls is a bit lighter or heavier (you do not know) than the others.
At least how many times do you have to use an old balance-type pair of scales to identify this ball?
### Our Solution:
It is enough to use the pair of scales just 3 times.
We know of two possible solutions:
Solution 1
Let's mark the balls using numbers from 1 to 12 and these special symbols:
x? means I know nothing about ball number x;
xL means that this ball is maybe lighter then the others;
xH means that this ball is maybe heavier then the others;
x. means this ball is "normal".
At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.
If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10? 11? on the right pan.
If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.
If the left pan is heavier, I know that 12 is normal and 9L 10L 11L. I weigh 9L and 10L.
If they are the same weight, then ball 11 is lighter then all other balls.
If they are not the same weight, then the lighter ball is the one up.
If the right pan is heavier, then 9H 10H and 11H and the procedure is similar to the former text.
If the left pan is heavier, then 1H 2H 3H 4H, 5L 6L 7L 8L and 9. 10. 11. 12. Now I lay on the left pan 1H 2H 3H 5L and on the right pan 4H 9. 10. 11.
If there is equilibrium, then the suspicious balls are 6L 7L and 8L. Identifying the wrong one is similar to the former case of 9L 10L 11L
If the left pan is lighter, then the wrong ball can be 5L or 4H. I compare for instance 1. and 4H. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).
If the left pan is heavier, then all balls are normal except for 1H 2H and 3H. Identifying the wrong ball among 3 balls was described earlier.
Solution 2
This solution was provided by Charles Naumann. His method also solves it with just three weighings:
Label the balls 1-12
First Weighing:
Left: 1 2 3 4
Right: 5 6 7 8
Off: 9 10 11 12
Record the heavier side (L, R, or B)
Second Weighing:
Left: 1 2 5 9
Right: 3 4 10 11
Off: 6 7 8 12
Record the heavier side (L, R or B)
Third Weighing:
Left: 3 7 9 10
Right: 1 4 6 12
Off: 2 5 8 11
Record the heavier side (L, R, B)
There are 27 (3^3) possible combination of scale readings. A complete sorted list of the scale reading appears below. Note that only 24 of the 27 readings should be possible given the original problem statement. The algorithm was designed so that if all three scale readings are the same, an error is flagged indicating that the scale is stuck.
BBB Error! There is not a single light or heavy ball (or scale is stuck).
BBL Ball #12 is light
BBR Ball #12 is heavy
BLB Ball #11 is light
BLL Ball #9 is heavy
BLR Ball #10 is light
BRB Ball #11 is heavy
BRL Ball #10 is heavy
BRR Ball #9 is light
LBB Ball #8 is light
LBL Ball #6 is light
LBR Ball #7 is light
LLL Error! Scale is stuck!
LLB Ball #2 is heavy
LLR Ball #1 is heavy
LRB Ball #5 is light
LRL Ball #3 is heavy
LRR Ball #4 is heavy
RBB Ball #8 is heavy
RBL Ball #7 is heavy
RBR Ball #6 is heavy
RLB Ball #5 is heavy
RLL Ball #4 is light
RLR Ball #3 is light
RRB Ball #2 is light
RRL Ball #1 is light
RRR Error! Scale is stuck!
See this puzzle without solution
Discuss this puzzle at the Math is Fun Forum | 1,079 | 3,459 | {"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-2018-09 | longest | en | 0.91198 |
https://betterworld2016.org/what-type-of-image-does-a-concave-lens-form/ | 1,656,661,942,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00608.warc.gz | 186,562,153 | 4,968 | As a convex lens develops different species of images relying on the position of object, an in similar way a concave lens additionally forms different types of images when the object is inserted at
1. At infinity 2. In between infinity and optical centre
1. Formation of image by a concave lens when the thing is placed at infinity
When an item is put at infinity, the 2 rays AO and BD running parallel to the principal axis get refracted at allude O and D respectively and also get diverged follow me the directions OX and also DY respectively. These diverged rays i.e. OX and also DY shows up to crossing each various other at the principal emphasis of the concave lens ~ extending ago by dotted lines. Therefore, in instance of concave lens once the object is placed at infinity the photo is formed at the major focus, highly reduced (point sized), virtual and erect.
You are watching: What type of image does a concave lens form?
Why concave lens is referred to as diverging lens?
A concave lens is dubbed diverging lens since of its capability to diverge a parallel beam that light.
2. When the thing is placed in between infinity and also optical centre
When an object abdominal is placed between infinity and also optical center of a concave lens, a ray of irradiate AO i beg your pardon is parallel to the primary axis diverges along the direction OX after ~ refraction and appears to come from the principal focus F along the direction OF. When the other ray of irradiate AC goes directly through the optical centre C the the concave lens without any deviation along the direction CY. Together it is clear from the figure (given below) that both the refracted rays i.e. OX and also CY space diverging in nature, so this rays show up to crossing each other at suggest A’ ~ above the left next of the lens on developing back. Hence, the photo A’B’ developed in this instance is a virtual image which is created at the exact same side of the lens, between the optical centre and also focus. Additionally the image formed is diminished and also erect.
Table of summary of Image created by a Concave Lens
S No. Position that Object Position the Image Size of Image Nature that Image 1. At infinity At the focus Highly diminished, suggest sized Virtual and erect 2. Between the infinity and also optical centre Between focus and also optical centre Diminished Virtual and also erect
Differences between a Convex Lens and also a Concave Lens
S No. Convex lens Concave lens 1. A convex lens is special in the middle and thin at the edges. A concave lens is slim in the middle and also thick in ~ the edges. 2. It is converging in nature. It is diverging in nature. 3. A convex lens has a actual focus. A concave lens has actually a online image. 4. It is used for convey of hypermetropia and presbyopia.See more: How Many Factors Does 60 Have, All Factors Of 60 And Multiples Of 60 It is supplied for correction of myopia. | 650 | 2,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-27 | latest | en | 0.952111 |
http://en.m.wikibooks.org/wiki/Algorithms/Find_maximum | 1,398,244,573,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00402-ip-10-147-4-33.ec2.internal.warc.gz | 108,309,802 | 7,789 | # Algorithms/Find maximum
Top, Chapters: 1, 2, 3, 4, 5, 6, 7, 8, 9, A
This is a source for teaching someone for different language of programming language. Find max is the topic.
## Method OrangeEdit
``` if the tb length is 1 then
output the zero is index of array tb;
else if the tb length is 0 then
output 0;
define variable call o_intMax ( integer type);
o_intMax become tb[indexInd];
looping from 1 to (tb's length - 1) save the index into variable indexInt
if tb[indexInt] > maxInt then
maxInt become tb[indexInt];
end if
end loop
```
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php c Python
## Method AppleEdit
``` if input_array size is 0 then
return 0
end if
if input_array size is 1 then
return input_array[0]
end if
if input_array size is 2 then
if element zero is bigger than element one then
return element zero
else
return element one
end if
end if
create a array call aryA and the size is (input_array size divide 2)
create a array call aryB and the size is (input_array size - aryA size)
looping from zero to (input_array size / 2),
aryA[loopIndex] = input_array[loopindex]
loopindex = loopindex + 1;
end looping
define integer variable call intK, initial it the zero
looping from (aryA size) to input_array size
aryB[intK] = input_array[loopIndex]
loopindex = loopindex + 1
intK = intK + 1;
end loop
create a variable call intM
create a variable call intN
intM = run findMax(aryA);
intN = run findMax(aryB);
if intM > intN then
return intM
else
return intN
end if
```
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php Python
## Method PearEdit
### AlgorithmEdit
Method findMax(i_aryTab,int i_start,int i_end)
``` if ( i_end - start equal one then
if ( i_aryTab[i_end] > i_aryTab[i_start] )
return i_aryTab[i_end]
else
return i_aryTab[i_start]
end if
define variable call intJ
intJ :=( (i_end - i_start ) + 1 ) / 2 - 1
define variable call oMaxN
define variable call oMaxM
oMaxN = findMax(i_aryTab,i_start,i_start+intJ)
oMaxM = findMax(i_aryTab,i_start + intJ + 1,i_end)
if ( oMaxN > oMax M )
return oMaxN
else
return oMaxM
```
Method findMaxHandler(i_aryTab)
``` return findMax(i_aryTab,0,i_aryTab.length - 1)
```
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php Python
## Method 4Edit
### AlgorithmEdit
Method : findMax
``` make all the element of input_array = mulitple itself -1;
run findMin(input_array) put result into o_Max;
return o_Max * -1;
```
Method : findMin
``` put to element[0] into min
for index = 1 to arraysize -1
if ( element[index] < min ) then
min = elemnt[index];
end if
end loop
return min
```
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php Python
## Method 5Edit
### AlgorithmEdit
Method : sorting
``` define p_blnFound As boolean type;
p_blnFound := true;
while ( p_blnFound is true ) do
begin
p_blnFound := false;
for loop from zero to( array size - 2 ) do
if ( inputArray[index] bigger than inputArray[index+1] ) then
swap inputArray[index] and inputArray[index + 1];
p_blnFound := true;
end if
end for loop
end while loop
return input_array;
```
Method : findMax
``` inputArray = sorting(inputArray);
return the last element of inputArray;
```
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php Python
## Method 6Edit
### AlgorithmEdit
Method : findMaxFourEle ( intA,intB,intC,intD)
``` int max = intA;
if intB > max then
begin
max = intB;
end if
if intC > max then
begin
max = intC;
end if
if intD > max then
begin
max = intD;
end if
return max;
```
Method: findMaxThreeEle(intA,intB,intC)
``` return findMaxFourEle(intA,intB,intC,intC);
```
NOTE: This method can be used with Method Pear
Java Vbscript Java script JScript Visual basic pascal oracle sql Perl Php Python
More data will be coming soon...
Top, Chapters: 1, 2, 3, 4, 5, 6, 7, 8, 9, A | 1,163 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2014-15 | longest | en | 0.398356 |
https://cp-wiki.vercel.app/tutorial/leetcode/WC303/ | 1,686,210,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00725.warc.gz | 210,862,310 | 10,180 | # # Leetcode 第303场周赛题解
## # Problem A - 第一个出现两次的字母 (opens new window)
### # 方法一：模拟
• 时间复杂度 $\mathcal{O}(\min(|S|,|\Sigma|))$
• 空间复杂度 $\mathcal{O}(\min(|S|,|\Sigma|))$
class Solution:
def repeatedCharacter(self, s: str) -> str:
cnt = collections.Counter()
for ch in s:
cnt[ch] += 1
if cnt[ch] == 2:
return ch
return ''
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## # Problem B - 相等行列对 (opens new window)
### # 方法一：暴力
• 时间复杂度 $\mathcal{O}(N^2)$
• 空间复杂度 $\mathcal{O}(N^2)$
class Solution:
def equalPairs(self, grid: List[List[int]]) -> int:
n = len(grid)
rows = collections.Counter()
cols = collections.Counter()
for i in range(n):
row = tuple(grid[i])
col = tuple([grid[j][i] for j in range(n)])
rows[row] += 1
cols[col] += 1
ans = 0
for row in rows:
ans += rows[row] * cols[row]
return ans
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## # Problem C - 设计食物评分系统 (opens new window)
### # 方法一：哈希表+平衡树
• 时间复杂度：修改操作 $\mathcal{O}(\log N)$，查询操作 $\mathcal{O}(1)$
• 空间复杂度 $\mathcal{O}(N)$
class FoodRatings {
unordered_map<string, set<pair<int, string>>> r;
unordered_map<string, int> f;
unordered_map<string, string> c;
public:
FoodRatings(vector<string>& foods, vector<string>& cuisines, vector<int>& ratings) {
int n = foods.size();
for (int i = 0; i < n; ++i) {
f[foods[i]] = ratings[i];
c[foods[i]] = cuisines[i];
r[cuisines[i]].emplace(-ratings[i], foods[i]);
}
}
void changeRating(string food, int newRating) {
r[c[food]].erase(make_pair(-f[food], food));
f[food] = newRating;
r[c[food]].emplace(-newRating, food);
}
string highestRated(string cuisine) {
return r[cuisine].begin()->second;
}
};
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## # Problem D - 优质数对的数目 (opens new window)
### # 方法一：按二进制表示中 1 的个数分组统计 + 前缀和
• 时间复杂度 $\mathcal{O}(N)$
• 空间复杂度 $\mathcal{O}(N)$
const int K = 30;
class Solution {
public:
long long countExcellentPairs(vector<int>& nums, int k) {
int n = nums.size();
vector<unordered_set<int>> s(K + 1);
vector<int> fcnt(K + 1);
for (int num : nums)
s[__builtin_popcount(num)].insert(num);
for (int i = K - 1; i >= 0; --i)
fcnt[i] = (int)s[i].size() + fcnt[i + 1];
long long ans = 0;
for (int i = 0; i <= K; ++i) {
int req = min(max(0, k - i), K);
ans += 1LL * (int)s[i].size() * fcnt[req];
}
return ans;
}
};
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22 | 910 | 2,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "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} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.20407 |
https://mathmatik.com/math-answers/least-common-measure/fractions-5th-problem-solving.html | 1,524,157,135,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937015.7/warc/CC-MAIN-20180419165443-20180419185443-00461.warc.gz | 655,072,873 | 12,233 | ## We Promise to Make your Math Frustrations Go Away!
Try the Free Math Solver or Scroll down to Tutorials!
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Number of inequalities to solve: 23456789
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• formula percentage | 933 | 3,927 | {"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-2018-17 | latest | en | 0.880123 |
https://www.r-bloggers.com/2016/01/precision-in-mcmc/ | 1,725,804,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00529.warc.gz | 916,846,711 | 22,647 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
While browsing Images des Mathématiques, I came across this article [in French] that studies the impact of round-off errors on number representations in a dynamical system and checked how much this was the case for MCMC algorithms like the slice sampler (recycling some R code from Monte Carlo Statistical Methods). By simply adding a few signif(…,dig=n) in the original R code. And letting the precision n vary.
“…si on simule des trajectoires pendant des intervalles de temps très longs, trop longs par rapport à la précision numérique choisie, alors bien souvent, les résultats des simulations seront complètement différents de ce qui se passe en réalité…” Pierre-Antoine Guihéneuf
Rather unsurprisingly (!), using a small enough precision (like two digits on the first row) has a visible impact on the simulation of a truncated normal. Moving to three digits seems to be sufficient in this example… One thing this tiny experiment reminds me of is the lumpability property of Kemeny and Snell. A restriction on Markov chains for aggregated (or discretised) versions to be ergodic or even Markov. Also, in 2000, Laird Breyer, Gareth Roberts and Jeff Rosenthal wrote a Statistics and Probability Letters paper on the impact of round-off errors on geometric ergodicity. However, I presume [maybe foolishly!] that the result stated in the original paper, namely that there exists an infinite number of precision digits for which the dynamical system degenerates into a small region of the space does not hold for MCMC. Maybe foolishly so because the above statement means that running a dynamical system for “too” long given the chosen precision kills the intended stationary properties of the system. Which I interpret as getting non-ergodic behaviour when exceeding the period of the uniform generator. More or less.
Filed under: Books, R, Statistics, University life Tagged: aperiodicity, CNRS, dynamical systems, Images des Mathématiques, MCMC algorithms, Metropolis-Hastings algorithm, Monte Carlo Statistical Methods, pseudo-random generator, round-off error, slice sampler | 476 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.680011 |
https://www.clutchprep.com/physics/practice-problems/141862/a-two-springs-with-spring-constants-k1-and-k2-are-connected-in-parallel-as-shown | 1,632,593,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00226.warc.gz | 732,087,722 | 28,285 | Spring Force (Hooke's Law) Video Lessons
Concept
# Problem: Two springs with spring constants k1 and k2 are connected in parallel, as shown in Fig. 4.18. What is the effective spring constant, keff? In other words, if the mass is displaced by x, find the keff for which the force equals F = -keffx.
###### FREE Expert Solution
Hooke's law:
$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{k}}{\mathbf{·}}{\mathbf{∆}}{\mathbf{x}}}$
Δx is the same for parallel springs.
Δx = Δx1 = Δx2
88% (366 ratings)
###### Problem Details
Two springs with spring constants kand k2 are connected in parallel, as shown in Fig. 4.18. What is the effective spring constant, keff? In other words, if the mass is displaced by x, find the keff for which the force equals F = -keffx. | 225 | 765 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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} | 3.78125 | 4 | CC-MAIN-2021-39 | latest | en | 0.834541 |
usefulinformationpk.com | 1,548,196,540,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00397.warc.gz | 237,176,977 | 4,641 | # How to measure your plot through internet
Internet utilities
## Measure land from anywhere in the world
Now you can measure land from anywhere in the world. For this purpose do follow the following instructions.
2. Once Google Earth is installed now you can measure any land or plot from anywhere in the world.
3. Open Google Earth and first of all find your plot, house or land with the help of search option (top left).
4. Now click on Show Ruler option from the top line of the interface as guided in the image below.
5. By clicking the ruler option a dialogue box shall appear. Being in the Line tab select your Map length as Feet.
6. Now put the cursor on the one corner of your plot and right click there and move it to the other corner. The measurement shall be calculated automatically in the dialogue box.
7. Now measure all four sides of your plot in the same way and note all the measurements with you.
## How to calculate marlas and kinals
If you want to know about how many marlas or kinals are there in your plot, land or house just follow the following formula.
## The formula
Multiply length with width (in feet) and divide it into 272.
If the length and width of your plot is 74 feet than
Multiply length (74) with width (74): 74 X 74 = 5476
Now divide 5476 to 272: 5476 / 272= 20.13
Note: Remember 272 are the square feet in one marla
It means your plot is of 20 marlas or one kinal.
You have to calculate every plot according to this formula. Remember if two sides of length or width are difference then calculate their average (For average add two measurements and divided them into 2).
Find the length of any road
In order to calculate the length of any road open Google Earth, click on Ruler icon at the top, select Path in the dialogue box and give length in Kilometers. Now if you click on any two points of the road it shall calculate its distance in kilometers. You can select other lengths like; miles, yards, feet etc. | 449 | 1,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-04 | longest | en | 0.898993 |
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#### Resources tagged with Mathematical reasoning & proof similar to Six Times Five:
Filter by: Content type:
Age range:
Challenge level:
### There are 162 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Even So
##### Age 11 to 14 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Neighbourly Addition
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours' house numbers. Can you explain the patterns I noticed?
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Always the Same
##### Age 11 to 14 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Adding All Nine
##### Age 11 to 14 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### More Mathematical Mysteries
##### Age 11 to 14 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### Cycle It
##### Age 11 to 14 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### 1 Step 2 Step
##### Age 11 to 14 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Tis Unique
##### Age 11 to 14 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Aba
##### Age 11 to 14 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Power Mad!
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Eleven
##### Age 11 to 14 Challenge Level:
Replace each letter with a digit to make this addition correct.
### 9 Weights
##### Age 11 to 14 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Cross-country Race
##### Age 11 to 14 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Is it Magic or Is it Maths?
##### Age 11 to 14 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### A Long Time at the Till
##### Age 14 to 18 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Clocked
##### Age 11 to 14 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Not Necessarily in That Order
##### Age 11 to 14 Challenge Level:
Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation?
### N000ughty Thoughts
##### Age 14 to 16 Challenge Level:
How many noughts are at the end of these giant numbers?
### Mod 3
##### Age 14 to 16 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Sticky Numbers
##### Age 11 to 14 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### Sixational
##### Age 14 to 18 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Pyramids
##### Age 11 to 14 Challenge Level:
What are the missing numbers in the pyramids?
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Never Prime
##### Age 14 to 16 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Unit Fractions
##### Age 11 to 14 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### Children at Large
##### Age 11 to 14 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### Chocolate Maths
##### Age 11 to 14 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Pattern of Islands
##### Age 11 to 14 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Always Perfect
##### Age 14 to 16 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### A Chordingly
##### Age 11 to 14 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Con Tricks
##### Age 11 to 14
Here are some examples of 'cons', and see if you can figure out where the trick is.
### Go Forth and Generalise
##### Age 11 to 14
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Disappearing Square
##### Age 11 to 14 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### For What?
##### Age 14 to 16 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
### Advent Calendar 2011 - Secondary
##### Age 11 to 18 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Logic
##### Age 7 to 14
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### More Number Pyramids
##### Age 11 to 14 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Unit Interval
##### Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Tessellating Hexagons
##### Age 11 to 14 Challenge Level:
Which hexagons tessellate?
### Concrete Wheel
##### Age 11 to 14 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### Königsberg
##### Age 11 to 14 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Triangle Inequality
##### Age 11 to 14 Challenge Level:
ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm.
### Why 24?
##### Age 14 to 16 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
### DOTS Division
##### Age 14 to 16 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Ratty
##### Age 11 to 14 Challenge Level:
If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation? | 2,418 | 9,988 | {"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-2019-43 | latest | en | 0.84556 |
https://www.coursehero.com/file/5654645/Practice6ans/ | 1,498,482,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320736.82/warc/CC-MAIN-20170626115614-20170626135614-00633.warc.gz | 837,265,319 | 328,069 | Practice6ans
# Practice6ans - Practice Questions(with answers Quiz#6...
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Practice Questions (with answers) Quiz #6 material Consider the plot below to answer questions 1 and 2. 1. Which of the following orbitals could the probability density shown above represent? a. 3 p b. 2 s c. 3 s d. both b. and c. e. both a. and c. 2. At what distance from the nucleus do the nodes occur in this orbital? a. approximately 2.5 pm and 6.5 pm b. approximately 132 pm and 344 pm c. approximately 212 pm and 582 pm d. approximately 4.5 pm and 11 pm e. approximately zero pm from the nucleus
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Consider the following plots to answer questions 3-5. 3. Which of the following must be true about the curves A, B, and C plotted above? a. Curves A, B, and C represent orbitals that have different energies b. Curves A, B, and C represent orbitals that are degenerate c. All of the curves must represent orbitals with l > 0 d. both a. and c. e. all of a., b., and c. 4. Which of the following orbitals could curve A above represent? a. 2p b. 2s c. 1s d. both a. and b. e. both b. and c. A B C
5. Which of the following must be true about the curves A, B, and C plotted above? a. The orbitals represented by A, B, and C have only one phase. b. Curve C could represent an orbital with n = 3. c. The orbitals represented by A, B, and C have l > 0. d. both b. and c. e. all of a., b., and c. 6. The nucleus of a hydrogen atom is located at the origin on a set of x , y , and z axes. Suppose the probability of finding the 2 s electron at the point ( x , y , z ) = (d, 0, 0) is 0.01. Which of the following must be true? a. The probability of finding the electron at ( x , y , z ) = (0, 0, d) is 0.01. b. The probability of finding the electron at ( x , y , z ) = (0, 0, d) is zero. c. The probability of finding the electron at ( x , y , z ) = (2d, 0, 0) is equal to 0.02. d. both a. and c. e. both b. and c. 7. The nucleus of a hydrogen atom is located at the origin on a set of x , y , and z axes. Suppose the probability of finding the 2 p z electron at the point ( x , y , z ) = (0, 0, 0.05) is equal to 0.001. Which of the following must be true? a. The probability of the finding the electron at ( x , y , z ) = (0, 0.05, 0) is equal to 0.001. b. The probability of finding the electron at ( x , y , z ) = (0.05, 0, 0) is equal to 0.001. c. The probability of finding the electron at the point ( x , y , z ) = (0.05, 0.05, 0) is less than 0.001. d. both a. and b. e. all of a., b., and c.
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8. Consider the following sets of quantum numbers for the electron in a hydrogen atom: I. n = 2, l = 1 II. n = 3 III. n = 3, l = 3 IV. n = 4, l = 2 Which of the following must be true?
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## This note was uploaded on 11/03/2009 for the course CHEM 1403 taught by Professor Leornardfine during the Fall '07 term at Columbia.
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Practice6ans - Practice Questions(with answers Quiz#6...
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Ask a homework question - tutors are online | 1,015 | 3,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-26 | longest | en | 0.908127 |
http://aa.springer.de/papers/8329001/2300001/sc3.htm | 1,653,368,792,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00486.warc.gz | 771,841 | 7,067 | Astron. Astrophys. 329, 1-13 (1998)
## 3. Spatial distribution of gravitational potential
As the first step of our consideration we need to reveal and to describe the properties of the simulated potential distribution. We need also to introduce and to test some quantitative characteristics of the spatial distribution of the potential field.
The spatial distribution of random field can be conveniently characterized by various mean values that are combinations of moments of the power spectrum (some examples of such characteristics were given by Bardeen et al. 1986 and Sahni & Coles 1995). It is important to specify whether they are given for untruncated power spectra which is important for theoretical considerations and comparison with observations, or together with various window functions what is important for comparisons with simulations which operate with truncated power spectra.
### 3.1. Qualitative description
The main properties of the potential distribution can be demonstrated by the evolution of the spatial distribution of the potential in a slice plotted in Fig. 2 for CDM75 at and at . The larger features of the potential are unchanged while on small scales the structure of potential is modified strongly and all the wrinkles in the equipotential lines are smoothed out with time. The lines come from the intersection of the plane under consideration with the 2-dimensional equipotential surfaces in 3-dimensional space.
Fig. 2. Slices of the gravitational potential from CDM75 (left at , right at ). The contour numbers are the values of the potential in units of (km/s)2
The matter concentrations in denser clumps influences locally the potential. This can be clearly seen in the most negative regions of the potential in Fig. 2, where some very deep potential wells evolve with time. The deep hole in the centre of Fig. 2b is especially prominent.
The isolines of potential distribution in Fig. 2 demonstrate also the well known fact, that the shapes of these isolines are more or less elliptical for high potential peaks and deep wells, and they become more and more complicated near the zero level. These properties are typical for random Gaussian fields which we now describe quantitatively.
### 3.2. Quantitative characteristics of the spatial potential distribution
Let us consider the potential along a random straight line. Thus we transform the complicated three-dimensional problem to the much simpler one-dimensional one. A 1D potential distribution is plotted in Fig. 3 for CDM75 and CDM200. The CDM75 profiles in Fig. 3a correspond to the cuts Mpc of Fig. 2, so that the "blob" at Mpc of Fig. 2a is included. This feature is seen in Fig. 3a as a tangent to . Fig. 3 demonstrates that in general, excluding the high dense clumps, the potential becomes smoother with time. Indeed, here the potential contains only one large wavelength giving raise to one maximum and one minimum. It is interesting that both the CDM75 and the CDM200 simulation contain just one basic wavelength of the potential perturbations which is an characteristic of the realized range of the CDM potential perturbations.
Fig. 3. One-dimensional potential distribution in CDM75 & CDM200
In the following we call regions with positive potential ( ) -regions. During cosmic evolution the density in these regions becomes lower than the mean density (under dense region - UDR). In -regions ( ) the density becomes higher than the mean density (over dense region - ODR). This can be clearly seen in Fig. 4 where the mean value of the density contrast in both regions is shown separately. Note, that the density contrast increases in -regions only by a factor of about 1.4 for CDM200. In the more evolved CDM25 simulation, we find over/underdensities of 2.2 and 0.17 for and , respectively. More detailed properties of and regions are given later on (Table 2). There we will show that the density contrast arises from a small mass flow through the boundary between and regions, whereas the respective volumes change only by a very small amount.
Fig. 4. Evolution of the density contrast in the regions of positive ( ) and negative ( ) potential for CDM200
The 1D approach allows us to introduce a simple quantitative characteristic of the potential distribution, the mean separation (mean free path) between zero points of the potential, . Demianski & Doroshkevich (1992) found a relation between and the power spectrum ,
where k is the comoving wave number. The value is closely linked with the correlation length of the potential (see, e.g., Bardeen et al. 1986). Equation ( 1) can be used for simulations with truncated power spectra, however, it can obviously not be used for spectra with Harrison-Zeldovich asymptote at , because the upper integral is divergent and leads to an infinite correlation length of the potential. Doroshkevich et al. (1997a) have derived a general relation which can be simplified for broad band power spectra to the equation
the solution of which gives the correct value both for the truncated and the total power spectrum. For the standard CDM and BSI matter dominated power spectra with we obtain
Note, that due to the scaling of the transfer function with the resulting of our models scales with .
The comparison of these results obtained for the theoretical spectra with similar estimates for simulations provides a simplest quantitative characteristic of the representativity of the simulations with respect to the large scale spatial potential distribution because both expressions (1) and (2) are sensitive to small k -values and to the size of the simulation box. Hence, it allows to estimate the impact of the computational box size on the matter evolution.
In simulations the mean separation of points along a random straight line, , can be found directly. Thus, if is the set of measured distances between levels and N is their total number, then the mean separation and dispersion are given by
In Table 1 the results are listed for CDM and BSI simulations at redshift , and together with the theoretical values of calculated from Eq. (2) for the truncated spectra used for simulations. The upper and lower limits of the integral were taken from the box size and the cell size.
Table 1. Theoretical and measured values of the mean separation and their dispersion for potential levels
Table 2. Parameters of the ODR and UDR, defined as regions where and , respectively for redshifts and . The columns 25-0 denotes the comparison of the density field at with respect to the the initial potential field at (for the columns 25 and 25-0 are identical). is the fraction of mass, is the fraction of space and is the density (in units of the critical density)
The dispersion in Table 1 characterizes the actual distance distribution rather than the errors in the measurements of . The reason for the large dispersion becomes clear from frequency distributions of the measured values (see Figs. 5 and 6 for CDM25 and CDM200, respectively). They are produced by 3963 random straight lines cutting the actual contour levels of the potential in the simulation box.
Fig. 5. Frequency distribution of measured distances between potential levels in the Mpc simulations. The dashed lines are the mean from Table 1, and the dotted lines are the theoretical values
Fig. 6. Frequency distribution of measured distances between potential levels in the Mpc simulations. The dashed lines are the mean from Table 1, and the dotted lines are the theoretical values . Fits to the histogram with the Poisson distribution are included as dotted lines.
In CDM25 and BSI25 simulations the different distances between levels are almost equal abundant, but we see already a trend towards smaller distances. This becomes more evident in the bigger simulations. Actually, the histogram is well described by a Poisson distribution in case of the CDM200 and BSI200 simulations (see the dotted line in Fig. 6).
From a theoretical point of view, we expect a Poisson distribution of SLSS elements (and also of ) along a random straight line for distances where any correlation becomes negligible (White 1979, Buryak et al. 1991, Borgani 1996). However, we see that for the small box sizes the measured distribution is far from Poissonian. For distances ( ) this is caused by the limitations of the simulations regarding SLSS elements because the number of long wave harmonics in the perturbations are strongly limited. However, in case of smaller scales the potential distribution is correlated and these correlations depend on the power spectrum. The CDM200 model shows the best indications of a Poissonian distribution because it contains the most SLSS elements in all the simulations. This can be easily seen from the values of in CDM200, which shows an average of 3 to 4 intersections with levels per box length, while on the other hand the Mpc simulations have only about two.
In this respect the CDM25 and BSI25 models are similar to the scale free models with a power index between and (see Fig. 1). Only for the models CDM200 and BSI200 deviations from the power law becomes important, so that we have to deal with a real broad band power spectrum. As we could see, this is also reflected by the potential characteristics. Thus, both for =75 Mpc and =25 Mpc the values are close to half the box size, 0.5 . In all simulations besides CDM200, the realized values are much smaller than the values obtained for the untruncated power spectra. This means that only the CDM simulation in the larger computational boxes realistically reproduces the potential distribution, and therefore it can expected to describe the matter evolution on the SLSS scale, comparable with the observed matter distribution. In contrast, any simulations in smaller boxes have more or less methodical character.
A common feature of all histograms (see also Table 1) is the slow increase of during evolution due to non-linear effects whereas is defined by the initial power spectrum. However, remains near to during the whole evolution. The slightly faster evolution of CDM75 and CDM25 should be ascribed to the late evolutionary stage of these simulations. Indeed, some structure elements have been destroyed at , and the matter distribution starts to transform into a system of isolated clumps (see also the discussion in Doroshkevich et al. 1997a).
### 3.3. Potential-potential correlation
The simple but important quantitative characteristic of the potential distribution is the two-point autocorrelation function. It allows us to obtain the characteristics of potential distribution averaged over the entire simulation.
For the primordial Harrison-Zeldovich spectrum the `gravitational potential' cannot be used directly because the dispersion of the potential diverges in the limit . Therefore, here one has to use the 'potential difference' between two points. This implies some modifications of the theoretical description (see, e.g., Demianski & Doroshkevich 1997). However, in case of simulations this problem disappears due to the finite box size. Thus, we can use here the `potential' without any restriction.
Thus, the standard correlation function of the potential in points and can be written as following:
where . For CDM-like power spectra this relation can be fitted as following (Demianski & Doroshkevich, 1997)
where the value is the mean separation of the zero potential along a random straight line defined by Eq. ( 2) and
is the dispersion of the potential perturbations.
In Fig. 7 the evolution of the power spectrum is depicted for CDM25 at redshifts . At high redshifts the power spectrum has some unphysical fluctuations at short wavelengths, caused by the finite resolution of the simulation. As upper limit in the integration of the power spectrum, we take the Nyquist frequency. Later the evolution increases strongly the short wave amplitude relatively to the linear law. In our case the long wave spectral amplitude decreases slightly due to the special initial realization.
Fig. 7. The evolution of the power spectrum with redshift for CDM25. The dashed lines are the corresponding power spectra calculated by linear theory. The horizontal bar indicates box size and Nyquist frequency.
The function has been calculated for the CDM200 and CDM25 models using the power spectra reconstructed at several redshifts (z = 25, 2 & 0) from the simulation. The results are presented in Fig. 8 together with the theoretical values given by Eq. ( 5). The different behaviour of theoretical and measured values at large distances characterizes the different potential distribution for truncated power spectra realized in simulations and the theoretical CDM spectra, i.e. it is caused by the impact of the finite box size. This conclusion is consistent to the former discussion of the one-dimendional analysis of separations of zero-points of the potential since this separation concerns the length scales larger than half of the simulation box.
Fig. 8. The normalized correlation function of potential vs. comoving distance for two models and two redshifts. The dotted lines present the simulation results and the dashed line the expectation of Eq. (5).
© European Southern Observatory (ESO) 1998
Online publication: November 24, 1997 | 2,756 | 13,303 | {"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} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.911422 |
https://www.physicsforums.com/threads/tests-of-efe.816403/ | 1,508,332,335,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00229.warc.gz | 997,899,361 | 20,821 | # Tests of EFE
Tags:
1. May 29, 2015
### CycoFin
There are several tests of general relativity:
- gravitional redshift
- deflaction of light
- perihelion precession of Mercury
- Shapiro delay
- Lense-Thirring precession
- binary pulsars
Now comes the part that problems me. These all are based on vacuum solution of Einstein field equations.
With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.
So is there any real tests of EFE, other than Cosmology? (which have it's problems as we all know)
Also looking for good explanations why left side of EFE must equal right side.
2. May 29, 2015
### Mentz114
The field equations follow from extrememizing an action as in any field theory.
Your concerns about vacuum solutions are groundless. The vacuum solutions have zero Ricci tensor but still possess non-zero Weyl curvature which
describes the field. It is very technical and if you study hard for a few years you will understand.
3. May 29, 2015
### Staff: Mentor
You can always simplify any equation to that form if you choose. This is a complete non issue.
All of the tests you mention are non-trivial solutions to the EFE.
4. May 29, 2015
### CycoFin
Can you then write EFE using only Weyl tensors? Is it so that basically Riemann tensor = Ricci tensor (trace part) + Weyl tensor(everything else)?
I disagree, the vacuum solution is the Schwarzschild metric which have 0 = 0
5. May 29, 2015
### Mentz114
The only way to change the EFE is to change the action being extremized.
A vacuum solution is created by solving the differential equations
$G(g)^{\mu\nu} =0$
where $g$ is a function of the coordinates. We thus find a valid metric, which gives us $0=0$, which is what we wanted.
See here http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution
6. May 29, 2015
### CycoFin
Yes, deriving the schwarzschild solution we assume that rigth side must be zero because vacuum and then find metric that gives also left side zero (R = 0).
But this 0 = 0 does not prove that EFE is correct hypothesis. One can put anything one wants to the right side, just make sure it is zero (e.g. number of magnetic monopoles).
I ask again, is there any experimental tests that have rigth side of EFE not zero?
7. May 29, 2015
### Mentz114
That is rubbish. You have to put the correct metric in. The metric is the solution. Your understanding of logic and mathematics is flawed.
The Schwarzschild metric satisfies all the experimental tests that have been tried - so your assumption is flawed in any case.
The equation $x^2-2ax+a^2=0$ has a solution $x=a$ which gives $0=0$. What is wrong with that logic ?
Last edited by a moderator: May 29, 2015
8. May 29, 2015
### CycoFin
Nothing is wrong with that logic.
You have proved that x2-2ax+a2 = 0 is true (have solution) but you have not proved that x2-2ax+a2 = b when b is not 0.
I agree that the Schwarschild metric satisfies all the experimental tests that have been tried and those I listed on first post.
I just don't understand how Schwardschild metric can "prove" EFE if equation reads 0 = 0.
9. May 29, 2015
### Mentz114
Well, you need to understand what solving an equation means. It is like a set of scales. The scales still work when they have nothing to weight. They are saying $0=0$.
Take the quadratic equation I wrote. I could have written $x^2-2ax = -a^2$. Now put in the solution $x=a$. Now the equation reads $-a^2=-a^2$.
We know the solution is correct when both sides are the same.
I don't think I can help any further with this.
10. May 29, 2015
### Staff: Mentor
No, it isn't. The EFE is a second order partial differential equation (more precisely, it's a system of 10 of them in the general case). So the RHS being zero just means it's a homogeneous second order partial differential equation, which certainly has solutions other than the trivial one. See here for an overview:
http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)
11. May 29, 2015
### John F. Gogo
So, you wrote this for wiki? Cool.
12. May 29, 2015
### Staff: Mentor
If you mean, did I write the Wikipedia page I linked to, no, I didn't. I just linked to it as a reference.
13. May 29, 2015
### Staff: Mentor
So what? This is true of any solution of any equation. You can always simplify it to 0=0:
Given A=B
A-B=0 by subtraction
0=0 by substitution
It is a completely useless complaint because it is always true for any equation. If you over-simplify you can always get 0=0. Your query about experiments or observations with a non-vacuum metric is a valid question, but this bit about 0=0 is not.
14. May 29, 2015
### atyy
Conceptually, these are not tests of the pure vacuum solutions, because in addition to the vacuum solution, geodesic motion is assumed. The geodesic motion is derived by assuming the presence of matter: http://arxiv.org/abs/0806.3293.
What problems does cosmology have? Are you thinking about dark matter?
For general relativity with matter, there seems to still be research going in these areas;
http://www.einstein-online.info/spotlights/hydrodynamics_realm
http://arxiv.org/abs/1210.4921
http://arxiv.org/abs/1206.2503
15. May 29, 2015
### Staff: Mentor
Just to clarify, this is referring to geodesic motion of bodies that are not "test bodies", because they have significant self-gravity, correct? For example, geodesic motion of the Earth in the gravitational field of the Sun.
This "presence of matter" is still very different from the "matter" that is present in cosmological solutions. In the latter, the matter significantly affects the spacetime geometry everywhere. In the former, the matter only significantly affects the spacetime geometry (in the sense of requiring a nonzero stress-energy tensor in the local solution of the EFE) in isolated regions corresponding to the central gravitating body producing the overall geometry (e.g., the Sun), and the self-gravitating bodies undergoing geodesic motion in this geometry (e.g., the Earth). Everywhere else, the solution is a vacuum solution, and that vacuum solution is what is used to predict the results of the classic solar system test experiments. (For example, nobody uses a nonzero stress-energy tensor for the Sun to predict the bending of light by the Sun; they just use the Schwarzschild geometry of the vacuum region exterior to the Sun.)
16. May 29, 2015
### atyy
I only meant that there are no test bodies, so although we treat the earth in the gravitational field of the sun as a test body, we do it by assuming a solution in which matter is present, and derive the treatment of the earth as a test body as a very good approximation.
17. May 30, 2015
### CycoFin
First I like to thank to get my thread back. I know my questions are very edge of the this forum rules. I will try not to go over.
I will try to make my point by some simple examles.
One can write equation:
apples = oranges
This is true if set both apples and oranges to zero. Equation reads 0 = 0. But surely it is not true if there are nonzero items of either.
One can write equation:
polar bears = constant * pink elephants.
Assume we make some experimetal tests on Afrika. Because there is no polar bears and neither pink elephants on Afrika, we are happy that test results agree with our equation which reads 0 = 0. We know there is polar bears on north pole but we have not yet made any tests because north pole is so far away from Afrika.
Can we now say from our test results that our polar bear/pink elephant equation is correct? No we don't. We have only tested one point (0=0).
This is the reason I asked is there any experimental test results where we have EFE something else than 0 = 0.
Last edited: May 30, 2015
18. May 30, 2015
### atyy
@CycoFin, as DaleSpam and others have said, your objection makes no sense. Every equation can be written in the form U = 0, including the full Maxwell's equations with charges and currents, and Newton's laws etc. It is true that there are regimes of GR that remain untested. However, you can consider the vacuum solutions of GR analogous to Maxwell's equations without charges. Even in the absence of charge, Maxwell's equations predict interesting phenomena such as electrotromagnetic waves that travel over great distances.
19. May 30, 2015
### atyy
In fact, the equation can be rewritten as
apples - oranges = 0.
Also, the equation can be true if there are nonzero items of either - it means that the number of apples and oranges is the same.
20. May 30, 2015
### CycoFin
Now I am little bit lost. You say that even we have RHS stress-energy T zero, LHS Einstein tensor G is not???
I know there are several diffrent vacuum solutions with different metrics but it does not change that in tensor point of view EFE reads 0 = 0 (these are zero tensors, not zeros) | 2,220 | 8,847 | {"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} | 3.140625 | 3 | CC-MAIN-2017-43 | longest | en | 0.927952 |
https://maimelatct.wordpress.com/2014/05/30/pressure-temperature-relation/ | 1,518,936,487,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811794.67/warc/CC-MAIN-20180218062032-20180218082032-00458.warc.gz | 708,952,716 | 23,350 | ## Pressure-temperature relation
Posted: May 30, 2014 in Uncategorized
The pressure of a gas is directly proportional to its temperature, if the volume is kept constant (Figure Figure 7). Recall that as the temperature of a gas increases, so does the kinetic energy of the particles in the gas. This causes the particles in the gas to move more rapidly and to collide with each other and with the side of the container more often. Since pressure is a measure of these collisions, the pressure of the gas increases with an increase in temperature. The pressure of the gas will decrease if its temperature decreases.
Tip:
You may see this law referred to as Gay-Lussac’s law or as Amontons’ law. Many scientists were working on the same problems at the same time and it is often difficult to know who actually discovered a particular law.
Chipa Thomas Maimela‘s insight:
In the same way that we have done for the other gas laws, we can describe the relationship between temperature and pressure using symbols, as follows:
T∝p,
therefore:
p=kT
Rearranging this we get:
and that, provided the amount of gas stays the same (and the volume also stays the same):
See on everythingscience.co.za | 264 | 1,198 | {"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-2018-09 | latest | en | 0.941007 |
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```Common-Mode Current is in Your Future!
Feature
Fig. 1: Differential-mode current in a PCB.
mode current flows out of the source, down the
trace, through the load, and back to the source
through the " ground " plane.
III. Common-Mode Current
But there is another kind of current that does seem
like magic. It is called common-mode current. It
should be hyphenated because " common current "
and " mode current " do not make any sense, we
need to join those first two words together when
they describe a particular type of current. What? A
particular type of current?
Common-mode current is usually referred to as an
" unintended " or " antenna " current. In contrast to
differential-mode current, common-mode current
flows in the same direction along a group of
conductors. Its presence, and its path from source to
load and back, is not obvious and usually not shown
on a circuit schematic. Figures 2 and 3 illustrate
differential-mode current and common-mode
current, respectively.
In Figure 4 we see the common-mode current
flowing out one wire, and returning back to the
(induced) source as a displacement current. The
displacement current flows between the two wires
because there is capacitance between the two
wires. This loop, consisting of part common-mode
conduction current plus displacement current is the
same model we use to describe the flow of current
along a dipole antenna as shown in Figure 5.
Fig. 4: Common-mode current in a PCB.
Fig. 5: Common-mode circuit model.
Fig. 2: Differential-mode
current in a simple circuit.
Fig. 3: Common-mode
current in a simple circuit.
Where does it come from? In PCB design, a time
changing current flowing on an external trace can
magnetically couple a second loop, just like in a
purpose-built transformer. The induced voltage then
pushes a common-mode current out of the wires
attached to the PCB " ground " plane. Common-mode
current is still a current, so it must flow in a loop.
The presence of either common-mode current or
displacement current often indicate the presence of
far-field wave propagation. When we do not want
this to happen, we call it a radiated EMI problem,
which can cause our electronic system to fail
mandatory U.S. (FCC CFR Part 15) and international
(CISPR 11, 24, etc.) standards that limit radio wave
" pollution " . We do not want our switch-mode power
supply or microprocessor PCB to interfere with
communications systems, such as in an ambulance,
helicopter, or spacecraft.
IV. Measurement of Common-Mode Current
Finally, it is relatively easy to measure actual
common-mode current in physical wiring. In Figure
6, we see that if a current probe were placed around
HKN.ORG
13
```
https://hkn.ieee.org/
# The Bridge - Issue 2, 2022
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https://www.nxtbook.com/nxtbooks/ieee/bridge_issue1_2023
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue3_2022
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue2_2022
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue1_2022
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue3_2021
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue2_2021
https://www.nxtbook.com/nxtbooks/ieee/bridge_issue1_2021
https://www.nxtbook.com/nxtbooks/ieee/bridge_2020_issue3
https://www.nxtbook.com/nxtbooks/ieee/bridge_2020_issue2
https://www.nxtbook.com/nxtbooks/ieee/bridge_2020_issue1
https://www.nxtbook.com/nxtbooks/ieee/bridge_2019_issue3
https://www.nxtbook.com/nxtbooks/ieee/bridge_2019_issue2
https://www.nxtbook.com/nxtbooks/ieee/bridge_2019_issue1
https://www.nxtbook.com/nxtbooks/ieee/bridge_2018_issue3
https://www.nxtbook.com/nxtbooks/ieee/bridge_2018_issue2
https://www.nxtbook.com/nxtbooks/ieee/bridge_2018_issue1
https://www.nxtbookmedia.com | 1,591 | 5,110 | {"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-2023-50 | latest | en | 0.916569 |
https://stats.stackexchange.com/questions/635432/an-example-of-a-random-variable-y-in-l-dagger-2-having-more-than-one-linear-c | 1,717,055,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00117.warc.gz | 464,504,546 | 39,098 | # An example of a random variable $y\in L^\dagger_2$ having more than one linear combination, $y = \Sigma_{i}\alpha_i x_i = \Sigma_{i}\beta_i x_i$
In the answer for the following exercise:
Let $$\{x_1,...,x_n\}$$ be a finite collection of random variables with $$E(x_i^2) \lt \infty$$ ($$i = 1,..., n$$). Show that the set of all linear combinations $$\Sigma_{i=1}^{n} \alpha_i x_i$$ constitutes a vector space, which we denote by $$L^\dagger_2$$,
there's a statement that I don't really understand, that is:
The difficulty is that a random variable $$y = \Sigma_{i}\alpha_i x_i$$ might also be expressible as $$y = \Sigma_{i}\beta_i x_i$$, where $$(\alpha_1,...,\alpha_n) \ne (\beta_1,...,\beta_n)$$.
For simplicity, suppose that $$n=2$$ so we have only two random variables $$x_1$$ and $$x_2$$ in the collection. How could I construct an example where $$y=\alpha_1 x_1 + \alpha_2 x_2=\beta_1 x_1 + \beta_2 x_2$$ where $$(\alpha_1, \alpha_2)\ne (\beta_1, \beta_2)$$?
• for your n=2 example, you would be able to do this exactly when $x_1 = a x_2$ for some constant $a$ (such as 1). In generally, you will be able to multiply express a linear combination in terms of its coefficients when the vectors constituting it are linearly dependent. Dec 21, 2023 at 18:59
• As emphasized by the context of the exercise, it isn't about random variables: it's purely about linear algebra. Let $x_2=-x_1,$ for instance. Then $\sum \alpha_i x_i = \alpha_1x_1+\alpha_2x_2=(\alpha_1-\alpha_2+\gamma)x_1+\gamma x_2$ for any real number $\gamma.$
– whuber
Dec 21, 2023 at 20:21
• BTW, it's hard to see how that statement is any difficulty at all. All you need to do in this exercise is apply the closure property, which is that any linear combination of linear combinations of the $x_i$ is a linear combination of them.
– whuber
Dec 21, 2023 at 20:23
• @JohnMadden and whuber Thanks for your comments! It's simpler than I thought it was. Dec 22, 2023 at 2:47
• @TranKhanh as I mentioned, there is no such case for N=2. Can you let me know what you don't like about my N=3 case? (note that in my example we have $0=x_3-\frac{x_1+x_2}{2}$). Dec 23, 2023 at 14:54 | 692 | 2,149 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.876712 |
http://datasciencehack.com/blog/category/main-category/natural-language-processing/ | 1,675,546,480,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00718.warc.gz | 10,265,224 | 52,664 | ## Prerequisites for understanding RNN at a more mathematical level
Writing the A gentle introduction to the tiresome part of understanding RNN Article Series on recurrent neural network (RNN) is nothing like a creative or ingenious idea. It is quite an ordinary topic. But still I am going to write my own new article on this ordinary topic because I have been frustrated by lack of sufficient explanations on RNN for slow learners like me.
I think many of readers of articles on this website at least know that RNN is a type of neural network used for AI tasks, such as time series prediction, machine translation, and voice recognition. But if you do not understand how RNNs work, especially during its back propagation, this blog series is for you.
After reading this articles series, I think you will be able to understand RNN in more mathematical and abstract ways. But in case some of the readers are allergic or intolerant to mathematics, I tried to use as little mathematics as possible.
Ideal prerequisite knowledge:
• Some understanding on densely connected layers (or fully connected layers, multilayer perception) and how their forward/back propagation work.
• Some understanding on structure of Convolutional Neural Network.
*In this article “Densely Connected Layers” is written as “DCL,” and “Convolutional Neural Network” as “CNN.”
### 1, Difficulty of Understanding RNN
I bet a part of difficulty of understanding RNN comes from the variety of its structures. If you search “recurrent neural network” on Google Image or something, you will see what I mean. But that cannot be helped because RNN enables a variety of tasks.
Another major difficulty of understanding RNN is understanding its back propagation algorithm. I think some of you found it hard to understand chain rules in calculating back propagation of densely connected layers, where you have to make the most of linear algebra. And I have to say backprop of RNN, especially LSTM, is a monster of chain rules. I am planing to upload not only a blog post on RNN backprop, but also a presentation slides with animations to make it more understandable, in some external links.
In order to avoid such confusions, I am going to introduce a very simplified type of RNN, which I call a “simple RNN.” The RNN displayed as the head image of this article is a simple RNN.
### 2, How Neurons are Connected
How to connect neurons and how to activate them is what neural networks are all about. Structures of those neurons are easy to grasp as long as that is about DCL or CNN. But when it comes to the structure of RNN, many study materials try to avoid showing that RNNs are also connections of neurons, as well as DCL or CNN(*If you are not sure how neurons are connected in CNN, this link should be helpful. Draw a random digit in the square at the corner.). In fact the structure of RNN is also the same, and as long as it is a simple RNN, and it is not hard to visualize its structure.
Even though RNN is also connections of neurons, usually most RNN charts are simplified, using blackboxes. In case of simple RNN, most study material would display it as the chart below.
But that also cannot be helped because fancier RNN have more complicated connections of neurons, and there are no longer advantages of displaying RNN as connections of neurons, and you would need to understand RNN in more abstract way, I mean, as you see in most of textbooks.
I am going to explain details of simple RNN in the next article of this series.
### 3, Neural Networks as Mappings
If you still think that neural networks are something like magical spider webs or models of brain tissues, forget that. They are just ordinary mappings.
If you have been allergic to mathematics in your life, you might have never heard of the word “mapping.” If so, at least please keep it in mind that the equation , which most people would have seen in compulsory education, is a part of mapping. If you get a value x, you get a value y corresponding to the x.
But in case of deep learning, x is a vector or a tensor, and it is denoted in bold like . If you have never studied linear algebra , imagine that a vector is a column of Excel data (only one column), a matrix is a sheet of Excel data (with some rows and columns), and a tensor is some sheets of Excel data (each sheet does not necessarily contain only one column.)
CNNs are mainly used for image processing, so their inputs are usually image data. Image data are in many cases (3, hight, width) tensors because usually an image has red, blue, green channels, and the image in each channel can be expressed as a height*width matrix (the “height” and the “width” are number of pixels, so they are discrete numbers).
The convolutional part of CNN (which I call “feature extraction part”) maps the tensors to a vector, and the last part is usually DCL, which works as classifier/regressor. At the end of the feature extraction part, you get a vector. I call it a “semantic vector” because the vector has information of “meaning” of the input image. In this link you can see maps of pictures plotted depending on the semantic vector. You can see that even if the pictures are not necessarily close pixelwise, they are close in terms of the “meanings” of the images.
In the example of a dog/cat classifier introduced by François Chollet, the developer of Keras, the CNN maps (3, 150, 150) tensors to 2-dimensional vectors, (1, 0) or (0, 1) for (dog, cat).
Wrapping up the points above, at least you should keep two points in mind: first, DCL is a classifier or a regressor, and CNN is a feature extractor used for image processing. And another important thing is, feature extraction parts of CNNs map images to vectors which are more related to the “meaning” of the image.
Importantly, I would like you to understand RNN this way. An RNN is also just a mapping.
*I recommend you to at least take a look at the beautiful pictures in this link. These pictures give you some insight into how CNN perceive images.
### 4, Problems of DCL and CNN, and needs for RNN
Taking an example of RNN task should be helpful for this topic. Probably machine translation is the most famous application of RNN, and it is also a good example of showing why DCL and CNN are not proper for some tasks. Its algorithms is out of the scope of this article series, but it would give you a good insight of some features of RNN. I prepared three sentences in German, English, and Japanese, which have the same meaning. Assume that each sentence is divided into some parts as shown below and that each vector corresponds to each part. In machine translation we want to convert a set of the vectors into another set of vectors.
Then let’s see why DCL and CNN are not proper for such task.
• The input size is fixed: In case of the dog/cat classifier I have mentioned, even though the sizes of the input images varies, they were first molded into (3, 150, 150) tensors. But in machine translation, usually the length of the input is supposed to be flexible.
• The order of inputs does not mater: In case of the dog/cat classifier the last section, even if the input is “cat,” “cat,” “dog” or “dog,” “cat,” “cat” there’s no difference. And in case of DCL, the network is symmetric, so even if you shuffle inputs, as long as you shuffle all of the input data in the same way, the DCL give out the same outcome . And if you have learned at least one foreign language, it is easy to imagine that the orders of vectors in sequence data matter in machine translation.
*It is said English language has phrase structure grammar, on the other hand Japanese language has dependency grammar. In English, the orders of words are important, but in Japanese as long as the particles and conjugations are correct, the orders of words are very flexible. In my impression, German grammar is between them. As long as you put the verb at the second position and the cases of the words are correct, the orders are also relatively flexible.
### 5, Sequence Data
We can say DCL and CNN are not useful when you want to process sequence data. Sequence data are a type of data which are lists of vectors. And importantly, the orders of the vectors matter. The number of vectors in sequence data is usually called time steps. A simple example of sequence data is meteorological data measured at a spot every ten minutes, for instance temperature, air pressure, wind velocity, humidity. In this case the data is recorded as 4-dimensional vector every ten minutes.
But this “time step” does not necessarily mean “time.” In case of natural language processing (including machine translation), which you I mentioned in the last section, the numberings of each vector denoting each part of sentences are “time steps.”
And RNNs are mappings from a sequence data to another sequence data.
In case of the machine translation above, the each sentence in German, English, and German is expressed as sequence data , and machine translation is nothing but mappings between these sequence data.
*At least I found a paper on the RNN’s capability of universal approximation on many-to-one RNN task. But I have not found any papers on universal approximation of many-to-many RNN tasks. Please let me know if you find any clue on whether such approximation is possible. I am desperate to know that.
### 6, Types of RNN Tasks
RNN tasks can be classified into some types depending on the lengths of input/output sequences (the “length” means the times steps of input/output sequence data).
If you want to predict the temperature in 24 hours, based on several time series data points in the last 96 hours, the task is many-to-one. If you sample data every ten minutes, the input size is 96*6=574 (the input data is a list of 574 vectors), and the output size is 1 (which is a value of temperature). Another example of many-to-one task is sentiment classification. If you want to judge whether a post on SNS is positive or negative, the input size is very flexible (the length of the post varies.) But the output size is one, which is (1, 0) or (0, 1), which denotes (positive, negative).
*The charts in this section are simplified model of RNN used for each task. Please keep it in mind that they are not 100% correct, but I tried to make them as exact as possible compared to those in other study materials.
Music/text generation can be one-to-many tasks. If you give the first sound/word you can generate a phrase.
Next, let’s look at many-to-many tasks. Machine translation and voice recognition are likely to be major examples of many-to-many tasks, but here name entity recognition seems to be a proper choice. Name entity recognition is task of finding proper noun in a sentence . For example if you got two sentences “He said, ‘Teddy bears on sale!’ ” and ‘He said, “Teddy Roosevelt was a great president!” ‘ judging whether the “Teddy” is a proper noun or a normal noun is name entity recognition.
Machine translation and voice recognition, which are more popular, are also many-to-many tasks, but they use more sophisticated models. In case of machine translation, the inputs are sentences in the original language, and the outputs are sentences in another language. When it comes to voice recognition, the input is data of air pressure at several time steps, and the output is the recognized word or sentence. Again, these are out of the scope of this article but I would like to introduce the models briefly.
Machine translation uses a type of RNN named sequence-to-sequence model (which is often called seq2seq model). This model is also very important for other natural language processes tasks in general, such as text summarization. A seq2seq model is divided into the encoder part and the decoder part. The encoder gives out a hidden state vector and it used as the input of the decoder part. And decoder part generates texts, using the output of the last time step as the input of next time step.
Voice recognition is also a famous application of RNN, but it also needs a special type of RNN.
*To be honest, I don’t know what is the state-of-the-art voice recognition algorithm. The example in this article is a combination of RNN and a collapsing function made using Connectionist Temporal Classification (CTC). In this model, the output of RNN is much longer than the recorded words or sentences, so a collapsing function reduces the output into next output with normal length.
You might have noticed that RNNs in the charts above are connected in both directions. Depending on the RNN tasks you need such bidirectional RNNs. I think it is also easy to imagine that such networks are necessary. Again, machine translation is a good example.
And interestingly, image captioning, which enables a computer to describe a picture, is one-to-many-task. As the output is a sentence, it is easy to imagine that the output is “many.” If it is a one-to-many task, the input is supposed to be a vector.
Where does the input come from? I mentioned that the last some layers in of CNN are closely connected to how CNNs extract meanings of pictures. Surprisingly such vectors, which I call a “semantic vectors” is the inputs of image captioning task (after some transformations, depending on the network models).
I think this articles includes major things you need to know as prerequisites when you want to understand RNN at more mathematical level. In the next article, I would like to explain the structure of a simple RNN, and how it forward propagate.
* I make study materials on machine learning, sponsored by DATANOMIQ. I do my best to make my content as straightforward but as precise as possible. I include all of my reference sources. If you notice any mistakes in my materials, please let me know (email: yasuto.tamura@datanomiq.de). And if you have any advice for making my materials more understandable to learners, I would appreciate hearing it.
## Über die Integration symbolischer Inferenz in tiefe neuronale Netze
Tiefe neuronale Netze waren in den letzten Jahren eine enorme Erfolgsgeschichte. Viele Fortschritte im Bereich der KI, wie das Erkennen von Objekten, die fließende Übersetzung natürlicher Sprache oder das Spielen von GO auf Weltklasseniveau, basieren auf tiefen neuronalen Netzen. Über die Grenzen dieses Ansatzes gab es jedoch nur wenige Berichte. Eine dieser Einschränkungen ist die Unfähigkeit, aus einer kleinen Anzahl von Beispielen zu lernen. Tiefe neuronale Netze erfordern in der Regel eine Vielzahl von Trainingsbeispielen, während der Mensch aus nur einem einzigen Beispiel lernen kann. Wenn Sie eine Katze einem Kind zeigen, das noch nie zuvor eine gesehen hat, kann es eine weitere Katze anhand dieser einzigen Instanz erkennen. Tiefe neuronale Netze hingegen benötigen Hunderttausende von Bildern, um zu erlernen, wie eine Katze aussieht. Eine weitere Einschränkung ist die Unfähigkeit, Rückschlüsse aus bereits erlerntem Allgemeinwissen zu ziehen. Beim Lesen eines Textes neigen Menschen dazu, weitreichende Rückschlüsse auf mögliche Interpretationen des Textes zu ziehen. Der Mensch ist dazu in der Lage, weil er Wissen aus sehr unterschiedlichen Bereichen abrufen und auf den Text anwenden kann.
Diese Einschränkungen deuten darauf hin, dass in tiefen neuronalen Netzen noch etwas Grundsätzliches fehlt. Dieses Etwas ist die Fähigkeit, symbolische Bezüge zu Entitäten in der realen Welt herzustellen und sie in Beziehung zueinander zu setzen. Symbolische Inferenz in Form von formaler Logik ist seit Jahrzehnten der Kern der klassischen KI, hat sich jedoch als spröde und komplex in der Anwendung erwiesen. Gibt es dennoch keine Möglichkeit, tiefe neuronale Netze so zu verbessern, dass sie in der Lage sind, symbolische Informationen zu verarbeiten? Tiefe neuronale Netzwerke wurden von biologischen neuronalen Netzwerken wie dem menschlichen Gehirn inspiriert. Im Wesentlichen sind sie ein vereinfachtes Modell der Neuronen und Synapsen, die die Grundbausteine des Gehirns ausmachen. Eine solche Vereinfachung ist, dass statt mit zeitlich begrenzten Aktionspotenzialen nur mit einem Aktivierungswert gearbeitet wird. Aber was ist, wenn es nicht nur wichtig ist, ob ein Neuron aktiviert wird, sondern auch, wann genau. Was wäre, wenn der Zeitpunkt, zu dem ein Neuron feuert, einen relationalen Kontext herstellt, auf den sich diese Aktivierung bezieht? Nehmen wir zum Beispiel ein Neuron, das für ein bestimmtes Wort steht. Wäre es nicht sinnvoll, wenn dieses Neuron jedes Mal ausgelöst würde, wenn das Wort in einem Text erscheint? In diesem Fall würde das Timing der Aktionspotenziale eine wichtige Rolle spielen. Und nicht nur das Timing einer einzelnen Aktivierung, sondern auch das Timing aller eingehenden Aktionspotenziale eines Neurons relativ zueinander wäre wichtig. Dieses zeitliche Muster kann verwendet werden, um eine Beziehung zwischen diesen Eingangsaktivierungen herzustellen. Wenn beispielsweise ein Neuron, das ein bestimmtes Wort repräsentiert, eine Eingabesynapse für jeden Buchstaben in diesem Wort hat, ist es wichtig, dass das Wort Neuron nur dann ausgelöst wird, wenn die Buchstabenneuronen in der richtigen Reihenfolge zueinander abgefeuert wurden. Konzeptionell könnten diese zeitlichen Unterschiede als Relationen zwischen den Eingangssynapsen eines Neurons modelliert werden. Diese Relationen definieren auch den Zeitpunkt, zu dem das Neuron selbst im Verhältnis zu seinen Eingangsaktivierungen feuert. Aus praktischen Gründen kann es sinnvoll sein, der Aktivierung eines Neurons mehrere Slots zuzuordnen, wie z.B. den Anfang und das Ende eines Wortes. Andernfalls müssten Anfang und Ende eines Wortes als zwei getrennte Neuronen modelliert werden. Diese Relationen sind ein sehr mächtiges Konzept. Sie ermöglichen es, die hierarchische Struktur von Texten einfach zu erfassen oder verschiedene Bereiche innerhalb eines Textes miteinander in Beziehung zu setzen. In diesem Fall kann sich ein Neuron auf eine sehr lokale Information beziehen, wie z.B. einen Buchstaben, oder auf eine sehr weitreichende Information, wie z.B. das Thema eines Textes.
Eine weitere Vereinfachung im Hinblick auf biologische neuronale Netze besteht darin, dass mit Hilfe einer Aktivierungsfunktion die Feuerrate eines einzelnen Neurons angenähert wird. Zu diesem Zweck nutzen klassische neuronale Netze die Sigmoidfunktion. Die Sigmoidfunktion ist jedoch symmetrisch bezüglich großer positiver oder negativer Eingangswerte, was es sehr schwierig macht, ausssagenlogische Operationen mit Neuronen mit der Sigmoidfunktion zu modellieren. Spiking-Netzwerke hingegen haben einen klaren Schwellenwert und ignorieren alle Eingangssignale, die unterhalb dieses Schwellenwerts bleiben. Daher ist die ReLU-Funktion oder eine andere asymmetrische Funktion eine deutlich bessere Annäherung für die Feuerrate. Diese Asymmetrie ist auch für Neuronen unerlässlich, die relationale Informationen verarbeiten. Das Neuron, das ein bestimmtes Wort repräsentiert, muss nämlich für alle Zeitpunkte, an denen das Wort nicht vorkommt, völlig inaktiv bleiben.
Ebenfalls vernachlässigt wird in tiefen neuronalen Netzwerken die Tatsache, dass verschiedene Arten von Neuronen in der Großhirnrinde vorkommen. Zwei wichtige Typen sind die bedornte Pyramidenzelle, die in erster Linie eine exzitatorische Charakteristik aufweist, und die nicht bedornte Sternzelle, die eine hemmende aufweist. Die inhibitorischen Neuronen sind besonders, weil sie es ermöglichen, negative Rückkopplungsschleifen aufzubauen. Solche Rückkopplungsschleifen finden sich normalerweise nicht in einem tiefen neuronalen Netzwerk, da sie einen inneren Zustand in das Netzwerk einbringen. Betrachten wir das folgende Netzwerk mit einem hemmenden Neuron und zwei exzitatorischen Neuronen, die zwei verschiedene Bedeutungen des Wortes “August” darstellen.
Beide Bedeutungen schließen sich gegenseitig aus, so dass das Netzwerk nun zwei stabile Zustände aufweist. Diese Zustände können von weiteren Eingangssynapsen der beiden exzitatorischen Neuronen abhängen. Wenn beispielsweise das nächste Wort nach dem Wort ‘August’ ein potenzieller Nachname ist, könnte eine entsprechende Eingabesynapse für das Entitätsneuron August-(Vorname) das Gewicht dieses Zustands erhöhen. Es ist nun wahrscheinlicher, dass das Wort “August” als Vorname und nicht als Monat eingestuft wird. Aber bedenken Sie, dass beide Zustände evaluiert werden müssen. In größeren Netzwerken können viele Neuronen durch negative oder positive Rückkopplungsschleifen verbunden sein, was zu einer großen Anzahl von stabilen Zuständen im Netzwerk führen kann.
Aus diesem Grund ist ein effizienter Optimierungsprozess erforderlich, der den besten Zustand in Bezug auf eine Zielfunktion ermittelt. Diese Zielfunktion könnte darin bestehen, die Notwendigkeit der Unterdrückung stark aktivierter Neuronen zu minimieren. Diese Zustände haben jedoch den enormen Vorteil, dass sie es erlauben, unterschiedliche Interpretationen eines bestimmten Textes zu berücksichtigen. Es ist eine Art Denkprozess, in dem verschiedene Interpretationen bewertet werden und die jeweils stärkste als Ergebnis geliefert wird. Glücklicherweise lässt sich die Suche nach einem optimalen Lösungszustand recht gut optimieren.
Der Grund, warum wir in diesen Rückkopplungsschleifen hemmende Neuronen benötigen, ist, dass sonst alle gegenseitig unterdrückenden Neuronen vollständig miteinander verbunden sein müssten. Das würde zu einer quadratisch zunehmenden Anzahl von Synapsen führen.
Durch die negativen Rückkopplungsschleifen, d.h. durch einfaches Verbinden einer negativen Synapse mit einem ihrer Vorläuferneuronen, haben wir plötzlich den Bereich der nichtmonotonen Logik betreten. Die nichtmonotone Logik ist ein Teilgebiet der formalen Logik, in dem Implikationen nicht nur zu einem Modell hinzugefügt, sondern auch entfernt werden. Es wird davon ausgegangen, dass eine nichtmonotone Logik erforderlich ist, um Schlussfolgerungen für viele Common Sense Aufgaben ziehen zu können. Eines der Hauptprobleme der nichtmonotonen Logik ist, dass sie oft nicht entscheiden kann, welche Schlussfolgerungen sie ziehen soll und welche eben nicht. Einige skeptische oder leichtgläubige Schlussfolgerungen sollten nur gezogen werden, wenn keine anderen Schlussfolgerungen wahrscheinlicher sind. Hier kommt die gewichtete Natur neuronaler Netze zum Tragen. In neuronalen Netzen können nämlich eher wahrscheinliche Zustände weniger wahrscheinliche Zustände unterdrücken.
### Beispielimplementierung innerhalb des Aika-Frameworks
An dieser Stelle möchte ich noch einmal das Beispielneuron für das Wort ‘der’ vom Anfang aufgreifen. Das Wort-Neuron besteht aus drei Eingabesynapsen, die sich jeweils auf die einzelnen Buchstaben des Wortes beziehen. Über die Relationen werden die Eingabesynapsen nun zueinander in eine bestimmte Beziehung gesetzt, so dass das Wort ‘der’ nur erkannt wird, wenn alle Buchstaben in der korrekten Reihenfolge auftreten.
Als Aktivierungsfunktion des Neurons wird hier der im negativen Bereich abgeschnittene (rectified) hyperbolische Tangens verwendet. Dieser hat gerade bei einem UND-verknüpfenden Neuron den Vorteil, dass er selbst bei sehr großen Werten der gewichteten Summe auf den Wert 1 begrenzt ist. Alternativ kann auch die ReLU-Funktion (Rectified Linear Unit) verwendet werden. Diese eignet sich insbesondere für ODER-verknüpfende Neuronen, da sie die Eingabewerte unverzerrt weiterleitet.
Im Gegensatz zu herkömmlichen neuronalen Netzen gibt es hier mehrere Bias Werte, einen für das gesamte Neuron (in diesem Fall auf 5.0 gesetzt) und einen für jede Synapse. Intern werden diese Werte zu einem gemeinsamen Bias aufsummiert. Es ist schon klar, dass dieses Aufteilen des Bias nicht wirklich gut zu Lernregeln wie der Delta-Rule und dem Backpropagation passt, allerdings eignen sich diese Lernverfahren eh nur sehr begrenzt für diese Art von neuronalem Netzwerk. Als Lernverfahren kommen eher von den natürlichen Mechanismen Langzeit-Potenzierung und Langzeit-Depression inspirierte Ansätze in Betracht.
### Fazit
Obwohl tiefe neuronale Netze bereits einen langen Weg zurückgelegt haben und mittlerweile beeindruckende Ergebnisse liefern, kann es sich doch lohnen, einen weiteren Blick auf das Original, das menschliche Gehirn und seine Schaltkreise zu werfen. Wenn eine so inhärent komplexe Struktur wie das menschliche Gehirn als Blaupause für ein neuronales Modell verwendet werden soll, müssen vereinfachende Annahmen getroffen werden. Allerdings ist bei diesem Prozess Vorsicht geboten, da sonst wichtige Aspekte des Originals verloren gehen können.
### Referenzen
1. Der Aika-Algorithm
Lukas Molzberger
2. Neuroscience: Exploring the Brain
Mark F. Bear, Barry W. Connors, Michael A. Paradiso
3. Neural-Symbolic Learning and Reasoning: A Survey and Interpretation
Tarek R. Besold, Artur d’Avila Garcez, Sebastian Bader; Howard Bowman, Pedro Domingos, Pascal Hitzler, Kai-Uwe Kuehnberger, Luis C. Lamb, ; Daniel Lowd, Priscila Machado Vieira Lima, Leo de Penning, Gadi Pinkas, Hoifung Poon, Gerson Zaverucha
4. Deep Learning: A Critical Appraisal
Gary Marcus
5. Nonmonotonic Reasoning
Gerhard Brewka, Ilkka Niemela, Mirosław Truszczynski
## Einstieg in Natural Language Processing – Teil 2: Preprocessing von Rohtext mit Python
Dies ist der zweite Artikel der Artikelserie Einstieg in Natural Language Processing.
In diesem Artikel wird das so genannte Preprocessing von Texten behandelt, also Schritte die im Bereich des NLP in der Regel vor eigentlichen Textanalyse durchgeführt werden.
### Tokenizing
Um eingelesenen Rohtext in ein Format zu überführen, welches in der späteren Analyse einfacher ausgewertet werden kann, sind eine ganze Reihe von Schritten notwendig. Ganz allgemein besteht der erste Schritt darin, den auszuwertenden Text in einzelne kurze Abschnitte – so genannte Tokens – zu zerlegen (außer man bastelt sich völlig eigene Analyseansätze, wie zum Beispiel eine Spracherkennung anhand von Buchstabenhäufigkeiten ect.).
Was genau ein Token ist, hängt vom verwendeten Tokenizer ab. So bringt NLTK bereits standardmäßig unter anderem BlankLine-, Line-, Sentence-, Word-, Wordpunkt- und SpaceTokenizer mit, welche Text entsprechend in Paragraphen, Zeilen, Sätze, Worte usw. aufsplitten. Weiterhin ist mit dem RegexTokenizer ein Tool vorhanden, mit welchem durch Wahl eines entsprechenden Regulären Ausdrucks beliebig komplexe eigene Tokenizer erstellt werden können.
Üblicherweise wird ein Text (evtl. nach vorherigem Aufsplitten in Paragraphen oder Sätze) schließlich in einzelne Worte und Interpunktionen (Satzzeichen) aufgeteilt. Hierfür kann, wie im folgenden Beispiel z. B. der WordTokenizer oder die diesem entsprechende Funktion word_tokenize() verwendet werden.
### Stemming & Lemmatizing
Andere häufig durchgeführte Schritte sind Stemming sowie Lemmatizing. Hierbei werden die Suffixe der einzelnen Tokens des Textes mit Hilfe eines Stemmers in eine Form überführt, welche nur den Wortstamm zurücklässt. Dies hat den Zweck verschiedene grammatikalische Formen des selben Wortes (welche sich oft in ihrer Endung unterscheiden (ich gehe, du gehst, er geht, wir gehen, …) ununterscheidbar zu machen. Diese würden sonst als mehrere unabhängige Worte in die darauf folgende Analyse eingehen.
Neben bereits fertigen Stemmern bietet NLTK auch für diesen Schritt die Möglichkeit sich eigene Stemmer zu programmieren. Da verschiedene Stemmer Suffixe nach unterschiedlichen Regeln entfernen, sind nur die Wortstämme miteinander vergleichbar, welche mit dem selben Stemmer generiert wurden!
Im forlgenden Beispiel werden verschiedene vordefinierte Stemmer aus dem Paket NLTK auf den bereits oben verwendeten Beispielsatz angewendet und die Ergebnisse der gestemmten Tokens in einer Art einfachen Tabelle ausgegeben:
Sehr ähnlich den Stemmern arbeiten Lemmatizer: Auch ihre Aufgabe ist es aus verschiedenen Formen eines Wortes die jeweilige Grundform zu bilden. Im Unterschied zu den Stemmern ist das Lemma eines Wortes jedoch klar als dessen Grundform definiert.
### Vokabular
Auch das Vokabular, also die Menge aller verschiedenen Worte eines Textes, ist eine informative Kennzahl. Bezieht man die Größe des Vokabulars eines Textes auf seine gesamte Anzahl verwendeter Worte, so lassen sich hiermit Aussagen zu der Diversität des Textes machen.
Außerdem kann das auftreten bestimmter Worte später bei der automatischen Einordnung in Kategorien wichtig werden: Will man beispielsweise Nachrichtenmeldungen nach Themen kategorisieren und in einem Text tritt das Wort „DAX“ auf, so ist es deutlich wahrscheinlicher, dass es sich bei diesem Text um eine Meldung aus dem Finanzbereich handelt, als z. B. um das „Kochrezept des Tages“.
Dies mag auf den ersten Blick trivial erscheinen, allerdings können auch mit einfachen Modellen, wie dem so genannten „Bag-of-Words-Modell“, welches nur die Anzahl des Auftretens von Worten prüft, bereits eine Vielzahl von Informationen aus Texten gewonnen werden.
Das reine Vokabular eines Textes, welcher in der Variable “rawtext” gespeichert ist, kann wie folgt in der Variable “vocab” gespeichert werden. Auf die Ausgabe wurde in diesem Fall verzichtet, da diese im Falle des oben als Beispiel gewählten Satzes den einzelnen Tokens entspricht, da kein Wort öfter als ein Mal vorkommt.
### Stopwords
Unter Stopwords werden Worte verstanden, welche zwar sehr häufig vorkommen, jedoch nur wenig Information zu einem Text beitragen. Beispiele in der beutschen Sprache sind: der, und, aber, mit, …
Sowohl NLTK als auch cpaCy bringen vorgefertigte Stopwordsets mit.
Vorsicht: NLTK besitzt eine Stopwordliste, welche erst in ein Set umgewandelt werden sollte um die lookup-Zeiten kurz zu halten – schließlich muss jedes einzelne Token des Textes auf das vorhanden sein in der Stopworditerable getestet werden!
### POS-Tagging
POS-Tagging steht für „Part of Speech Tagging“ und entspricht ungefähr den Aufgaben, die man noch aus dem Deutschunterricht kennt: „Unterstreiche alle Subjekte rot, alle Objekte blau…“. Wichtig ist diese Art von Tagging insbesondere, wenn man später tatsächlich strukturiert Informationen aus dem Text extrahieren möchte, da man hierfür wissen muss wer oder was als Subjekt mit wem oder was als Objekt interagiert.
Obwohl genau die selben Worte vorkommen, bedeutet der Satz „Die Katze frisst die Maus.“ etwas anderes als „Die Maus frisst die Katze.“, da hier Subjekt und Objekt aufgrund ihrer Reihenfolge vertauscht sind (Stichwort: Subjekt – Prädikat – Objekt ).
Weniger wichtig ist dieser Schritt bei der Kategorisierung von Dokumenten. Insbesondere bei dem bereits oben erwähnten Bag-of-Words-Modell, fließen POS-Tags überhaupt nicht mit ein.
## Und weil es so schön einfach ist: Die obigen Schritte mit spaCy
Die obigen Methoden und Arbeitsschritte, welche Texte die in natürlicher Sprache geschrieben sind, allgemein computerzugänglicher und einfacher auswertbar machen, können beliebig genau den eigenen Wünschen angepasst, einzeln mit dem Paket NLTK durchgeführt werden. Dies zumindest einmal gemacht zu haben, erweitert das Verständnis für die funktionsweise einzelnen Schritte und insbesondere deren manchmal etwas versteckten Komplexität. (Wie muss beispielsweise ein Tokenizer funktionieren der den Satz “Schwierig ist z. B. dieser Satz.” korrekt in nur einen Satz aufspaltet, anstatt ihn an jedem Punkt welcher an einem Wortende auftritt in insgesamt vier Sätze aufzuspalten, von denen einer nur aus einem Leerzeichen besteht?) Hier soll nun aber, weil es so schön einfach ist, auch das analoge Vorgehen mit dem Paket spaCy beschrieben werden:
Dieser kurze Codeabschnitt liest den an spaCy übergebenen Rohtext in ein spaCy Doc-Object ein und führt dabei automatisch bereits alle oben beschriebenen sowie noch eine Reihe weitere Operationen aus. So stehen neben dem immer noch vollständig gespeicherten Originaltext, die einzelnen Sätze, Worte, Lemmas, Noun-Chunks, Named Entities, Part-of-Speech-Tags, ect. direkt zur Verfügung und können.über die Methoden des Doc-Objektes erreicht werden. Des weiteren liegen auch verschiedene weitere Objekte wie beispielsweise Vektoren zur Bestimmung von Dokumentenähnlichkeiten bereits fertig vor.
Die Folgende Übersicht soll eine kurze (aber noch lange nicht vollständige) Übersicht über die automatisch von spaCy generierten Objekte und Methoden zur Textanalyse geben:
Diese „Vollautomatisierung“ der Vorabschritte zur Textanalyse hat jedoch auch seinen Preis: spaCy geht nicht gerade sparsam mit Ressourcen wie Rechenleistung und Arbeitsspeicher um. Will man einen oder einige Texte untersuchen so ist spaCy oft die einfachste und schnellste Lösung für das Preprocessing. Anders sieht es aber beispielsweise aus, wenn eine bestimmte Analyse wie zum Beispiel die Einteilung in verschiedene Textkategorien auf eine sehr große Anzahl von Texten angewendet werden soll. In diesem Fall, sollte man in Erwägung ziehen auf ressourcenschonendere Alternativen wie zum Beispiel gensim auszuweichen.
Wer beim lesen genau aufgepasst hat, wird festgestellt haben, dass ich im Abschnitt POS-Tagging im Gegensatz zu den anderen Abschnitten auf ein kurzes Codebeispiel verzichtet habe. Dies möchte ich an dieser Stelle nachholen und dabei gleich eine Erweiterung des Pakets spaCy vorstellen: displaCy.
Displacy bietet die Möglichkeit, sich Zusammenhänge und Eigenschaften von Texten wie Named Entities oder eben POS-Tagging graphisch im Browser anzeigen zu lassen.
Nach ausführen des obigen Codes erhält man eine Ausgabe die wie folgt aussieht:
Nun öffnet man einen Browser und ruft die URL ‘http://127.0.0.1:5000’ auf (Achtung: localhost anstatt der IP funktioniert – warum auch immer – mit displacy nicht). Im Browser sollte nun eine Seite mit einem SVG-Bild geladen werden, welches wie folgt aussieht
Die Abbildung macht deutlich was POS-Tagging genau ist und warum es von Nutzen sein kann wenn man Informationen aus einem Text extrahieren will. Jedem Word (Token) ist eine Wortart zugeordnet und die Beziehung der einzelnen Worte durch Pfeile dargestellt. Dies ermöglicht es dem Computer zum Beispiel in dem Satzteil “der grüne Apfel”, das Adjektiv “grün” auf das Nomen “Apfel” zu beziehen und diesem somit als Eigenschaft zuzuordnen.
Nachdem dieser Artikel wichtige Schritte des Preprocessing von Texten beschrieben hat, geht es im nächsten Artikel darum was man an Texten eigentlich analysieren kann und welche Analysemöglichkeiten die verschiedenen für Python vorhandenen Module bieten.
## Einstieg in Natural Language Processing – Teil 1: Natürliche vs. Formale Sprachen
Dies ist Artikel 1 von 4 der Artikelserie Einstieg in Natural Language Processing – Artikelserie.
Versuche und erste Ansätze, Maschinen beizubringen menschliche Sprache zu verstehen, gibt es bereits seit den 50er Jahren. Trotz der jahrzehntelangen Forschung und Entwicklung gelingt dies bis heute nicht umfassend. Woran liegt dies?
Um diese Frage zu beantworten, hilft es, sich die Unterschiede zwischen „natürlichen“, also sich selbstständig entwickelnden, typischerweise von Menschen gesprochenen Sprachen und den von Computern interpretieren formalen Sprachen klar zu machen. Formale Sprachen, wie zum Beispiel Python zum Ausführen der Codebeispiele in dieser Artikelserie, HTML (Hyper Text Markup Language) zur Darstellung von Webseiten und andere typische Programmier- und Skriptsprachen, sind üblicherweise sehr streng strukturiert.
Alle diese Sprachen weisen eine Reihe von Gemeinsamkeiten auf, welche es Computern einfach machen, sie korrekt zu interpretieren (also den Informationsinhalt zu “verstehen”). Das vermutlich auffälligste Merkmal formaler Sprachen ist eine relativ strikte Syntax, welche (wenn überhaupt) nur geringe Abweichungen von einem Standard erlaubt. Wie penibel die jeweilige Syntax oft einzuhalten ist, wird am ehesten deutlich, wenn diese verletzt wird:
Solche so genannten “Syntax Error” gehören daher zu den häufigsten Fehlern beim Schreiben von Quellcode.
Ganz anders dagegen sieht es in der Kommunikation mit natürlichen Sprachen aus. Zwar fördert falsche Komma-Setzung in der Regel nicht die Leserlichkeit eines Textes, jedoch bleibt dieser in der Regel trotzdem verständlich. Auch macht es keinen Unterschied ob ich sage „Es ist heiß heute.“ oder „Heute ist es heiß.“. Genau wie in der deutschen Sprache funktioniert dieses Beispiel auch im Englischen sowie in anderen natürlichen Sprachen. Insbesondere Spanisch ist ein Beispiel für eine Sprache mit extrem variabler Satzstellung. Jedoch kann in anderen Fällen eine andere Reihenfolge der selben Worte deren Bedeutung auch verändern. So ist „Ist es heute heiß?“ ganz klar eine Frage, obwohl exakt die selben Worte wie in den Beispielsätzen oben vorkommen.
Ein weiterer wichtiger, hiermit verwandter Unterschied ist, dass es bei formalen Sprachen in der Regel einen Ausdruck gibt, welcher eine spezifische Bedeutung besitzt, während es in natürlichen Sprachen oft viele Synonyme gibt, die ein und dieselbe Sache (oder zumindest etwas sehr ähnliches) ausdrücken. Ein wahrer boolscher Wert wird in Python als
geschrieben. Es gibt keine andere Möglichkeit, diesen Wert auszudrücken (zumindest nicht ohne irgend eine Art von Operatoren wie das Doppelgleichheitszeichen zu benutzen und damit z. B. “0 == 0” zu schreiben). Anders hingegen zum Beispiel in der Deutschen Sprache: Wahr, richtig, korrekt, stimmt, ja,
Um einen Vorstellung davon zu bekommen, wie verbreitet Synonyme in natürlichen Sprachen sind, lässt sich die Internetseite https://www.openthesaurus.de verwenden. Beispielshalber findet man dutzende Synonyme für das Wort „schnell“ hier: https://www.openthesaurus.de/synonyme/schnell
Eine weitere große Schwierigkeit, welche in den meisten natürlichen Sprachen und nahezu allen Arten von Texten zu finden ist, stellen verschiedene grammatikalische Formen eines Wortes dar. So sind die Worte bin, wäre, sind, waren, wirst, werden… alles Konjugationen desselben Verbs, nämlich sein. Eine durchaus beeindruckende Übersicht über die verwirrende Vielfalt von Konjugationen dieses kleinen Wörtchens, findet sich unter: https://www.verbformen.de/konjugation/sein.htm.
Dieses Problem wird um so schwerwiegender, da viele Verben, insbesondere die am häufigsten genutzten, sehr unregelmäßige Konjugationsformen besitzen und damit keiner generellen Regel folgen. Daher ist computerintern oft ein Mapping für jede mögliche Konjugationsform bei vielen Verben die einzige Möglichkeit, an die Grundform zu kommen (mehr dazu in Teil 3 dieser Artikelserie).
Die Liste der sprachlichen Schwierigkeiten beim computergestützten Auswerten natürlicher Sprache ließe sich an diesem Punkt noch beliebig weiter fortsetzen:
• Rechtschreibfehler
• falsche Grammatik
• Smileys
• der „Substantivverkettungswahn“ im Deutschen
• mehrdeutige Worte und Abkürzungen
• abwegige Redewendungen (z. B. “ins Gras beißen”)
• Ironie
• und, und, und …
Ob und welche Rolle jede dieser Schwierigkeiten im einzelnen spielt, hängt natürlich sehr stark von den jeweiligen Texten ab und kann nicht pauschalisiert werden – ein typischer Chatverlauf wird ganz andere Probleme bereithalten als ein Wikipedia-Artikel. Wie man einige dieser Probleme in der Praxis vereinfachen oder sogar lösen kann und welche Ansätze und Methoden zur Verfügung stehen und regelmäßig zur Anwendung kommen wird im nächsten Teil dieser Artikelserie an praktischen Codebeispielen genauer unter die Lupe genommen.
## NLTK vs. Spacy – Eine kurze Übersicht
Möchte man einen (oder auch einige) Text(e) mit den Methoden des natural language processings untersuchen um die darin verwendete Sprache auswerten oder nach bestimmten Informationen suchen, so sind insbesondere die Pakete NLTK und spaCy zu empfehlen (bei sehr vielen Texten sieht das schon wieder anders aus und wird am Ende der Artikelserie mit dem Paket gensim vorgestellt); beide bieten eine unglaubliche Vielzahl von Analysemöglichkeiten, vorgefertigten Wortsets, vortrainierte Stemmer und Lemmatiser, POS Tagger und, und, und…
Ist man vor allem an den Ergebnissen der Analyse selbst interessiert, so bietet sich spaCy an, da hier bereits mit wenigen Zeilen Code viele interessante Informationen generiert werden können.
Wer dagegen gerne selber bastelt oder wissen möchte wie die einzelnen Tools und Teilschritte genau funktionieren oder sich seine eigenen Stemmer, Tagger ect. trainieren will, ist vermutlich mit NLTK besser beraten. Zwar ist hier oft mehr Quellcode für das gleiche Ergebnis notwendig, allerdings kann das Preprocessing der Texte hierbei relativ einfach exakt den eigenen Vorstellungen angepasst werden. Zudem bietet NLTK eine Vielzahl von Beispieltexten und bereits fertig getagte Daten, mit welchen eigene Tagger trainiert und getestet werden können.
## Einstieg in Natural Language Processing – Artikelserie
Unter Natural Language Processing (NLP) versteht man ein Teilgebiet der Informatik bzw. der Datenwissenschaft, welches sich mit der Analyse und Auswertung , aber auch der Synthese natürlicher Sprache befasst. Mit natürlichen Sprachen werden Sprachen wie zum Beispiel Deutsch, Englisch oder Spanisch bezeichnet, welche nicht geplant entworfen wurden, sondern sich über lange Zeit allein durch ihre Benutzung entwickelt haben. Anders ausgedrückt geht es um die Schnittstelle zwischen unserer im Alltag verwendeten und für uns Menschen verständlichen Sprache auf der einen, und um deren computergestützte Auswertung auf der anderen Seite.
Diese Artikelserie soll eine Einführung in die Thematik des Natural Language Processing sein, dessen Methoden, Möglichkeiten, aber auch der Grenzen . Im einzelnen werden folgende Themen näher behandelt:
1. Artikel – Natürliche vs. Formale Sprachen
2. Artikel – Preprocessing von Rohtext mit Python
3. Artikel – Möglichkeiten/Methoden der Textanalyse an Beispielen (erscheint demnächst…)
4. Artikel – NLP, was kann es? Und was nicht? (erscheint demnächst…)
Zur Verdeutlichung der beschriebenen Zusammenhänge und Methoden und um Interessierten einige Ideen für mögliche Startpunkte aufzuzeigen, werden im Verlauf der Artikelserie an verschiedenen Stellen Codebeispiele in der Programmiersprache Python vorgestellt.
Von den vielen im Internet zur Verfügung stehenden Python-Paketen zum Thema NLP, werden in diesem Artikel insbesondere die drei Pakete NLTK, Gensim und Spacy verwendet.
## Aika: Ein semantisches neuronales Netzwerk
Wenn es darum geht Informationen aus natürlichsprachigen Texten zu extrahieren, stehen einem verschiedene Möglichkeiten zur Verfügung. Eine der ältesten und wohl auch am häufigsten genutzten Möglichkeiten ist die der regulären Ausdrücke. Hier werden exakte Muster definiert und in einem Textstring gematcht. Probleme bereiten diese allerdings, wenn kompliziertere semantische Muster gefunden werden sollen oder wenn verschiedene Muster aufeinander aufbauen oder miteinander interagieren sollen. Gerade das ist aber der Normalfall bei der Verarbeitung von natürlichem Text. Muster hängen voneinander ab, verstärken oder unterdrücken sich gegenseitig.
Prädestiniert um solche Beziehungen abzubilden wären eigentlich künstliche neuronale Netze. Diese haben nur das große Manko, dass sie keine strukturierten Informationen verarbeiten können. Neuronale Netze bringen von sich aus keine Möglichkeit mit, die relationalen Beziehungen zwischen Worten oder Phrasen zu verarbeiten. Ein weiteres Problem neuronaler Netze ist die Verarbeitung von Feedback-Schleifen, bei denen einzelne Neuronen von sich selbst abhängig sind. Genau diese Probleme versucht der Aika Algorithmus (www.aika-software.org) zu lösen.
Der Aika Algorithmus ist als Open Source Java-Bibliothek implementiert und dient dazu semantische Informationen in Texten zu erkennen und zu verarbeiten. Da semantische Informationen sehr häufig mehrdeutig sind, erzeugt die Bibliothek für jede dieser Bedeutungen eine eigene Interpretation und wählt zum Schluss die am höchsten gewichtete aus. Aika kombiniert dazu aktuelle Ideen und Konzepte aus den Bereichen des maschinellen Lernens und der künstlichen Intelligenz, wie etwa künstliche neuronale Netze, Frequent Pattern Mining und die auf formaler Logik basierenden Expertensysteme. Aika basiert auf der heute gängigen Architektur eines künstlichen neuronalen Netzwerks (KNN) und nutzt diese, um sprachliche Regeln und semantische Beziehungen abzubilden.
## Die Knackpunkte: relationale Struktur und zyklische Abhängigkeiten
Das erste Problem: Texte haben eine von Grund auf relationale Struktur. Die einzelnen Worte stehen über ihre Reihenfolge in einer ganz bestimmten Beziehung zueinander. Gängige Methoden, um Texte für die Eingabe in ein KNN auszuflachen, sind beispielsweise Bag-of-Words oder Sliding-Window. Mittlerweile haben sich auch rekurrente neuronale Netze etabliert, die das gesamte Netz in einer Schleife für jedes Wort des Textes mehrfach hintereinander schalten. Aika geht hier allerdings einen anderen Weg. Aika propagiert die relationalen Informationen, also den Textbereich und die Wortposition, gemeinsam mit den Aktivierungen durch das Netzwerk. Die gesamte relationale Struktur des Textes bleibt also erhalten und lässt sich jederzeit zur weiteren Verarbeitung nutzen.
Das zweite Problem ist, dass bei der Verarbeitung von Text häufig nicht klar ist, in welcher Reihenfolge einzelne Informationen verarbeitet werden müssen. Wenn wir beispielsweise den Namen „August Schneider“ betrachten, können sowohl der Vor- als auch der Nachname in einem anderen Zusammenhang eine völlig andere Bedeutung annehmen. August könnte sich auch auf den Monat beziehen. Und genauso könnte Schneider eben auch den Beruf des Schneiders meinen. Einfache Regeln, um hier dennoch den Vor- und den Nachnamen zu erkennen, wären: „Wenn das nachfolgende Wort ein Nachname ist, handelt es sich bei August um einen Vornamen“ und „Wenn das vorherige Wort ein Vorname ist, dann handelt es sich bei Schneider um einen Nachnamen“. Das Problem dabei ist nur, dass unsere Regeln nun eine zyklische Abhängigkeit beinhalten. Aber ist das wirklich so schlimm? Aika erlaubt es, genau solche Feedback-Schleifen abzubilden. Wobei die Schleifen sowohl positive, als auch negative Gewichte haben können. Negative rekurrente Synapsen führen dazu, dass zwei sich gegenseitig ausschließende Interpretationen entstehen. Der Trick ist nun zunächst nur Annahmen zu treffen, also etwa dass es sich bei dem Wort „Schneider“ um den Beruf handelt und zu schauen wie das Netzwerk auf diese Annahme reagiert. Es bedarf also einer Evaluationsfunktion und einer Suche, die die Annahmen immer weiter variiert, bis schließlich eine optimale Interpretation des Textes gefunden ist. Genau wie schon der Textbereich und die Wortposition werden nun auch die Annahmen gemeinsam mit den Aktivierungen durch das Netzwerk propagiert.
## Die zwei Ebenen des Aika Algorithmus
Aber wie lassen sich diese Informationen mit den Aktivierungen durch das Netzwerk propagieren, wo doch der Aktivierungswert eines Neurons für gewöhnlich nur eine Fließkommazahl ist? Genau hier liegt der Grund, weshalb Aika unter der neuronalen Ebene mit ihren Neuronen und kontinuierlich gewichteten Synapsen noch eine diskrete Ebene besitzt, in der es eine Darstellung aller Neuronen in boolscher Logik gibt. Aika verwendet als Aktivierungsfunktion die obere Hälfte der Tanh-Funktion. Alle negativen Werte werden auf 0 gesetzt und führen zu keiner Aktivierung des Neurons. Es gibt also einen klaren Schwellenwert, der zwischen aktiven und inaktiven Neuronen unterscheidet. Anhand dieses Schwellenwertes lassen sich die Gewichte der einzelnen Synapsen in boolsche Logik übersetzen und entlang der Gatter dieser Logik kann nun ein Aktivierungsobjekt mit den Informationen durch das Netzwerk propagiert werden. So verbindet Aika seine diskrete bzw. symbolische Ebene mit seiner subsymbolischen Ebene aus kontinuierlichen Synapsen-Gewichten.
Die Logik Ebene in Aika erlaubt außerdem einen enormen Effizienzgewinn im Vergleich zu einem herkömmlichen KNN, da die gewichtete Summe von Neuronen nur noch für solche Neuronen berechnet werden muss, die vorher durch die Logikebene aktiviert wurden. Im Falle eines UND-verknüpfenden Neurons bedeutet das, dass das Aktivierungsobjekt zunächst mehrere Ebenen einer Lattice-Datenstruktur aus UND-Knoten durchlaufen muss, bevor das eigentliche Neuron berechnet und aktiviert werden kann. Diese Lattice-Datenstruktur stammt aus dem Bereich des Frequent Pattern Mining und enthält in einem gerichteten azyklischen Graphen alle Teilmuster eines beliebigen größeren Musters. Ein solches Frequent Pattern Lattice kann in zwei Richtungen betrieben werden. Zum Einen können damit bereits bekannte Muster gematcht werden, und zum Anderen können auch völlig neue Muster damit erzeugt werden.
Da es schwierig ist Netze mit Millionen von Neuronen im Speicher zu halten, nutzt Aika das Provider Architekturpattern um selten verwendete Neuronen oder Logikknoten in einen externen Datenspeicher (z.B. eine Mongo DB) auszulagern, und bei Bedarf nachzuladen.
## Ein Beispielneuron
Hier soll nun noch beispielhaft gezeigt werden wie ein Neuron innerhalb des semantischen Netzes angelegt werden kann. Zu beachten ist, dass Neuronen sowohl UND- als auch ODER-Verknüpfungen abbilden können. Das Verhalten hängt dabei alleine vom gewählten Bias ab. Liegt der Bias bei 0.0 oder einem nur schwach negativen Wert reicht schon die Aktivierung eines positiven Inputs aus um auch das aktuelle Neuron zu aktivieren. Es handelt sich dann um eine ODER-Verknüpfung. Liegt der Bias hingegen tiefer im negativen Bereich dann müssen mitunter mehrere positive Inputs gleichzeitig aktiviert werden damit das aktuelle Neuron dann auch aktiv wird. Jetzt handelt es sich dann um eine UND-Verknüpfung. Der Bias Wert kann der initNeuron einfach als Parameter übergeben werden. Um jedoch die Berechnung des Bias zu erleichtern bietet Aika bei den Inputs noch den Parameter BiasDelta an. Der Parameter BiasDelta nimmt einen Wert zwischen 0.0 und 1.0 entgegen. Bei 0.0 wirkt sich der Parameter gar nicht aus. Bei einem höheren Wert hingegen wird er mit dem Betrag des Synapsengewichts multipliziert und von dem Bias abgezogen. Der Gesamtbias lautet in diesem Beispiel also -55.0. Die beiden positiven Eingabesynapsen müssen also aktiviert werden und die negative Eingabesynapse darf nicht aktiviert werden, damit dieses Neuron selber aktiv werden kann. Das Zusammenspiel von Bias und Synpasengewichten ist aber nicht nur für die Aktivierung eines Neurons wichtig, sondern auch für die spätere Auswahl der finalen Interpretation. Je stärker die Aktivierungen innerhalb einer Interpretation aktiv sind, desto höher wird diese Interpretation gewichtet.
Um eine beliebige Graphstruktur abbilden zu können, trennt Aika das Anlegen der Neuronen von der Verknüpfung mit anderen Neuronen. Mit createNeuron(“E-Schneider (Nachname)”) wird also zunächst einmal ein unverknüpftes Neuron erzeugt, das dann über die initNeuron Funktion mit den Eingabeneuronen wortSchneiderNeuron, kategorieVornameNeuron und unterdrueckendesNeuron verknüpft wird. Über den Parameter RelativeRid wird hier angegeben auf welche relative Wortposition sich die Eingabesynapse bezieht. Die Eingabesynpase zu der Kategorie Vorname bezieht sich also mit -1 auf die vorherige Wortposition. Der Parameter Recurrent gibt an ob es sich bei dieser Synpase um eine Feedback-Schleife handelt. Über den Parameter RangeMatch wird angegeben wie sich der Textbereich, also die Start- und die Endposition zwischen der Eingabe- und der Ausgabeaktivierung verhält. Bei EQUALS sollen die Bereiche also genau übereinstimmen, bei CONTAINED_IN reicht es hingegen wenn der Bereich der Eingabeaktivierung innerhalb des Bereichs der Ausgabeaktivierung liegt. Dann kann noch über den Parameter RangeOutput angegeben werden, dass der Bereich der Eingabeaktivierung an die Ausgabeaktivierung weiterpropagiert werden soll.
## Fazit
Mit Aika können sehr flexibel umfangreiche semantische Modelle erzeugt und verarbeitet werden. Aus Begriffslisten verschiedener Kategorien, wie etwa: Vor- und Nachnamen, Orten, Berufen, Strassen, grammatikalischen Worttypen usw. können automatisch Neuronen generiert werden. Diese können dann dazu genutzt werden, Worte und Phrasen zu erkennen, einzelnen Begriffen eine Bedeutung zuzuordnen oder die Kategorie eines Begriffs zu bestimmen. Falls in dem zu verarbeitenden Text mehrdeutige Begriffe oder Phrasen auftauchen, kann Aika für diese jeweils eigene Interpretationen erzeugen und gewichten. Die sinnvollste Interpretation wird dann als Ergebnis zurück geliefert. | 13,102 | 52,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.952564 |
http://algebra-help.com/algebra-help-factor/angle-complements/long-division-explained.html | 1,519,373,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814538.66/warc/CC-MAIN-20180223075134-20180223095134-00268.warc.gz | 14,841,146 | 6,909 | # Algebrator can start solving your homework in the next 5 minutes
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• solving solutions online calulator | 1,073 | 4,700 | {"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 | latest | en | 0.849297 |
https://gsebsolutions.in/gseb-solutions-class-8-science-chapter-13/ | 1,719,280,206,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00416.warc.gz | 245,124,003 | 53,248 | # GSEB Solutions Class 8 Science Chapter 13 Sound
Gujarat BoardĀ GSEB Textbook Solutions Class 8 Science Chapter 13 Sound Textbook Questions and Answers, Notes Pdf.
## Gujarat Board Textbook Solutions Class 8 Science Chapter 13 Sound
### Gujarat Board Class 8 Science Sound Textbook Questions and Answers
Choose the correct answer
Question 1.
Sound can travel through:
(a) gases only
(b) solids only
(c) liquids only
(d) solids, liquids, and gases.
Answer:
(d) solids, liquids, and gases.
Question 2.
Voice of which of the following is likely to have a minimum frequency?
(a) Baby girl
(b) Baby boy
(c) A man
(d) A woman
Answer:
(a) Baby girl.
Question 3.
In the following statements, tick āTā against those which are true, and āFā against those which are false:
(a) Sound cannot travel in a vacuum.
(b) The number of oscillations per second of a vibrating object is called its time period.
(c) If the amplitude of vibration is large, the sound is feeble.
(d) For human ears, the audible range is 20 Hz to 20,000 Hz.
(e) The lower the frequency of vibration, the higher is the pitch.
(f) Unwanted or unpleasant sound is termed as music.
(g) Noise pollution may cause partial hearing impairment.
Answer:
(a) True
(b) False
(c) False
(b) True
(e) False
(f)False
(g) True.
Question 4.
in the blanks with suitable words:
(a) Time taken by an object to complete one oscillation is called ………
(b) Loudness is determined by the ………… of vibration.
(c) The unit of frequency is ………..
(d) Unwanted sound is called ……….
(e) The shrillness of a sound is determined by the ……….. of vibration.
Answer:
(a) time period
(b) amplitude
(c) hertz
(d) noise
(e) frequency
Question 5.
A pendulum oscillates 40 times in 4 seconds. Find its time period and frequency.
Answer:
Number of oscillations = 40
Total time = 4 seconds
Time taken to complete one oscillation 4 1
= $$\frac {4}{40}$$ = $$\frac {1}{10}$$ = 0.1 second
So time period = 0.1 second.
Question 6.
The sound from a mosquito is produced when it vibrates its wings at an average rate of 500 vibrations per second, what is the time period of the vibration?
Answer:
Total vibrations = 500
Time taken = 1 second
Time taken to complete one vibration
= $$\frac {1}{500}$$ = 0.002 seconds.
So time period = 0.002 seconds.
Question 7.
Identify the part which vibrates to produce sound in the following instruments.
(a) Dholak
(b) Sitar
(c) Flute
Answer:
(a) Stretched membrane
(b) Strings
(c) Air column.
Question 8.
What is the difference between noise and music? Can music become noise sometimes?
Answer:
The sound which is unpleasant for our ears is called noise while music is the sound that is pleasant for our ears. Music becomes noise sometimes when it crosses the bearable range of sound for our ears.
Question 9.
List sources of noise pollution in your surroundings.
Answer:
Sources of noise pollution: Honking of horns, loudspeakers, loud sounds of machines in factories, loud sounds of T.V., radio, domestic appliances etc.
Question 10.
Explain in what way noise pollution is harmful to humans.
Answer:
Harmful effects of noise pollution:
• It causes deafness
• It causes mental illness
• It causes headache and high blood pressure
Question 11.
Your parents are going to buy a house. They have been offered one on the roadside and another three lanes away from the roadside. Which house would you suggest to your parents to buy? Explain your answer.
Answer:
I would suggest my parents buy the house three lanes away from the roadside. This house would safeguard our health and peace of mind.
Question 12.
Sketch larynx and explain its function in your own words.
Answer:
The larynx is also known as the voice box. It has vocal cords which have air column vibrating in them, which cause sound in humans.
Question 13.
Lightning and thunder take place in the sky at the same time and at the same distance from us. Lightning is seen earlier and thunder is heard later. Can you explain why?
Answer:
The speed of light is more than that of sound. Due to more speed, the light reaches us before the sound does. So lightning is seen earlier and thunder is heard later. | 1,024 | 4,130 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.867483 |
https://www.unitconverters.net/length/megaparsec-to-foot.htm | 1,725,778,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00167.warc.gz | 993,143,075 | 4,295 | Home / Length Conversion / Convert Megaparsec to Foot
# Convert Megaparsec to Foot
Please provide values below to convert megaparsec [Mpc] to foot [ft], or vice versa.
From: megaparsec To: foot
### Megaparsec to Foot Conversion Table
Megaparsec [Mpc]Foot [ft]
0.01 Mpc1.0123614111811E+21 ft
0.1 Mpc1.0123614111811E+22 ft
1 Mpc1.0123614111811E+23 ft
2 Mpc2.0247228223622E+23 ft
3 Mpc3.0370842335433E+23 ft
5 Mpc5.0618070559055E+23 ft
10 Mpc1.0123614111811E+24 ft
20 Mpc2.0247228223622E+24 ft
50 Mpc5.0618070559055E+24 ft
100 Mpc1.0123614111811E+25 ft
1000 Mpc1.0123614111811E+26 ft
### How to Convert Megaparsec to Foot
1 Mpc = 1.0123614111811E+23 ft
1 ft = 9.8778952749031E-24 Mpc
Example: convert 15 Mpc to ft:
15 Mpc = 15 × 1.0123614111811E+23 ft = 1.5185421167717E+24 ft | 335 | 782 | {"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 | 0.425848 |
https://socratic.org/questions/58cfd7cf11ef6b36bba3dffe | 1,708,727,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00494.warc.gz | 546,864,081 | 6,177 | # What energy is required to melt a block of ice at 0 ""^@C, and raise the temperature of the water to 10 ""^@C?
##### 1 Answer
Mar 26, 2017
You need to quote.........(i)
#### Explanation:
You need to quote.........$\text{(i) the latent heat of fusion of ice}$, and this for the transition ice to water at $0$ ""^@C....
And $\text{(ii) the specific heat of water.........}$
Of course, we could look these up ourselves, but why should we if you can't be bothered? | 130 | 468 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.891724 |
https://ultimatescholar.com/2022/09/07/university-of-phoenix-comparing-means-worksheet-john-is-interested-in-determining-if-frequency-of-exercise-affects-pulse-rate-john-randomly-samples-indivi/ | 1,669,875,364,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00605.warc.gz | 625,654,840 | 12,078 | # University of Phoenix Comparing Means Worksheet John is interested in determining if frequency of exercise affects pulse rate. John randomly samples indivi
University of Phoenix Comparing Means Worksheet John is interested in determining if frequency of exercise affects pulse rate. John randomly samples individuals at a local gym to ask if they will participate in his study. 55 individuals agree, and they are divided into three groups of exercisers: 1 = high frequency, 2 = moderate frequency, 3 = low frequency. Next, John measures their pulse after their workout. Do pulse rates differ among individuals who exercise with high frequency versus those who exercise moderately versus those who exercise with low frequency?
Note: Make sure you adjust the data so that there are three exercise variables (one for each level).
Conduct a one-way ANOVA to investigate the relationship between exercise frequency (Exercise) and pulse rate (Pulse1). Conduct Tukey’s HSD post hoc tests, if necessary. On the output, identify the following:
a. F ratio for the group effect
b. Sums of squares for the exercise effect
c. Mean for moderate exercisers
d. P value for exercise effect Height Weight Age Gender Smokes Alcohol Exercise Ran Pulse1 Pulse2
173
57 18
2
2
1
2
2
86
88
179
58 19
2
2
1
2
1
82
150
167
62 18
2
2
1
1
1
96
176
195
84 18
1
2
1
1
2
71
73
173
64 18
2
2
1
3
2
90
88
184
74 22
1
2
1
3
1
78
141
162
57 20
2
2
1
2
2
68
72
169
55 18
2
2
1
2
2
71
77
164
56 19
2
2
1
1
2
68
68
168
60 23
1
2
1
2
1
88
150
170
75 20
1
2
1
1
1
76
88
178
58 19
1
2
2
3
2
74
76
170
68 22
1
1
1
2
2
70
71
187
59 18
1
2
1
1
2
78
82
180
72 18
1
2
1
2
2
69
67
185
110 22
1
2
1
3
2
77
73
170
56 19
1
2
2
3
2
64
63
180
70 18
1
2
1
2
1
80
146
166
56 21
2
1
2
2
2
83
79
155
50 19
2
2
2
2
2
78
79
175
60 19
1
2
2
3
2
88
86
140
50 34
2
2
2
3
1
70
98
163
55 20
2
2
2
3
2
78
74
182
75 26
1
1
1
2
2
80
76
176
59 19
1
2
2
2
2
68
69
177
74 18
2
2
2
2
1
70
96
170
60 18
1
2
1
2
2
62
59
172
60 21
2
2
2
3
2
81
79
189
60 19
1
2
1
2
1
78
168
178
56 21
2
2
1
2
1
86
150
175
75 20
1
2
1
2
1
59
92
180
85 19
1
1
1
2
1
68
125
160
57 19
2
2
2
2
1
75
130
164
66 23
2
2
2
3
1
74
168
175
65 19
1
2
1
2
1
60
104
163
55 20
2
2
2
2
1
70
119
185
90 18
1
2
2
3
1
80
140
169
68 19
1
2
2
2
2
58
58
165
63 18
2
2
1
2
2
84
84
155
49 18
2
2
1
2
2
104
92
175
66 20
1
2
1
2
2
66
68
178
63 23
1
2
1
3
2
84
90
184
65 21
1
1
2
2
2
65
67
170
60 19
2
2
1
2
2
80
80
162
60 19
2
2
1
2
2
66
60
164
46 18
2
2
2
2
2
104
96
171
182
174
167
157
183
167
171
182
70
85
60
70
41
73
75
67
63
26
20
19
22
20
20
20
18
20
2
1
2
1
2
1
2
2
1
2
1
2
1
2
2
2
2
2
2
1
1
1
2
1
1
1
1
2
3
3
3
2
2
2
3
1
2
2
1
2
1
2
2
2
1
76
70
66
92
70
63
65
76
56
76
68
89
84
95
65
67
74
110
Variable
Height
Weight
Age
Gender
Smokes
Alcohol
Exercise
Description
Height (cm)
Weight (kg)
Age (years)
Sex (1 = male, 2 = female)
Regular smoker? (1 = yes, 2 = no)
Regular drinker? (1 = yes, 2 = no)
Frequency of exercise (1 = high, 2 = moderate, 3 = low)
Whether the student ran or sat between the first and
Ran
second pulse measurements (1 = ran, 2 = sat)
Pulse1 First pulse measurement (rate per minute)
Pulse2 Second pulse measurement (rate per minute)
Comparing Means Worksheet
PSYCH/625 Version 5
University of Phoenix Material
Comparing Means Worksheet
The f ollowing questions require that you access Microsof t ® Excel® for analysis. The team must work
together to solve these questions. The data set f or this assignment is located in the Pulse Rate Dataset.
1.
John is interested in determining if f requency of exercise af f ects pulse rate. John randomly samples
individuals at a local gym to ask if they will participate in his study. 55 individuals agree, and they are
divided into three groups of exercisers: 1 = high f requency, 2 = moderate f requency, 3 = low
f requency. Next, John measures their pulse af ter their workout. Do pulse rates dif f er among
individuals who exercise with high f requency versus those who exercise moderately versus those
who exercise with low f requency?
Note: Make sure you adjust the data so that there are three exercise variables (one f or each level).
Conduct a one-way ANOVA to investigate the relationship between exercise f requency (Exercise)
and pulse rate (Pulse1). Conduct Tukey’s HSD post hoc tests, if necessary. On the output, identify
the f ollowing:
a. F ratio f or the group ef f ect
b. Sums of squares f or the exercise ef f ect
c. Mean f or moderate exercisers
d. P value f or exercise ef f ect
2.
Write a 125- to 175-word summary of the results, f ormatted according to APA guidelines. | 1,831 | 4,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-49 | longest | en | 0.713123 |
https://www.realestatelicense.org/viewtopic.php?id=2c6aaa-horizontal-component-of-earth%27s-magnetic-field | 1,656,163,990,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00651.warc.gz | 1,020,296,777 | 6,008 | For example magnetic north drifts gradually over the … fl Secular variations are slow changes in the Earth’s magnetic ßeld with time. The horizontal component of the Earth’s magnetic field B h . fond the resultant intensity of earth - 17766855 Electric currents in the Earth. Causes of the Earth’s magnetism: Some important factors which may be the cause of Earth’s magnetism are: Magnetic masses in the Earth. Electric currents in the upper regions of the atmosphere. At a certain location, the horizontal component of the earth’s magnetic field is , due north. The horizontal component of earth’s magnetic field at place is 0.36× 10-4 weber/m². If the angle of dip at that placed is 60° then the value of vertical component of earth’s magnetic field will be (in wb/m²) These are explained on the next page. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. In temperate northern latitudes, this field is… THE EARTH’S MAGNETIC FIELD (2) General features fl The Earth’s magnetic ßeld can be mapped by means of isomagnetic charts. Physics 121 Lab: Finding the horizontal component of the magnetic field of the Earth Introduction The Earth’s magnetic field exhibits changes with time and is not uniform across the surface of the earth. It is the component of the earths magnet field which aligns the compass needle to the Magnetic North Pole. Find the speed of the proton I know this one is so simple but I still cannot figure it out. horizontal and vertical components of earth is magnetic field at a place are 0.22 Tesla and 0.38 tesla respectively. Let, Total magnetic field is B and, B H be the horizontal component of the magnetic field.. At given place, Horizontal component of magnetic field is given by, B H = B cos60 ∴ B = 2B H At equator dip angle = 0 0 ∴ B’ H = I cos0 = I = 2B H It is not completely clear how the earth’s magnetic field is produced. Solution for Earth's magnetic field near the ground is typically 0.50 G (0.50 gauss), where 1G = 10 4 tesla. Radiations from the Sun. | 480 | 2,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-27 | latest | en | 0.863228 |
http://www.gradesaver.com/textbooks/math/geometry/elementary-geometry-for-college-students-5th-edition/chapter-1-section-1-5-introduction-to-geometric-proof-exercises-page-44/14 | 1,481,452,602,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544672.33/warc/CC-MAIN-20161202170904-00330-ip-10-31-129-80.ec2.internal.warc.gz | 496,782,138 | 40,162 | ## Elementary Geometry for College Students (5th Edition)
Published by Brooks Cole
# Chapter 1 - Section 1.5 - Introduction to Geometric Proof - Exercises: 14
#### Answer
m$\angle$1=m$\angle$2
#### Work Step by Step
This conclusion can be made using the definition an angle bisector
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 105 | 449 | {"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} | 2.640625 | 3 | CC-MAIN-2016-50 | longest | en | 0.827794 |
https://questioncove.com/updates/4f5fbb33e4b038e07180526b | 1,638,347,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00635.warc.gz | 550,341,066 | 4,530 | Mathematics
OpenStudy (anonymous):
Mr. Hauseman has 17 students in his class, three of which are freshman and the others are sophomores and juniors. He is going to randomly draw two students to be partners. He calculates the probablility of drawing a junior and a freshman to be 9136. How many juniors are in the class? A.) 6 b.)7 c.)8 d.)9 | 91 | 341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.956806 |
https://support.khanacademy.org/hc/pt-br/profiles/398579337272-gunank312 | 1,606,213,218,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00629.warc.gz | 511,999,205 | 10,885 | Como podemos ajudar?
# gunank312
• gunank312 criou uma publicação,
When one is using khan academy, it is conceived to be in “elementary” level or something, if there is to be more content on the related topics, for example, some trigonometric proofs are not “legit...
• gunank312 comentou,
• gunank312 criou uma publicação,
### linear approximation of a rational function
hello, i have question on : https://www.khanacademy.org/video/linear-approximation-example?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI_CxIIVXNlckRhdGEiHGthaWRfNTE0MDMyNDAyODgyNDMwNTM3NzY2MzYMCxIIRmVlZGJh...
• gunank312 criou uma publicação,
### product rule, squeeze theorom
hello, in the calculus topic, in proving the product rule, at 1:46, why is (f(x)g(x))=(f(x+h)g(x+h)-f(x)g(x))/h? i am a little confused here, as wouldn't the "equation" change, as its not exactly a...
• gunank312 comentou,
I got my doubt clarified, denominator to becomes x plus delta x minus x giving delta x... Thanks
• gunank312 criou uma publicação,
### Question in a video
For the proof of derivative of ln(x) is 1/x, at 00:40, why is the denominator simply delta x and not x? | 348 | 1,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.816863 |
https://tutorbin.com/subject/catia-homework-help | 1,708,945,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00445.warc.gz | 587,209,207 | 17,078 | # catia homework help
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• Q1:1. Applications - CATIA Exercises: Exercise 1: From page 3.10-4 b) Ø16.5 Ø21.5 118 Exercise 3.2-b REQUIREMENTS: Create a 3D model Save as: 34 Ta Tex TH 54.5 。 10 4 115 Ho 25 30° FORES jar 143 R6 FT Til Lin 30 -38- Ø10 LØ5 75 2X M 10 x 1.5 Exercise 3.2-b.CATDrawing YOU WILL HAVE TO UPLOAD 2 CAD FILES (CATPart and .CATDrawing) 45 19 80 Exercise 3.2-b.CATPart Create a detailed drawing of the part Include all required views, as shown. ➤ Assign the specified dimensions only. Add border and title block (use CATIA template #2, and modify it as shown during the lecture). Save as:See Answer
• Q2:Exercise 2: REQUIREMENTS: Create a 3D model > Save as: Exercise 2-Block.CATPart Create a detailed drawing the part shown > Units: millimeters ➤ Respect proportions ➤ Include only the necessary views: 2 standard views (front and top), and auxiliary views to show the true size and shape of the oblique surfaces. Do not dimension Add border and title block (use CATIA template #2, and modify it as shown during the lecture). Save as: Exercise 2-Block.CATDrawing YOU WILL HAVE TO UPLOAD 2 CAD FILES (CATPart and .CATDrawing) ➤See Answer
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## TutorBin helping students around the globe
TutorBin believes that distance should never be a barrier to learning. Over 500000+ orders and 100000+ happy customers explain TutorBin has become the name that keeps learning fun in the UK, USA, Canada, Australia, Singapore, and UAE. | 692 | 2,860 | {"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-2024-10 | latest | en | 0.830303 |
http://garsia.math.yorku.ca/~nantel/teaching/Old/index_Algebra6121_2017.html | 1,516,617,097,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00767.warc.gz | 138,312,260 | 3,804 | # Graduate Level Applied Algebra (Math 6121)
This is a new course created in 2015.
## Nantel Bergeron
tel: 736-2100 Ext 33968
e-mail: bergeron@mathstat.yorku.ca
Office: 2029 TEL BUILDING.
## Books
Algebra, T. W. Hungerford, GTM Springer. (recommended but not required)
Abstract Algebra, Dummit and Foote, Willey. (Highly recommended but not required)
An Introduction to Computational Algebraic Geometry and Commutative Algebra, D. A. Cox, J. Little and D. O'shea, UTM Springer (Highly recommended but not required)
## I plan as follow:
I added some reference DF=Dummit and foote(third edition), and S=Sagan(second Edition). This could be useful if you want to work on some exercises. Just pick some in those chapter. Remember that it is important to work the math in order to learn it.
• Introduction Why applied algebra
• Linear Algebra (Recall crash course, Graduate level): [DF Chapter 11] (Sept 7,12,14)
• THM For any fin. Gen. vector space V (over C)
• V has a (ordered) basis B
• dim(V) = |B| =n is well defined
• L: V ----> C^n where L(v)=[v] is an isomorphism
• THM For any linear transformation T: V ---> W, and fixed bases in V and W, There is a unique matrix M=[T] such that LoT=MoL
• All questions about T can be answered using algorithms on matrix [T]
• End(V) = Mat(nxn) and Aut(V)=Gl(n)
• Direct sums and tensor products have corresponding operations on bases and linear transformations.
• Group Theory and representation Theory [DF Chapter 1,2,3,4,5,6]
• Recall : Groups, morphisms, subgroups, G-sets (and G-morphisms), Isomorphism Theorems and quotient groups. (Sept 19, 21, 26)
• Jordan-Holder Theorem (Sept 28,Oct 3)
• Sylow Theorem (Oct 5,10,12)
• Representation of finite groups and characters (over C) (Oct 17,19)
• Midterm (Oct 24): include Linear algebra, Basic groups theory, Jordan-Holder, Sylow Theorem;
Basic Representation Theory: definition of G-module, Representation, invariant subspace (G-submodule), G-morphism.
Lots of my presentation is out of the book The Symmetric Group , Bruce Sagan, Springer GTM 203, (2001). [S Chapter 1]
• Maske's Theorem
• Schur's lemma
• Structure of the space of G-endomorphisms
• Structure of the inner space of characters on G
• THM the number of irreducible representations for G equal the number of conjugacy classes of G
• Preliminary notions in ring [DF Chapter 7,8,9]
• Euclidian domain
• Principal ideal domain
• Unique Factorization domain
• Polynomial rings
• Grobner basis with emphasis on algorithmic aspect and computational geometry
• solving polynomial system of equations (with some application to robotics and computational geometry)
--------------------------------------------------------------------------------------------------
--- If time allows, Module over PID... it is cover in more details in Math 6122.
--------------------------------------------------------------------------------------------------
• Modules over PID (Advanced linear algebra) [DF Chapter 10,12]
• Chinese Remainder Theorem
• Classification of finitely generated modules over PID
• Classification of finitely generated abelian groups
• rational canonical form
• Jordan canonical form
## Proposed Presentation, Project and Applications:
My list is not inclusive and is just suggestions
• Pick a Theorem above and present it in class (let me know in advance to coordinate when to present it)
• The book Combinatorial Species and Tree-like Structures, Encyclopedia of Math. , Cambridge Univ. Press, (1998), contain many possible Presentation, Project and Applications related to action of groups:
• Polya Theory
• Combinatorial Enumeration
• Species
• and more ...
• The Book Group representations in probability and statistics , Persi Diaconis, Institute of Math. Stat. Lecture Notes, Vol. 11 (1988), contain many possible Presentation, Project and Applications related to representation of groups:
• Discrete Fourrier Transform
• Markov Chain
• Sampling in groups
• and more ...
• The book The Symmetric Group , Bruce Sagan, Springer GTM 203, (2001), contain many possible Presentation, Project and Applications related representation theory and the symmetric group:
• program the Young Natural Representation
• Present symmetric functions
• Introduce Robinson-Schensted algorithm and its consequences (What was the original application of Schensted?)
• and more ...
## Evaluation:
Students will be evaluated on five aspects (which are parts of the life of any living mathematician). The final grade will be base on the average of the best three.
• 1 Project/Homework (working on an extended project or working on exercises)
• 2 Midterms (Writing exams).
• 3 Oral Presentation (Presenting some special topic or long proofs).
• 4 Comprehensive exam (writing the comprehensive exam at the end, Note that for some of you it is one of your Ph. D. requirements).
5 Participation in class (Being there, asking questions, being curious, etc.) is ALSO an important aspect of the evaluation. It may help increase any of your average above by up to 10%.
Nantel Bergeron
Office: 2029 TEL Building
tel: 416-736-2100 x 33968
email address: bergeron at yorku dot ca
Department of Mathematics and Statistics.
2029 TEL Building
York University
North York, Ontario M3J 1P3, Canada
To Department's Public Page
last revised Sept. 2015 | 1,268 | 5,289 | {"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-2018-05 | latest | en | 0.84895 |
https://programmer.help/blogs/learning-diary-of-slam-beginners.html | 1,581,978,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00527.warc.gz | 548,054,818 | 11,562 | # Learning diary of slam beginners
Learning SLAM for the first time, because one note cannot be used in ubuntu, I decided to use CSDN to record my learning notes. The purpose of publishing is to share and exchange progress with you. Most of the notes come from the experience of the big guys. I will also attach a link. If there is infringement, please let me know.
The main steps to create g2o are as follows:
## 1. Create a linear solver
The form of the incremental equation we require is: H △ X=-b. generally, the method we think of is to directly inverse, that is, △ X=-H^-1*b. But when the dimension of H is large, it is very difficult to invert the matrix, so we need some special methods to invert the matrix.
Linear solver cholmod: use sparse cholesky decomposition method. Inherited from LinearSolverCCS
LinearSolverCSparse: use CSparse method. Inherited from LinearSolverCCS
LinearSolverPCG: using the preconditioned reconcile gradient method, inherited from LinearSolver
LinearSolverDense: use the deny Cholesky decomposition method. Inherited from LinearSolver
LinearSolverEigen: the dependency is only eigen, which is solved by sparse Cholesky in eigen. The performance is similar to that of CSparse. Inherited from LinearSolver
## 2. Create BlockSolver. It is initialized with the linear solver defined above.
The BlockSolver contains the LinearSolver internally, which is initialized with the LinearSolver defined above. It is defined in the following folder:
```g2o/g2o/core/block_solver.h
```
BlockSolver can be defined in two ways
1. One is the solver of the specified fixed variable, which is defined as follows
``` using BlockSolverPL = BlockSolver< BlockSolverTraits<p, l> >;
```
Where p represents the dimension of post (note that it must be the minimum representation under manifold), and l represents the dimension of landmark
2. The other is variable size solver, which is defined as follows
```using BlockSolverX = BlockSolverPL<Eigen::Dynamic, Eigen::Dynamic>;
```
At the end of block ﹣ solver. H, several commonly used types are predefined, as follows:
```Block solver ﹣ 6 ﹣ 3: indicates that the post is 6-dimensional and the observation point is 3-dimensional. BA for 3D SLAM
Blocksolver? 7? 3: there is an additional scale based on blocksolver? 6? 3
Block solver ﹣ 3 ﹣ 2: indicates that post is 3D and observation point is 2D
```
## 3. Create the total solver. And select one from GN, LM and dogleg, and then initialize it with BlockSolver
In the directory of g2o/g2o/core /, there are three optimization methods for Solver: Gauss Newton method, LM (Levenberg Marquardt) method and Dogleg method, as shown in the figure below, which also match the previous figureThe GN, LM and Doglet algorithms inherit from the same class: OptimizationWithHessian.
Optimization algorithm with Hessian, in turn, inherits from optimization algorithm.
In short, at this stage, we can choose three methods:
```g2o::OptimizationAlgorithmGaussNewton
g2o::OptimizationAlgorithmLevenberg
g2o::OptimizationAlgorithmDogleg
```
## 4. Create the ultimate large boss sparse optimizer and use the defined solver as the solution method.
Create sparse optimizer
```g2o::SparseOptimizer optimizer;
```
Use the solver defined above as the solution method:
```SparseOptimizer::setAlgorithm(OptimizationAlgorithm* algorithm)
```
setVerbose is used to set the output information of optimization process
```SparseOptimizer::setVerbose(bool verbose)
```
## 5. Defines the vertices and edges of a graph. And add to SparseOptimizer
First, let's look at the ① class related to vertex in the above figure: HyperGraph::Vertex, which is in the path of
```g2o/core/hyper_graph.h
```
This HyperGraph::Vertex is an abstract vertex and must be used by derivation. As shown in the figure below
Then we look at the ② class in the G 2O class structure diagram. We see that HyperGraph::Vertex is inherited by the class optimizeablegraph, which is defined in
```g2o/core/optimizable_graph.h
```
We found the vertex definition and found that, as expected, the optimizeablegraph inherits from HyperGraph, as shown in the following figure
However, this OptimizableGraph::Vertex is also very low-level, and it will be extended when it is used. Therefore, a general template suitable for most situations is provided in g2o. It is the ③ class corresponding to the structure diagram of G 2O class
BaseVertex<D, T>
Its path is:
```g2o/core/base_vertex.h
```
D is of type int, representing the minimum dimension of vertex. For example, if rotation in 3D space is 3-dimensional, then D = 3
T is the data type of the vertex to be estimated. For example, if the Quaternion is used to express three-dimensional rotation, t is the Quaternion type
How do I define vertices myself?
Some commonly used vertex types are defined in g2o itself:
```VertexSE2 : public BaseVertex<3, SE2> //2D pose Vertex, (x,y,theta)
VertexSE3 : public BaseVertex<6, Isometry3> //6d vector (x,y,z,qx,qy,qz) (note that we leave out the w part of the quaternion)
VertexPointXY : public BaseVertex<2, Vector2>
VertexPointXYZ : public BaseVertex<3, Vector3>
VertexSBAPointXYZ : public BaseVertex<3, Vector3>
// Internal and external parameterization and exponential mapping of Se 3 vertices
VertexSE3Expmap : public BaseVertex<6, SE3Quat>
// SBACam Vertex, (x,y,z,qw,qx,qy,qz),(x,y,z,qx,qy,qz) (note that we leave out the w part of the quaternion.
// Suppose qw is positive, otherwise there is an ambiguous rotation in qx, qy, qz
VertexCam : public BaseVertex<6, SBACam>
// Sim3 Vertex, (x,y,z,qw,qx,qy,qz),7d vector,(x,y,z,qx,qy,qz) (note that we leave out the w part of the quaternion.
VertexSim3Expmap : public BaseVertex<7, Sim3>
```
Redefining a vertex generally requires rewriting the following functions:
```virtual bool read(std::istream& is);
virtual bool write(std::ostream& os) const;
virtual void oplusImpl(const number_t* update);
virtual void setToOriginImpl();
```
Read, write: read disk and save disk functions respectively. In general, if there is no need to perform read / write operation, just declare it
setToOriginImpl: vertex reset function, which sets the original value of the optimized variable.
oplusImpl: vertex update function. A very important function, mainly used for the calculation of delta △ x in the optimization process. After we calculate the increment according to the increment equation, we adjust the estimated value through this function, so we must pay attention to the content of this function.
Let's take a simple example. It comes from the curve fitting in Lecture 14. The sources are as follows
```ch6/g2o_curve_fitting/main.cpp
```
//Vertex and template parameters of curve model: optimize variable dimension and data type
```class CurveFittingVertex: public g2o::BaseVertex<3, Eigen::Vector3d>
{
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW
virtual void setToOriginImpl() // Reset
{
_estimate << 0,0,0;
}
virtual void oplusImpl( const double* update ) // To update
{
_estimate += Eigen::Vector3d(update);
}
// Save and read: leave blank
virtual bool read( istream& in ) {}
virtual bool write( ostream& out ) const {}
};
```
The initial value of the vertex in the code is set to 0. When updating, the update amount is added directly because of X + △ x (because the vertex type is Eigen::Vector3d, which belongs to vector, and can be updated by adding)
But some examples don't work, such as the following complex point example: Lie algebra represents vertex se3expmap
From g2o official website
```g2o/types/sba/types_six_dof_expmap.h
```
```//brief SE3 Vertex parameterized internally with a transformation matrix and externally with its exponential map
class G2O_TYPES_SBA_API VertexSE3Expmap : public BaseVertex<6, SE3Quat>{
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW
VertexSE3Expmap();
bool write(std::ostream& os) const;
virtual void setToOriginImpl() {
_estimate = SE3Quat();
}
virtual void oplusImpl(const number_t* update_) {
Eigen::Map<const Vector6> update(update_);
setEstimate(SE3Quat::exp(update)*estimate()); //Regeneration mode
}
};
```
The 6 and se3quat in this are:
The first parameter 6 represents the optimization variable dimension of internal storage, which is a six dimensional Lie algebra
The second parameter is the type of optimization variable, which uses the camera pose type defined by g2o: SE3Quat.
Here you can see the details of g2o/types/slam3d/se3quat.h
It uses quaternion to express rotation, and then adds displacement to store position and pose. At the same time, it supports operations on Lie algebra, such as log function, update function and so on
How do I add vertices to a graph?
It's easy to add vertices to the graph. Let's first look at the first curve fitting example, setEstimate(type) function to set the initial value; setId(int) defines the node number
``` // Add vertex to graph
CurveFittingVertex* v = new CurveFittingVertex();
v->setEstimate( Eigen::Vector3d(0,0,0) );
v->setId(0);
```
This is an example of adding vertex point XYZ, which is easy to understand
```/ch7/pose_estimation_3d2d.cpp
```
``` int index = 1;
for ( const Point3f p:points_3d ) // landmarks
{
g2o::VertexSBAPointXYZ* point = new g2o::VertexSBAPointXYZ();
point->setId ( index++ );
point->setEstimate ( Eigen::Vector3d ( p.x, p.y, p.z ) );
point->setMarginalized ( true );
}
```
A preliminary understanding of the edge of G 2O
g2o/g2o/core/hyper_graph.h
g2o/g2o/core/optimizable_graph.h
g2o/g2o/core/base_edge.h
BaseUnaryEdge, BaseBinaryEdge and BaseMultiEdge represent one, two and multiple edge respectively.
One edge can be understood as an edge connecting only one vertex, two edges can be understood as an edge connecting two vertices, and multiple edges can be understood as an edge connecting more than three vertices
Their main parameters are: D, E, VertexXi, VertexXj, respectively representing:
D is of type int, indicating the dimension of the measured value
E is the data type of the measurement
Vertex Xi and vertex XJ represent different vertex types respectively
For example, if we use an edge to represent the re projection error of a 3D point projected onto an image plane, we can set the input parameters as follows:
``` BaseBinaryEdge<2, Vector2D, VertexSBAPointXYZ, VertexSE3Expmap>
```
The first 2 is that the measurement value is 2D, that is, the difference of image pixel coordinates x and Y. the corresponding measurement value type is Vector2D, and the two vertices, that is, the optimization variables are three-dimensional vertex vertex sapointxyz and Lie group pose vertex sexe3expmap respectively
When defining an edge, we usually need to copy some important member functions. The edge mainly has the following important member functions
```virtual bool read(std::istream& is);
virtual bool write(std::ostream& os) const;
virtual void computeError();
virtual void linearizeOplus();
```
Here's a brief explanation
Read, write: read disk and save disk functions respectively. In general, if there is no need to perform read / write operation, just declare it
Computererror function: it is very important to use the value of the current vertex to calculate the error between the measured value and the actual measured value
linearizeOplus function: it is very important that under the current vertex value, the partial derivative of the error to the optimization variable, i.e. Jacobian
In addition to the above member functions, several important member variables and functions are also explained:
```_measurement: store observations
_Error: stores the error computed by the computeError() function
_Vertices []: for storing vertex information, such as binary edges, the size of [vertices [] is 2, and the storage order is related to the set int (0 or 1) when setVertex(int, vertex) is called
setId(int): to define the number of the edge (determines the position in the H matrix)
setMeasurement(type) function to define observations
setVertex(int, vertex) to define vertices
setInformation() to define the inverse of covariance matrix
```
How to customize the edges of g2o?
Basically, the edge in G 2O is defined as follows:
```class myEdge: public g2o::BaseBinaryEdge<errorDim, errorType, Vertex1Type, Vertex2Type>
{
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW
myEdge(){}
virtual bool write(ostream& out) const {}
virtual void computeError() override
{
// ...
_error = _measurement - Something;
}
virtual void linearizeOplus() override
{
_jacobianOplusXi(pos, pos) = something;
// ...
/*
_jocobianOplusXj(pos, pos) = something;
...
*/
}
private:
// data
}
```
We can find that the most important two functions are computeError(), linearizeOplus()
This is a unary side. It mainly defines the error function. As shown below, you can see that this example is basically a lost extension of the above example
```// Error model template parameters: observation dimension, type, connection vertex type
class CurveFittingEdge: public g2o::BaseUnaryEdge<1,double,CurveFittingVertex>
{
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW
CurveFittingEdge( double x ): BaseUnaryEdge(), _x(x) {}
// Calculation curve model error
void computeError()
{
const CurveFittingVertex* v = static_cast<const CurveFittingVertex*> (_vertices[0]);
const Eigen::Vector3d abc = v->estimate();
_error(0,0) = _measurement - std::exp( abc(0,0)*_x*_x + abc(1,0)*_x + abc(2,0) ) ;
}
virtual bool read( istream& in ) {}
virtual bool write( ostream& out ) const {}
public:
double _x; // x value, y value is "measurement"
};
```
Here's a complicated example: the PnP problem of 3D-2D points, which is to minimize the re projection error. This problem is very common. The most common binary edge is used, and the basic edge related code is almost all in one
```//Inheriting the BaseBinaryEdge class, the observation value is 2D, the type is Vector2D, and the vertices are 3D points and Lie group pose respectively
class G2O_TYPES_SBA_API EdgeProjectXYZ2UV : public BaseBinaryEdge<2, Vector2D, VertexSBAPointXYZ, VertexSE3Expmap>{
public:
EIGEN_MAKE_ALIGNED_OPERATOR_NEW;
//1. Default initialization
EdgeProjectXYZ2UV();
//2. Calculation error
void computeError() {
//Pose of Li group camera v1
const VertexSE3Expmap* v1 = static_cast<const VertexSE3Expmap*>(_vertices[1]);
// Vertex v2
const VertexSBAPointXYZ* v2 = static_cast<const VertexSBAPointXYZ*>(_vertices[0]);
//Camera parameters
const CameraParameters * cam
= static_cast<const CameraParameters *>(parameter(0));
//Error calculation, measured value minus estimated value, i.e. re projection error OBS cam
//The estimated value is calculated by T*p. the camera coordinates are obtained, and then the pixel coordinates are obtained by camera2pixel() function.
Vector2D obs(_measurement);
_error = obs-cam->cam_map(v1->estimate().map(v2->estimate()));
//Error = observation projection
}
//3. Calculation method of linear increment function, i.e. Jacobian matrix J
virtual void linearizeOplus();
//4. Camera parameters
CameraParameters * _cam;
bool write(std::ostream& os) const;
};
```
How do I add edges to a graph?
The following code is from GitHub, which is still an example of curve fitting
slambook/ch6/g2o_curve_fitting/main.cpp
```// Add edge to graph
for ( int i=0; i<N; i++ )
{
CurveFittingEdge* edge = new CurveFittingEdge( x_data[i] );
edge->setId(i);
edge->setVertex( 0, v ); // Set connected vertices
edge->setMeasurement( y_data[i] ); // Observational value
edge->setInformation( Eigen::Matrix<double,1,1>::Identity()*1/(w_sigma*w_sigma) ); // Information matrix: the inverse of covariance matrix
}
```
For this curve fitting, the observed value is the actual observed data point. For visual SLAM, it is usually the coordinate of feature points we observed. Here is an example. This example is a little more complicated than the one just mentioned, because it is a binary edge, and it needs to connect two vertices with an edge
Code from GitHub
slambook/ch7/pose_estimation_3d2d.cpp
```index = 1;
for ( const Point2f p:points_2d )
{
g2o::EdgeProjectXYZ2UV* edge = new g2o::EdgeProjectXYZ2UV();
edge->setId ( index );
edge->setVertex ( 0, dynamic_cast<g2o::VertexSBAPointXYZ*> ( optimizer.vertex ( index ) ) );
edge->setVertex ( 1, pose );
edge->setMeasurement ( Eigen::Vector2d ( p.x, p.y ) );
edge->setParameterId ( 0,0 );
edge->setInformation ( Eigen::Matrix2d::Identity() );
index++;
}
```
The p in the setMeasurement function here comes from the vector points 2D, that is, the image coordinates (x,y) of the feature points.
setVertex has two vertices of type 0 and vertexsapointxyz, and one is 1 and pose.
_vertices[0] corresponds to vertex of vertexsapointxyz type, i.e. 3D point, and "vertices[1] corresponds to vertex of VertexSE3Expmap type, i.e. pose
## 6. Set optimization parameters to start optimization.
Set the initialization, iterations, save results, etc. of SparseOptimizer.
Initialization
```SparseOptimizer::initializeOptimization(HyperGraph::EdgeSet& eset)
```
Set the number of iterations, and then you start to perform the graph optimization.
```SparseOptimizer::optimize(int iterations, bool online)
```
---------------
```typedef g2o::BlockSolver< g2o::BlockSolverTraits<3,1> > Block; // The optimization variable dimension of each error item is 3, and the error value dimension is 1
// Step 1: create a linear solver
Block::LinearSolverType* linearSolver = new g2o::LinearSolverDense<Block::PoseMatrixType>();
// Step 2: create BlockSolver. And initialized with the linear solver defined above
Block* solver_ptr = new Block( linearSolver );
// Step 3: create the total solver. And select one from GN, LM and dogleg, and then initialize it with BlockSolver
g2o::OptimizationAlgorithmLevenberg* solver = new g2o::OptimizationAlgorithmLevenberg( solver_ptr );
// Step 4: create the ultimate big boss sparse optimizer
g2o::SparseOptimizer optimizer; // Graph model
optimizer.setAlgorithm( solver ); // Set solver
optimizer.setVerbose( true ); // Turn on debug output
// Step 5: define the vertices and edges of the graph. And add to SparseOptimizer
CurveFittingVertex* v = new CurveFittingVertex(); //Add vertex to graph
v->setEstimate( Eigen::Vector3d(0,0,0) );
v->setId(0);
for ( int i=0; i<N; i++ ) // Add edge to graph
{
CurveFittingEdge* edge = new CurveFittingEdge( x_data[i] );
edge->setId(i);
edge->setVertex( 0, v ); // Set connected vertices
edge->setMeasurement( y_data[i] ); // Observational value
edge->setInformation( Eigen::Matrix<double,1,1>::Identity()*1/(w_sigma*w_sigma) ); // Information matrix: the inverse of covariance matrix
}
// Step 6: set optimization parameters and start optimization
optimizer.initializeOptimization();
optimizer.optimize(100);
```
reference material:
14 lectures on visual SLAM by Gao Xiang
https://blog.csdn.net/electech6/article/details/88018481
https://www.cnblogs.com/CV-life/p/10286037.html
https://www.cnblogs.com/CV-life/archive/2019/03/13/10525579.html
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Tags: github Ubuntu
Posted on Fri, 14 Feb 2020 02:00:26 -0500 by grantf | 4,780 | 19,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-10 | longest | en | 0.885565 |
https://www.ademcetinkaya.com/2023/03/gsrm-gsr-ii-meteora-acquisition-corp.html | 1,709,589,859,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00558.warc.gz | 643,442,460 | 61,851 | Outlook: GSR II Meteora Acquisition Corp. Class A Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Wait until speculative trend diminishes
Time series to forecast n: 06 Mar 2023 for (n+1 year)
Methodology : Modular Neural Network (Emotional Trigger/Responses Analysis)
## Abstract
GSR II Meteora Acquisition Corp. Class A Common Stock prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation1,2,3,4 and it is concluded that the GSRM stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
## Key Points
1. Fundemental Analysis with Algorithmic Trading
2. What are the most successful trading algorithms?
3. Stock Rating
## GSRM Target Price Prediction Modeling Methodology
We consider GSR II Meteora Acquisition Corp. Class A Common Stock Decision Process with Modular Neural Network (Emotional Trigger/Responses Analysis) where A is the set of discrete actions of GSRM stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Pearson Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Emotional Trigger/Responses Analysis)) X S(n):→ (n+1 year) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$
n:Time series to forecast
p:Price signals of GSRM stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## GSRM Stock Forecast (Buy or Sell) for (n+1 year)
Sample Set: Neural Network
Stock/Index: GSRM GSR II Meteora Acquisition Corp. Class A Common Stock
Time series to forecast n: 06 Mar 2023 for (n+1 year)
According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for GSR II Meteora Acquisition Corp. Class A Common Stock
1. As with all fair value measurements, an entity's measurement method for determining the portion of the change in the liability's fair value that is attributable to changes in its credit risk must make maximum use of relevant observable inputs and minimum use of unobservable inputs.
2. Conversely, if the critical terms of the hedging instrument and the hedged item are not closely aligned, there is an increased level of uncertainty about the extent of offset. Consequently, the hedge effectiveness during the term of the hedging relationship is more difficult to predict. In such a situation it might only be possible for an entity to conclude on the basis of a quantitative assessment that an economic relationship exists between the hedged item and the hedging instrument (see paragraphs B6.4.4–B6.4.6). In some situations a quantitative assessment might also be needed to assess whether the hedge ratio used for designating the hedging relationship meets the hedge effectiveness requirements (see paragraphs B6.4.9–B6.4.11). An entity can use the same or different methods for those two different purposes.
3. An entity shall apply the impairment requirements in Section 5.5 retrospectively in accordance with IAS 8 subject to paragraphs 7.2.15 and 7.2.18–7.2.20.
4. An example of a fair value hedge is a hedge of exposure to changes in the fair value of a fixed-rate debt instrument arising from changes in interest rates. Such a hedge could be entered into by the issuer or by the holder.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
GSR II Meteora Acquisition Corp. Class A Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. GSR II Meteora Acquisition Corp. Class A Common Stock prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation1,2,3,4 and it is concluded that the GSRM stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Wait until speculative trend diminishes
### GSRM GSR II Meteora Acquisition Corp. Class A Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCaa2B1
Balance SheetBaa2C
Leverage RatiosB2Caa2
Cash FlowBa3Caa2
Rates of Return and ProfitabilityBaa2B2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 75 out of 100 with 515 signals.
## References
1. J. Harb and D. Precup. Investigating recurrence and eligibility traces in deep Q-networks. In Deep Reinforcement Learning Workshop, NIPS 2016, Barcelona, Spain, 2016.
2. V. Borkar. Stochastic approximation: a dynamical systems viewpoint. Cambridge University Press, 2008
3. Belsley, D. A. (1988), "Modelling and forecast reliability," International Journal of Forecasting, 4, 427–447.
4. V. Borkar. An actor-critic algorithm for constrained Markov decision processes. Systems & Control Letters, 54(3):207–213, 2005.
5. E. Altman, K. Avrachenkov, and R. N ́u ̃nez-Queija. Perturbation analysis for denumerable Markov chains with application to queueing models. Advances in Applied Probability, pages 839–853, 2004
6. N. B ̈auerle and A. Mundt. Dynamic mean-risk optimization in a binomial model. Mathematical Methods of Operations Research, 70(2):219–239, 2009.
7. Mikolov T, Sutskever I, Chen K, Corrado GS, Dean J. 2013b. Distributed representations of words and phrases and their compositionality. In Advances in Neural Information Processing Systems, Vol. 26, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 3111–19. San Diego, CA: Neural Inf. Process. Syst. Found.
Frequently Asked QuestionsQ: What is the prediction methodology for GSRM stock?
A: GSRM stock prediction methodology: We evaluate the prediction models Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation
Q: Is GSRM stock a buy or sell?
A: The dominant strategy among neural network is to Wait until speculative trend diminishes GSRM Stock.
Q: Is GSR II Meteora Acquisition Corp. Class A Common Stock stock a good investment?
A: The consensus rating for GSR II Meteora Acquisition Corp. Class A Common Stock is Wait until speculative trend diminishes and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of GSRM stock?
A: The consensus rating for GSRM is Wait until speculative trend diminishes.
Q: What is the prediction period for GSRM stock?
A: The prediction period for GSRM is (n+1 year)
Stop Guessing, Start Winning.
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Click here to see what the AI recommends. | 1,967 | 8,061 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.793445 |
https://math.stackexchange.com/questions/814392/revisiting-algebra-for-the-proofs | 1,561,002,402,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00370.warc.gz | 511,769,048 | 35,177 | # Revisiting algebra for the proofs
As I was reading over Galois' wikipedia page, I noticed that he (Galois) read a book called "Réflexions sur la résolution algébrique des équations" at age 15, which I then read has to do with the Lagrange resolvent of polynomial equations, and from there I somehow got to the fundamental theorem of algebra blah blah blah you know how wikipedia is...
And to make a long story short, I realized that there seems to be a heck of a lot of very complex, interesting-looking algebra (especially on the proof side of things) that I didn't even get close to in Algebra 1 and 2 back before I even knew what a proof was. I'm just wondering whether it's worth it to go back and look over some of the "advanced algebra" the we glossed over in early high school.
I recently finished calc 1 and I'm currently learning some basic set theory/logic/proof-writing stuff (reading How to Prove it) because I started to get interested in proof construction.
I plan to read Spivak's this summer (I'll be a freshman in college next fall), but is it worth it to also go back to stuff the we only learned superficially in 9-11th grade or will I get into that in abstract algebra and whatnot in the years to come?
Sorry about how long this was… the background on my excursions on wikipedia was probably unnecessary.
• Lagrange's book is short and accessible. It is important historical background material for Galois theory, but nowadays primarily of historical interest. – André Nicolas May 30 '14 at 0:43
What you read in high school is necessary for college level math. Abstract algebra assumes you know the following: set theory, logic, proof techniques, functions and relations, induction, cardinal numbers, and number theory.
Abstract algebra covers groups, rings, fields, modules, vector spaces, and algebras. Hence, as I said above, functions and relations, which is what you learn in high school, is required for abstract algebra.
Most universities do not teach precalculus, assuming that the students are familiar and comfortable with it. It is important to know at least that much, since it'll help you in calculus of a single and of multi variables.
Good luck! | 488 | 2,193 | {"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-2019-26 | longest | en | 0.945329 |
https://chess.stackexchange.com/questions/14073/non-losing-strategy-for-white | 1,642,487,908,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00703.warc.gz | 230,663,639 | 33,960 | # Non-losing strategy for white
If you modify chess so that white gets to make one extra move within the first 10 moves, is it possible to show there is a non-losing strategy for white?
Student T gives an explicit answer, but you can also view it from a game-theoretical perspective.
Suppose normal chess is a draw (or a win for black). Then, white can play 1. Ng1-f3 and Nf3-g1, effectively becoming black and the rest of the game is normal, so it is a draw (or a win for white).
Suppose normal chess is a win for white. This will almost certainly involve moving a pawn two steps from its initial position within the first 10 moves, or moving a bishop, queen, rook or king. You can use your extra move to perform this move in two steps, e.g. e2-e3-e4 or Bf1-d3-e2.
In any case, White will not lose the game.
• That is a very elegant answer! Apr 4 '16 at 8:12
• This is a non-constructive proof. You must be a mathematician! Apr 4 '16 at 8:39
• This is not a proof. The knight manoeuvre cost two tempi, not only one. Thus it does not prove that Black has no winning strategy. Apr 4 '16 at 10:06
• @jknappen - the second tempo is 'paid' by the extra move White gets. And DagOskarMadsen, you are right. Apr 4 '16 at 10:38
Yes. 1.e4, then move the white Queen out to f3|g4|h5. Prepare for Qxf7+ then Qxe8 or Qxd7+ then Qxe8. Now white wins because the black king is gone.
• Say 1.e4 Nf6. Can you win Black's king by force now? Or are you just going to grab a piece? Can you prove that White can win or draw with an extra piece?
– bof
Apr 4 '16 at 9:03
• 2. e5 and your knight can't run (otherwise 3. e6 wins) so white can at least grab your knight without using his extra move. Apr 4 '16 at 12:17
• 2.Bc4 can give Black some trouble. 1.e4 Nf6 2.Bc4 (threatens 3.Qh5,Qxf7#) g6 3.e5 (threatens 4.e6,exf7# or 4.Qf3,Qxf7# if the knight moves anywhere but d5, where it is lost.) e6 4.exf6. Now if 4…Qxf6 5.Qf3,Qxf6 and if anything else then 5.Qg4 threatens 6.Qxe6,Qxf7# or even just 6.Qxe6,Qxe8#. 5…Be7 doesn't prevent mate on f7, and 5…Qe7 or 5…Qf6 loses the queen. Also 5…d5 just loses to 6.Bb5,Bxe8# Sacrificing the knight on d5 on move 2 or 3, followed by e6, seems best for Black, but probably still loses. Jul 11 '16 at 9:42 | 724 | 2,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-05 | latest | en | 0.903136 |
https://elliottwavegold.com/2014/12/gold-elliott-wave-technical-analysis-16th-december-2014/ | 1,725,718,730,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00638.warc.gz | 216,299,275 | 14,297 | Select Page
A little more downwards movement takes this move into its fifth day. I am swapping the Elliott wave counts over. Yesterday’s alternate is now the main preferred wave count. This is in line with GDX and Silver.
Summary: The bigger picture remains the same, but a second wave correction should complete over the next one to two days. I expect it to move lower to at least 1,172.
Click on charts to enlarge
Main Wave Count
Primary wave 4 is complete and primary wave 5 is unfolding. Primary wave 5 may only subdivide as an impulse or an ending diagonal. So far it looks most likely to be an impulse.
Within primary wave 5 intermediate wave (1) fits perfectly as an impulse. There is perfect alternation within intermediate wave (1): minor wave 2 is a deep zigzag lasting a Fibonacci five days and minor wave 4 is a shallow triangle lasting a Fibonacci eight days, 1.618 the duration of minor wave 2. Minor wave 3 is 9.65 longer than 1.618 the length of minor wave 1, and minor wave 5 is just 0.51 short of 0.618 the length of minor wave 1. I am confident this movement is one complete impulse.
Intermediate wave (2) is an incomplete expanded flat correction. Within it minor wave A is a double zigzag. The downwards wave labelled minor wave B has a corrective count of seven and subdivides perfectly as a zigzag. Minor wave B is a 172% correction of minor wave A. This is longer than the maximum common length for a B wave within a flat correction at 138%, but within the allowable range of twice the length of minor wave A. Minor wave C may not exhibit a Fibonacci ratio to minor wave A, and I think the target for it to end would best be calculated at minute degree. At this stage I would expect intermediate wave (2) to end close to the 0.618 Fibonacci ratio of intermediate wave (1) at 1,283.27.
Intermediate wave (1) lasted a Fibonacci 13 weeks. If intermediate wave (2) exhibits a Fibonacci duration it may be 13 weeks to be even with intermediate wave (1). Intermediate wave (2) is now in its tenth week.
So far within minor wave C the highest volume is on two up days. This supports the idea that at this stage the trend remains up.
See the most recent Historic Analysis to see the long term channel about this whole downwards movement. The channel does not copy over to the daily chart when I put the daily chart on an arithmetic scale, so this channel must be drawn on a weekly chart on a semi log scale. The upper edge of that channel may be where intermediate wave (2) finally ends. I would not expect the upper edge of this channel to be breached.
The target for primary wave 5 at this stage remains the same. At 956.97 it would reach equality in length with primary wave 1. However, if this target is wrong it may be too low. When intermediate waves (1) through to (4) within it are complete I will calculate the target at intermediate degree and if it changes it may move upwards. This is because waves following triangles tend to be more brief and weak than otherwise expected. A perfect example is on this chart: minor wave 5 to end intermediate wave (1) was particularly short and brief after the triangle of minor wave 4.
Within minor wave C minute wave i subdivides perfectly as a leading contracting diagonal. When leading diagonals unfold in first wave positions they are normally followed by very deep second wave corrections. There is a nice example here on the daily chart: at the top left of the chart minor wave 1 was a leading contracting diagonal and it was followed by a deep 65% zigzag for minor wave 2. I will expect minute wave ii to be deep, at least to the 0.618 Fibonacci ratio at 1,172. When it is over then a third wave up should begin.
Minute wave ii may not move beyond the start of minute wave i below 1,138.19.
Intermediate wave (2) may not move beyond the start of intermediate wave (1) above 1,345.22. I have confidence this price point will not be passed because the structure of primary wave 5 is incomplete because downwards movement from the end of the triangle of primary wave 4 does not fit well as either a complete impulse nor an ending diagonal.
To see a prior example of an expanded flat correction for Gold on the daily chart, and an explanation of this structure, go here.
I would expect minute wave ii to be deep, and so I expect it is not over. I would not want to label this as complete at the low of 1,187. That would be far too shallow for a second wave correction following a first wave leading diagonal.
Minute wave ii so far looks like it may be unfolding as a double zigzag. However, there are more than several structures it could complete as. Over the next one or two days the wave count at the hourly chart level may change as more of this correction unfolds and the structure becomes clearer.
Within the first zigzag subminuette wave a subdivides okay as an impulse; the only problem here is the disproportion between micro waves 2 and 4. However, all subdivisions fit.
Within subminuette wave a micro wave B fits well as a running contracting triangle, and this is supported by MACD sitting right at the zero line while it unfolds. I am confident there is a triangle in this position. It is for this reason that I see minuette wave (w) as a zigzag.
Within subminuette wave c it fits best as an ending expanding diagonal. I have checked on the five minute chart to see if micro waves 1, 3 and 5 fit as zigzags, and this is how they look best. Although the trend lines of this structure look odd, the wave lengths are all expanding and all the sub waves subdivide as zigzags as they should within an ending diagonal. Within diagonals the fourth wave should overlap back into first wave price territory.
Subminuette wave c is 2.72 short of 1.618 the length of subminuette wave a.
If this analysis to this point is correct, then minute wave ii is most likely to be a double zigzag and not a double combination. Their purposes are quite different: double combinations exist to take up time and move price sideways, while double zigzags exist to deepen a correction. Considering the expected depth of minute wave ii a double zigzag would fit the purpose. I expect minuette wave (y) to move price lower.
If minuette wave (y) is a zigzag then within it subminuette wave b may not move beyond the start of subminuette wave a above 1,223.57. Subminuette wave a subdivides perfectly as a five wave impulse on the five minute chart.
I do not think that subminuette wave b is over at this point. It should move price sideways and / or higher. Because I do not know where subminuette wave b has ended I cannot use the ratio between subminuette waves a and c to calculate a target for the zigzag of minuette wave (y) to end for you. I should be able to do that tomorrow. The most likely length for subminuette wave c will be equality with subminuette wave a at 36.
Minute wave ii may not move beyond the start of minute wave i below 1,138.19.
At any time any movement at all below 1,186.29 would invalidate the alternate wave count below and so provide confidence in this main wave count.
Alternate Wave Count
I almost do not want to publish this wave count today, because it has the “wrong” look. I consider it to now be extremely unlikely. I am concerned that publishing it gives it too much weight.
If minor wave C is within the middle of a third wave of a third wave of a third wave, then this current second wave correction should be brief and shallow. It is neither. This second wave correction has now lasted one day longer than the second wave correction one degree higher: minuette wave (ii) lasted four days and the current correction for subminuette wave ii is extremely deep and has lasted now five days. This gives this structure a strange look, and does not look like the middle of a third wave anymore.
Subminuette wave ii may not move beyond the start of subminuette wave i below 1,186.29. While price remains above this point this wave count will remain technically possible, but still unlikely.
Second Alternate Wave Count
I have published this wave count before, and will publish it again today in response to a member’s query. I consider it to have an extremely low probability.
Of all the markets I have analysed daily over the last six or so years it is Gold which exhibits the most typical looking Elliott wave structures, probably because of the huge volume of this truly global market. It is often so typical it looks like a textbook perfect case. I have learned to give a lot of weight to the “right look” for this particular market, and to consider alternates and change my wave count earlier rather than later if a wave count changes from the right look.
For this wave count intermediate wave (2) looks to be too brief and too shallow. It lasted only 11 days and is only 45% of intermediate wave (1). That is the first indication this wave count may be wrong.
Minor wave 1 does not subdivide clearly as a five wave impulse, and looks better as a three on the daily chart.
Now minor wave 2 within a larger degree third wave of intermediate wave (3) is very deep and lasted longer than intermediate wave (2) one degree higher. It is 86% of minor wave 1 and has lasted twice the duration of intermediate wave (2) at 22 days.
The black channel is a base channel about intermediate waves (1) and (2). A lower degree second wave correction should not breach a base channel drawn about a first and second wave one or more degrees higher. Minor wave 2 clearly and strongly breaches this channel. This further reduces the probability of this wave count.
I will probably not publish this wave count again. Publishing it gives it too much weight.
This analysis is published about 4:05 p.m. EST. | 2,173 | 9,714 | {"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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.918617 |
https://rmazing.wordpress.com/tag/confidence-interval/ | 1,606,722,498,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00255.warc.gz | 458,863,341 | 23,980 | ## Introducing ‘propagate’
August 31, 2013
With this post, I want to introduce the new ‘propagate’ package on CRAN.
It has one single purpose: propagation of uncertainties (“error propagation”). There is already one package on CRAN available for this task, named ‘metRology’ (http://cran.r-project.org/web/packages/metRology/index.html).
‘propagate’ has some additional functionality that some may find useful. The most important functions are:
* propagate: A general function for the calculation of uncertainty propagation by first-/second-order Taylor expansion and Monte Carlo simulation including covariances. Input data can be any symbolic/numeric differentiable expression and data based on replicates, summaries (mean & s.d.) or sampled from a distribution. Uncertainty propagation is based completely on matrix calculus accounting for full covariance structure. Monte Carlo simulation is conducted using multivariate normal or t-distributions with covariance structure. The second-order Taylor approximation is the new aspect, because it is not based on the assumption of linearity around $f(x)$ but uses a second-order polynomial to account for nonlinearities, making heavy use of numerical or symbolical Hessian matrices. Interestingly, the second-order approximation gives results quite similar to the MC simulations!
* plot.propagate: Graphing error propagation with the histograms of the MC simulations and MC/Taylor-based confidence intervals.
* predictNLS: The propagate function is used to calculate the propagated error to the fitted values of a nonlinear model of type nls or nlsLM. Please refer to my post here: https://rmazing.wordpress.com/2013/08/26/predictnls-part-2-taylor-approximation-confidence-intervals-for-nls-models/.
* makeGrad, makeHess, numGrad, numHess are functions to create symbolical or numerical gradient and Hessian matrices from an expression containing first/second-order partial derivatives. These can then be evaluated in an environment with evalDerivs.
* fitDistr: This function fits 21 different continuous distributions by (weighted) NLS to the histogram or kernel density of the Monte Carlo simulation results as obtained by propagate or any other vector containing large-scale observations. Finally, the fits are sorted by ascending AIC.
* random samplers for 15 continuous distributions under one hood, some of them previously unavailable:
Skewed-normal distribution, Generalized normal distributionm, Scaled and shifted t-distribution, Gumbel distribution, Johnson SU distribution, Johnson SB distribution, 3P Weibull distribution, 4P Beta distribution, Triangular distribution, Trapezoidal distribution, Curvilinear Trapezoidal distribution, Generalized trapezoidal distribution, Laplacian distribution, Arcsine distribution, von Mises distribution.
Most of them sample from the inverse cumulative distribution function, but 11, 12 and 15 use a vectorized version of “Rejection Sampling” giving roughly 100000 random numbers/s.
An example (without covariance for simplicity): $\mu_a = 5, \sigma_a = 0.1, \mu_b = 10, \sigma_b = 0.1, \mu_x = 1, \sigma_x = 0.1$
$f(x) = a^{bx}$:
>DAT <- data.frame(a = c(5, 0.1), b = c(10, 0.1), x = c(1, 0.1)) >EXPR <- expression(a^b*x) >res <- propagate(EXPR, DAT)
Results from error propagation: Mean.1 Mean.2 sd.1 sd.2 2.5% 97.5% 9765625 10067885 2690477 2739850 4677411 15414333
Results from Monte Carlo simulation: Mean sd Median MAD 2.5% 97.5% 10072640 2826027 9713207 2657217 5635222 16594123
The plot reveals the resulting distribution obtained from Monte Carlo simulation:
>plot(res)
Seems like a skewed distributions. We can now use fitDistr to find out which comes closest:
> fitDistr(res$resSIM) Fitting Normal distribution...Done. Fitting Skewed-normal distribution...Done. Fitting Generalized normal distribution...Done. Fitting Log-normal distribution...Done. Fitting Scaled/shifted t- distribution...Done. Fitting Logistic distribution...Done. Fitting Uniform distribution...Done. Fitting Triangular distribution...Done. Fitting Trapezoidal distribution...Done. Fitting Curvilinear Trapezoidal distribution...Done. Fitting Generalized Trapezoidal distribution...Done. Fitting Gamma distribution...Done. Fitting Cauchy distribution...Done. Fitting Laplace distribution...Done. Fitting Gumbel distribution...Done. Fitting Johnson SU distribution...........10.........20.........30.........40.........50 .........60.........70.........80.Done. Fitting Johnson SB distribution...........10.........20.........30.........40.........50 .........60.........70.........80.Done. Fitting 3P Weibull distribution...........10.........20.......Done. Fitting 4P Beta distribution...Done. Fitting Arcsine distribution...Done. Fitting von Mises distribution...Done.$aic Distribution AIC 4 Log-normal -4917.823 16 Johnson SU -4861.960 15 Gumbel -4595.917 19 4P Beta -4509.716 12 Gamma -4469.780 9 Trapezoidal -4340.195 1 Normal -4284.706 5 Scaled/shifted t- -4283.070 6 Logistic -4266.171 3 Generalized normal -4264.102 14 Laplace -4144.870 13 Cauchy -4099.405 2 Skewed-normal -4060.936 11 Generalized Trapezoidal -4032.484 10 Curvilinear Trapezoidal -3996.495 8 Triangular -3970.993 7 Uniform -3933.513 20 Arcsine -3793.793 18 3P Weibull -3783.041 21 von Mises -3715.034 17 Johnson SB -3711.034
Log-normal wins, which makes perfect sense after using an exponentiation function... Have fun with the package. Comments welcome! Cheers, Andrej
17 Comments | General, R Internals | Tagged: confidence interval, first-order, fitting, Monte Carlo, nls, nonlinear, predict, second-order, Taylor approximation | Permalink Posted by anspiess
predictNLS (Part 2, Taylor approximation): confidence intervals for ‘nls’ models August 26, 2013 Initial Remark: Reload this page if formulas don’t display well! As promised, here is the second part on how to obtain confidence intervals for fitted values obtained from nonlinear regression via nls or nlsLM (package ‘minpack.lm’). I covered a Monte Carlo approach in https://rmazing.wordpress.com/2013/08/14/predictnls-part-1-monte-carlo-simulation-confidence-intervals-for-nls-models/, but here we will take a different approach: First- and second-order Taylor approximation around $f(x)$: $f(x) \approx f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2$. Using Taylor approximation for calculating confidence intervals is a matter of propagating the uncertainties of the parameter estimates obtained from vcov(model) to the fitted value. When using first-order Taylor approximation, this is also known as the “Delta method”. Those familiar with error propagation will know the formula $\displaystyle \sum_{i=1}^2 \rm{j_i}^2 \sigma_i^2 + 2\sum_{i=1\atop i \neq k}^n\sum_{k=1\atop k \neq i}^n \rm{j_i j_k} \sigma_{ik}$. Heavily underused is the matrix notation of the famous formula above, for which a good derivation can be found at http://www.nada.kth.se/~kai-a/papers/arrasTR-9801-R3.pdf: $\sigma_y^2 = \nabla_x\mathbf{C}_x\nabla_x^T$, where $\nabla_x$ is the gradient vector of first-order partial derivatives and $\mathbf{C}_x$ is the variance-covariance matrix. This formula corresponds to the first-order Taylor approximation. Now the problem with first-order approximations is that they assume linearity around $f(x)$. Using the “Delta method” for nonlinear confidence intervals in R has been discussed in http://thebiobucket.blogspot.de/2011/04/fit-sigmoid-curve-with-confidence.html or http://finzi.psych.upenn.edu/R/Rhelp02a/archive/42932.html. For highly nonlinear functions we need (at least) a second-order polynomial around $f(x)$ to realistically estimate the surrounding interval (red is linear approximation, blue is second-order polynomial on a sine function around $x = 5$): Interestingly, there are also matrix-like notations for the second-order mean and variance in the literature (see http://dml.cz/dmlcz/141418 or http://iopscience.iop.org/0026-1394/44/3/012/pdf/0026-1394_44_3_012.pdf): Second-order mean: $\rm{E}[y] = f(\bar{x}_i) + \frac{1}{2}\rm{tr}(\mathbf{H}_{xx}\mathbf{C}_x)$. Second-order variance: $\sigma_y^2 = \nabla_x\mathbf{C}_x\nabla_x^T + \frac{1}{2}\rm{tr}(\mathbf{H}_{xx}\mathbf{C}_x\mathbf{H}_{xx}\mathbf{C}_x)$, where $\mathbf{H}_{xx}$ is the Hessian matrix of second-order partial derivatives and $tr(\cdot)$ is the matrix trace (sum of diagonals). Enough theory, for wrapping this all up we need three utility functions: 1) numGrad for calculating numerical first-order partial derivatives. numGrad <- function(expr, envir = .GlobalEnv) { f0 <- eval(expr, envir) vars <- all.vars(expr) p <- length(vars) x <- sapply(vars, function(a) get(a, envir)) eps <- 1e-04 d <- 0.1 r <- 4 v <- 2 zero.tol <- sqrt(.Machine$double.eps/7e-07) h0 <- abs(d * x) + eps * (abs(x) < zero.tol) D <- matrix(0, length(f0), p) Daprox <- matrix(0, length(f0), r) for (i in 1:p) { h <- h0 for (k in 1:r) { x1 <- x2 <- x x1 <- x1 + (i == (1:p)) * h f1 <- eval(expr, as.list(x1)) x2 <- x2 - (i == (1:p)) * h f2 <- eval(expr, envir = as.list(x2)) Daprox[, k] <- (f1 - f2)/(2 * h[i]) h <- h/v } for (m in 1:(r - 1)) for (k in 1:(r - m)) { Daprox[, k] <- (Daprox[, k + 1] * (4^m) - Daprox[, k])/(4^m - 1) } D[, i] <- Daprox[, 1] } return(D) } 2) numHess for calculating numerical second-order partial derivatives. numHess <- function(expr, envir = .GlobalEnv) { f0 <- eval(expr, envir) vars <- all.vars(expr) p <- length(vars) x <- sapply(vars, function(a) get(a, envir)) eps <- 1e-04 d <- 0.1 r <- 4 v <- 2 zero.tol <- sqrt(.Machine$double.eps/7e-07) h0 <- abs(d * x) + eps * (abs(x) < zero.tol) Daprox <- matrix(0, length(f0), r) Hdiag <- matrix(0, length(f0), p) Haprox <- matrix(0, length(f0), r) H <- matrix(NA, p, p) for (i in 1:p) { h <- h0 for (k in 1:r) { x1 <- x2 <- x x1 <- x1 + (i == (1:p)) * h f1 <- eval(expr, as.list(x1)) x2 <- x2 - (i == (1:p)) * h f2 <- eval(expr, envir = as.list(x2)) Haprox[, k] <- (f1 - 2 * f0 + f2)/h[i]^2 h <- h/v } for (m in 1:(r - 1)) for (k in 1:(r - m)) { Haprox[, k] <- (Haprox[, k + 1] * (4^m) - Haprox[, k])/(4^m - 1) } Hdiag[, i] <- Haprox[, 1] } for (i in 1:p) { for (j in 1:i) { if (i == j) { H[i, j] <- Hdiag[, i] } else { h <- h0 for (k in 1:r) { x1 <- x2 <- x x1 <- x1 + (i == (1:p)) * h + (j == (1:p)) * h f1 <- eval(expr, as.list(x1)) x2 <- x2 - (i == (1:p)) * h - (j == (1:p)) * h f2 <- eval(expr, envir = as.list(x2)) Daprox[, k] <- (f1 - 2 * f0 + f2 - Hdiag[, i] * h[i]^2 - Hdiag[, j] * h[j]^2)/(2 * h[i] * h[j]) h <- h/v } for (m in 1:(r - 1)) for (k in 1:(r - m)) { Daprox[, k] <- (Daprox[, k + 1] * (4^m) - Daprox[, k])/(4^m - 1) } H[i, j] <- H[j, i] <- Daprox[, 1] } } } return(H) } And a small function for the matrix trace: tr <- function(mat) sum(diag(mat), na.rm = TRUE) 1) and 2) are modified versions of the genD function in the “numDeriv” package that can handle expressions. Now we need the predictNLS function that wraps it all up: predictNLS <- function( object, newdata, interval = c("none", "confidence", "prediction"), level = 0.95, ... ) { require(MASS, quietly = TRUE) interval <- match.arg(interval) ## get right-hand side of formula RHS <- as.list(object$call$formula)[[3]] EXPR <- as.expression(RHS) ## all variables in model VARS <- all.vars(EXPR) ## coefficients COEF <- coef(object) ## extract predictor variable predNAME <- setdiff(VARS, names(COEF)) ## take fitted values, if 'newdata' is missing if (missing(newdata)) { newdata <- eval(object$data)[predNAME] colnames(newdata) <- predNAME } ## check that 'newdata' has same name as predVAR if (names(newdata)[1] != predNAME) stop("newdata should have name '", predNAME, "'!") ## get parameter coefficients COEF <- coef(object) ## get variance-covariance matrix VCOV <- vcov(object) ## augment variance-covariance matrix for 'mvrnorm' ## by adding a column/row for 'error in x' NCOL <- ncol(VCOV) ADD1 <- c(rep(0, NCOL)) ADD1 <- matrix(ADD1, ncol = 1) colnames(ADD1) <- predNAME VCOV <- cbind(VCOV, ADD1) ADD2 <- c(rep(0, NCOL + 1)) ADD2 <- matrix(ADD2, nrow = 1) rownames(ADD2) <- predNAME VCOV <- rbind(VCOV, ADD2) NR <- nrow(newdata) respVEC <- numeric(NR) seVEC <- numeric(NR) varPLACE <- ncol(VCOV) outMAT <- NULL ## define counter function counter <- function (i) { if (i%%10 == 0) cat(i) else cat(".") if (i%%50 == 0) cat("\n") flush.console() } ## calculate residual variance r <- residuals(object) w <- weights(object) rss <- sum(if (is.null(w)) r^2 else r^2 * w) df <- df.residual(object) res.var <- rss/df ## iterate over all entries in 'newdata' as in usual 'predict.' functions for (i in 1:NR) { counter(i) ## get predictor values and optional errors predVAL <- newdata[i, 1] if (ncol(newdata) == 2) predERROR <- newdata[i, 2] else predERROR <- 0 names(predVAL) <- predNAME names(predERROR) <- predNAME ## create mean vector meanVAL <- c(COEF, predVAL) ## create augmented variance-covariance matrix ## by putting error^2 in lower-right position of VCOV newVCOV <- VCOV newVCOV[varPLACE, varPLACE] <- predERROR^2 SIGMA <- newVCOV ## first-order mean: eval(EXPR), first-order variance: G.S.t(G) MEAN1 <- try(eval(EXPR, envir = as.list(meanVAL)), silent = TRUE) if (inherits(MEAN1, "try-error")) stop("There was an error in evaluating the first-order mean!") GRAD <- try(numGrad(EXPR, as.list(meanVAL)), silent = TRUE) if (inherits(GRAD, "try-error")) stop("There was an error in creating the numeric gradient!") VAR1 <- GRAD %*% SIGMA %*% matrix(GRAD) ## second-order mean: firstMEAN + 0.5 * tr(H.S), ## second-order variance: firstVAR + 0.5 * tr(H.S.H.S) HESS <- try(numHess(EXPR, as.list(meanVAL)), silent = TRUE) if (inherits(HESS, "try-error")) stop("There was an error in creating the numeric Hessian!") valMEAN2 <- 0.5 * tr(HESS %*% SIGMA) valVAR2 <- 0.5 * tr(HESS %*% SIGMA %*% HESS %*% SIGMA) MEAN2 <- MEAN1 + valMEAN2 VAR2 <- VAR1 + valVAR2 ## confidence or prediction interval if (interval != "none") { tfrac <- abs(qt((1 - level)/2, df)) INTERVAL <- tfrac * switch(interval, confidence = sqrt(VAR2), prediction = sqrt(VAR2 + res.var)) LOWER <- MEAN2 - INTERVAL UPPER <- MEAN2 + INTERVAL names(LOWER) <- paste((1 - level)/2 * 100, "%", sep = "") names(UPPER) <- paste((1 - (1- level)/2) * 100, "%", sep = "") } else { LOWER <- NULL UPPER <- NULL } RES <- c(mu.1 = MEAN1, mu.2 = MEAN2, sd.1 = sqrt(VAR1), sd.2 = sqrt(VAR2), LOWER, UPPER) outMAT <- rbind(outMAT, RES) } cat("\n") rownames(outMAT) <- NULL return(outMAT) } With all functions at hand, we can now got through the same example as used in the Monte Carlo post: DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) > predictNLS(fm1DNase1, newdata = data.frame(conc = 5), interval = "confidence") . mu.1 mu.2 sd.1 sd.2 2.5% 97.5% [1,] 1.243631 1.243288 0.03620415 0.03620833 1.165064 1.321511 The errors/confidence intervals are larger than with the MC approch (who knows why?) but it is very interesting to see how close the second-order corrected mean (1.243288) comes to the mean of the simulated values from the Monte Carlo approach (1.243293)! The two approach (MC/Taylor) will be found in the predictNLS function that will be part of the “propagate” package in a few days at CRAN… Cheers, Andrej 12 Comments | General, R Internals | Tagged: confidence interval, first-order, fitting, Monte Carlo, nls, nonlinear, predict, second-order, Taylor approximation | Permalink Posted by anspiess predictNLS (Part 1, Monte Carlo simulation): confidence intervals for ‘nls’ models August 14, 2013 Those that do a lot of nonlinear fitting with the nls function may have noticed that predict.nls does not have a way to calculate a confidence interval for the fitted value. Using confint you can obtain the error of the fit parameters, but how about the error in fitted values? ?predict.nls says: “At present se.fit and interval are ignored.” What a pity… This is largely to the fact that confidence intervals for nonlinear fits are not easily calculated and under some debate, see http://r.789695.n4.nabble.com/plotting-confidence-bands-from-predict-nls-td3505012.html or http://thr3ads.net/r-help/2011/05/1053390-plotting-confidence-bands-from-predict.nls. In principle, since calculating the error in the fitted values is a matter of “error propagation”, two different approaches can be used: 1) Error propagation using approximation by (first-order) Taylor expansion around $f(x)$, 2) Error propagation using Monte Carlo simulation. Topic 1) will be subject of my next post, today we will stick with the MC approach. When calculating the error in the fitted values, we need to propagate the error of all variables, i.e. the error in all predictor variables $x_m$ and the error of the fit parameters $\beta$, to the response $y = f(x_m, \beta)$. Often (as in the ‘Examples’ section of nls), there is no error in the $x_m$-values. The errors of the fit parameters $\beta$ are obtained, together with their correlations, in the variance-covariance matrix $\Sigma$ from vcov(object). A Monte Carlo approach to nonlinear error propagation does the following: 1) Use as input $\mu_m$ and $\sigma_m^2$ of all predictor variables and the vcov matrix $\Sigma$ of the fit parameters $\beta$. 2) For each variable $m$, we create $n$ samples from a multivariate normal distribution using the variance-covariance matrix: $x_{m, n} \sim \mathcal{N}(\mu, \Sigma)$. 3) We evaluate the function on each simulated variable: $y_n = f(x_{m, n}, \beta)$ 4) We calculate statistics (mean, s.d., median, mad) and quantile-based confidence intervals on the vector $y_n$. This is exactly what the following function does: It takes an nls object, extracts the variables/parameter values/parameter variance-covariance matrix, creates an “augmented” covariance matrix (with the variance/covariance values from the parameters and predictor variables included, the latter often being zero), simulates from a multivariate normal distribution (using mvrnorm of the ‘MASS’ package), evaluates the function (object$call$formula) on the values and finally collects statistics. Here we go: predictNLS <- function( object, newdata, level = 0.95, nsim = 10000, ... ) { require(MASS, quietly = TRUE) ## get right-hand side of formula RHS <- as.list(object$call$formula)[[3]] EXPR <- as.expression(RHS) ## all variables in model VARS <- all.vars(EXPR) ## coefficients COEF <- coef(object) ## extract predictor variable predNAME <- setdiff(VARS, names(COEF)) ## take fitted values, if 'newdata' is missing if (missing(newdata)) { newdata <- eval(object$data)[predNAME] colnames(newdata) <- predNAME } ## check that 'newdata' has same name as predVAR if (names(newdata)[1] != predNAME) stop("newdata should have name '", predNAME, "'!") ## get parameter coefficients COEF <- coef(object) ## get variance-covariance matrix VCOV <- vcov(object) ## augment variance-covariance matrix for 'mvrnorm' ## by adding a column/row for 'error in x' NCOL <- ncol(VCOV) ADD1 <- c(rep(0, NCOL)) ADD1 <- matrix(ADD1, ncol = 1) colnames(ADD1) <- predNAME VCOV <- cbind(VCOV, ADD1) ADD2 <- c(rep(0, NCOL + 1)) ADD2 <- matrix(ADD2, nrow = 1) rownames(ADD2) <- predNAME VCOV <- rbind(VCOV, ADD2) ## iterate over all entries in 'newdata' as in usual 'predict.' functions NR <- nrow(newdata) respVEC <- numeric(NR) seVEC <- numeric(NR) varPLACE <- ncol(VCOV) ## define counter function counter <- function (i) { if (i%%10 == 0) cat(i) else cat(".") if (i%%50 == 0) cat("\n") flush.console() } outMAT <- NULL for (i in 1:NR) { counter(i) ## get predictor values and optional errors predVAL <- newdata[i, 1] if (ncol(newdata) == 2) predERROR <- newdata[i, 2] else predERROR <- 0 names(predVAL) <- predNAME names(predERROR) <- predNAME ## create mean vector for 'mvrnorm' MU <- c(COEF, predVAL) ## create variance-covariance matrix for 'mvrnorm' ## by putting error^2 in lower-right position of VCOV newVCOV <- VCOV newVCOV[varPLACE, varPLACE] <- predERROR^2 ## create MC simulation matrix simMAT <- mvrnorm(n = nsim, mu = MU, Sigma = newVCOV, empirical = TRUE) ## evaluate expression on rows of simMAT EVAL <- try(eval(EXPR, envir = as.data.frame(simMAT)), silent = TRUE) if (inherits(EVAL, "try-error")) stop("There was an error evaluating the simulations!") ## collect statistics PRED <- data.frame(predVAL) colnames(PRED) <- predNAME FITTED <- predict(object, newdata = data.frame(PRED)) MEAN.sim <- mean(EVAL, na.rm = TRUE) SD.sim <- sd(EVAL, na.rm = TRUE) MEDIAN.sim <- median(EVAL, na.rm = TRUE) MAD.sim <- mad(EVAL, na.rm = TRUE) QUANT <- quantile(EVAL, c((1 - level)/2, level + (1 - level)/2)) RES <- c(FITTED, MEAN.sim, SD.sim, MEDIAN.sim, MAD.sim, QUANT[1], QUANT[2]) outMAT <- rbind(outMAT, RES) } colnames(outMAT) <- c("fit", "mean", "sd", "median", "mad", names(QUANT[1]), names(QUANT[2])) rownames(outMAT) <- NULL cat("\n") return(outMAT) } The input is an ‘nls’ object, a data.frame ‘newdata’ of values to be predicted with the value $x_{new}$ in the first column and (optional) “errors-in-x” (as $\sigma$) in the second column. The number of simulations can be tweaked with nsim as well as the alpha-level for the confidence interval. The output is $f(x_{new}, \beta)$ (fitted value), $\mu(y_n)$ (mean of simulation), $\sigma(y_n)$ (s.d. of simulation), $median(y_n)$ (median of simulation), $mad(y_n)$ (mad of simulation) and the lower/upper confidence interval. Ok, let’s go to it (taken from the ‘?nls’ documentation): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) ## usual predict.nls has no confidence intervals implemented predict(fm1DNase1, newdata = data.frame(conc = 5), interval = "confidence") [1] 1.243631 attr(,"gradient") Asym xmid scal [1,] 0.5302925 -0.5608912 -0.06804642 In the next post we will see how to use the gradient attribute to calculate a first-order Taylor expansion around $f(x)$… However, predictNLS gives us the error and confidence interval at $x = 5$: predictNLS(fm1DNase1, newdata = data.frame(conc = 5)) . fit mean sd median mad 2.5% 97.5% [1,] 1.243631 1.243293 0.009462893 1.243378 0.009637439 1.224608 1.261575 Interesting to see, how close the mean of the simulation comes to the actual fitted value… We could also add some error in $x$ to propagate to $y$: > predictNLS(fm1DNase1, newdata = data.frame(conc = 5, error = 0.1)) . fit mean sd median mad 2.5% 97.5% [1,] 1.243631 1.243174 0.01467673 1.243162 0.01488567 1.214252 1.272103 Have fun. If anyone know how to calculate a “prediction interval” (maybe quantile regression) give me hint… Cheers, Andrej 12 Comments | General, R Internals | Tagged: confidence interval, fitting, Monte Carlo, nls, nonlinear | Permalink Posted by anspiess
Search Recent Posts Monte Carlo-based prediction intervals for nonlinear regression Linear regression with random error giving EXACT predefined parameter estimates Introducing: Orthogonal Nonlinear Least-Squares Regression in R Error propagation based on interval arithmetics I’ll take my NLS with weights, please… Archives May 2018 January 2016 January 2015 September 2014 January 2014 August 2013 July 2013 April 2013 February 2013 January 2013 November 2012 July 2012 June 2012 Categories General nonlinear regression R Internals Uncategorized Meta Register Log in Entries feed Comments feed WordPress.com Blogroll R-bloggers __ATA.cmd.push(function() { __ATA.initDynamicSlot({ id: 'atatags-286348-5fc4a3c2234e1', location: 140, formFactor: '003', label: { text: 'Advertisements', }, creative: { reportAd: { text: 'Report this ad', }, privacySettings: { text: 'Privacy settings', } } }); }); Create a free website or blog at WordPress.com.
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https://mathoverflow.net/questions/442548/do-compactly-generated-spaces-have-a-more-direct-definition | 1,721,855,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00033.warc.gz | 356,439,069 | 26,286 | # Do compactly generated spaces have a more direct definition?
Is there an elementary way to define Haussdorf-compactly generated weakly Hausdorff topological spaces in a way that does not need defining topological space first?
Weakly Hausdorff sequential spaces have an alternative cryptomorphic axiomatization as Hausdorff subsequential spaces, which are arguably simpler to define than topological spaces, and have the property that dropping the Hausdorff condition turns it into a quasitopos.
One would hope then that compactly generated k-Hausdorff spaces have a similarly well behaved definition that can skip the usual presentation of things, since the category is very natural. Is there such a presentation?
• Could you explain how defining `Hausdorff subsequential spaces' is simpler than defining topological spaces? Commented Mar 12, 2023 at 13:52
• @Tyrone because morphisms of sequential spaces are characterized as ones preserving an (infinitary) operation of a limit; whereas continuous maps cannot be defined without referring to elements and involve a very ill-behaved notion of preimage, or an even nastier notion of induced mapping on set of subests. Commented Mar 13, 2023 at 12:49
• @DenisT that’s not simpler. Commented Mar 13, 2023 at 14:54
• @FernandoMuro it is "arguably" simpler as in, "there are situations where this is simpler" as explained by Denis T. Some results, for e.g. Manes theorem or Tychonov theorem are much more natural when you think in terms of limits as an operation, instead of open sets. One could also argue that most students encounter the notion of limits long before the notion of open sets, so maybe axiomatizing the notion of limits is more natural than axiomatizing the notion of open sets. Not that I'm not saying this is the case - this is obviously opinion based, but you can't deny this can be argued. Commented Mar 13, 2023 at 18:07
• One way to view it as simpler is that topological spaces formulated in terms of filters or nets are much more complex than subsequential/sequential spaces formulated in terms of sequences. The open set formulation of topological spaces is simple, but the open set formulation of topological spaces with nice properties isn't usually as simple and you often end up with an ugly category unless you add a bunch of seemingly arbitrary conditions Commented Mar 13, 2023 at 18:39
Sorry, I misremembered this. There are more quasiseparated condensed sets than just the weak Hausdorff $$k$$-spaces. See p. 15, right after Prop 2.9 here. I don't know if there's a way to define weak Hausdorff $$k$$-spaces in terms of condensed sets alone.
Weakly Hausdorff k-spaces are the quasiseparated objects in the category of condensed sets (see Thm 2.16 here). Quasiseparated is a categorical notion which is standard in topos theory.
In turn, the category of condensed sets can be defined as
• sheaves on compact Hausdorff spaces (= algebras for the ultrafilter monad on $$Set$$ = opposite category of commutative unital $$C^\ast$$ algebras), or as
• sheaves on totally disconnected compact Hausdorff spaces (= profinite sets), or as
• sheaves on extremally disconnected spaces (=idempotent completion of the Kleisli category for the ultrafilter monad), or as
• sheaves on free extremally disconnected spaces (=the Kleisli category for the ultrafilter monad).
Here, the ultrafilter monad is the unique monad whose underlying functor is the ultrafilter functor $$\beta : Set \to Set$$, carrying a set to the set of ultrafilters thereon. The topology with respect to which we take sheaves is some additional data in the first two descriptions, but in the last two descriptions it's very straightforward: a sheaf is a presheaf carrying coproducts to products.
Anyway, that gives a few possible definitions which don't require you to know what a topological space is.
• I've never encountered this "categorical notion which is standard in topos theory". (-: Can you give a reference? Commented Mar 16, 2023 at 3:07
• @MikeShulman I think it should be in SGA4-VII somewhere. Definition is an obvious one mimicking situation for T1 spaces: one says that an object of a topos is quasicompact if every covering (effective epi from coproduct) admits a finite subcovering, and X being quasiseparated means that (binary) fibered products over X preserve quasicompactness. Commented Mar 16, 2023 at 13:03
• @MikeShulman Denis T has given the definition; I know I first encountered the concept when reading about how to recover a coherent pretopos from a coherent topos, since a coherent object is one which is quasicompact and quasiseparated. I just looked in Makkai and Reyes Def 9.2.1 and they don't give a name to quasiseparatedness though. But presumably it is defined in SGA. Commented Mar 16, 2023 at 14:13
• Ah, this sounds like the notion that is called stable in D3.3 of the Elephant? And since non-algebraic-geometers generally drop the "quasi" from "compact", is it also common to drop it from "separated"? Commented Mar 16, 2023 at 19:45 | 1,225 | 5,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.944179 |
https://de.maplesoft.com/support/help/errors/view.aspx?path=GraphTheory/TuttePolynomial&L=G | 1,670,205,312,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00560.warc.gz | 212,061,022 | 51,250 | AcyclicPolynomial - Maple Help
GraphTheory
TuttePolynomial
compute Tutte polynomial
ChromaticPolynomial
compute chromatic polynomial
FlowPolynomial
compute flow polynomial
RankPolynomial
compute rank polynomial
AcyclicPolynomial
compute acyclic polynomial
ReliabilityPolynomial
compute reliability polynomial
Calling Sequence TuttePolynomial(H, x, y) ChromaticPolynomial(G, t) FlowPolynomial(G, x) RankPolynomial(G, x, y) AcyclicPolynomial(G, p) ReliabilityPolynomial(H, p)
Parameters
H - undirected graph G - undirected unweighted graph t,x,y,p - variables or values
Description
TuttePolynomial
• The TuttePolynomial command returns a bivariate polynomial in x and y when x and y are variables or the evaluation of the bivariate polynomial when x or y are values.
• The Tutte polynomial, T(G;x,y), is a generalization of the chromatic polynomial and is defined as follows:
If G has no edges then T(G;x,y) = 1.
Let e be any edge in G and let G-e and G/e denote the graph G with e removed and with e contracted, respectively.
If e is a loop then T(G;x,y) = y*T(G-e;x,y)
If e is a bridge (cut-edge) then T(G;x,y) = x*T(G/e;x,y)
If e is not a bridge nor a loop then T(G;x,y) = T(G-e;x,y) + T(G/e;x,y)
• The ChromaticPolynomial, FlowPolynomial, RankPolynomial, and ReliabilityPolynomial are functions of the Tutte polynomial. They are all NP-hard to compute in general.
ChromaticPolynomial
• The ChromaticPolynomial command returns a polynomial, P(G,t), which is the number of proper vertex colorings of G using no more than t colors, where t is a non-negative integer.
The chromatic polynomial, P(G;t), for a graph G with n vertices, and k connected components, can be expressed in terms of the Tutte polynomial T(G;x,y), as follows:
P(G;t) = (-1)^(n-k)*t^k*T(G;1-t,0)
• P(G,t) has been determined for certain classes of graphs. Fast codes for the special cases for the complete graph, its complement, trees and cycles have been included in the command.
FlowPolynomial and RankPolynomial
• The FlowPolynomial command returns a polynomial in x when x is a variable or the evaluation of the polynomial when x is a value. The value of this polynomial at a positive integer k gives the number of nowhere-zero flows on G with edge labels chosen from the integers modulo k.
The flow polynomial, Q(G;x), for a graph G with n vertices, m edges, and k connected components, can be expressed in terms of the Tutte polynomial, T(G;x,y), as follows:
Q(G;x) = (-1)^(m-n+k)*T(G;0,1-x)
• The RankPolynomial command returns a bivariate polynomial in x and y when x and y are variables or the evaluation of the polynomial when x or y are values.
S(G;x,y) = T(G;x+1,y+1) where S(G;x,y) is the rank polynomial of G.
AcyclicPolynomial and ReliabilityPolynomial
• The AcyclicPolynomial command returns a polynomial in p when p is a variable or the evaluation of the polynomial when p is a value. The value of this polynomial at a value $0\le p\le 1$ gives the probability that G is acyclic when each edge operates with probability p.
• The ReliabilityPolynomial command returns a polynomial in p when p is a variable or the evaluation of the polynomial when p is a value. The value of this polynomial at a value $0\le p\le 1$ gives the probability that G is connected when each edge fails with probability p. For example, if G is a tree on n vertices, then if any edge fails G will become disconnected. Hence the reliability polynomial for a tree is (1-p)^(n-1). It can be computed as follows.
If the graph G is not connected, then its reliability polynomial is 0.
If e is a loop in G then R(G;p) = R(G-e;p)
If e is a bridge (isthmus) then R(G;p) = (1-p)*T(G/e;p)
If e is not a bridge nor a loop then R(G;p) = p*R(G-e;p) + (1-p)*R(G/e;p)
• If G has n vertices and m edges, the reliability polynomial R(G;p) is related to the Tutte polynomial T(G;x,y) as follows
R(G;p) = T(G;1,1/p)*p^(n-1)*(1-p)^(m-n+1)
Notes
• The TuttePolynomial and ReliabilityPolynomial commands accept a weighted graph H as input. The edge weights must be positive integers and they are interpreted as multiple edges.
• The computation of the Tutte polynomial is NP-hard. We compute the Tutte polynomial using edge deletion and contraction and we remember the Tutte polynomial for each connected subgraph computed. By processing edges in a canonical ordering this enables us to identify subgraphs already seen without using a general graph isomorphism test. See references [2] and [4] Monagan in the References section.
Examples
> $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
> $\mathrm{with}\left(\mathrm{SpecialGraphs}\right):$
> $\mathrm{with}\left(\mathrm{RandomGraphs}\right):$
TuttePolynomial
> $G≔\mathrm{CompleteGraph}\left(4\right)$
${G}{≔}{\mathrm{Graph 1: an undirected unweighted graph with 4 vertices and 6 edge\left(s\right)}}$ (1)
> $\mathrm{TuttePolynomial}\left(G,'x','y'\right)$
${{x}}^{{3}}{+}{{y}}^{{3}}{+}{3}{}{{x}}^{{2}}{+}{4}{}{x}{}{y}{+}{3}{}{{y}}^{{2}}{+}{2}{}{x}{+}{2}{}{y}$ (2)
> $\mathrm{TuttePolynomial}\left(G,'x',1\right)$
${{x}}^{{3}}{+}{3}{}{{x}}^{{2}}{+}{6}{}{x}{+}{6}$ (3)
We can verify the recurrence relation
> $G≔\mathrm{PetersenGraph}\left(\right)$
${G}{≔}{\mathrm{Graph 2: an undirected unweighted graph with 10 vertices and 15 edge\left(s\right)}}$ (4)
> $f≔\mathrm{TuttePolynomial}\left(G,'x','y'\right)$
${f}{≔}{{x}}^{{9}}{+}{6}{}{{x}}^{{8}}{+}{21}{}{{x}}^{{7}}{+}{56}{}{{x}}^{{6}}{+}{12}{}{{x}}^{{5}}{}{y}{+}{{y}}^{{6}}{+}{114}{}{{x}}^{{5}}{+}{70}{}{{x}}^{{4}}{}{y}{+}{30}{}{{x}}^{{3}}{}{{y}}^{{2}}{+}{15}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{10}{}{x}{}{{y}}^{{4}}{+}{9}{}{{y}}^{{5}}{+}{170}{}{{x}}^{{4}}{+}{170}{}{{x}}^{{3}}{}{y}{+}{105}{}{{x}}^{{2}}{}{{y}}^{{2}}{+}{65}{}{x}{}{{y}}^{{3}}{+}{35}{}{{y}}^{{4}}{+}{180}{}{{x}}^{{3}}{+}{240}{}{{x}}^{{2}}{}{y}{+}{171}{}{x}{}{{y}}^{{2}}{+}{75}{}{{y}}^{{3}}{+}{120}{}{{x}}^{{2}}{+}{168}{}{x}{}{y}{+}{84}{}{{y}}^{{2}}{+}{36}{}{x}{+}{36}{}{y}$ (5)
> $e≔\mathrm{Edges}\left(G\right)\left[1\right]$
${e}{≔}\left\{{1}{,}{2}\right\}$ (6)
> $\mathrm{Gminuse}≔\mathrm{DeleteEdge}\left(G,e,'\mathrm{inplace}'=\mathrm{false}\right)$
${\mathrm{Gminuse}}{≔}{\mathrm{Graph 3: an undirected unweighted graph with 10 vertices and 14 edge\left(s\right)}}$ (7)
> $\mathrm{Gcontracte}≔\mathrm{Contract}\left(G,e,'\mathrm{inplace}'=\mathrm{false}\right)$
${\mathrm{Gcontracte}}{≔}{\mathrm{Graph 4: an undirected unweighted graph with 9 vertices and 14 edge\left(s\right)}}$ (8)
> $\mathrm{expand}\left(f-\left(\mathrm{TuttePolynomial}\left(\mathrm{Gminuse},'x','y'\right)+\mathrm{TuttePolynomial}\left(\mathrm{Gcontracte},'x','y'\right)\right)\right)$
${0}$ (9)
ChromaticPolynomial
> $P≔\mathrm{ChromaticPolynomial}\left(G,'x'\right)$
${P}{≔}{x}{}\left({x}{-}{1}\right){}\left({x}{-}{2}\right){}\left({{x}}^{{7}}{-}{12}{}{{x}}^{{6}}{+}{67}{}{{x}}^{{5}}{-}{230}{}{{x}}^{{4}}{+}{529}{}{{x}}^{{3}}{-}{814}{}{{x}}^{{2}}{+}{775}{}{x}{-}{352}\right)$ (10)
• This must be zero since the Petersen graph is not 2-colorable
> $\mathrm{eval}\left(P,x=2\right)$
${0}$ (11)
> $\mathrm{eval}\left(P,x=3\right)$
${120}$ (12)
• We can verify the recurrence relation
> $\mathrm{expand}\left(P-\left(\mathrm{ChromaticPolynomial}\left(\mathrm{Gminuse},'x'\right)-\mathrm{ChromaticPolynomial}\left(\mathrm{Gcontracte},'x'\right)\right)\right)$
${0}$ (13)
> $K≔\mathrm{CompleteGraph}\left(10\right)$
${K}{≔}{\mathrm{Graph 5: an undirected unweighted graph with 10 vertices and 45 edge\left(s\right)}}$ (14)
• We can convince ourselves that this graph needs 10 colors and can be colored 10! ways with 10 colors
> $\mathrm{ChromaticPolynomial}\left(K,'t'\right)$
${t}{}\left({t}{-}{1}\right){}\left({t}{-}{2}\right){}\left({t}{-}{3}\right){}\left({t}{-}{4}\right){}\left({t}{-}{5}\right){}\left({t}{-}{6}\right){}\left({t}{-}{7}\right){}\left({t}{-}{8}\right){}\left({t}{-}{9}\right)$ (15)
> $\mathrm{ChromaticPolynomial}\left(K,10\right)-10!$
${0}$ (16)
> $\mathrm{ChromaticPolynomial}\left(\mathrm{RandomTree}\left(100\right),'t'\right)$
${t}{}{\left({t}{-}{1}\right)}^{{99}}$ (17)
FlowPolynomial and RankPolynomial
> $Q≔\mathrm{FlowPolynomial}\left(G,x\right)$
${Q}{≔}{{x}}^{{6}}{-}{15}{}{{x}}^{{5}}{+}{95}{}{{x}}^{{4}}{-}{325}{}{{x}}^{{3}}{+}{624}{}{{x}}^{{2}}{-}{620}{}{x}{+}{240}$ (18)
> $\mathrm{eval}\left(Q,x=4\right)$
${0}$ (19)
> $\mathrm{eval}\left(Q,x=5\right)$
${240}$ (20)
> $T≔\mathrm{TuttePolynomial}\left(G,0,1-x\right)$
${T}{≔}{{x}}^{{6}}{-}{15}{}{{x}}^{{5}}{+}{95}{}{{x}}^{{4}}{-}{325}{}{{x}}^{{3}}{+}{624}{}{{x}}^{{2}}{-}{620}{}{x}{+}{240}$ (21)
> $n≔\mathrm{NumberOfVertices}\left(G\right)$
${n}{≔}{10}$ (22)
> $m≔\mathrm{NumberOfEdges}\left(G\right)$
${m}{≔}{15}$ (23)
> $k≔\mathrm{nops}\left(\mathrm{ConnectedComponents}\left(G\right)\right)$
${k}{≔}{1}$ (24)
> $\mathrm{expand}\left(Q-{\left(-1\right)}^{m-n+k}T\right)$
${0}$ (25)
> $\mathrm{K4}≔\mathrm{CompleteGraph}\left(4\right)$
${\mathrm{K4}}{≔}{\mathrm{Graph 6: an undirected unweighted graph with 4 vertices and 6 edge\left(s\right)}}$ (26)
> $S≔\mathrm{RankPolynomial}\left(\mathrm{K4},x,y\right)$
${S}{≔}{{x}}^{{3}}{+}{{y}}^{{3}}{+}{6}{}{{x}}^{{2}}{+}{4}{}{x}{}{y}{+}{6}{}{{y}}^{{2}}{+}{15}{}{x}{+}{15}{}{y}{+}{16}$ (27)
• The number of subgraphs
> $\mathrm{eval}\left(S,\left\{x=1,y=1\right\}\right)$
${64}$ (28)
• The number of acyclic subgraphs
> $\mathrm{eval}\left(S,\left\{x=1,y=0\right\}\right)$
${38}$ (29)
• The number of subgraphs whose rank = rank(G)
> $\mathrm{eval}\left(S,\left\{x=0,y=1\right\}\right)$
${38}$ (30)
• The number of maximum spanning forests
> $\mathrm{eval}\left(S,\left\{x=0,y=0\right\}\right)$
${16}$ (31)
ReliabilityPolynomial and AcyclicPolynomial
> $P≔\mathrm{Graph}\left(\left\{\left\{1,2\right\},\left\{2,3\right\}\right\}\right)$
${P}{≔}{\mathrm{Graph 7: an undirected unweighted graph with 3 vertices and 2 edge\left(s\right)}}$ (32)
> $R≔\mathrm{ReliabilityPolynomial}\left(P,p\right)$
${R}{≔}{\left({1}{-}{p}\right)}^{{2}}$ (33)
> $\mathrm{AcyclicPolynomial}\left(P,p\right)$
${1}$ (34)
• The reliability of a connected network should increase if we add an edge. The difference Q-P below is positive for $0.
> $\mathrm{AddEdge}\left(P,\left\{1,3\right\}\right)$
${\mathrm{Graph 7: an undirected unweighted graph with 3 vertices and 3 edge\left(s\right)}}$ (35)
> $Q≔\mathrm{ReliabilityPolynomial}\left(P,p\right)$
${Q}{≔}\left({2}{}{p}{+}{1}\right){}{\left({1}{-}{p}\right)}^{{2}}$ (36)
> $\mathrm{factor}\left(Q-R\right)$
${2}{}{\left({-}{1}{+}{p}\right)}^{{2}}{}{p}$ (37)
> $\mathrm{expand}\left(\mathrm{AcyclicPolynomial}\left(P,p\right)\right)$
${-}{{p}}^{{3}}{+}{1}$ (38)
• Multiple edges may be specified as edge weights.
> $G≔\mathrm{Graph}\left(\left\{\left[\left\{1,2\right\},2\right],\left[\left\{1,3\right\},1\right],\left[\left\{2,3\right\},1\right]\right\}\right)$
${G}{≔}{\mathrm{Graph 8: an undirected weighted graph with 3 vertices and 3 edge\left(s\right)}}$ (39)
> $\mathrm{ReliabilityPolynomial}\left(G,p\right)$
$\left({2}{}{{p}}^{{2}}{+}{2}{}{p}{+}{1}\right){}{\left({1}{-}{p}\right)}^{{2}}$ (40)
• The following graph represents the Arpanet, the early internet, in late 1970.
> $V≔\left[\mathrm{MIT},\mathrm{LINCOLN},\mathrm{CASE},\mathrm{CMU},\mathrm{HARVARD},\mathrm{BBN},\mathrm{UCSB},\mathrm{UCLA},\mathrm{STANFORD},\mathrm{SRI},\mathrm{RAND},\mathrm{UTAH},\mathrm{SDC}\right]:$
> $\mathrm{Arpanet}≔\mathrm{Graph}\left(V,\mathrm{Trail}\left(\mathrm{UCSB},\mathrm{UCLA},\mathrm{RAND},\mathrm{SDC},\mathrm{UTAH},\mathrm{SRI},\mathrm{STANFORD},\mathrm{UCLA},\mathrm{SRI},\mathrm{UCSB}\right),\left\{\left\{\mathrm{BBN},\mathrm{RAND}\right\},\left\{\mathrm{MIT},\mathrm{UTAH}\right\}\right\},\mathrm{Trail}\left(\mathrm{MIT},\mathrm{LINCOLN},\mathrm{CASE},\mathrm{CMU},\mathrm{HARVARD},\mathrm{BBN},\mathrm{MIT}\right)\right)$
${\mathrm{Arpanet}}{≔}{\mathrm{Graph 9: an undirected unweighted graph with 13 vertices and 17 edge\left(s\right)}}$ (41)
> $\mathrm{SetVertexPositions}\left(\mathrm{Arpanet},\left[\left[1.0,1.0\right],\left[0.9,1.2\right],\left[0.5,1.1\right],\left[0.6,0.8\right],\left[1.0,0.6\right],\left[1.0,0.8\right],\left[-1.1,0.1\right],\left[-0.8,0.3\right],\left[-0.6,0.5\right],\left[-0.8,0.7\right],\left[-0.8,-0.1\right],\left[-0.3,0.9\right],\left[-0.5,0.2\right]\right]\right)$
> $\mathrm{DrawGraph}\left(\mathrm{Arpanet}\right)$
> $R≔\mathrm{ReliabilityPolynomial}\left(\mathrm{Arpanet},p\right)$
${R}{≔}\left({280}{}{{p}}^{{5}}{+}{310}{}{{p}}^{{4}}{+}{186}{}{{p}}^{{3}}{+}{63}{}{{p}}^{{2}}{+}{12}{}{p}{+}{1}\right){}{\left({1}{-}{p}\right)}^{{12}}$ (42)
• Which edge (link) should we add to the Arpanet to improve the reliability the most? Let's try adding the edge from Stanford to CMU.
> $H≔\mathrm{AddEdge}\left(\mathrm{Arpanet},\left\{\mathrm{CMU},\mathrm{STANFORD}\right\},\mathrm{inplace}=\mathrm{false}\right)$
${H}{≔}{\mathrm{Graph 10: an undirected unweighted graph with 13 vertices and 18 edge\left(s\right)}}$ (43)
> $S≔\mathrm{ReliabilityPolynomial}\left(H,p\right)$
${S}{≔}\left({976}{}{{p}}^{{6}}{+}{1118}{}{{p}}^{{5}}{+}{703}{}{{p}}^{{4}}{+}{276}{}{{p}}^{{3}}{+}{72}{}{{p}}^{{2}}{+}{12}{}{p}{+}{1}\right){}{\left({1}{-}{p}\right)}^{{12}}$ (44)
• We can compare the reliability polynomials visually for $0\le p\le 1$ then compute the improvement by computing the area of the enclosed curves as an integral.
> $\mathrm{plot}\left(\left[R,S\right],p=0..1,\mathrm{color}=\left[\mathrm{blue},\mathrm{red}\right]\right)$
> $\mathrm{improvement}≔\mathrm{int}\left(S-R,p=0..1\right)$
${\mathrm{improvement}}{≔}\frac{{443879}}{{10581480}}$ (45)
> $\mathrm{evalf}\left(\mathrm{improvement}\right)$
${0.04194866881}$ (46)
References
[1] Bollobás, Béla. Modern Graph Theory. Graduate Texts in Mathematics, 184, Springer-Verlag, New York, 1998.
[2] Farr, Khatarinejad, Khodadad, and Monagan. A Graph Theory Package for Maple. Proceedings of the 2005 Maple Conference, Maplesoft, July 2005: 260-271.
[3] Haggard, Pearce, and Royle. "Computing Tutte Polynomials." TOMS. Vol. 37(24). (2010): 1-17.
[4] Monagan. "A new edge selection heuristic for computing Tutte polynomials." Proceedings of FPSAC 2012.
Compatibility
• The GraphTheory[ReliabilityPolynomial] command was introduced in Maple 17.
• For more information on Maple 17 changes, see Updates in Maple 17. | 5,307 | 14,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 103, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.673075 |
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| By Jensenmath9
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Jensenmath9
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Quizzes Created: 7 | Total Attempts: 3,884
Questions: 6 | Attempts: 382
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• 1.
### Here are the number of points scored by Crosby in each of his NHL seasons: 37, 56, 66, 72, 102, 103, 109, 120 Calculate the Interquartile range
45
Explanation
The interquartile range is a measure of the spread or variability of a dataset. It is calculated by finding the difference between the upper quartile and the lower quartile. In this case, the lower quartile is the median of the first four numbers (56 and 66) and the upper quartile is the median of the last four numbers (103 and 109). So, the interquartile range is 109 - 56 = 45.
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• 2.
### Here are the number of points scored by Crosby in each of his NHL seasons: 37, 56, 66, 72, 102, 103, 109, 120 Calculate the standard deviation (round to 1 decimal place)
27.5
Explanation
The standard deviation is a measure of how spread out the data is from the mean. To calculate the standard deviation, we first find the mean of the data set, which is the sum of all the numbers divided by the total number of numbers. In this case, the mean is 81.9. Then, we subtract the mean from each data point, square the result, and sum all the squared differences. Next, we divide the sum by the total number of numbers minus 1, and finally, take the square root of the result. In this case, the standard deviation is 27.5.
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• 3.
0.94
• 4.
### The mean of a set of 4 number is 19. What would the mean be if each of the numbers was decreased by 11?
• A.
8
• B.
11
• C.
19
• D.
30
A. 8
Explanation
If each of the numbers in the set is decreased by 11, the new set of numbers would be:
-3 (8 minus 11)
0 (11 minus 11)
8 (19 minus 11)
19 (30 minus 11)
To find the mean of this new set, we add up all the numbers and divide by the total count. In this case, the sum of the new set is 24, and since there are 4 numbers, the mean would be 24 divided by 4, which is 6.
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• 5.
### The mean of a set of 5 numbers is 26. What would the mean be if one of the numbers was increased by 10?
• A.
10
• B.
26
• C.
28
• D.
36
C. 28
Explanation
If one of the numbers in the set is increased by 10, the new sum of the set would be the original sum plus 10. Since the mean is calculated by dividing the sum by the number of values, the new mean would be the new sum divided by 5 (since there are still 5 numbers in the set). Therefore, the new mean would be (original sum + 10) / 5, which simplifies to (130 + 10) / 5 = 140 / 5 = 28.
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Jensenmath9 | 908 | 3,378 | {"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} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.911196 |
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posted by .
∫1(3e^x+3x^2 )dx
0
• calculus -
I guess you mean integral from 0 to one
[ 3 e^x + x^3 ] at x = 1 - at x = 0
at x = 1
3 e^1 + 1 = 3e+1
at x = 0
3 e^0 + 0 = 3
so in the end
3 e - 2 | 109 | 209 | {"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} | 3.453125 | 3 | CC-MAIN-2017-26 | latest | en | 0.889899 |
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### Author Topic: Split from "The law of conservation of mass?" (Read 3464 times)
#### Pmb
• Neilep Level Member
• Posts: 1838
• Physicist
##### Split from "The law of conservation of mass?"
« on: 18/10/2011 16:31:06 »
Mod Note - this thread split from original question to allow discussion of off-topic matter
Hi Kat
Quote from: katblakeslee
The law of conservation of mass simply states that matter can not be created nor destroyed,...
Let us define matter as it was defined by Einstein as well as in other modern text books. Einstein wrote n his 1906 review paper on General Relativity. In the test The Principle of Relativity on page 14. Read section 14 The Field Equations of of Gravitation in the absence of matter. It reads
Quote
We make a distinction hearafter between "gravitational field" and "matter" in this way, that we denote everything but the gravitational field as "matter". Our use of the therefore includes not only in the ordinary sense, but the electric field as well.
You might be interested in the following web page of mine at
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
Quote
... but when we look at any object it had to have come from somewhere ..
Have you considered that it may have come from nowhere? Near the beginning of the Big-Bang event it seems clear to me that the universe started out as a quantum singularity which eventually led to the big bang. It's possible that the universe started out with zero total energy. As the Big-Bang progressed, matter was created in two forms, i.e. matter which had positive energy and that with negative energy, each form having matter.
Best wishes Kat
Pete
« Last Edit: 19/10/2011 10:43:05 by imatfaal »
#### yor_on
• Naked Science Forum GOD!
• Posts: 11697
• Thanked: 1 times
• (Ah, yes:) *a table is always good to hide under*
##### Re: Split from "The law of conservation of mass?"
« Reply #1 on: 18/10/2011 21:08:20 »
It's a interesting point Pmb, as if we was at the receiving end of a prism splitting light into different 'types'. Or maybe we are inside a 'fault line' of something else, where our SpaceTime 'doesn't exist', where it comes from?
If we assume it to be 'energy' then we don't have 'negative and positive' energy, because that would break the conservation laws as I see it. It would be very near the way I think 'real antimatter' should work, just silently disappearing in contact with matter, leaving no 'residue' of any kind definable, not even as entropy's 'work done'. What we call 'anti matter' today is not of that kind, when it meets matter we get a lot of 'energy' released.
« Last Edit: 18/10/2011 21:10:32 by yor_on »
#### Pmb
• Neilep Level Member
• Posts: 1838
• Physicist
##### Re: Split from "The law of conservation of mass?"
« Reply #2 on: 18/10/2011 21:35:43 »
It's a interesting point Pmb, as if we was at the receiving end of a prism splitting light into different 'types'. Or maybe we are inside a 'fault line' of something else, where our SpaceTime 'doesn't exist', where it comes from?
If we assume it to be 'energy' then we don't have 'negative and positive' energy, because that would break the conservation laws as I see it. It would be very near the way I think 'real antimatter' should work, just silently disappearing in contact with matter, leaving no 'residue' of any kind definable, not even as entropy's 'work done'. What we call 'anti matter' today is not of that kind, when it meets matter we get a lot of 'energy' released.
Have you ever browsed through an article I wrote on this subject? It has yet to be published, but that will happen when it happens.
There are two;
http://arxiv.org/abs/0709.0687
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
Best wishes
Pete
#### yor_on
• Naked Science Forum GOD!
• Posts: 11697
• Thanked: 1 times
• (Ah, yes:) *a table is always good to hide under*
##### Re: Split from "The law of conservation of mass?"
« Reply #3 on: 19/10/2011 02:17:36 »
#### MikeS
• Neilep Level Member
• Posts: 1044
##### Re: Split from "The law of conservation of mass?"
« Reply #4 on: 19/10/2011 08:11:58 »
Hi Kat
Quote
... but when we look at any object it had to have come from somewhere ..
Have you considered that it may have come from nowhere? Near the beginning of the Big-Bang event it seems clear to me that the universe started out as a quantum singularity which eventually led to the big bang. It's possible that the universe started out with zero total energy. As the Big-Bang progressed, matter was created in two forms, i.e. matter which had positive energy and that with negative energy, each form having matter.
Best wishes Kat
Pete
Pete
Presumably you are talking about the creation of matter and antimatter? Both are created from energy (photons) but photons are their own antiparticle, so I don't understand what you mean by negative energy?
Granted, matter and anti-matter (if they gravitationally repel) could be thought of as cancelling.
The big bang accounts for the surplus of matter over antimatter by allowing a very slight lop sided creation favouring matter. This next bit is not a mainstream idea but I mention it as it fits in with the idea of the universe being a free lunch. What if, the creation of matter and antimatter were equal and what wasn't mutually annihilated gravitationally sorted to become two sister universes. Each going different directions in time relative to the other. That would mean that regardless of the time the universes had existed the total time (as a pair) that they had existed would always be zero. That scenario seems to fit in with quantum mechanics nicely.
#### JP
• Neilep Level Member
• Posts: 3366
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##### Split from "The law of conservation of mass?"
« Reply #5 on: 19/10/2011 14:26:32 »
Mike, the negative energy comes about because Einstein's equation for the relationship between energy, momentum and mass in special relativity is (setting c=1 to make it look simple):
E2=m2+p2,
where E is energy, m is (rest or invariant) mass and p is momentum.
To solve for the energy, you take a square root, and that can give positive or negative solutions. Dirac was the first to really work out that these negative energy particles were antimatter--no one had actually seen antimatter, so he was just working out the math. Later on, people saw antimatter and verified that his math was describing real particles by using negative energy.
#### yor_on
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##### Split from "The law of conservation of mass?"
« Reply #6 on: 19/10/2011 15:03:19 »
So we set a scale, and then we measure. Under that 0 we call it negative, above we call it positive. Does that mean that energy comes in two qualities?
Not as I see it. If you from a anti particle annihilating with a particle gets a 'positive energy' (as in radiation), sufficient to cover both particles energy/momentum then that last energy is of only one quality.
And that's also what the conservation laws and entropy demands.
« Last Edit: 19/10/2011 15:28:45 by yor_on »
#### yor_on
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##### Split from "The law of conservation of mass?"
« Reply #7 on: 19/10/2011 15:05:42 »
The real mystery is 'work done' here.
#### Pmb
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##### Split from "The law of conservation of mass?"
« Reply #8 on: 19/10/2011 16:22:46 »
Hi Kat
Quote
... but when we look at any object it had to have come from somewhere ..
Have you considered that it may have come from nowhere? Near the beginning of the Big-Bang event it seems clear to me that the universe started out as a quantum singularity which eventually led to the big bang. It's possible that the universe started out with zero total energy. As the Big-Bang progressed, matter was created in two forms, i.e. matter which had positive energy and that with negative energy, each form having matter.
Best wishes Kat
Pete
Pete
Presumably you are talking about the creation of matter and antimatter? Both are created from energy (photons) but photons are their own antiparticle, so I don't understand what you mean by negative energy?
Granted, matter and anti-matter (if they gravitationally repel) could be thought of as cancelling.
The big bang accounts for the surplus of matter over antimatter by allowing a very slight lop sided creation favouring matter. This next bit is not a mainstream idea but I mention it as it fits in with the idea of the universe being a free lunch. What if, the creation of matter and antimatter were equal and what wasn't mutually annihilated gravitationally sorted to become two sister universes. Each going different directions in time relative to the other. That would mean that regardless of the time the universes had existed the total time (as a pair) that they had existed would always be zero. That scenario seems to fit in with quantum mechanics nicely.
Actually I was thinking about positive energy and negative energy. The positive energy comes from matter, the negative energy comes from the gravitational field. I'm not 100% sure about this though. I thinkk I read it in either a cosmology text or an article.
#### MikeS
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##### Split from "The law of conservation of mass?"
« Reply #9 on: 20/10/2011 07:35:48 »
JP
Thanks
So negative energy is something that can appear by taking the square root, ok but I think I am correct in believing that we don't actually know of anything that is negative energy other than the gravitational field?
Pmb
I agree that gravity can be thought of as negative energy and it is easy to see why the negative energy of gravity would cancel the positive energy contained in matter. However that still leaves energy that is not contained in matter and it is difficult to see how negative gravitational energy can cancel that?
#### Pmb
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##### Split from "The law of conservation of mass?"
« Reply #10 on: 20/10/2011 15:44:32 »
JP
Thanks
So negative energy is something that can appear by taking the square root, ok but I think I am correct in believing that we don't actually know of anything that is negative energy other than the gravitational field?
Pmb
I agree that gravity can be thought of as negative energy and it is easy to see why the negative energy of gravity would cancel the positive energy contained in matter. However that still leaves energy that is not contained in matter and it is difficult to see how negative gravitational energy can cancel that?
I'm starting a self study of cosmology so by next year I should know much more about things like this.
Note - Please take notice that using educated guesses here.The universe is so enormous that the energy of the negative gravity is enourmous. Same can be said about the energy in matter (including the EM fields).
#### yor_on
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##### Split from "The law of conservation of mass?"
« Reply #11 on: 20/10/2011 20:49:44 »
To me it's a mathematical function. SpaceTime does not have a 'negative energy' because if it had the conservation laws would be wrong. Neither would an outcome between a anti particle and a particle reflect both energies (as a 'positive' function/radiation of them).
Which either makes me believe that my thoughts about what 'anti' something should be seen as is wrong, and so can't exist, or that the definition we use is one describing 'symmetries', but not 'negative energy'.
==
What would happen if we could run 'clocks' backward?
With 'energy'?
It wouldn't turn 'negative', but it would reverse.
« Last Edit: 21/10/2011 20:14:27 by yor_on »
#### MikeS
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##### Split from "The law of conservation of mass?"
« Reply #12 on: 22/10/2011 07:38:51 »
JP
Thanks
So negative energy is something that can appear by taking the square root, ok but I think I am correct in believing that we don't actually know of anything that is negative energy other than the gravitational field?
Pmb
I agree that gravity can be thought of as negative energy and it is easy to see why the negative energy of gravity would cancel the positive energy contained in matter. However that still leaves energy that is not contained in matter and it is difficult to see how negative gravitational energy can cancel that?
I'm starting a self study of cosmology so by next year I should know much more about things like this.
Note - Please take notice that using educated guesses here.The universe is so enormous that the energy of the negative gravity is enourmous. Same can be said about the energy in matter (including the EM fields).
Pmb
What I was getting at is gravity is created by mass, so presumably the gravitational negative energy is balanced by the positive energy contained in matter. That still leaves a surplus of positive energy that is not contained in matter. For the universe to have zero net energy what balances this surplus of positive energy?
#### Pmb
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##### Split from "The law of conservation of mass?"
« Reply #13 on: 22/10/2011 22:31:33 »
JP
Thanks
So negative energy is something that can appear by taking the square root, ok but I think I am correct in believing that we don't actually know of anything that is negative energy other than the gravitational field?
Pmb
I agree that gravity can be thought of as negative energy and it is easy to see why the negative energy of gravity would cancel the positive energy contained in matter. However that still leaves energy that is not contained in matter and it is difficult to see how negative gravitational energy can cancel that?
It's the gravitational field that the negative gravity comes from, on a uiversal level. Even on the human level, if you were at rest on earth then you youself have negative gravity.
Quote
I'm starting a self study of cosmology so by next year I should know much more about things like this.
Good man my friend. I too am doing the same thing. What text are you using?
Pmb
[quote[
What I was getting at is gravity is created by mass, ...
The gravitational field which is the cause of the universal acceleration, and that might not be caused by matter as we know it.
[quote
...so presumably the gravitational negative energy is balanced by the positive energy contained in matter.
Why? More matter then the greater the negative pat of the field.
Newtonian wise it's like this. If we let g[/g] represent the gravitational field then g = -Mn/d^2. The negative sign represents the negstive part of the megative field.
#### The Naked Scientists Forum
##### Split from "The law of conservation of mass?"
« Reply #13 on: 22/10/2011 22:31:33 » | 3,624 | 14,974 | {"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} | 2.640625 | 3 | CC-MAIN-2016-44 | longest | en | 0.936248 |
https://paroj.github.io/gltut/Positioning/Tut06%20Translation.html | 1,542,823,108,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039749562.99/warc/CC-MAIN-20181121173523-20181121195523-00169.warc.gz | 699,366,055 | 5,600 | ## Translation
The simplest space transformation operation is translation. Indeed, we have not only seen this transform before, it has been used in all of the tutorials with a perspective projection. Recall this line from the vertex shaders:
vec4 cameraPos = position + vec4(offset.x, offset.y, 0.0, 0.0);
This is a translation transformation: it is used to position the origin point of the initial space relative to the destination space. Since all of the coordinates in a space are relative to the origin point of that space, all a translation needs to do is add a vector to all of the coordinates in that space. The vector added to these values is the location of where the user wants the origin point relative to the destination coordinate system.
Here is a more concrete example. Let us say that an object which in its model space is near its origin. This means that, if we want to see that object in front of the camera, we must position the origin of the model in front of the camera. If the extent of the model is only [-1, 1] in model space, we can ensure that the object is visible by adding this vector to all of the model space coordinates: (0, 0, -3). This puts the origin of the model at that position in camera space.
Translation is ultimately just that simple. So let's make it needlessly complex. And the best tool for doing that: matrices. Oh, we could just use a 3D uniform vector to pass an offset to do the transformation. But matrices have hidden benefits we will explore very soon.
All of our position vectors are 4D vectors, with a final W coordinate that is always 1.0. In Tutorial 04, we took advantage of this with our perspective transformation matrix. The equation for the Z coordinate needed an additive term, so we put that term in the W column of the transformation matrix. Matrix multiplication causes the value in the W column to be multiplied by the W coordinate of the vector (which is 1) and added to the sum of the other terms.
But how do we keep the matrix from doing something to the other terms? We only want this matrix to apply an offset to the position. We do not want to have it modify the position in some other way.
This is done by modifying an identity matrix. An identity matrix is a matrix that, when performing matrix multiplication, will return the matrix (or vector) it was multiplied with. It is sort of like the number 1 with regular multiplication: 1*X = X. The 4x4 identity matrix looks like this:
Equation 6.2. Identity Matrix
$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$
To modify the identity matrix into one that is suitable for translation, we simply put the offset into the W column of the identity matrix.
Equation 6.3. Translation Matrix
$\begin{array}{c}\text{Translation}=\left(x,y,z\right)\\ \left[\begin{array}{cccc}1& 0& 0& x\\ 0& 1& 0& y\\ 0& 0& 1& z\\ 0& 0& 0& 1\end{array}\right]\end{array}$
The tutorial project cleverly titled Translation performs translation operations.
This tutorial renders 3 of the same object, all in different positions. One of the objects is positioned in the center of the screen, and the other two's positions orbit it at various speeds.
Because of the prevalence of matrix math, this is the first tutorial that uses the GLM math library. So let's take a look at the shader program initialization code to see it in action.
Example 6.1. Translation Shader Initialization
void InitializeProgram()
{
"PosColorLocalTransform.vert"));
"ColorPassthrough.frag"));
positionAttrib = glGetAttribLocation(theProgram, "position");
colorAttrib = glGetAttribLocation(theProgram, "color");
modelToCameraMatrixUnif = glGetUniformLocation(theProgram,
"modelToCameraMatrix");
cameraToClipMatrixUnif = glGetUniformLocation(theProgram,
"cameraToClipMatrix");
float fzNear = 1.0f; float fzFar = 45.0f;
cameraToClipMatrix[0].x = fFrustumScale;
cameraToClipMatrix[1].y = fFrustumScale;
cameraToClipMatrix[2].z = (fzFar + fzNear) / (fzNear - fzFar);
cameraToClipMatrix[2].w = -1.0f;
cameraToClipMatrix[3].z = (2 * fzFar * fzNear) / (fzNear - fzFar);
glUseProgram(theProgram);
glUniformMatrix4fv(cameraToClipMatrixUnif, 1, GL_FALSE,
glm::value_ptr(cameraToClipMatrix));
glUseProgram(0);
}
GLM takes a unique approach for a vector/matrix math library. It attempts to emulate GLSL's approach to vector operations where possible. It uses C++ operator overloading to effectively emulate GLSL. In many cases, GLM-based expressions would compile in GLSL.
The matrix cameraToClipMatrix is defined as a glm::mat4, which has the same properties as a GLSL mat4. Array indexing of a mat4, whether GLM or GLSL, returns the zero-based column of the matrix as a vec4.
The glm::value_ptr function is used to get a direct pointer to the matrix data, in column-major order. This is useful for uploading data to OpenGL, as shown with the call to glUniformMatrix4fv.
With the exception of getting a second uniform location (for our model transformation matrix), this code functions exactly as it did in previous tutorials.
There is one important note: fFrustumScale is not 1.0 anymore. Until now, the relative sizes of objects were not particularly meaningful. Now that we are starting to deal with more complex objects that have a particular scale, picking a proper field of view for the perspective projection is very important.
The new fFrustumScale is computed with this code:
Example 6.2. Frustum Scale Computation
float CalcFrustumScale(float fFovDeg)
{
const float degToRad = 3.14159f * 2.0f / 360.0f;
return 1.0f / tan(fFovRad / 2.0f);
}
const float fFrustumScale = CalcFrustumScale(45.0f);
The function CalcFrustumScale computes the frustum scale based on a field-of-view angle in degrees. The field of view in this case is the angle between the forward direction and the direction of the farmost-extent of the view.
This project, and many of the others in this tutorial, uses a fairly complex bit of code to manage the transform matrices for the various object instances. There is an Instance object for each actual object; it has a function pointer that is used to compute the object's offset position. The Instance object then takes that position and computes a transformation matrix, based on the current elapsed time, with this function:
Example 6.3. Translation Matrix Generation
glm::mat4 ConstructMatrix(float fElapsedTime)
{
glm::mat4 theMat(1.0f);
theMat[3] = glm::vec4(CalcOffset(fElapsedTime), 1.0f);
return theMat;
}
The glm::mat4 constructor that takes only a single value constructs what is known as a diagonal matrix. That is a matrix with all zeros except for along the diagonal from the upper-left to the lower-right. The values along that diagonal will be the value passed to the constructor. An identity matrix is just a diagonal matrix with 1 as the value along the diagonal.
This function simply replaces the W column of that identity matrix with the offset value.
This all produces the following:
Fork me on GitHub | 1,694 | 6,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 3.796875 | 4 | CC-MAIN-2018-47 | longest | en | 0.926868 |
http://mathhelpforum.com/calculus/8321-slope-predicator-function-4-step-process.html | 1,527,177,819,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00308.warc.gz | 186,965,075 | 10,481 | 1. ## slope-predicator function/4-step process?
Given $\displaystyle f(x)=\frac{2}{x-1}$ , use the four step process to find a slope-predicator function m(x). Then write an equation for the line tangent to the curve at the point $\displaystyle x = 0$.
2. I am not sure what you are asking.
If you want the tangent at $\displaystyle x=0$.
Use the slope-intercept form.
$\displaystyle y-y_0=m(x-x_0)$
Where,
$\displaystyle m=f'(0)$
And,
$\displaystyle x_0=0$ but then $\displaystyle y_0=\frac{2}{x_0-1}=-2$.
Also,
$\displaystyle f'(x)=-2(x-1)^{-2}$
At, $\displaystyle x=0$ we have,
$\displaystyle f'(0)=-2(0-1)^{-2}=-2$
Thus,
$\displaystyle y+2=-2(x-0)$
Thus,
$\displaystyle y=-2x-2$
Is equation of tangent line.
3. My text explains it as follows:
1. Write the definition of the derivative
2. Substitute the expresions f(x+h) and f(x) as determined by the paticular function of f.
3. Simplify the result by algebraic methods until it is possible to....
4. Apply appropriate limit laws to finally evaluate the limit.
4. Hello, FLTR!
Someone is inventing a new language ?!
Given $\displaystyle f(x) \:=\:\frac{2}{x-1}$, use the four-step process
to find a slope-predicator function m(x). . Why can't they say "derivative"?
Then write an equation for the line tangent to the curve at the point $\displaystyle x = 0$.
I assume your difficulty is with the algebra in the four-step process.
We have: .$\displaystyle f(x+h) - f(x)\;=\;\frac{2}{x+h-1} - \frac{2}{x-1}$
Get a common denominator:
. . $\displaystyle = \;\frac{2}{x+h-1}\cdot\frac{x-1}{x-1} \,- \,\frac{2}{x-1}\cdot\frac{x+h-1}{x+h-1} \;=\;\frac{2(x-1) - 2(x+h-1)}{(x-1)(x+h-1)}$
. . $\displaystyle = \;\frac{2x - 2 - 2x - 2h + 2}{(x-1)(x+h-1)} \;=\;\frac{-2h}{(x-1)(x+h-1)}$
Divide by $\displaystyle h\!:\;\;\frac{f+h) - f(x)}{h}\;=\;\frac{-2\not{h}}{\not{h}(x-1)(x+h-1)} \;= \;\frac{-2}{(x-1)(x+h-1)}$
Take limits: .$\displaystyle f'(x) \;= \;\lim_{h\to0}\frac{f(x+h) - f(x)}{h} \;=\;\lim_{h\to0}\frac{-2}{(x-1)(x+h-1)}$
Therefore: .$\displaystyle \boxed{f'(x)\;=\;\frac{-2}{(x-1)^2}}$
I assume you can finish the problem now . . . | 769 | 2,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.637889 |
http://www.exampleproblems.com/wiki/index.php/Cross_product | 1,657,160,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00420.warc.gz | 83,603,870 | 12,083 | # Cross product
In mathematics, the cross product is a binary operation on vectors in a three dimensional vector space. It is also known as the vector product or outer product. It differs from the dot product in that it results in a pseudovector rather than in a scalar. Its main use lies in the fact that the cross product of two vectors is orthogonal to both of them.
## Definition
The cross product of the two vectors a and b is denoted by a × b (in longhand some mathematicians write ab to avoid confusion with the letter x - this should not be confused with the logical and operator, $\displaystyle \and$ ). It can be defined by
$\displaystyle \mathbf{a} \times \mathbf{b} = \mathbf\hat{n} \left| \mathbf{a} \right| \left| \mathbf{b} \right| \sin \theta$
where θ is the measure of the angle between a and b (0° ≤ θ ≤ 180°) on the plane defined by the span of the vectors, and n is a unit vector perpendicular to both a and b.
The problem with this definition is that there are two unit vectors perpendicular to both a and b: if n is perpendicular, then so is −n.
Which vector is the "correct" one by convention depends upon the orientation of the vector space—i.e., on the handedness of the given orthogonal coordinate system (i, j, k). The cross product a × b is defined in such a way that (a, b, a × b) becomes right-handed if (i, j, k) is right-handed, or left-handed if (i, j, k) is left-handed.
An easy way to compute the direction of the resultant vector is the "right-hand rule." If the coordinate system is right-handed, one simply points the forefinger in the direction of the first operand and the middle finger in the direction of the second operand. Then, the resultant vector is coming out of the thumb.
Because the cross product depends on the choice of coordinate system, its result is referred to as a pseudovector. Fortunately, in nature cross products tend to come in pairs, so that the “handedness” of the coordinate system is undone by a second cross product.
The cross product can be represented graphically, with respect to a right-handed coordinate system, as shown in the picture below.
File:Crossproduct.png
## Properties
### Geometric meaning
The length of the cross product, can be interpreted as the area of the parallelogram having a and b as sides:
$\displaystyle |a \times b|$
This means that the magnitude of the triple product gives the volume V of the parallelepiped formed by a, b, and c:
$\displaystyle V = |\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})|.$
### Algebraic properties
The cross product is anticommutative,
a × b = -b × a,
a × (b + c) = a × b + a × c,
and compatible with scalar multiplication so that
(ra) × b = a × (rb) = r(a × b).
It is not associative, but satisfies the Jacobi identity:
a × (b × c) + b × (c × a) + c × (a × b) = 0.
The distributivity, linearity and Jacobi identity show that R3 together with vector addition and cross product forms a Lie algebra.
Further, two non-zero vectors a and b are parallel iff a × b = 0.
### Associativity
The vector cross product is an example of a non-associative map. In general
$\displaystyle \left(\mathbf{A}\times\mathbf{B}\right)\times\mathbf{C}\neq \mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right).$
To see this, consider the case where $\displaystyle \mathbf{A}$ and $\displaystyle \mathbf{B}$ are parallel to one another. Then the left hand side is zero, and the right hand side is (in general) non-zero.
### Matrix notation
The unit vectors i, j, and k from the given orthogonal coordinate system satisfy the following equalities:
i × j = k j × k = i k × i = j.
With these rules, the coordinates of the cross product of two vectors can be computed easily, without the need to determine any angles: Let
a = a1i + a2j + a3k = [a1, a2, a3]
and
b = b1i + b2j + b3k = [b1, b2, b3].
Then
a × b = [a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1].
The above component notation can also be written formally as the determinant of a matrix:
$\displaystyle \mathbf{a}\times\mathbf{b}=\det \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{bmatrix}.$
The determinant of three vectors can be recovered as
det (a, b, c) = a · (b × c).
Intuitively, the cross product can be described by Sarrus's scheme. Consider the table
$\displaystyle \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} & \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 & a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 & b_1 & b_2 & b_3 \end{matrix}$
For the first three unit vectors, multiply the elements on the diagonal to the right (e.g. the first diagonal would contain i, a2, and b3). For the last three unit vectors, multiply the elements on the diagonal to the left and then negate the product (e.g. the last diagonal would contain k, a2, and b1). The cross product would be defined by the sum of these products:
$\displaystyle \mathbf{i}(a_2b_3) + \mathbf{j}(a_3b_1) + \mathbf{k}(a_1b_2) - \mathbf{i}(a_3b_2) - \mathbf{j}(a_1b_3) - \mathbf{k}(a_2b_1).$
Although written here in terms of coordinates, it follows from the geometrical definition above that the cross product is invariant under rotations about the axis defined by $\displaystyle \mathbf{a}\times\mathbf{b}$ , and inverses under swapping $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}.$
The cross product can also be described in terms of quaternions. Notice for instance that the above given cross product relations among i, j, and k agree with the multiplicative relations among the quaternions i, j, and k. In general, if we represent a vector [a1, a2, a3] as the quaternion a1i + a2j + a3k, we obtain the cross product of two vectors by taking their product as quaternions and deleting the real part of the result (the real part will be the negative of the dot product of the two vectors). More about the connection between quaternion multiplication, vector operations and geometry can be found at [[quaternions and spatial rotation]].
### Lagrange's formula
While this is not strictly a property of the cross-product, it is an identity involving the cross-product which is very useful. It is written as
a × (b × c) = b(a · c) − c(a · b),
which is easier to remember as “BAC minus CAB”. This formula is very useful in simplifying vector calculations in physics. A special case regarding gradients and useful in vector calculus, is
$\displaystyle \begin{matrix} \nabla \times (\nabla \times \mathbf{f}) &=& \nabla (\nabla \cdot \mathbf{f} ) - (\nabla \cdot \nabla) \mathbf{f} \\ &=& \mbox{grad }(\mbox{div } \mathbf{f} ) - \mbox{laplacian } \mathbf{f}. \end{matrix}$
This is a special case of the more general Laplace-de Rham operator $\displaystyle \Delta = d \delta + \delta d$ .
Another useful identity of Lagrange is
$\displaystyle |a \times b|^2 + |a \cdot b|^2 = |a|^2 |b|^2.$
This is a special case of the multiplicativity $\displaystyle |vw| = |v| |w|$ of the norm in the quaternion algebra.
## Applications
The cross product occurs in the formula for the vector operator curl. It is also used to describe the Lorentz force experienced by a moving electrical charge in a magnetic field. The definitions of torque and angular momentum also involve the cross product.
The cross product can also be used to calculate the normal for a triangle or polygon.
Given a point p and a line through a and b in a plane, all with z coordinate zero, then the z component of (p-a) × (b-a) will be positive or negative, depending on which side of the line p is.
## Higher dimensions
A cross product for 7-dimensional vectors can be obtained in the same way by using the octonions instead of the quaternions. See seven dimensional cross product for the main article.
In general dimension, there is no direct analogue of the binary cross product. There is however the wedge product, which has similar properties, except that the wedge product of two vectors is now a 2-vector instead of an ordinary vector. The cross product can be interpreted as the wedge product in three dimensions after using Hodge duality to identify 2-vectors with vectors.
One can also construct an n-ary analogue of the cross product in Rn+1 given by
$\displaystyle \bigwedge(\mathbf{v}_1,\cdots,\mathbf{v}_n)= \begin{vmatrix} v_1{}^1 &\cdots &v_1{}^{n+1}\\ \vdots &\ddots &\vdots\\ v_n{}^1 & \cdots &v_n{}^{n+1}\\ \mathbf{e}_1 &\cdots &\mathbf{e}_{n+1} \end{vmatrix}.$
This formula is identical in structure to the determinant formula for the normal cross product in R3 except that the row of basis vectors is the last row in the determinant rather than the first. The reason for this is to ensure that the ordered vectors (v1,...,vn,Λ(v1,...,vn)) have a positive orientation with respect to (e1,...,en+1). If n is even, this modification leaves the value unchanged, so this convention agrees with the normal definition of the binary product. In the case that n is odd, however, the distinction must be kept. This n-ary form enjoys many of the same properties as the vector cross product: it is alternating and linear in its arguments, it is perpendicular to each argument, and its magnitude gives the hypervolume of the region bounded by the arguments. And just like the vector cross product, it can be defined in a coordinate independent way as the Hodge dual of the wedge product of the arguments.
The wedge product and dot product can be combined to form the Clifford product.
In the context of multilinear algebra, it is possible to define a generalized cross product in terms of parity such that the generalized cross product between two vectors of dimension n is a tensor of rank n−2. This is a different concept than what is discussed above. | 2,597 | 9,698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 18} | 4.40625 | 4 | CC-MAIN-2022-27 | latest | en | 0.922398 |
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# Rotation matrix
In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space. For example, using the convention below, the matrix
${\displaystyle R={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}}$
rotates points in the xy-plane counterclockwise through an angle θ with respect to the x axis about the origin of a two-dimensional Cartesian coordinate system. To perform the rotation on a plane point with standard coordinates v = (x, y), it should be written as a column vector, and multiplied by the matrix R:
${\displaystyle R\mathbf {v} \ =\ {\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}\ =\ {\begin{bmatrix}x\cos \theta -y\sin \theta \\x\sin \theta +y\cos \theta \end{bmatrix}}.}$
If x and y are the endpoint coordinates of a vector, where x is cosine and y is sine, then the above equations become the trigonometric summation angle formulae. Indeed, a rotation matrix can be seen as the trigonometric summation angle formulae in matrix form. One way to understand this is say we have a vector at an angle 30° from the x axis, and we wish to rotate that angle by a further 45°. We simply need to compute the vector endpoint coordinates at 75°.
The examples in this article apply to active rotations of vectors counterclockwise in a right-handed coordinate system (y counterclockwise from x) by pre-multiplication (R on the left). If any one of these is changed (such as rotating axes instead of vectors, a passive transformation), then the inverse of the example matrix should be used, which coincides with its transpose.
Since matrix multiplication has no effect on the zero vector (the coordinates of the origin), rotation matrices describe rotations about the origin. Rotation matrices provide an algebraic description of such rotations, and are used extensively for computations in geometry, physics, and computer graphics. In some literature, the term rotation is generalized to include improper rotations, characterized by orthogonal matrices with a determinant of −1 (instead of +1). These combine proper rotations with reflections (which invert orientation). In other cases, where reflections are not being considered, the label proper may be dropped. The latter convention is followed in this article.
Rotation matrices are square matrices, with real entries. More specifically, they can be characterized as orthogonal matrices with determinant 1; that is, a square matrix R is a rotation matrix if and only if RT = R−1 and det R = 1. The set of all orthogonal matrices of size n with determinant +1 forms a group known as the special orthogonal group SO(n), one example of which is the rotation group SO(3). The set of all orthogonal matrices of size n with determinant +1 or −1 forms the (general) orthogonal group O(n).
## In two dimensions
A counterclockwise rotation of a vector through angle θ. The vector is initially aligned with the x-axis.
In two dimensions, the standard rotation matrix has the following form:
${\displaystyle R(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}.}$
This rotates column vectors by means of the following matrix multiplication,
${\displaystyle {\begin{bmatrix}x'\\y'\\\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}{\begin{bmatrix}x\\y\\\end{bmatrix}}.}$
Thus, the new coordinates (x′, y′) of a point (x, y) after rotation are
{\displaystyle {\begin{aligned}x'&=x\cos \theta -y\sin \theta \,\\y'&=x\sin \theta +y\cos \theta \,\end{aligned}}.}
### Examples
For example, when the vector
${\displaystyle \mathbf {\hat {x}} ={\begin{bmatrix}1\\0\\\end{bmatrix}}}$
is rotated by an angle θ, its new coordinates are
${\displaystyle {\begin{bmatrix}\cos \theta \\\sin \theta \\\end{bmatrix}},}$
and when the vector
${\displaystyle \mathbf {\hat {y}} ={\begin{bmatrix}0\\1\\\end{bmatrix}}}$
is rotated by an angle θ, its new coordinates are
${\displaystyle {\begin{bmatrix}-\sin \theta \\\cos \theta \\\end{bmatrix}}.}$
### Direction
The direction of vector rotation is counterclockwise if θ is positive (e.g. 90°), and clockwise if θ is negative (e.g. −90°). Thus the clockwise rotation matrix is found as
${\displaystyle R(-\theta )={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \\\end{bmatrix}}.}$
The two-dimensional case is the only non-trivial (i.e. not one-dimensional) case where the rotation matrices group is commutative, so that it does not matter in which order multiple rotations are performed. An alternative convention uses rotating axes,[1] and the above matrices also represent a rotation of the axes clockwise through an angle θ.
### Non-standard orientation of the coordinate system
A rotation through angle θ with non-standard axes.
If a standard right-handed Cartesian coordinate system is used, with the x-axis to the right and the y-axis up, the rotation R(θ) is counterclockwise. If a left-handed Cartesian coordinate system is used, with x directed to the right but y directed down, R(θ) is clockwise. Such non-standard orientations are rarely used in mathematics but are common in 2D computer graphics, which often have the origin in the top left corner and the y-axis down the screen or page.[2]
See below for other alternative conventions which may change the sense of the rotation produced by a rotation matrix.
### Common rotations
Particularly useful are the matrices
${\displaystyle {\begin{bmatrix}0&-1\\[3pt]1&0\\\end{bmatrix}},\quad {\begin{bmatrix}-1&0\\[3pt]0&-1\\\end{bmatrix}},\quad {\begin{bmatrix}0&1\\[3pt]-1&0\\\end{bmatrix}}}$
for 90°, 180°, and 270° counter-clockwise rotations.
A 180° rotation (middle) followed by a positive 90° rotation (left) is equivalent to a single negative 90° (positive 270°) rotation (right). Each of these figures depicts the result of a rotation relative to an upright starting position (bottom left) and includes the matrix representation of the permutation applied by the rotation (center right), as well as other related diagrams. See "Permutation notation" on Wikiversity for details.
### Relationship with complex plane
Since
${\displaystyle {\begin{bmatrix}0&1\\-1&0\end{bmatrix}}^{2}\ =\ {\begin{bmatrix}-1&0\\0&-1\end{bmatrix}}\ =-I,}$
the matrices of the shape
${\displaystyle {\begin{bmatrix}x&y\\-y&x\end{bmatrix}}}$
form a ring isomorphic to the field of the complex numbers ${\displaystyle \mathbb {C} }$. Under this isomorphism, the rotation matrices correspond to circle of the unit complex numbers, the complex numbers of modulus 1.
If one identify ${\displaystyle \mathbb {R} ^{2}}$ with ${\displaystyle \mathbb {C} }$ through the linear isomorphism ${\displaystyle (a,b)\mapsto a+ib,}$ the action of a matrix of the above form on vectors of ${\displaystyle \mathbb {R} ^{2}}$ corresponds to the multiplication by the complex number x + iy, and rotations correspond to multiplication by complex numbers of modulus 1.
As every rotation matrix can be written
${\displaystyle {\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}},}$
the above correspondence associates such a matrix with the complex number
${\displaystyle \cos t+i\sin t=e^{it}}$
(this last equality is Euler's formula).
## In three dimensions
A positive 90° rotation around the y-axis (left) after one around the z-axis (middle) gives a 120° rotation around the main diagonal (right).
In the top left corner are the rotation matrices, in the bottom right corner are the corresponding permutations of the cube with the origin in its center.
### Basic rotations
A basic rotation (also called elemental rotation) is a rotation about one of the axes of a coordinate system. The following three basic rotation matrices rotate vectors by an angle θ about the x-, y-, or z-axis, in three dimensions, using the right-hand rule—which codifies their alternating signs. (The same matrices can also represent a clockwise rotation of the axes.[nb 1])
{\displaystyle {\begin{alignedat}{1}R_{x}(\theta )&={\begin{bmatrix}1&0&0\\0&\cos \theta &-\sin \theta \\[3pt]0&\sin \theta &\cos \theta \\[3pt]\end{bmatrix}}\\[6pt]R_{y}(\theta )&={\begin{bmatrix}\cos \theta &0&\sin \theta \\[3pt]0&1&0\\[3pt]-\sin \theta &0&\cos \theta \\\end{bmatrix}}\\[6pt]R_{z}(\theta )&={\begin{bmatrix}\cos \theta &-\sin \theta &0\\[3pt]\sin \theta &\cos \theta &0\\[3pt]0&0&1\\\end{bmatrix}}\end{alignedat}}}
For column vectors, each of these basic vector rotations appears counterclockwise when the axis about which they occur points toward the observer, the coordinate system is right-handed, and the angle θ is positive. Rz, for instance, would rotate toward the y-axis a vector aligned with the x-axis, as can easily be checked by operating with Rz on the vector (1,0,0):
${\displaystyle R_{z}(90^{\circ }){\begin{bmatrix}1\\0\\0\\\end{bmatrix}}={\begin{bmatrix}\cos 90^{\circ }&-\sin 90^{\circ }&0\\\sin 90^{\circ }&\quad \cos 90^{\circ }&0\\0&0&1\\\end{bmatrix}}{\begin{bmatrix}1\\0\\0\\\end{bmatrix}}={\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\\\end{bmatrix}}{\begin{bmatrix}1\\0\\0\\\end{bmatrix}}={\begin{bmatrix}0\\1\\0\\\end{bmatrix}}}$
This is similar to the rotation produced by the above-mentioned two-dimensional rotation matrix. See below for alternative conventions which may apparently or actually invert the sense of the rotation produced by these matrices.
### General rotations
Other rotation matrices can be obtained from these three using matrix multiplication. For example, the product
${\displaystyle R=R_{z}(\alpha )\,R_{y}(\beta )\,R_{x}(\gamma )={\overset {\text{yaw}}{\begin{bmatrix}\cos \alpha &-\sin \alpha &0\\\sin \alpha &\cos \alpha &0\\0&0&1\\\end{bmatrix}}}{\overset {\text{pitch}}{\begin{bmatrix}\cos \beta &0&\sin \beta \\0&1&0\\-\sin \beta &0&\cos \beta \\\end{bmatrix}}}{\overset {\text{roll}}{\begin{bmatrix}1&0&0\\0&\cos \gamma &-\sin \gamma \\0&\sin \gamma &\cos \gamma \\\end{bmatrix}}}}$
${\displaystyle R={\begin{bmatrix}\cos \alpha \cos \beta &\cos \alpha \sin \beta \sin \gamma -\sin \alpha \cos \gamma &\cos \alpha \sin \beta \cos \gamma +\sin \alpha \sin \gamma \\\sin \alpha \cos \beta &\sin \alpha \sin \beta \sin \gamma +\cos \alpha \cos \gamma &\sin \alpha \sin \beta \cos \gamma -\cos \alpha \sin \gamma \\-\sin \beta &\cos \beta \sin \gamma &\cos \beta \cos \gamma \\\end{bmatrix}}}$
represents a rotation whose yaw, pitch, and roll angles are α, β and γ, respectively. More formally, it is an intrinsic rotation whose Tait–Bryan angles are α, β, γ, about axes z, y, x, respectively. Similarly, the product
${\displaystyle R=R_{z}(\gamma )\,R_{y}(\beta )\,R_{x}(\alpha )={\begin{bmatrix}\cos \gamma &-\sin \gamma &0\\\sin \gamma &\cos \gamma &0\\0&0&1\\\end{bmatrix}}{\begin{bmatrix}\cos \beta &0&\sin \beta \\0&1&0\\-\sin \beta &0&\cos \beta \\\end{bmatrix}}{\begin{bmatrix}1&0&0\\0&\cos \alpha &-\sin \alpha \\0&\sin \alpha &\cos \alpha \\\end{bmatrix}}}$
represents an extrinsic rotation whose (improper) Euler angles are α, β, γ, about axes x, y, z.
These matrices produce the desired effect only if they are used to premultiply column vectors, and (since in general matrix multiplication is not commutative) only if they are applied in the specified order (see Ambiguities for more details).
### Conversion from rotation matrix and to axis–angle
Every rotation in three dimensions is defined by its axis (a vector along this axis is unchanged by the rotation), and its angle — the amount of rotation about that axis (Euler rotation theorem).
There are several methods to compute the axis and angle from a rotation matrix (see also axis–angle representation). Here, we only describe the method based on the computation of the eigenvectors and eigenvalues of the rotation matrix. It is also possible to use the trace of the rotation matrix.
#### Determining the axis
A rotation R around axis u can be decomposed using 3 endomorphisms P, (IP), and Q (click to enlarge).
Given a 3 × 3 rotation matrix R, a vector u parallel to the rotation axis must satisfy
${\displaystyle R\mathbf {u} =\mathbf {u} ,}$
since the rotation of u around the rotation axis must result in u. The equation above may be solved for u which is unique up to a scalar factor unless R = I.
Further, the equation may be rewritten
${\displaystyle R\mathbf {u} =I\mathbf {u} \implies \left(R-I\right)\mathbf {u} =0,}$
which shows that u lies in the null space of RI.
Viewed in another way, u is an eigenvector of R corresponding to the eigenvalue λ = 1. Every rotation matrix must have this eigenvalue, the other two eigenvalues being complex conjugates of each other. It follows that a general rotation matrix in three dimensions has, up to a multiplicative constant, only one real eigenvector.
One way to determine the rotation axis is by showing that:
{\displaystyle {\begin{aligned}0&=R^{\mathsf {T}}0+0\\&=R^{\mathsf {T}}\left(R-I\right)\mathbf {u} +\left(R-I\right)\mathbf {u} \\&=\left(R^{\mathsf {T}}R-R^{\mathsf {T}}+R-I\right)\mathbf {u} \\&=\left(I-R^{\mathsf {T}}+R-I\right)\mathbf {u} \\&=\left(R-R^{\mathsf {T}}\right)\mathbf {u} \end{aligned}}}
Since (RRT) is a skew-symmetric matrix, we can choose u such that
${\displaystyle [\mathbf {u} ]_{\times }=\left(R-R^{\mathsf {T}}\right).}$
The matrix–vector product becomes a cross product of a vector with itself, ensuring that the result is zero:
${\displaystyle \left(R-R^{\mathsf {T}}\right)\mathbf {u} =[\mathbf {u} ]_{\times }\mathbf {u} =\mathbf {u} \times \mathbf {u} =0\,}$
Therefore, if
${\displaystyle R={\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\\end{bmatrix}},}$
then
${\displaystyle \mathbf {u} ={\begin{bmatrix}h-f\\c-g\\d-b\\\end{bmatrix}}.}$
The magnitude of u computed this way is ||u|| = 2 sin θ, where θ is the angle of rotation.
This does not work if R is symmetric. Above, if RRT is zero, then all subsequent steps are invalid. In this case, it is necessary to diagonalize R and find the eigenvector corresponding to an eigenvalue of 1.
#### Determining the angle
To find the angle of a rotation, once the axis of the rotation is known, select a vector v perpendicular to the axis. Then the angle of the rotation is the angle between v and Rv.
A more direct method, however, is to simply calculate the trace: the sum of the diagonal elements of the rotation matrix. Care should be taken to select the right sign for the angle θ to match the chosen axis:
${\displaystyle \operatorname {tr} (R)=1+2\cos \theta ,}$
from which follows that the angle's absolute value is
${\displaystyle |\theta |=\arccos \left({\frac {\operatorname {tr} (R)-1}{2}}\right).}$
#### Rotation matrix from axis and angle
The matrix of a proper rotation R by angle θ around the axis u = (ux, uy, uz), a unit vector with u2
x
+ u2
y
+ u2
z
= 1
, is given by:[3]
${\displaystyle R={\begin{bmatrix}\cos \theta +u_{x}^{2}\left(1-\cos \theta \right)&u_{x}u_{y}\left(1-\cos \theta \right)-u_{z}\sin \theta &u_{x}u_{z}\left(1-\cos \theta \right)+u_{y}\sin \theta \\u_{y}u_{x}\left(1-\cos \theta \right)+u_{z}\sin \theta &\cos \theta +u_{y}^{2}\left(1-\cos \theta \right)&u_{y}u_{z}\left(1-\cos \theta \right)-u_{x}\sin \theta \\u_{z}u_{x}\left(1-\cos \theta \right)-u_{y}\sin \theta &u_{z}u_{y}\left(1-\cos \theta \right)+u_{x}\sin \theta &\cos \theta +u_{z}^{2}\left(1-\cos \theta \right)\end{bmatrix}}.}$
A derivation of this matrix from first principles can be found in section 9.2 here.[4] The basic idea to derive this matrix is dividing the problem into few known simple steps.
1. First rotate the given axis and the point such that the axis lies in one of the coordinate planes (xy, yz or zx)
2. Then rotate the given axis and the point such that the axis is aligned with one of the two coordinate axes for that particular coordinate plane (x, y or z)
3. Use one of the fundamental rotation matrices to rotate the point depending on the coordinate axis with which the rotation axis is aligned.
4. Reverse rotate the axis-point pair such that it attains the final configuration as that was in step 2 (Undoing step 2)
5. Reverse rotate the axis-point pair which was done in step 1 (undoing step 1)
This can be written more concisely as
${\displaystyle R=(\cos \theta )\,I+(\sin \theta )\,[\mathbf {u} ]_{\times }+(1-\cos \theta )\,(\mathbf {u} \otimes \mathbf {u} ),}$
where [u]× is the cross product matrix of u; the expression uu is the outer product, and I is the identity matrix. Alternatively, the matrix entries are:
${\displaystyle R_{jk}={\begin{cases}\cos ^{2}{\frac {\theta }{2}}+\sin ^{2}{\frac {\theta }{2}}\left(2u_{j}^{2}-1\right),\quad &{\text{if }}j=k\\2u_{j}u_{k}\sin ^{2}{\frac {\theta }{2}}-\varepsilon _{jkl}u_{l}\sin \theta ,\quad &{\text{if }}j\neq k\end{cases}}}$
where εjkl is the Levi-Civita symbol with ε123 = 1. This is a matrix form of Rodrigues' rotation formula, (or the equivalent, differently parametrized Euler–Rodrigues formula) with[nb 2]
${\displaystyle \mathbf {u} \otimes \mathbf {u} =\mathbf {u} \mathbf {u} ^{\mathsf {T}}={\begin{bmatrix}u_{x}^{2}&u_{x}u_{y}&u_{x}u_{z}\\[3pt]u_{x}u_{y}&u_{y}^{2}&u_{y}u_{z}\\[3pt]u_{x}u_{z}&u_{y}u_{z}&u_{z}^{2}\end{bmatrix}},\qquad [\mathbf {u} ]_{\times }={\begin{bmatrix}0&-u_{z}&u_{y}\\[3pt]u_{z}&0&-u_{x}\\[3pt]-u_{y}&u_{x}&0\end{bmatrix}}.}$
In ${\displaystyle \mathbb {R} ^{3}}$ the rotation of a vector x around the axis u by an angle θ can be written as:
${\displaystyle R_{\mathbf {u} }(\theta )\mathbf {x} =\mathbf {u} (\mathbf {u} \cdot \mathbf {x} )+\cos \left(\theta \right)(\mathbf {u} \times \mathbf {x} )\times \mathbf {u} +\sin \left(\theta \right)(\mathbf {u} \times \mathbf {x} )}$
If the 3D space is right-handed and θ > 0, this rotation will be counterclockwise when u points towards the observer (Right-hand rule). Explicitly, with ${\displaystyle ({\boldsymbol {\alpha }},{\boldsymbol {\beta }},\mathbf {u} )}$ a right-handed orthonormal basis,
${\displaystyle R_{\mathbf {u} }(\theta ){\boldsymbol {\alpha }}=\cos \left(\theta \right){\boldsymbol {\alpha }}+\sin \left(\theta \right){\boldsymbol {\beta }},\quad R_{\mathbf {u} }(\theta ){\boldsymbol {\beta }}=-\sin \left(\theta \right){\boldsymbol {\alpha }}+\cos \left(\theta \right){\boldsymbol {\beta }},\quad R_{\mathbf {u} }(\theta )\mathbf {u} =\mathbf {u} .}$
Note the striking merely apparent differences to the equivalent Lie-algebraic formulation below.
## Properties
For any n-dimensional rotation matrix R acting on ${\displaystyle \mathbb {R} ^{n},}$
• ${\displaystyle R^{\mathsf {T}}=R^{-1}}$ (The rotation is an orthogonal matrix)
It follows that:
• ${\displaystyle \det R=\pm 1}$
A rotation is termed proper if det R = 1, and improper (or a roto-reflection) if det R = –1. For even dimensions n = 2k, the n eigenvalues λ of a proper rotation occur as pairs of complex conjugates which are roots of unity: λ = e±j for j = 1, ..., k, which is real only for λ = ±1. Therefore, there may be no vectors fixed by the rotation (λ = 1), and thus no axis of rotation. Any fixed eigenvectors occur in pairs, and the axis of rotation is an even-dimensional subspace.
For odd dimensions n = 2k + 1, a proper rotation R will have an odd number of eigenvalues, with at least one λ = 1 and the axis of rotation will be an odd dimensional subspace. Proof:
${\displaystyle {\begin{array}{rcl}\det \left(R-I\right)&=&\det \left(R^{\mathsf {T}}\right)\det \left(R-I\right)\ =\ \det \left(R^{\mathsf {T}}R-R^{\mathsf {T}}\right)\ =\ \det \left(I-R^{\mathsf {T}}\right)\\&=&\det(I-R)\ =\ \left(-1\right)^{n}\det \left(R-I\right)\ =\ -\det \left(R-I\right).\end{array}}}$
Here I is the identity matrix, and we use det(RT) = det(R) = 1, as well as (−1)n = −1 since n is odd. Therefore, det(RI) = 0, meaning there is a null vector v with (R – I)v = 0, that is Rv = v, a fixed eigenvector. There may also be pairs of fixed eigenvectors in the even-dimensional subspace orthogonal to v, so the total dimension of fixed eigenvectors is odd.
For example, in 2-space n = 2, a rotation by angle θ has eigenvalues λ = e and λ = e, so there is no axis of rotation except when θ = 0, the case of the null rotation. In 3-space n = 3, the axis of a non-null proper rotation is always a unique line, and a rotation around this axis by angle θ has eigenvalues λ = 1, e, e. In 4-space n = 4, the four eigenvalues are of the form e±, e±. The null rotation has θ = φ = 0. The case of θ = 0, φ ≠ 0 is called a simple rotation, with two unit eigenvalues forming an axis plane, and a two-dimensional rotation orthogonal to the axis plane. Otherwise, there is no axis plane. The case of θ = φ is called an isoclinic rotation, having eigenvalues e± repeated twice, so every vector is rotated through an angle θ.
The trace of a rotation matrix is equal to the sum of its eigenvalues. For n = 2, a rotation by angle θ has trace 2 cos θ. For n = 3, a rotation around any axis by angle θ has trace 1 + 2 cos θ. For n = 4, and the trace is 2(cos θ + cos φ), which becomes 4 cos θ for an isoclinic rotation.
## Geometry
In Euclidean geometry, a rotation is an example of an isometry, a transformation that moves points without changing the distances between them. Rotations are distinguished from other isometries by two additional properties: they leave (at least) one point fixed, and they leave "handedness" unchanged. In contrast, a translation moves every point, a reflection exchanges left- and right-handed ordering, a glide reflection does both, and an improper rotation combines a change in handedness with a normal rotation.
If a fixed point is taken as the origin of a Cartesian coordinate system, then every point can be given coordinates as a displacement from the origin. Thus one may work with the vector space of displacements instead of the points themselves. Now suppose (p1, ..., pn) are the coordinates of the vector p from the origin O to point P. Choose an orthonormal basis for our coordinates; then the squared distance to P, by Pythagoras, is
${\displaystyle d^{2}(O,P)=\|\mathbf {p} \|^{2}=\sum _{r=1}^{n}p_{r}^{2}}$
which can be computed using the matrix multiplication
${\displaystyle \|\mathbf {p} \|^{2}={\begin{bmatrix}p_{1}\cdots p_{n}\end{bmatrix}}{\begin{bmatrix}p_{1}\\\vdots \\p_{n}\end{bmatrix}}=\mathbf {p} ^{\mathsf {T}}\mathbf {p} .}$
A geometric rotation transforms lines to lines, and preserves ratios of distances between points. From these properties it can be shown that a rotation is a linear transformation of the vectors, and thus can be written in matrix form, Qp. The fact that a rotation preserves, not just ratios, but distances themselves, is stated as
${\displaystyle \mathbf {p} ^{\mathsf {T}}\mathbf {p} =(Q\mathbf {p} )^{\mathsf {T}}(Q\mathbf {p} ),}$
or
{\displaystyle {\begin{aligned}\mathbf {p} ^{\mathsf {T}}I\mathbf {p} &{}=\left(\mathbf {p} ^{\mathsf {T}}Q^{\mathsf {T}}\right)(Q\mathbf {p} )\\&{}=\mathbf {p} ^{\mathsf {T}}\left(Q^{\mathsf {T}}Q\right)\mathbf {p} .\end{aligned}}}
Because this equation holds for all vectors, p, one concludes that every rotation matrix, Q, satisfies the orthogonality condition,
${\displaystyle Q^{\mathsf {T}}Q=I.}$
Rotations preserve handedness because they cannot change the ordering of the axes, which implies the special matrix condition,
${\displaystyle \det Q=+1.}$
Equally important, it can be shown that any matrix satisfying these two conditions acts as a rotation.
## Multiplication
The inverse of a rotation matrix is its transpose, which is also a rotation matrix:
{\displaystyle {\begin{aligned}\left(Q^{\mathsf {T}}\right)^{\mathsf {T}}\left(Q^{\mathsf {T}}\right)&=QQ^{\mathsf {T}}=I\\\det Q^{\mathsf {T}}&=\det Q=+1.\end{aligned}}}
The product of two rotation matrices is a rotation matrix:
{\displaystyle {\begin{aligned}\left(Q_{1}Q_{2}\right)^{\mathsf {T}}\left(Q_{1}Q_{2}\right)&=Q_{2}^{\mathsf {T}}\left(Q_{1}^{\mathsf {T}}Q_{1}\right)Q_{2}=I\\\det \left(Q_{1}Q_{2}\right)&=\left(\det Q_{1}\right)\left(\det Q_{2}\right)=+1.\end{aligned}}}
For n > 2, multiplication of n × n rotation matrices is generally not commutative.
{\displaystyle {\begin{aligned}Q_{1}&={\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}}&Q_{2}&={\begin{bmatrix}0&0&1\\0&1&0\\-1&0&0\end{bmatrix}}\\Q_{1}Q_{2}&={\begin{bmatrix}0&-1&0\\0&0&1\\-1&0&0\end{bmatrix}}&Q_{2}Q_{1}&={\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}}.\end{aligned}}}
Noting that any identity matrix is a rotation matrix, and that matrix multiplication is associative, we may summarize all these properties by saying that the n × n rotation matrices form a group, which for n > 2 is non-abelian, called a special orthogonal group, and denoted by SO(n), SO(n,R), SOn, or SOn(R), the group of n × n rotation matrices is isomorphic to the group of rotations in an n-dimensional space. This means that multiplication of rotation matrices corresponds to composition of rotations, applied in left-to-right order of their corresponding matrices.
## Ambiguities
Alias and alibi rotations
The interpretation of a rotation matrix can be subject to many ambiguities.
In most cases the effect of the ambiguity is equivalent to the effect of a rotation matrix inversion (for these orthogonal matrices equivalently matrix transpose).
Alias or alibi (passive or active) transformation
The coordinates of a point P may change due to either a rotation of the coordinate system CS (alias), or a rotation of the point P (alibi). In the latter case, the rotation of P also produces a rotation of the vector v representing P. In other words, either P and v are fixed while CS rotates (alias), or CS is fixed while P and v rotate (alibi). Any given rotation can be legitimately described both ways, as vectors and coordinate systems actually rotate with respect to each other, about the same axis but in opposite directions. Throughout this article, we chose the alibi approach to describe rotations. For instance,
${\displaystyle R(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}}}$
represents a counterclockwise rotation of a vector v by an angle θ, or a rotation of CS by the same angle but in the opposite direction (i.e. clockwise). Alibi and alias transformations are also known as active and passive transformations, respectively.
Pre-multiplication or post-multiplication
The same point P can be represented either by a column vector v or a row vector w. Rotation matrices can either pre-multiply column vectors (Rv), or post-multiply row vectors (wR). However, Rv produces a rotation in the opposite direction with respect to wR. Throughout this article, rotations produced on column vectors are described by means of a pre-multiplication. To obtain exactly the same rotation (i.e. the same final coordinates of point P), the equivalent row vector must be post-multiplied by the transpose of R (i.e. wRT).
Right- or left-handed coordinates
The matrix and the vector can be represented with respect to a right-handed or left-handed coordinate system. Throughout the article, we assumed a right-handed orientation, unless otherwise specified.
Vectors or forms
The vector space has a dual space of linear forms, and the matrix can act on either vectors or forms.
## Decompositions
### Independent planes
Consider the 3 × 3 rotation matrix
${\displaystyle Q={\begin{bmatrix}0.36&0.48&-0.80\\-0.80&0.60&0.00\\0.48&0.64&0.60\end{bmatrix}}.}$
If Q acts in a certain direction, v, purely as a scaling by a factor λ, then we have
${\displaystyle Q\mathbf {v} =\lambda \mathbf {v} ,}$
so that
${\displaystyle \mathbf {0} =(\lambda I-Q)\mathbf {v} .}$
Thus λ is a root of the characteristic polynomial for Q,
{\displaystyle {\begin{aligned}0&{}=\det(\lambda I-Q)\\&{}=\lambda ^{3}-{\tfrac {39}{25}}\lambda ^{2}+{\tfrac {39}{25}}\lambda -1\\&{}=(\lambda -1)\left(\lambda ^{2}-{\tfrac {14}{25}}\lambda +1\right).\end{aligned}}}
Two features are noteworthy. First, one of the roots (or eigenvalues) is 1, which tells us that some direction is unaffected by the matrix. For rotations in three dimensions, this is the axis of the rotation (a concept that has no meaning in any other dimension). Second, the other two roots are a pair of complex conjugates, whose product is 1 (the constant term of the quadratic), and whose sum is 2 cos θ (the negated linear term). This factorization is of interest for 3 × 3 rotation matrices because the same thing occurs for all of them. (As special cases, for a null rotation the "complex conjugates" are both 1, and for a 180° rotation they are both −1.) Furthermore, a similar factorization holds for any n × n rotation matrix. If the dimension, n, is odd, there will be a "dangling" eigenvalue of 1; and for any dimension the rest of the polynomial factors into quadratic terms like the one here (with the two special cases noted). We are guaranteed that the characteristic polynomial will have degree n and thus n eigenvalues. And since a rotation matrix commutes with its transpose, it is a normal matrix, so can be diagonalized. We conclude that every rotation matrix, when expressed in a suitable coordinate system, partitions into independent rotations of two-dimensional subspaces, at most n/2 of them.
The sum of the entries on the main diagonal of a matrix is called the trace; it does not change if we reorient the coordinate system, and always equals the sum of the eigenvalues. This has the convenient implication for 2 × 2 and 3 × 3 rotation matrices that the trace reveals the angle of rotation, θ, in the two-dimensional space (or subspace). For a 2 × 2 matrix the trace is 2 cos θ, and for a 3 × 3 matrix it is 1 + 2 cos θ. In the three-dimensional case, the subspace consists of all vectors perpendicular to the rotation axis (the invariant direction, with eigenvalue 1). Thus we can extract from any 3 × 3 rotation matrix a rotation axis and an angle, and these completely determine the rotation.
### Sequential angles
The constraints on a 2 × 2 rotation matrix imply that it must have the form
${\displaystyle Q={\begin{bmatrix}a&-b\\b&a\end{bmatrix}}}$
with a2 + b2 = 1. Therefore, we may set a = cos θ and b = sin θ, for some angle θ. To solve for θ it is not enough to look at a alone or b alone; we must consider both together to place the angle in the correct quadrant, using a two-argument arctangent function.
Now consider the first column of a 3 × 3 rotation matrix,
${\displaystyle {\begin{bmatrix}a\\b\\c\end{bmatrix}}.}$
Although a2 + b2 will probably not equal 1, but some value r2 < 1, we can use a slight variation of the previous computation to find a so-called Givens rotation that transforms the column to
${\displaystyle {\begin{bmatrix}r\\0\\c\end{bmatrix}},}$
zeroing b. This acts on the subspace spanned by the x- and y-axes. We can then repeat the process for the xz-subspace to zero c. Acting on the full matrix, these two rotations produce the schematic form
${\displaystyle Q_{xz}Q_{xy}Q={\begin{bmatrix}1&0&0\\0&\ast &\ast \\0&\ast &\ast \end{bmatrix}}.}$
Shifting attention to the second column, a Givens rotation of the yz-subspace can now zero the z value. This brings the full matrix to the form
${\displaystyle Q_{yz}Q_{xz}Q_{xy}Q={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}},}$
which is an identity matrix. Thus we have decomposed Q as
${\displaystyle Q=Q_{xy}^{-1}Q_{xz}^{-1}Q_{yz}^{-1}.}$
An n × n rotation matrix will have (n − 1) + (n − 2) + ⋯ + 2 + 1, or
${\displaystyle \sum _{k=1}^{n-1}k={\frac {n(n-1)}{2}}}$
entries below the diagonal to zero. We can zero them by extending the same idea of stepping through the columns with a series of rotations in a fixed sequence of planes. We conclude that the set of n × n rotation matrices, each of which has n2 entries, can be parameterized by n(n−1)/2 angles.
xzxw xzyw xyxw xyzw yxyw yxzw yzyw yzxw zyzw zyxw zxzw zxyw xzxb yzxb xyxb zyxb yxyb zxyb yzyb xzyb zyzb xyzb zxzb yxzb
In three dimensions this restates in matrix form an observation made by Euler, so mathematicians call the ordered sequence of three angles Euler angles. However, the situation is somewhat more complicated than we have so far indicated. Despite the small dimension, we actually have considerable freedom in the sequence of axis pairs we use; and we also have some freedom in the choice of angles. Thus we find many different conventions employed when three-dimensional rotations are parameterized for physics, or medicine, or chemistry, or other disciplines. When we include the option of world axes or body axes, 24 different sequences are possible. And while some disciplines call any sequence Euler angles, others give different names (Cardano, Tait–Bryan, roll-pitch-yaw) to different sequences.
One reason for the large number of options is that, as noted previously, rotations in three dimensions (and higher) do not commute. If we reverse a given sequence of rotations, we get a different outcome. This also implies that we cannot compose two rotations by adding their corresponding angles. Thus Euler angles are not vectors, despite a similarity in appearance as a triplet of numbers.
### Nested dimensions
A 3 × 3 rotation matrix such as
${\displaystyle Q_{3\times 3}={\begin{bmatrix}\cos \theta &-\sin \theta &{\color {CadetBlue}0}\\\sin \theta &\cos \theta &{\color {CadetBlue}0}\\{\color {CadetBlue}0}&{\color {CadetBlue}0}&{\color {CadetBlue}1}\end{bmatrix}}}$
suggests a 2 × 2 rotation matrix,
${\displaystyle Q_{2\times 2}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}},}$
is embedded in the upper left corner:
${\displaystyle Q_{3\times 3}=\left[{\begin{matrix}Q_{2\times 2}&\mathbf {0} \\\mathbf {0} ^{\mathsf {T}}&1\end{matrix}}\right].}$
This is no illusion; not just one, but many, copies of n-dimensional rotations are found within (n + 1)-dimensional rotations, as subgroups. Each embedding leaves one direction fixed, which in the case of 3 × 3 matrices is the rotation axis. For example, we have
{\displaystyle {\begin{aligned}Q_{\mathbf {x} }(\theta )&={\begin{bmatrix}{\color {CadetBlue}1}&{\color {CadetBlue}0}&{\color {CadetBlue}0}\\{\color {CadetBlue}0}&\cos \theta &-\sin \theta \\{\color {CadetBlue}0}&\sin \theta &\cos \theta \end{bmatrix}},\\[8px]Q_{\mathbf {y} }(\theta )&={\begin{bmatrix}\cos \theta &{\color {CadetBlue}0}&\sin \theta \\{\color {CadetBlue}0}&{\color {CadetBlue}1}&{\color {CadetBlue}0}\\-\sin \theta &{\color {CadetBlue}0}&\cos \theta \end{bmatrix}},\\[8px]Q_{\mathbf {z} }(\theta )&={\begin{bmatrix}\cos \theta &-\sin \theta &{\color {CadetBlue}0}\\\sin \theta &\cos \theta &{\color {CadetBlue}0}\\{\color {CadetBlue}0}&{\color {CadetBlue}0}&{\color {CadetBlue}1}\end{bmatrix}},\end{aligned}}}
fixing the x-axis, the y-axis, and the z-axis, respectively. The rotation axis need not be a coordinate axis; if u = (x,y,z) is a unit vector in the desired direction, then
{\displaystyle {\begin{aligned}Q_{\mathbf {u} }(\theta )&={\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}\sin \theta +\left(I-\mathbf {u} \mathbf {u} ^{\mathsf {T}}\right)\cos \theta +\mathbf {u} \mathbf {u} ^{\mathsf {T}}\\[8px]&={\begin{bmatrix}\left(1-x^{2}\right)c_{\theta }+x^{2}&-zs_{\theta }-xyc_{\theta }+xy&ys_{\theta }-xzc_{\theta }+xz\\zs_{\theta }-xyc_{\theta }+xy&\left(1-y^{2}\right)c_{\theta }+y^{2}&-xs_{\theta }-yzc_{\theta }+yz\\-ys_{\theta }-xzc_{\theta }+xz&xs_{\theta }-yzc_{\theta }+yz&\left(1-z^{2}\right)c_{\theta }+z^{2}\end{bmatrix}}\\[8px]&={\begin{bmatrix}x^{2}(1-c_{\theta })+c_{\theta }&xy(1-c_{\theta })-zs_{\theta }&xz(1-c_{\theta })+ys_{\theta }\\xy(1-c_{\theta })+zs_{\theta }&y^{2}(1-c_{\theta })+c_{\theta }&yz(1-c_{\theta })-xs_{\theta }\\xz(1-c_{\theta })-ys_{\theta }&yz(1-c_{\theta })+xs_{\theta }&z^{2}(1-c_{\theta })+c_{\theta }\end{bmatrix}},\end{aligned}}}
where cθ = cos θ, sθ = sin θ, is a rotation by angle θ leaving axis u fixed.
A direction in (n + 1)-dimensional space will be a unit magnitude vector, which we may consider a point on a generalized sphere, Sn. Thus it is natural to describe the rotation group SO(n + 1) as combining SO(n) and Sn. A suitable formalism is the fiber bundle,
${\displaystyle SO(n)\hookrightarrow SO(n+1)\to S^{n},}$
where for every direction in the base space, Sn, the fiber over it in the total space, SO(n + 1), is a copy of the fiber space, SO(n), namely the rotations that keep that direction fixed.
Thus we can build an n × n rotation matrix by starting with a 2 × 2 matrix, aiming its fixed axis on S2 (the ordinary sphere in three-dimensional space), aiming the resulting rotation on S3, and so on up through Sn−1. A point on Sn can be selected using n numbers, so we again have n(n − 1)/2 numbers to describe any n × n rotation matrix.
In fact, we can view the sequential angle decomposition, discussed previously, as reversing this process. The composition of n − 1 Givens rotations brings the first column (and row) to (1,0,...,0), so that the remainder of the matrix is a rotation matrix of dimension one less, embedded so as to leave (1, 0, ..., 0) fixed.
### Skew parameters via Cayley's formula
When an n × n rotation matrix Q, does not include a −1 eigenvalue, thus none of the planar rotations which it comprises are 180° rotations, then Q + I is an invertible matrix. Most rotation matrices fit this description, and for them it can be shown that (QI)(Q + I)−1 is a skew-symmetric matrix, A. Thus AT = −A; and since the diagonal is necessarily zero, and since the upper triangle determines the lower one, A contains 1/2n(n − 1) independent numbers.
Conveniently, IA is invertible whenever A is skew-symmetric; thus we can recover the original matrix using the Cayley transform,
${\displaystyle A\mapsto (I+A)(I-A)^{-1},}$
which maps any skew-symmetric matrix A to a rotation matrix. In fact, aside from the noted exceptions, we can produce any rotation matrix in this way. Although in practical applications we can hardly afford to ignore 180° rotations, the Cayley transform is still a potentially useful tool, giving a parameterization of most rotation matrices without trigonometric functions.
In three dimensions, for example, we have (Cayley 1846)
{\displaystyle {\begin{aligned}&{\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}\mapsto \\[3pt]\quad {\frac {1}{1+x^{2}+y^{2}+z^{2}}}&{\begin{bmatrix}1+x^{2}-y^{2}-z^{2}&2xy-2z&2y+2xz\\2xy+2z&1-x^{2}+y^{2}-z^{2}&2yz-2x\\2xz-2y&2x+2yz&1-x^{2}-y^{2}+z^{2}\end{bmatrix}}.\end{aligned}}}
If we condense the skew entries into a vector, (x,y,z), then we produce a 90° rotation around the x-axis for (1, 0, 0), around the y-axis for (0, 1, 0), and around the z-axis for (0, 0, 1). The 180° rotations are just out of reach; for, in the limit as x → ∞, (x, 0, 0) does approach a 180° rotation around the x axis, and similarly for other directions.
### Decomposition into shears
For the 2D case, a rotation matrix can be decomposed into three shear matrices (Paeth 1986):
{\displaystyle {\begin{aligned}R(\theta )&{}={\begin{bmatrix}1&-\tan {\frac {\theta }{2}}\\0&1\end{bmatrix}}{\begin{bmatrix}1&0\\\sin \theta &1\end{bmatrix}}{\begin{bmatrix}1&-\tan {\frac {\theta }{2}}\\0&1\end{bmatrix}}\end{aligned}}}
This is useful, for instance, in computer graphics, since shears can be implemented with fewer multiplication instructions than rotating a bitmap directly. On modern computers, this may not matter, but it can be relevant for very old or low-end microprocessors.
A rotation can also be written as two shears and scaling (Daubechies & Sweldens 1998):
{\displaystyle {\begin{aligned}R(\theta )&{}={\begin{bmatrix}1&0\\\tan \theta &1\end{bmatrix}}{\begin{bmatrix}1&-\sin \theta \cos \theta \\0&1\end{bmatrix}}{\begin{bmatrix}\cos \theta &0\\0&{\frac {1}{\cos \theta }}\end{bmatrix}}\end{aligned}}}
## Group theory
Below follow some basic facts about the role of the collection of all rotation matrices of a fixed dimension (here mostly 3) in mathematics and particularly in physics where rotational symmetry is a requirement of every truly fundamental law (due to the assumption of isotropy of space), and where the same symmetry, when present, is a simplifying property of many problems of less fundamental nature. Examples abound in classical mechanics and quantum mechanics. Knowledge of the part of the solutions pertaining to this symmetry applies (with qualifications) to all such problems and it can be factored out of a specific problem at hand, thus reducing its complexity. A prime example – in mathematics and physics – would be the theory of spherical harmonics. Their role in the group theory of the rotation groups is that of being a representation space for the entire set of finite-dimensional irreducible representations of the rotation group SO(3). For this topic, see Rotation group SO(3) § Spherical harmonics.
The main articles listed in each subsection are referred to for more detail.
### Lie group
The n × n rotation matrices for each n form a group, the special orthogonal group, SO(n). This algebraic structure is coupled with a topological structure inherited from GLn(${\displaystyle \mathbb {R} }$) in such a way that the operations of multiplication and taking the inverse are analytic functions of the matrix entries. Thus SO(n) is for each n a Lie group. It is compact and connected, but not simply connected. It is also a semi-simple group, in fact a simple group with the exception SO(4).[5] The relevance of this is that all theorems and all machinery from the theory of analytic manifolds (analytic manifolds are in particular smooth manifolds) apply and the well-developed representation theory of compact semi-simple groups is ready for use.
### Lie algebra
The Lie algebra so(n) of SO(n) is given by
${\displaystyle {\mathfrak {so}}(n)={\mathfrak {o}}(n)=\left\{X\in M_{n}(\mathbb {R} )\mid X=-X^{\mathsf {T}}\right\},}$
and is the space of skew-symmetric matrices of dimension n, see classical group, where o(n) is the Lie algebra of O(n), the orthogonal group. For reference, the most common basis for so(3) is
${\displaystyle L_{\mathbf {x} }=\left[{\begin{matrix}0&0&0\\0&0&-1\\0&1&0\end{matrix}}\right],\quad L_{\mathbf {y} }=\left[{\begin{matrix}0&0&1\\0&0&0\\-1&0&0\end{matrix}}\right],\quad L_{\mathbf {z} }=\left[{\begin{matrix}0&-1&0\\1&0&0\\0&0&0\end{matrix}}\right].}$
### Exponential map
Connecting the Lie algebra to the Lie group is the exponential map, which is defined using the standard matrix exponential series for eA[6] For any skew-symmetric matrix A, exp(A) is always a rotation matrix.[nb 3]
An important practical example is the 3 × 3 case. In rotation group SO(3), it is shown that one can identify every Aso(3) with an Euler vector ω = θu, where u = (x,y,z) is a unit magnitude vector.
By the properties of the identification su(2) ≅ ${\displaystyle \mathbb {R} ^{3}}$, u is in the null space of A. Thus, u is left invariant by exp(A) and is hence a rotation axis.
According to Rodrigues' rotation formula on matrix form, one obtains,
{\displaystyle {\begin{aligned}\exp(A)&=\exp {\bigl (}\theta (\mathbf {u} \cdot \mathbf {L} ){\bigr )}\\&=\exp \left(\left[{\begin{matrix}0&-z\theta &y\theta \\z\theta &0&-x\theta \\-y\theta &x\theta &0\end{matrix}}\right]\right)\\&=I+\sin \theta \ \mathbf {u} \cdot \mathbf {L} +(1-\cos \theta )(\mathbf {u} \cdot \mathbf {L} )^{2},\end{aligned}}}
where {\displaystyle {\begin{aligned}\mathbf {u} \cdot \mathbf {L} &=\left[{\begin{matrix}0&-z&y\\z&0&-x\\-y&x&0\end{matrix}}\right]\end{aligned}}.}
This is the matrix for a rotation around axis u by the angle θ. For full detail, see exponential map SO(3).
### Baker–Campbell–Hausdorff formula
The BCH formula provides an explicit expression for Z = log(eXeY) in terms of a series expansion of nested commutators of X and Y.[7] This general expansion unfolds as[nb 4]
${\displaystyle Z=C(X,Y)=X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}{\bigl [}X,[X,Y]{\bigr ]}-{\tfrac {1}{12}}{\bigl [}Y,[X,Y]{\bigr ]}+\cdots .}$
In the 3 × 3 case, the general infinite expansion has a compact form,[8]
${\displaystyle Z=\alpha X+\beta Y+\gamma [X,Y],}$
for suitable trigonometric function coefficients, detailed in the Baker–Campbell–Hausdorff formula for SO(3).
As a group identity, the above holds for all faithful representations, including the doublet (spinor representation), which is simpler. The same explicit formula thus follows straightforwardly through Pauli matrices; see the 2 × 2 derivation for SU(2). For the general n × n case, one might use Ref.[9]
### Spin group
The Lie group of n × n rotation matrices, SO(n), is not simply connected, so Lie theory tells us it is a homomorphic image of a universal covering group. Often the covering group, which in this case is called the spin group denoted by Spin(n), is simpler and more natural to work with.[10]
In the case of planar rotations, SO(2) is topologically a circle, S1. Its universal covering group, Spin(2), is isomorphic to the real line, R, under addition. Whenever angles of arbitrary magnitude are used one is taking advantage of the convenience of the universal cover. Every 2 × 2 rotation matrix is produced by a countable infinity of angles, separated by integer multiples of 2π. Correspondingly, the fundamental group of SO(2) is isomorphic to the integers, Z.
In the case of spatial rotations, SO(3) is topologically equivalent to three-dimensional real projective space, RP3. Its universal covering group, Spin(3), is isomorphic to the 3-sphere, S3. Every 3 × 3 rotation matrix is produced by two opposite points on the sphere. Correspondingly, the fundamental group of SO(3) is isomorphic to the two-element group, Z2.
We can also describe Spin(3) as isomorphic to quaternions of unit norm under multiplication, or to certain 4 × 4 real matrices, or to 2 × 2 complex special unitary matrices, namely SU(2). The covering maps for the first and the last case are given by
${\displaystyle \mathbb {H} \supset \{q\in \mathbb {H} :\|q\|=1\}\ni w+\mathbf {i} x+\mathbf {j} y+\mathbf {k} z\mapsto \left[{\begin{matrix}1-2y^{2}-2z^{2}&2xy-2zw&2xz+2yw\\2xy+2zw&1-2x^{2}-2z^{2}&2yz-2xw\\2xz-2yw&2yz+2xw&1-2x^{2}-2y^{2}\end{matrix}}\right]\in \mathrm {SO} (3),}$
and
${\displaystyle \mathrm {SU} (2)\ni \left[{\begin{matrix}\alpha &\beta \\-{\overline {\beta }}&{\overline {\alpha }}\end{matrix}}\right]\mapsto \left[{\begin{matrix}{\tfrac {1}{2}}(\alpha ^{2}-\beta ^{2}+{\overline {\alpha ^{2}}}-{\overline {\beta ^{2}}})&{\frac {i}{2}}(-\alpha ^{2}-\beta ^{2}+{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&-\alpha \beta -{\overline {\alpha }}{\overline {\beta }}\\{\tfrac {i}{2}}(\alpha ^{2}-\beta ^{2}-{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&{\frac {i}{2}}(\alpha ^{2}+\beta ^{2}+{\overline {\alpha ^{2}}}+{\overline {\beta ^{2}}})&-i(+\alpha \beta -{\overline {\alpha }}{\overline {\beta }})\\\alpha {\overline {\beta }}+{\overline {\alpha }}\beta &i(-\alpha {\overline {\beta }}+{\overline {\alpha }}\beta )&\alpha {\overline {\alpha }}-\beta {\overline {\beta }}\end{matrix}}\right]\in \mathrm {SO} (3).}$
For a detailed account of the SU(2)-covering and the quaternionic covering, see spin group SO(3).
Many features of these cases are the same for higher dimensions. The coverings are all two-to-one, with SO(n), n > 2, having fundamental group Z2. The natural setting for these groups is within a Clifford algebra. One type of action of the rotations is produced by a kind of "sandwich", denoted by qvq. More importantly in applications to physics, the corresponding spin representation of the Lie algebra sits inside the Clifford algebra. It can be exponentiated in the usual way to give rise to a 2-valued representation, also known as projective representation of the rotation group. This is the case with SO(3) and SU(2), where the 2-valued representation can be viewed as an "inverse" of the covering map. By properties of covering maps, the inverse can be chosen ono-to-one as a local section, but not globally.
### Infinitesimal rotations
The matrices in the Lie algebra are not themselves rotations; the skew-symmetric matrices are derivatives, proportional differences of rotations. An actual "differential rotation", or infinitesimal rotation matrix has the form
${\displaystyle I+A\,d\theta ,}$
where is vanishingly small and Aso(n), for instance with A = Lx,
${\displaystyle dL_{x}=\left[{\begin{matrix}1&0&0\\0&1&-d\theta \\0&d\theta &1\end{matrix}}\right].}$
The computation rules are as usual except that infinitesimals of second order are routinely dropped. With these rules, these matrices do not satisfy all the same properties as ordinary finite rotation matrices under the usual treatment of infinitesimals.[11] It turns out that the order in which infinitesimal rotations are applied is irrelevant. To see this exemplified, consult infinitesimal rotations SO(3).
## Conversions
We have seen the existence of several decompositions that apply in any dimension, namely independent planes, sequential angles, and nested dimensions. In all these cases we can either decompose a matrix or construct one. We have also given special attention to 3 × 3 rotation matrices, and these warrant further attention, in both directions (Stuelpnagel 1964).
### Quaternion
Given the unit quaternion q = w + xi + yj + zk, the equivalent left-handed (Post-Multiplied) 3 × 3 rotation matrix is
${\displaystyle Q={\begin{bmatrix}1-2y^{2}-2z^{2}&2xy-2zw&2xz+2yw\\2xy+2zw&1-2x^{2}-2z^{2}&2yz-2xw\\2xz-2yw&2yz+2xw&1-2x^{2}-2y^{2}\end{bmatrix}}.}$
Now every quaternion component appears multiplied by two in a term of degree two, and if all such terms are zero what is left is an identity matrix. This leads to an efficient, robust conversion from any quaternion – whether unit or non-unit – to a 3 × 3 rotation matrix. Given:
{\displaystyle {\begin{aligned}n&=w\times w+x\times x+y\times y+z\times z\\s&={\begin{cases}0&{\text{if }}n=0\\{\frac {2}{n}}&{\text{otherwise}}\end{cases}}\\wx&=s\times w\times x,&wy&=s\times w\times y,&wz&=s\times w\times z\\xx&=s\times x\times x,&xy&=s\times x\times y,&xz&=s\times x\times z\\yy&=s\times y\times y,&yz&=s\times y\times z,&zz&=s\times z\times z\end{aligned}}}
we can calculate
${\displaystyle Q={\begin{bmatrix}1-(yy+zz)&xy-wz&xz+wy\\xy+wz&1-(xx+zz)&yz-wx\\xz-wy&yz+wx&1-(xx+yy)\end{bmatrix}}}$
Freed from the demand for a unit quaternion, we find that nonzero quaternions act as homogeneous coordinates for 3 × 3 rotation matrices. The Cayley transform, discussed earlier, is obtained by scaling the quaternion so that its w component is 1. For a 180° rotation around any axis, w will be zero, which explains the Cayley limitation.
The sum of the entries along the main diagonal (the trace), plus one, equals 4 − 4(x2 + y2 + z2), which is 4w2. Thus we can write the trace itself as 2w2 + 2w2 − 1; and from the previous version of the matrix we see that the diagonal entries themselves have the same form: 2x2 + 2w2 − 1, 2y2 + 2w2 − 1, and 2z2 + 2w2 − 1. So we can easily compare the magnitudes of all four quaternion components using the matrix diagonal. We can, in fact, obtain all four magnitudes using sums and square roots, and choose consistent signs using the skew-symmetric part of the off-diagonal entries:
{\displaystyle {\begin{aligned}t&=Q_{xx}+Q_{yy}+Q_{zz}\quad ({\text{the trace of }}Q)\\r&={\sqrt {1+t}}\\w&={\tfrac {1}{2}}r\\x&=\operatorname {copysign} \left({\tfrac {1}{2}}{\sqrt {1+Q_{xx}-Q_{yy}-Q_{zz}}}\,,\left(Q_{zy}-Q_{yz}\right)\right)\\y&=\operatorname {copysign} \left({\tfrac {1}{2}}{\sqrt {1-Q_{xx}+Q_{yy}-Q_{zz}}}\,,\left(Q_{xz}-Q_{zx}\right)\right)\\z&=\operatorname {copysign} \left({\tfrac {1}{2}}{\sqrt {1-Q_{xx}-Q_{yy}+Q_{zz}}}\,,\left(Q_{yx}-Q_{xy}\right)\right)\end{aligned}}}
where copysign(p, q) is defined as p with the sign of q, that is
${\displaystyle \operatorname {copysign} (p,q)=\operatorname {sgn} (q)\,|p|.}$
Alternatively, use a single square root and division
{\displaystyle {\begin{aligned}t&=Q_{xx}+Q_{yy}+Q_{zz}\\r&={\sqrt {1+t}}\\s&={\tfrac {1}{2r}}\\w&={\tfrac {1}{2}}r\\x&=\left(Q_{zy}-Q_{yz}\right)s\\y&=\left(Q_{xz}-Q_{zx}\right)s\\z&=\left(Q_{yx}-Q_{xy}\right)s\end{aligned}}}
This is numerically stable so long as the trace, t, is not negative; otherwise, we risk dividing by (nearly) zero. In that case, suppose Qxx is the largest diagonal entry, so x will have the largest magnitude (the other cases are derived by cyclic permutation); then the following is safe.
{\displaystyle {\begin{aligned}r&={\sqrt {1+Q_{xx}-Q_{yy}-Q_{zz}}}\\s&={\tfrac {1}{2r}}\\w&=\left(Q_{zy}-Q_{yz}\right)s\\x&={\tfrac {1}{2}}r\\y&=\left(Q_{xy}+Q_{yx}\right)s\\z&=\left(Q_{zx}+Q_{xz}\right)s\end{aligned}}}
If the matrix contains significant error, such as accumulated numerical error, we may construct a symmetric 4 × 4 matrix,
${\displaystyle K={\frac {1}{3}}{\begin{bmatrix}Q_{xx}-Q_{yy}-Q_{zz}&Q_{yx}+Q_{xy}&Q_{zx}+Q_{xz}&Q_{yz}-Q_{zy}\\Q_{yx}+Q_{xy}&Q_{yy}-Q_{xx}-Q_{zz}&Q_{zy}+Q_{yz}&Q_{zx}-Q_{xz}\\Q_{zx}+Q_{xz}&Q_{zy}+Q_{yz}&Q_{zz}-Q_{xx}-Q_{yy}&Q_{xy}-Q_{yx}\\Q_{yz}-Q_{zy}&Q_{zx}-Q_{xz}&Q_{xy}-Q_{yx}&Q_{xx}+Q_{yy}+Q_{zz}\end{bmatrix}},}$
and find the eigenvector, (x,y,z,w), of its largest magnitude eigenvalue. (If Q is truly a rotation matrix, that value will be 1.) The quaternion so obtained will correspond to the rotation matrix closest to the given matrix (Bar-Itzhack 2000).
### Polar decomposition
If the n × n matrix M is nonsingular, its columns are linearly independent vectors; thus the Gram–Schmidt process can adjust them to be an orthonormal basis. Stated in terms of numerical linear algebra, we convert M to an orthogonal matrix, Q, using QR decomposition. However, we often prefer a Q closest to M, which this method does not accomplish. For that, the tool we want is the polar decomposition (Fan & Hoffman 1955; Higham 1989).
To measure closeness, we may use any matrix norm invariant under orthogonal transformations. A convenient choice is the Frobenius norm, ||QM||F, squared, which is the sum of the squares of the element differences. Writing this in terms of the trace, Tr, our goal is,
• Find Q minimizing Tr( (QM)T(QM) ), subject to QTQ = I.
Though written in matrix terms, the objective function is just a quadratic polynomial. We can minimize it in the usual way, by finding where its derivative is zero. For a 3 × 3 matrix, the orthogonality constraint implies six scalar equalities that the entries of Q must satisfy. To incorporate the constraint(s), we may employ a standard technique, Lagrange multipliers, assembled as a symmetric matrix, Y. Thus our method is:
• Differentiate Tr( (QM)T(QM) + (QTQI)Y ) with respect to (the entries of) Q, and equate to zero.
Consider a 2 × 2 example. Including constraints, we seek to minimize
{\displaystyle {\begin{aligned}&{\left(Q_{xx}-M_{xx}\right)^{2}+\left(Q_{xy}-M_{xy}\right)^{2}+\left(Q_{yx}-M_{yx}\right)^{2}+\left(Q_{yy}-M_{yy}\right)^{2}}\\&{\quad +\left(Q_{xx}^{2}+Q_{yx}^{2}-1\right)Y_{xx}+\left(Q_{xy}^{2}+Q_{yy}^{2}-1\right)Y_{yy}+2\left(Q_{xx}Q_{xy}+Q_{yx}Q_{yy}\right)Y_{xy}.}\end{aligned}}}
Taking the derivative with respect to Qxx, Qxy, Qyx, Qyy in turn, we assemble a matrix.
${\displaystyle {2{\begin{bmatrix}{Q_{xx}-M_{xx}+Q_{xx}Y_{xx}+Q_{xy}Y_{xy}}&{Q_{xy}-M_{xy}+Q_{xx}Y_{xy}+Q_{xy}Y_{yy}}\\{Q_{yx}-M_{yx}+Q_{yx}Y_{xx}+Q_{yy}Y_{xy}}&{Q_{yy}-M_{yy}+Q_{yx}Y_{xy}+Q_{yy}Y_{yy}}\end{bmatrix}}}}$
In general, we obtain the equation
${\displaystyle 0=2(Q-M)+2QY,}$
so that
${\displaystyle M=Q(I+Y)=QS,}$
where Q is orthogonal and S is symmetric. To ensure a minimum, the Y matrix (and hence S) must be positive definite. Linear algebra calls QS the polar decomposition of M, with S the positive square root of S2 = MTM.
${\displaystyle S^{2}=\left(Q^{\mathsf {T}}M\right)^{\mathsf {T}}\left(Q^{\mathsf {T}}M\right)=M^{\mathsf {T}}QQ^{\mathsf {T}}M=M^{\mathsf {T}}M}$
When M is non-singular, the Q and S factors of the polar decomposition are uniquely determined. However, the determinant of S is positive because S is positive definite, so Q inherits the sign of the determinant of M. That is, Q is only guaranteed to be orthogonal, not a rotation matrix. This is unavoidable; an M with negative determinant has no uniquely defined closest rotation matrix.
### Axis and angle
To efficiently construct a rotation matrix Q from an angle θ and a unit axis u, we can take advantage of symmetry and skew-symmetry within the entries. If x, y, and z are the components of the unit vector representing the axis, and
{\displaystyle {\begin{aligned}c&=\cos \theta \\s&=\sin \theta \\C&=1-c\end{aligned}}}
then
${\displaystyle Q(\theta )={\begin{bmatrix}xxC+c&xyC-zs&xzC+ys\\yxC+zs&yyC+c&yzC-xs\\zxC-ys&zyC+xs&zzC+c\end{bmatrix}}}$
Determining an axis and angle, like determining a quaternion, is only possible up to the sign; that is, (u, θ) and (−u, −θ) correspond to the same rotation matrix, just like q and q. Additionally, axis–angle extraction presents additional difficulties. The angle can be restricted to be from 0° to 180°, but angles are formally ambiguous by multiples of 360°. When the angle is zero, the axis is undefined. When the angle is 180°, the matrix becomes symmetric, which has implications in extracting the axis. Near multiples of 180°, care is needed to avoid numerical problems: in extracting the angle, a two-argument arctangent with atan2(sin θ, cos θ) equal to θ avoids the insensitivity of arccos; and in computing the axis magnitude in order to force unit magnitude, a brute-force approach can lose accuracy through underflow (Moler & Morrison 1983).
A partial approach is as follows:
{\displaystyle {\begin{aligned}x&=Q_{zy}-Q_{yz}\\y&=Q_{xz}-Q_{zx}\\z&=Q_{yx}-Q_{xy}\\r&={\sqrt {x^{2}+y^{2}+z^{2}}}\\t&=Q_{xx}+Q_{yy}+Q_{zz}\\\theta &=\operatorname {atan2} (r,t-1)\end{aligned}}}
The x-, y-, and z-components of the axis would then be divided by r. A fully robust approach will use a different algorithm when t, the trace of the matrix Q, is negative, as with quaternion extraction. When r is zero because the angle is zero, an axis must be provided from some source other than the matrix.
### Euler angles
Complexity of conversion escalates with Euler angles (used here in the broad sense). The first difficulty is to establish which of the twenty-four variations of Cartesian axis order we will use. Suppose the three angles are θ1, θ2, θ3; physics and chemistry may interpret these as
${\displaystyle Q(\theta _{1},\theta _{2},\theta _{3})=Q_{\mathbf {x} }(\theta _{1})Q_{\mathbf {y} }(\theta _{2})Q_{\mathbf {z} }(\theta _{3}),}$
while aircraft dynamics may use
${\displaystyle Q(\theta _{1},\theta _{2},\theta _{3})=Q_{\mathbf {z} }(\theta _{3})Q_{\mathbf {y} }(\theta _{2})Q_{\mathbf {x} }(\theta _{1}).}$
One systematic approach begins with choosing the rightmost axis. Among all permutations of (x,y,z), only two place that axis first; one is an even permutation and the other odd. Choosing parity thus establishes the middle axis. That leaves two choices for the left-most axis, either duplicating the first or not. These three choices gives us 3 × 2 × 2 = 12 variations; we double that to 24 by choosing static or rotating axes.
This is enough to construct a matrix from angles, but triples differing in many ways can give the same rotation matrix. For example, suppose we use the zyz convention above; then we have the following equivalent pairs:
(90°, 45°, −105°) ≡ (−270°, −315°, 255°) multiples of 360° (72°, 0°, 0°) ≡ (40°, 0°, 32°) singular alignment (45°, 60°, −30°) ≡ (−135°, −60°, 150°) bistable flip
Angles for any order can be found using a concise common routine (Herter & Lott 1993; Shoemake 1994).
The problem of singular alignment, the mathematical analog of physical gimbal lock, occurs when the middle rotation aligns the axes of the first and last rotations. It afflicts every axis order at either even or odd multiples of 90°. These singularities are not characteristic of the rotation matrix as such, and only occur with the usage of Euler angles.
The singularities are avoided when considering and manipulating the rotation matrix as orthonormal row vectors (in 3D applications often named the right-vector, up-vector and out-vector) instead of as angles. The singularities are also avoided when working with quaternions.
### Vector to vector formulation
In some instances it is interesting to describe a rotation by specifying how a vector is mapped into another through the shortest path (smallest angle). In ${\displaystyle \mathbb {R} ^{3}}$ this completely describes the associated rotation matrix. In general, given x, y${\displaystyle \mathbb {S} }$n, the matrix
${\displaystyle R:=I+yx^{\mathsf {T}}-xy^{\mathsf {T}}+{\frac {1}{1+\langle x,y\rangle }}\left(yx^{\mathsf {T}}-xy^{\mathsf {T}}\right)^{2}}$
belongs to SO(n + 1) and maps x to y.[12]
## Uniform random rotation matrices
We sometimes need to generate a uniformly distributed random rotation matrix. It seems intuitively clear in two dimensions that this means the rotation angle is uniformly distributed between 0 and 2π. That intuition is correct, but does not carry over to higher dimensions. For example, if we decompose 3 × 3 rotation matrices in axis–angle form, the angle should not be uniformly distributed; the probability that (the magnitude of) the angle is at most θ should be 1/π(θ − sin θ), for 0 ≤ θ ≤ π.
Since SO(n) is a connected and locally compact Lie group, we have a simple standard criterion for uniformity, namely that the distribution be unchanged when composed with any arbitrary rotation (a Lie group "translation"). This definition corresponds to what is called Haar measure. León, Massé & Rivest (2006) show how to use the Cayley transform to generate and test matrices according to this criterion.
We can also generate a uniform distribution in any dimension using the subgroup algorithm of Diaconis & Shashahani (1987). This recursively exploits the nested dimensions group structure of SO(n), as follows. Generate a uniform angle and construct a 2 × 2 rotation matrix. To step from n to n + 1, generate a vector v uniformly distributed on the n-sphere Sn, embed the n × n matrix in the next larger size with last column (0,...,0,1), and rotate the larger matrix so the last column becomes v.
As usual, we have special alternatives for the 3 × 3 case. Each of these methods begins with three independent random scalars uniformly distributed on the unit interval. Arvo (1992) takes advantage of the odd dimension to change a Householder reflection to a rotation by negation, and uses that to aim the axis of a uniform planar rotation.
Another method uses unit quaternions. Multiplication of rotation matrices is homomorphic to multiplication of quaternions, and multiplication by a unit quaternion rotates the unit sphere. Since the homomorphism is a local isometry, we immediately conclude that to produce a uniform distribution on SO(3) we may use a uniform distribution on S3. In practice: create a four-element vector where each element is a sampling of a normal distribution. Normalize its length and you have a uniformly sampled random unit quaternion which represents a uniformly sampled random rotation. Note that the aforementioned only applies to rotations in dimension 3. For a generalised idea of quaternions, one should look into Rotors.
Euler angles can also be used, though not with each angle uniformly distributed (Murnaghan 1962; Miles 1965).
For the axis–angle form, the axis is uniformly distributed over the unit sphere of directions, S2, while the angle has the nonuniform distribution over [0,π] noted previously (Miles 1965).
## Remarks
1. ^ Note that if instead of rotating vectors, it is the reference frame that is being rotated, the signs on the sin θ terms will be reversed. If reference frame A is rotated anti-clockwise about the origin through an angle θ to create reference frame B, then Rx (with the signs flipped) will transform a vector described in reference frame A coordinates to reference frame B coordinates. Coordinate frame transformations in aerospace, robotics, and other fields are often performed using this interpretation of the rotation matrix.
2. ^ Note that
${\displaystyle \mathbf {u} \otimes \mathbf {u} ={\bigl (}[\mathbf {u} ]_{\times }{\bigr )}^{2}+{\mathbf {I} }}$
so that, in Rodrigues' notation, equivalently,
${\displaystyle \mathbf {R} =\mathbf {I} +(\sin \theta )[\mathbf {u} ]_{\times }+(1-\cos \theta ){\bigl (}[\mathbf {u} ]_{\times }{\bigr )}^{2}.}$
3. ^ Note that this exponential map of skew-symmetric matrices to rotation matrices is quite different from the Cayley transform discussed earlier, differing to the third order,
${\displaystyle e^{2A}-{\frac {I+A}{I-A}}=-{\tfrac {2}{3}}A^{3}+\mathrm {O} \left(A^{4}\right).}$
Conversely, a skew-symmetric matrix A specifying a rotation matrix through the Cayley map specifies the same rotation matrix through the map exp(2 arctanh A).
4. ^ For a detailed derivation, see Derivative of the exponential map. Issues of convergence of this series to the right element of the Lie algebra are here swept under the carpet. Convergence is guaranteed when ||X|| + ||Y|| < log 2 and ||Z|| < log 2. If these conditions are not fulfilled, the series may still converge. A solution always exists since exp is onto[clarification needed] in the cases under consideration.
## Notes
1. ^ Swokowski, Earl (1979). Calculus with Analytic Geometry (Second ed.). Boston: Prindle, Weber, and Schmidt. ISBN 0-87150-268-2.
2. ^ W3C recommendation (2003). "Scalable Vector Graphics – the initial coordinate system".
3. ^ Taylor, Camillo J.; Kriegman, David J. (1994). "Minimization on the Lie Group SO(3) and Related Manifolds" (PDF). Technical Report No. 9405. Yale University.
4. ^ https://dspace.lboro.ac.uk/dspace-jspui/handle/2134/18050
5. ^
6. ^ (Wedderburn 1934, §8.02)
7. ^ Hall 2004, Ch. 3; Varadarajan 1984, §2.15
8. ^
9. ^ Curtright, T L; Fairlie, D B; Zachos, C K (2014). "A compact formula for rotations as spin matrix polynomials". SIGMA. 10: 084. arXiv:1402.3541. Bibcode:2014SIGMA..10..084C. doi:10.3842/SIGMA.2014.084. S2CID 18776942.
10. ^ Baker 2003, Ch. 5; Fulton & Harris 1991, pp. 299–315
11. ^ (Goldstein, Poole & Safko 2002, §4.8)
12. ^ Cid, Jose Ángel; Tojo, F. Adrián F. (2018). "A Lipschitz condition along a transversal foliation implies local uniqueness for ODEs". Electronic Journal of Qualitative Theory of Differential Equations. 13 (13): 1–14. arXiv:1801.01724. doi:10.14232/ejqtde.2018.1.13.
## References
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation. | 19,993 | 67,427 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 122, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-25 | latest | en | 0.816431 |
https://quant.stackexchange.com/questions/15945/arent-technical-indicators-calculated-on-adjusted-close-price | 1,719,168,133,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00535.warc.gz | 414,597,363 | 40,845 | # Aren't Technical Indicators calculated on Adjusted Close Price?
I would assume that day to day movement in stock can be accurately compared with the Adjusted Close Price and not simply the Close Price, taking into account Stock Splits Dividends etc. http://www.investopedia.com/ask/answers/06/adjustedclosingprice.asp
Eg: For INFOSYS (INFY, NSE) https://in.finance.yahoo.com/q/hp?s=INFY.NS&a=11&b=12&c=2002&d=11&e=20&f=2014&g=d
The 2 day SMA (Simple Moving Average) on 19 Dec 2014 is (1998.65 + 1965.90) / 2 = 1982.275 Which is what Yahoo Finance is giving me.
But the 2 Day SMA on 2nd Dec 2014 Should be =
(Adj Closing on 1st Dec + Adj Closing on 2nd Dec ) / 2 = (1087.46 + 2126.60)/2 = 1606.73
But it is infact (According to Y! Finance)
(Closing on 1st Dec + Closing on 2nd Dec ) / 2 = (2174.93 + 2126.60)/2 = 2150.765
So the 2:1 Split has no effect on the SMA? I have taken the example of 2 Day SMA for easy calculation but my question is for general MACD, Bollinger Bands, EMA etc.
Many people seem to point to bugs with Yahoo's implementation. I thought given Yahoo's reputation, I thought this was too far fetched but this almost confirms it
Any known bugs with Yahoo Finance adjusted close data ?
Still one thing remains, should one adjust for these splits manually then for Technical Indicators? If not wouldn't it cause a skew in the results?
• I am not sure how Yahoo or Google finance calculate their technical indicators but I would assume it is based off of the last traded price. For example, if you check out freestockcharts.com they calculate their technical indicators real-time (during normal market hours) based on the last traded price thanks to BATS. I would hope Yahoo corrects for splits etc
– Rime
Commented Dec 22, 2014 at 21:28
• @Rime This is exactly what I mean! I also hoped they adjust for splits (which they do in Adjusted Close price in CSVs) but for Tech indicators apparently they don't! Commented Dec 23, 2014 at 7:43
• A very warm welcome to QuantSE and thank you for your great question! Commented Dec 23, 2014 at 8:31
• @vonjd Thank You for your welcome :)! Unfortunately the question has not generated any views, hence the bounty! Commented Dec 23, 2014 at 8:33
• Whether Y!Finance is wrong is not a matter of opinion. Sometimes it just is wrong as pointed out by vonjd, Matt Wolf and chrisaycock. I believe the important lesson here is: don't rely on the data provided by Yahoo, it might be best but it's not perfect. Satisfaction guaranteed, or your money back. By the way: it seems to me that data quality is not the topic of this question and vonjd deserves his bounty. Commented Dec 27, 2014 at 10:55
As can be seen from this example from Yahoo!Finance this should not happen (click on "+ The adjusted close"):
https://help.yahoo.com/kb/finance/SLN2311.html?impressions=true
Another more complete example can be found here:
http://luminouslogic.com/how-to-normalize-historical-data-for-splits-dividends-etc.htm
So my explanation is that this is a glitch in the data... remember: Yahoo!Finance is free and they sometimes don't have the highest quality!
An alternative would be Google Finance but they don't provide adjusted data at all. Yet from these data here it can be seen that it must be a glitch in the Yahoo adjusted data: | 875 | 3,290 | {"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-2024-26 | latest | en | 0.931103 |
https://oeis.org/A192136 | 1,708,972,199,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00394.warc.gz | 442,627,838 | 4,456 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A192136 a(n) = (5*n^2 - 3*n + 2)/2. 4
1, 2, 8, 19, 35, 56, 82, 113, 149, 190, 236, 287, 343, 404, 470, 541, 617, 698, 784, 875, 971, 1072, 1178, 1289, 1405, 1526, 1652, 1783, 1919, 2060, 2206, 2357, 2513, 2674, 2840, 3011, 3187, 3368, 3554, 3745, 3941, 4142, 4348, 4559, 4775, 4996, 5222, 5453 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Binomial transform of [1,1,5,0,0,0,0,0,....]. - Johannes W. Meijer, Jul 07 2011 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA a(n) = (5*n^2 - 3*n + 2)/2. a(n) = 2*a(n-1) - a(n-2) + 5. a(n) = a(n-1) + 5*n - 4. a(n) = 5*binomial(n+2, 2) - 9*n - 4. a(n) = A000217(n+1) - A000217(n) + 5*A000217(n-1); triangular numbers. - Johannes W. Meijer, Jul 07 2011 O.g.f.: (1-x+5*x^2)/((1-x)^3). MATHEMATICA Table[(5n^2-3n+2)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 2, 8}, 50] (* Harvey P. Dale, Aug 08 2016 *) PROG (Magma) A192136:=func< n | (5*n^2-3*n+2)/2 >; [ A192136(n): n in [0..50] ]; // Klaus Brockhaus, Jun 27 2011 (PARI) a(n)=n*(5*n-3)/2+1 \\ Charles R Greathouse IV, Jun 17 2017 CROSSREFS Cf. A000124, A002522, A143689, A130883, A000217, A006137, A140063, A140064. Sequence in context: A109071 A196134 A256321 * A031327 A193389 A030504 Adjacent sequences: A192133 A192134 A192135 * A192137 A192138 A192139 KEYWORD nonn,easy AUTHOR Eric Werley, Jun 24 2011 STATUS approved
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Last modified February 26 12:54 EST 2024. Contains 370352 sequences. (Running on oeis4.) | 818 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | latest | en | 0.524183 |
https://bintanvictor.wordpress.com/2017/02/22/substring-search-boyer-moore/ | 1,611,517,730,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703550617.50/warc/CC-MAIN-20210124173052-20210124203052-00191.warc.gz | 230,330,855 | 29,361 | substring search – Boyer-Moore
I think single-string coding questions are evergreen. The insights and techniques might be …reusable?
Brute force search — shift by one, then attempt to match the entire substring. My code below is more efficient than brute-force. Mostly it shifts by one each time, but checks just one char.
Q: write a python function match(string haystack, string nonRegexNeedle), following the simplified Boyer-Moore algo:
Try matching at the left end. If mismatch, then identify the LEFT-most offending char (J) in haystack. Now locate the RIGHT-most position of J in needle (if not found then simply shift all the way past the J.) and right-shift the needle to align the two J.
Q: So when would we right-shift by one? Not sure. So I don’t know if my code below is 100% correct in all cases.
```def visualize(haystack, substring, offset):
print '\t--' + '0123456789 123456789 12345-----'
print '\t--' + haystack
print '\t--' + ' '*offset + substring
print '\t--' + '-------------------------------'
def match(haystack, substring):
offset = 0 #offset from start of haystack, where substring is now
while(True):
#now locate first offender in haytack
offenderOffset, offender = -9999, ''
for i, char in enumerate(substring):
offenderOffset = offset+i
if substring[i] != haystack[offenderOffset]:
offender = haystack[offenderOffset]
print offender + ' <-- spotted at offenderOffset = ', offenderOffset
visualize(haystack, substring, offset)
break;
if offender == '': return True
# now back-scan substring to locate offender, and then shift to align
# not really forward-scan haystack
while True:
offset += 1
if offset + len(substring) > len(haystack):
print 'Gave up complately:'
visualize(haystack, substring, offset)
return False
if offenderOffset - offset == -1:
print 'gave up on aligning: '
visualize(haystack, substring, offset)
break
if substring[offenderOffset - offset] == offender:
print 'aligned: '
visualize(haystack, substring, offset)
break # go back to start of outer while loop
print match('aborb bdb', 'abors')
``` | 518 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.521488 |
https://questions.llc/questions/22643 | 1,709,280,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00142.warc.gz | 478,531,952 | 7,477 | # Posted by COFFEE on Monday, July 9, 2007 at 9:10pm.
Solve the initial-value problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r^2+4r+6=0,
r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1)
r=(16 +/- Sqrt(-8))
r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2)
y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2
y'(0)=4, c2=4
y(x)=e^(8x)*(2*cos(Sqrt(2)x)+4*sin(Sqrt(2)x))
What am I doing wrong here? This is the answer that I came up with but it is incorrect. Thanks.
* Calculus - Second Order Differential Equations - bobpursley, Monday, July 9, 2007 at 10:09pm
your use of the quadratic formula is wrong. r= (-b +- sqrt (b^2 -4ac)/2a
you did not use -b.
--------------------------
y''+4y'+6y=0, y(0)=2, y'(0)=4
r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)(6))/2
r=-2 +/- sqrt(2)*i
y=e^-2x*(c1*cos(sqrt(2))x+c2*sin(sqrt(2))x)
y(0)=1*(c1+0)=2, c1=2
y'=(-1/2)e^-2x*(c1*(sin(sqrt(2)))/sqrt(2)-c2*(cos(sqrt(2)))/sqrt(2))
y'(0)=(-1/2)(0-1/sqrt(2)*c2)=4
c2=2/sqrt(2)
y(x)=e^-2x*(2cos(sqrt(2))x+(2/sqrt(2))sin(sqrt(x))x)
What is wrong with my solution? thanks.
## In your initial calculations, when you used the quadratic formula, you made a mistake. The correct quadratic formula is:
r = (-b ± sqrt(b^2 - 4ac)) / 2a
But in your solution, you missed the negative sign in front of the coefficient of the linear term. You wrote it as r=(16 ± Sqrt(-8)), but it should be r=(-4 ± Sqrt(16-4(1)(6))/2. This error affected the subsequent steps of your solution and led to the incorrect answer.
Let's go through the correct solution step by step:
1. Start with the given differential equation: y'' + 4y' + 6y = 0.
2. Find the roots of the characteristic equation: r^2 + 4r + 6 = 0.
Use the quadratic formula to solve for r:
r = (-b ± sqrt(b^2 - 4ac)) / 2a.
In this case, a = 1, b = 4, and c = 6.
Plugging these values into the quadratic formula, we get:
r = (-4 ± sqrt(16 - 4(1)(6))) / 2
r = (-4 ± sqrt(16 - 24)) / 2
r = (-4 ± sqrt(-8)) / 2
r = (-4 ± sqrt(8)i) / 2
r = -2 ± sqrt(2)i.
3. Write the general solution for y(x):
Using the complex roots, r = -2 ± sqrt(2)i, we can write the general solution as:
y(x) = e^(-2x)(c1*cos(sqrt(2)x) + c2*sin(sqrt(2)x)).
4. Apply the initial conditions to find the particular solution:
y(0) = 2:
Plugging x = 0 and y = 2 into the general solution, we get:
2 = e^(-2*0)(c1*cos(0) + c2*sin(0))
2 = e^0(c1 + 0)
2 = c1
Therefore, c1 = 2.
y'(0) = 4:
Plugging x = 0 and y' = 4 into the derivative of the general solution, we get:
4 = (-2)e^(-2*0)(2*(sin(sqrt(2)*0))/sqrt(2) - c2*(cos(sqrt(2)*0))/sqrt(2))
4 = -2(0 - c2/sqrt(2))
4 = 0 + c2/sqrt(2)
c2 = 4*sqrt(2).
5. Substitute the values of c1 and c2 back into the general solution:
y(x) = e^(-2x)(2*cos(sqrt(2)x) + (4*sqrt(2))*sin(sqrt(2)x)).
So, the correct solution to the initial-value problem is:
y(x) = e^(-2x)(2*cos(sqrt(2)x) + (4*sqrt(2))*sin(sqrt(2)x)).
I hope this clears up the confusion and helps you understand the correct solution. Remember to always double-check your calculations to avoid mistakes. | 1,161 | 2,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.787384 |
http://lukasz.dk/mirror/hsfortuna/backf/backf.htm | 1,558,444,264,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256381.7/warc/CC-MAIN-20190521122503-20190521144503-00006.warc.gz | 132,159,774 | 4,038 | PS2 Linux Programming
Back Face Culling
Introduction
This tutorial illustrates a method of using VU1 to cull a triangle from the render pipeline if it is facing away from the camera. The winding order of the vertices of the triangle (clockwise or counter clockwise) is used to determine whether the front or back face of the triangle is facing the camera. If the back is facing the camera the triangle is not rendered.
To illustrate the technique in operation, there is a rotating triangle rendered in the scene. If the triangle is facing the camera it is rendered with each vertex having a different colour. If the triangle is facing away from the camera it is rendered in a solid white colour.
Background
Figure 1
Figure 1 illustrates a triangle with three vertices V1, V2 and V3. The vector cross product (D) is taken between the vectors (V2-V1) and (V2-V1) and the vector C, from the vertex V1 to the camera position VC, is obtained. The vector dot product between C and D determines the orientation of the triangle relative to the camera. The orientation of the triangle is determined from the sign of the dot product.
Included within the VU1 macro file supplied by Sony (vcl_sml.i) there is a macro called TriangleWinding.
TriangleWinding result, vert1, vert2, vert3, eyepos
This macro has five parameters; vert1, vert2 and vert3 are the three triangle vertices and eyepos is the position of the camera. result is the output of the macro which is non-zero if the winding is clockwise. In essence, the macro executes the algorithm described above and “result” is used to determine the facing of the triangle relative to the camera.
VU1 Micro Code
In order to undertake back face culling each of the three vertices of the triangle to be rendered must be in the same space (World space) as the camera in order to perform the "winding" calculation. Therefore, at the start of the VU code loop, each vertex is transformed from local to world space using the World transformation matrix, then stored in a temporary register so that they are available for the culling calculation. This is achieved with the following micro-code.
; Transform the vertex by the World matrix (this is a macro from vcl_sml.i)
MatrixMultiplyVertex Vert, fWorld, Vert
; Store the vertex position in the temporary vertex buffer
; We use this for back face culling later
sq Vert, VertexBuffer(iVert)
Once all three triangle vertices have been transformed and stored, the back face culling calculation can be performed. This is achieved using the VU micro code below. Firstly the triangle vertices are read from the buffer into floating point registered. The TriangleWinding macro is then executed with the output "result" being non-zero if the winding is clockwise.
If the "result" is non-zero the back face culling code is executed. For the purpose of this illustration, all three vertex colours are set to white, but in reality the ADC bit of each vertex would be set to 1 (to not draw the vertex): note that this code is commented out in the listing below.
If the "result" is zero the triangle is drawn unmodified.
;--------------------------------
;Back Face Culling (Per Triangle)
;--------------------------------
; Read our buffered vertex positions
lq Vertex1, 2+VertexBuffer(vi00)
lq Vertex2, 1+VertexBuffer(vi00)
lq Vertex3, 0+VertexBuffer(vi00)
; if vi01 = 1, our face needs to be culled
TriangleWinding Result, Vertex1, Vertex2, Vertex3, fCameraPos
ibeq Result, vi00, Draw
; Cull the triangle here
; Note that in this case we don't use this cull code
; In our case we just set the vertex colour to white
sq Colour, StartVert-1(vi00)
sq Colour, StartVert+1(vi00)
sq Colour, StartVert+3(vi00)
b End_Cull
Draw:
; Set the ADC bit to 0 to draw vertex
End_Cull:
;----------------------------------------
The Example Application
In the example application a rotating triangle is shown on screen. When the front face of the triangle is facing the camera it is drawn with each vertex having a different colour. When the back face of the triangle is facing the camera it is drawn with all vertices being white.
Conclusions
This tutorial introduces a technique that can be used for back face culling within the graphics pipeline. As written the code is skeletal in nature and would need adaptation in order to incorporate the technique into a realistic application.
Dr Henry S Fortuna
University of Abertay Dundee | 1,004 | 4,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-22 | latest | en | 0.89108 |
https://youaremom.com/pre-pregnancy/calculate-date-conception/ | 1,550,876,283,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249406966.99/warc/CC-MAIN-20190222220601-20190223002601-00121.warc.gz | 1,028,503,837 | 15,755 | # How to Calculate the Date of Conception
· May 24, 2018
Calculating the date of conception is rather complicated. However, knowing this information is very valuable.
Gestational age is known as the period that covers all of conception. The process begins on the first day of your last menstruation.
Specialists believe that it’s very hard to determine the exact date of conception. In order to estimate the date, the age of the baby and an inexact day is taken into account.
For a more precise approximation, menstruation is used. After a period, ovulation begins either late or early.
There is a natural average to calculate your ovulation date. It should happen in the first 15 days after your period. But ovulation can happen on the 9th, 11th, 14th or 16th day, the timing isn’t known.
Specialists usually suggest that we should have in mind 6 days around the ovulation date. If a mother knows her cycle, it’s easier to get closer to the date.
However, even with this method, there can be a considerable margin of error. In this article we’ll describe other mechanisms to calculate the date of conception.
### Calculate the date of conception
Like we previously stated, the main index to calculate the conception date is the date of the mother’s last menstruation.
This calculation allows us to know the gestational age although it doesn’t tell us the precise date that conception occurred.
A woman who has a regular cycle can allow for more concrete details. However, this can still fail in certain occasions. In order to find out the date, other methods can be used. Take the following examples.
#### Ultrasounds
With as little as 5 weeks after a woman’s last period, the embryo can be measured. In order to perform this measurement, the ultrasound can be divided in specific fractions.
All babies have the same measurements in the first weeks no matter what their final weight and measurements will be. That’s why age approximation is so precise when it’s done in the first 18 weeks.
This method is useful when dealing with a woman whose menstrual cycle is irregular. It also helps clear up doubts when a woman is confused due to bleeding.
Some women confuse implantation bleeding for menstruation. This makes them think they’re not pregnant. Ultrasounds are never wrong.
#### Typical date estimation
In the case of a normal pregnancy in a woman with a regular menstrual cycle, it’s most common for conception to occur between the first 11–21 days after the last period.
We know that the exact date of ovulation is difficult to determine. That’s why this stays as an estimation.
#### Growth of the baby
The growth of the baby can be a double-edged mechanism to find out the baby’s age.
Measurements of the baby are usually similar according to the gestational age, however, unusual growth can occur. The baby may be smaller or larger than average.
“Having a baby is like falling in love again, as much with your husband as with your child” – anonymous.
There are situations when the size of the uterus (height) doesn’t correspond with the date of the last menstruation.
However, the evolution of the pregnancy is a great indicator to find out the date of conception. It’s also useful to estimate when the birth will be.
#### Exact conception date
In most cases, the exact conception date is unknown. This is due to circumstances previously mentioned. However, it’s possible to find out the exact gestational age.
When a woman receives artificial insemination procedures, the exact date of conception can even be determined ahead of time.
Some special cases allow for the knowledge of a bit more precise details. Of course there is still mystery surrounding conception.
Even with the development of special techniques, ovulation tests and other methods, we’re still far from knowing when the magical moment of conception occurs. | 800 | 3,865 | {"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-2019-09 | latest | en | 0.935352 |
www.stagnes-school.org | 1,495,783,976,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00072.warc.gz | 824,912,669 | 196,879 | Problem of the Week
Problem of the Week for May 22, 2017
4th Grade Pow due May 26, 2017
Monday: The cost to run a refrigerator is about \$63 each year. About how much will it have cost to run by the time it is 15 years old?
Tuesday: Fritz is in charge of buying uniform for his baseball team. One uniform costs \$20. There are 14 players on his team. How much will he spend for the uniforms?
Wednesday: A box of organges weighs 24 pounds, how much does 12 boxes of organges weigh? Use partial products to help.
Monday: The cost to run a refrigerator is about \$63 each year. About how much will it have cost to run by the time it is 15 years old?
Tuesday: Fritz is in charge of buying uniform for his baseball team. One uniform costs \$20. There are 14 players on his team. How much will he spend for the uniforms?
Wednesday: A box of organges weighs 24 pounds, how much does 12 boxes of organges weigh? Use partial products to help. | 234 | 936 | {"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} | 3.59375 | 4 | CC-MAIN-2017-22 | longest | en | 0.987668 |
http://physics.stackexchange.com/questions/tagged/conformal-field-theory?page=3&sort=newest&pagesize=50 | 1,469,828,549,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257832385.29/warc/CC-MAIN-20160723071032-00151-ip-10-185-27-174.ec2.internal.warc.gz | 189,799,485 | 26,786 | # Tagged Questions
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### In 1d criticality, what is the relation between the universality class and central charge?
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### Correlator $bc$ system [closed]
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### Connection between the M5 brane and (2, 0) superconformal field theory
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### Problem with OPE (from Polchinski) [closed]
I was reading Polchinski, Vol. 2 pag 12, while I found (10.3.12a): $$e^{iH(z)}e^{-iH(z)}=\frac{1}{2z} + i\partial H(0) + 2zT^H_B(0) + O(z^2).\tag{10.3.12a}$$ Now I tried to do the OPE, what I ...
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### Conformal blocks in 2D CFTs
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### Target Space Lorentz Invariance vs. World Sheet Weyl Invariance
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### Charged CFT observables and AdS/CFT
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### Is there a maximum number of fixed points that a QFT can have?
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### OPE in a general $d$-dimensional CFT
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### Calculating OPE of Graviton Vertex Operator [duplicate]
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### Mode operators in the Virasoro algebra
This questions concerns Exercise 2.11 in Polchinski. We are asked to compute the commutator $$L_{m}(L_{-m}|0;0\rangle) - L_{-m}(L_{m} |0;0\rangle)$$ By plugging the mode expansions, we use the ...
156 views | 1,258 | 4,791 | {"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} | 2.6875 | 3 | CC-MAIN-2016-30 | latest | en | 0.886052 |
http://essay911.org/102-comma-splices-and-run-on-sentences.html | 1,555,607,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517745.15/warc/CC-MAIN-20190418161426-20190418182414-00011.warc.gz | 64,878,128 | 5,933 | # COMMA SPLICES AND RUN-ON SENTENCES
COMMA SPLICES AND RUN-ON SENTENCES
A COMMA SPLICE is an error where two sentences (two main clauses) are joined together with only a comma:
We have enough money, we can buy a piano.
It is almost midnight, however, I'm not tired.
A RUN-ON SENTENCE is an error where two sentences (two main clauses) are joined together with no punctuation between them:
We have enough money we can buy a piano.
It is almost midnight however, I am not tired.
There are FOUR ways to correct a comma splice or run-on.
(1) Use a period.
We have enough money. We can buy a piano.
It is midnight. However, I'm not tired.
(2) Use a semi-colon.
We have enough money; we can buy a piano.
It is midnight; however, I'm not tired.
(3) Use a conjunction.
WHEN we have enough money, we can buy a piano.
We have, enough money, SO we can buy a piano.
AS SOON AS we have enough money, we can buy a piano.
It is midnight, BUT I'm not tired.
ALTHOUGH it is almost midnight, I am not tired.
(4) Change the sentence structure.
We have enough money to buy a piano.
Having enough money, we can buy a piano.
We have enough money for a piano.
I am never tired until long after midnight.
At midnight I'm not tired like other people.
DO NOT CORRECT A COMMA SPLICE BY ERASING THE COMMA. DO NOT CORRECT A RUN-ON BY INSERTING A COMMA. | 347 | 1,324 | {"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} | 3.1875 | 3 | CC-MAIN-2019-18 | longest | en | 0.953064 |
http://math.uww.edu/~mcfarlat/141/polyn13.htm | 1,512,977,469,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00284.warc.gz | 189,880,305 | 896 | Strategy:
Arrange polynomials in columns by degree.
Add the coefficients in each column.
Add: 3x3 + 4x2 - 5x + 7-x3 + 3x - 7 2x3 + 4x2 -2x
The term "+7 - 7 = 0" is not written | 74 | 188 | {"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-2017-51 | latest | en | 0.757467 |
https://howto.org/are-p-%E2%86%92-q-and-p-%E2%88%A8-q-logically-equivalent-68350/ | 1,695,674,274,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00544.warc.gz | 333,085,247 | 12,811 | ## What is logically equivalent to P → Q?
P→Q is logically equivalent to ¬P∨Q. … Example: “If a number is a multiple of 4, then it is even” is equivalent to, “a number is not a multiple of 4 or (else) it is even.”
## Are P → Q → R and P → Q → R logically equivalent?
No, compare (p⟹p)⟹r, which is equivalent to r, and p⟹(p⟹r), which is equivalent to p⟹r. It is not. Suppose p, q, r are false. Then p→q and q→r are true, so (p→q)→r is false and p→(q→r) is true.
## Are the statements P → Q ∨ R and P → Q ∨ P → are logically equivalent?
1.3. 24 Show that (p → q) ∨ (p → r) and p → (q ∨ r) are logically equivalent. By the definition of conditional statements on page 6, using the Com- mutativity Law, the hypothesis is equivalent to (q ∨ ¬p) ∨ (¬p ∨ r). … This means that the conditional from the second-to-last column the last column is always true (T).
## How do you know if two statements are logically equivalent?
To test for logical equivalence of 2 statements, construct a truth table that includes every variable to be evaluated, and then check to see if the resulting truth values of the 2 statements are equivalent.
## Is P → Q ∨ q → p a tautology?
Example: The proposition p ∨ ¬p is a tautology. 2. A proposition is said to be a contradiction if its truth value is F for any assignment of truth values to its components. … A proposition of the form “if p then q” or “p implies q”, represented “p → q” is called a conditional proposition.
## Is P → Q ∧ Q → P logically equivalent to P → Q ∨ Q ↔ P?
Look at the following two compound propositions: p → q and q ∨ ¬p. (p → q) and (q ∨ ¬p) are logically equivalent. So (p → q) ↔ (q ∨ ¬p) is a tautology. Thus: (p → q)≡ (q ∨ ¬p).
## What is the truth value of ∼ P ∨ Q ∧ P?
So because we don’t have statements on either side of the “and” symbol that are both true, the statment ~p∧q is false. So ~p∧q=F. Now that we know the truth value of everything in the parintheses (~p∧q), we can join this statement with ∨p to give us the final statement (~p∧q)∨p.
Truth Tables.
p q p∧q
T F F
F T F
F F F
## Is logically equivalent to?
Two expressions are logically equivalent provided that they have the same truth value for all possible combinations of truth values for all variables appearing in the two expressions. In this case, we write X≡Y and say that X and Y are logically equivalent.
## Is P → Q → P → Q → QA tautology Why or why not?
[p∧(p→q)]→q ≡ F is not true. Therefore [p∧(p→q)]→q is tautology.
## Are the statements P ∧ Q ∨ R and P ∧ Q ∨ R logically equivalent?
This particular equivalence is known as De Morgan’s Law. Since columns corresponding to p∨(q∧r) and (p∨q)∧(p∨r) match, the propositions are logically equivalent. This particular equivalence is known as the Distributive Law.
## Is P -> Q equivalent to Q -> p?
The conditional of q by p is “If p then q” or “p implies q” and is denoted by p q. It is false when p is true and q is false; otherwise it is true. … Suppose a conditional statement of the form “If p then q” is given. The converse is “If q then p.” Symbolically, the converse of p q is q p.
## What is logical equivalence examples?
Now, consider the following statement: If Ryan gets a pay raise, then he will take Allison to dinner. This means we can also say that If Ryan does not take Allison to dinner, then he did not get a pay raise is logically equivalent.
## Which of the following is logically equivalent to if/p then not q?
If p, then not q” is equivalent to “No p are q.” Example: “If something is a poodle, then it is a dog” is a round-about way of saying “All poodles are dogs.
## What is the negation of the statement P → Q ∨ R?
q ∨ r → pD. P ∧∼ q ∧∼ r.
## Which is the inverse of P → Q?
The inverse of p → q is ¬p → ¬q. If p and q are propositions, the biconditional “p if and only if q,” denoted by p ↔ q, is true if both p and q have the same truth values and is false if p and q have opposite truth values. The words if and only if are sometimes abbreviated iff.
## Which statement is denoted by if not p then not q?
Comparisons
name form description
implication if P then Q first statement implies truth of second
inverse if not P then not Q negation of both statements
converse if Q then P reversal of both statements
contrapositive if not Q then not P reversal and negation of both statements
## What does P only if Q mean?
Only if introduces a necessary condition: P only if Q means that the truth of Q is necessary, or required, in order for P to be true. That is, P only if Q rules out just one possibility: that P is true and Q is false.
## Is contrapositive the same as Contraposition?
As nouns the difference between contrapositive and contraposition. is that contrapositive is (logic) the inverse of the converse of a given proposition while contraposition is (logic) the statement of the form “if not q then not p”, given the statement “if p then q”.
## When the statements p and q both are true then the value of the statement P → Q ∧ Q is?
This has some significance in logic because if two propositions have the same truth table they are in a logical sense equal to each other – and we say that they are logically equivalent. So: ¬p∨(p∧q)≡p→q, or “Not p or (p and q) is equivalent to if p then q.”
Logically Equivalent Statements.
p q p→q
F F T
Mar 7, 2021
## What are logically equivalent statements?
Logical Equivalence. Definition. Two statement forms are called logically equivalent if, and only if, they have identical truth values for each possible substitution for their. statement variables. | 1,479 | 5,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.88559 |
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