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### Klein Bottle Intersection
```Date: 11/04/2002 at 09:23:53
From: Gilles Eveloy
Subject: Klein Bottle surface
I would like a detailed description of the intersection domain in a
Klein bottle.
To be more specific, does the external surface contain a hole to let
the handle cross it, or do we consider that if I consider a point A on
the external surface and a point B on the handle surface, I can draw
a continuous path that belongs to the Klein bottle surface ?
Thank you.
Gilles Eveloy
```
```
Date: 12/15/2002 at 17:31:23
From: Doctor Nitrogen
Subject: Re: Klein Bottle surface
Hi, Gilles:
There is no hole at all in the Klein bottle. There is only one surface
for the Klein bottle, and it is not divided up into an "inside" and an
"outside," although it seems that way from the view of our ordinary
space.
Here is a description of the topology for a Klein bottle.
Imagine a long, hollow circular cylinder, with circles for the ends.
Imagine that the ends are open, and that the cylinder is made of some
very pliable clear plastic. Suppose you could give the top opening a
little clockwise "twist" and you could give the bottom opening a
little counterclockwise "twist," and you could keep both ends twisted
that way and then try to join the ends. If you want both circular ends
to match, both clockwise or both counterclockwise, you obviously
cannot just join them together, because each end is oriented a
different way, one clockwise, the other counterclockwise. What a
topologist does to join them so that the two orientations match, is
"push" the top end directly "into" the cylinder's surface, but he does
this in a higher-dimensional space, not in our ordinary three-
dimensional space. The result is that he makes both ends match in
orientation (both clockwise) when he joins them in the higher-
dimensional space.
The result is the peculiar Klein bottle. The original cylinder had a
hollow inside and an outside. But the Klein bottle does not divide our
space into an inside and an outside as an ordinary bottle does. There
is no "interior" and "exterior" surface; there is only one surface, so
that if you were to connect your two points A and B, it might look as
if point B is located at some mysterious "interior" of the bottle, but
such is not the case....
You can also make a Klein bottle by "sewing" two Moebius strips
together. You can read about this and some more about Klein bottles
at:
Klein Bottle - Davide P. Cervone, The Geometry Center
http://www.math.union.edu/~dpvc/papers/RP2/Glossary/KleinBottle.html
Another surface that is strange like the Klein bottle is the "twisted
sphere," or the projective plane. This projective plane looks as if
it's infinitely long, but if you start walking straight ahead from
point A and keep going, you arrive at point A again, although the
plane seems infinite in extent.
I hope this helped answer the questions you had.
- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry
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http://mathforum.org/dr.math/ | 853 | 3,554 | {"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} | 3.65625 | 4 | CC-MAIN-2018-09 | longest | en | 0.942105 |
https://blog.csdn.net/a_flying_bird/article/details/78597092 | 1,537,326,578,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155817.29/warc/CC-MAIN-20180919024323-20180919044323-00555.warc.gz | 469,719,680 | 11,906 | # Question
False不是False，而是True！？
def foolish(value):
print 'Foolish!', value, type(value),
if value:
print "True"
else:
print "False"
if __name__ == "__main__":
foolish('False')
Foolish! False <type 'str'> True
# bool()小函数
#!/usr/bin/env python
import types
import string
import sys
# Transform string or int value to boolean.
def bool(a):
if type(a) == types.StringType:
tmp = a.lower()
if tmp == 'false':
return False
if tmp == 'true':
return True
raise ValueError, "Invalid bool value."
if type(a) == types.IntType:
if a == 0:
return False
if a == 1:
return True
raise ValueError, "Invalid bool value."
if type(a) == types.BooleanType:
return a
raise ValueError, "Invalid bool value."
def helper(value):
print value, type(value),
try:
print bool(value)
except Exception as e:
print e
def foolish(value):
print 'Foolish!', value, type(value),
if value:
print "True"
else:
print "False"
if __name__ == "__main__":
foolish('False')
foolish('True')
values = [None, True, False, 0, 1, 'True', 'false', 2, 't', (3, 1)]
for value in values:
helper(value)
if len(sys.argv) == 2:
helper(sys.argv[1])
# 运行结果
\$ ./test.py True
Foolish! False <type 'str'> True
Foolish! True <type 'str'> True
None <type 'NoneType'> Invalid bool value.
True <type 'bool'> True
False <type 'bool'> False
0 <type 'int'> False
1 <type 'int'> True
True <type 'str'> True
false <type 'str'> False
2 <type 'int'> Invalid bool value.
t <type 'str'> Invalid bool value.
(3, 1) <type 'tuple'> Invalid bool value.
True <type 'str'> True
• 评论
• 上一篇 | 469 | 1,535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-39 | latest | en | 0.207536 |
https://www.justcrackinterview.com/interviews/motion-control-engineering/ | 1,695,588,607,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.30/warc/CC-MAIN-20230924191454-20230924221454-00732.warc.gz | 932,674,435 | 19,530 | # Motion control engineering Interview Questions Answers, HR Interview Questions, Motion control engineering Aptitude Test Questions, Motion control engineering Campus Placements Exam Questions
Find best Interview questions and answer for Motion control engineering Job. Some people added Motion control engineering interview Questions in our Website. Check now and Prepare for your job interview. Interview questions are useful to attend job interviews and get shortlisted for job position. Find best Motion control engineering Interview Questions and Answers for Freshers and experienced. These questions can surely help in preparing for Motion control engineering interview or job.
All of the questions listed below were collected by students recently placed at Motion control engineering.
Ques:- January 1, 1992 was Wednesday. What day of the week was January 1, 1993?
A. Monday
B. Tuesday
C. Thursday
D. Friday
Friday
Ques:- to travel ‘m’ miles the time is ‘h’ hours,then what is the time taken to travel M miles.
speed=mile/hour
.•. time taken to travel M miles is
So, time =distance/speed
=M/m/h
=Mh/m hrs
Ques:- A man driving the car at twice the speed of auto one day he was driven car for 10 min. and car is failed. he left the car and took auto to go to the office .he spent 30 min. in the auto. what will be the time take by car to go office?
25
Ques:- A sheet of paper has statements numbered from 1 to 40. For each value of n from 1 to 40, statement n says “At least and of the statements on this sheet are true.” Which statements are true and which are false?
Ques:- What is the coolest hack you've ever written?
Ques:- Number of zeros in the product of the first 100 numbers?
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
Your donation keeps our website running smoothly and accessible to all. Support us today to ensure we continue to provide valuable information and resources.
Ques:- Are you interested in helping others?
Ques:- What are the disadvantages of the mobile.
Ques:- What would you do if you lost your current job without information ?
Ques:- How would you describe the corporate culture?
Ques:- What is the most important things to run businesssuccessfully? pls, answer at least three main porints.
Ques:- Do you think you are overqualified for this position?
Ques:- What goals have you achieved in your present career?And How did you plan to have achieved these goals?
Ques:- If x/y=4 and y is not '0' what % of x is 2x-y(a)150%(b)175%(c)200%(d)250%
Ques:- Took a typing test.
Ques:- What will be the growth rate after you join?
Ques:- What is product?
Ques:- A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B? | 708 | 2,886 | {"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-2023-40 | latest | en | 0.933488 |
https://db0nus869y26v.cloudfront.net/en/Flux_tube | 1,659,951,996,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00732.warc.gz | 200,540,515 | 21,369 | Diagram of a flux tube showing the magnetic field lines ${\displaystyle B}$ in the tube walls. The same amount of magnetic flux enters the tube through surface ${\displaystyle S_{1))$ as leaves the tube through surface ${\displaystyle S_{2))$
A flux tube is a generally tube-like (cylindrical) region of space containing a magnetic field, B, such that the cylindrical sides of the tube are everywhere parallel to the magnetic field lines. It is a graphical visual aid for visualizing a magnetic field. Since no magnetic flux passes through the sides of the tube, the flux through any cross section of the tube is equal, and the flux entering the tube at one end is equal to the flux leaving the tube at the other. Both the cross-sectional area of the tube and the magnetic field strength may vary along the length of the tube, but the magnetic flux inside is always constant.
As used in astrophysics, a flux tube generally means an area of space through which a strong magnetic field passes, in which the behavior of matter (usually ionized gas or plasma) is strongly influenced by the field. They are commonly found around stars, including the Sun, which has many flux tubes from tens to hundreds of kilometers in diameter.[1] Sunspots are also associated with larger flux tubes of 2500 km diameter.[1] Some planets also have flux tubes. A well-known example is the flux tube between Jupiter and its moon Io.
## Definition
The flux of a vector field passing through any closed orientable surface is the surface integral of the field over the surface. For example, for a vector field consisting of the velocity of a volume of liquid in motion, and an imaginary surface within the liquid, the flux is the volume of liquid passing through the surface per unit time.
A flux tube can be defined passing through any closed, orientable surface ${\displaystyle S_{1))$ in a vector field ${\displaystyle F}$, as the set of all points on the field lines passing through the boundary of ${\displaystyle S_{1))$. This set forms a hollow tube. The tube follows the field lines, possibly turning, twisting, and changing its cross sectional size and shape as the field lines converge or diverge. Since no field lines pass through the tube walls there is no flux through the walls of the tube, so all the field lines enter and leave through the end surfaces. Thus a flux tube divides all the field lines into two sets; those passing through the inside of the tube, and those outside. Consider the volume bounded by the tube and any two surfaces ${\displaystyle S_{1))$ and ${\displaystyle S_{2))$ intersecting it. If the field ${\displaystyle F}$ has sources or sinks within the tube the flux out of this volume will be nonzero. However, if the field is divergenceless (solenoidal, ${\displaystyle \operatorname {div} F=0}$) then from the divergence theorem the sum of the flux leaving the volume through these two surfaces will be zero, so the flux leaving through ${\displaystyle S_{2))$ will be equal to the flux entering through ${\displaystyle S_{1))$. In other words, the flux within the tube through any surface intersecting the tube is equal, the tube encloses a constant quantity of flux along its length. The strength (magnitude) of the vector field, and the cross sectional area of the tube varies along its length, but the surface integral of the field over any surface spanning the tube is equal.
Since from Maxwell's equations (specifically Gauss's law for magnetism) magnetic fields are divergenceless, magnetic flux tubes have this property, so flux tubes are mainly used as an aid in visualizing magnetic fields. However flux tubes can also be useful for visualizing other vector fields in regions of zero divergence, such as electric fields in regions where there are no charges and gravitational fields in regions where there is no mass.
In particle physics, the hadron particles that make up all matter, such as neutrons and protons, are composed of more basic particles called quarks, which are bound together by thin flux tubes of strong nuclear force field. The flux tube model is important in explaining the so-called color confinement mechanism, why quarks are never seen separately in particle experiments.
## Types
• Flux rope: Twisted magnetic flux tube.[1]
• Fibril field: Magnetic flux tube that does not have a magnetic field outside the tube.[1]
## History
Main article: Line of force
In 1861, James Clerk Maxwell gave rise to the concept of a flux tube inspired by Michael Faraday's work in electrical and magnetic behavior in his paper titled "On Physical Lines of Force".[2] Maxwell described flux tubes as:
"If upon any surface which cuts the lines of fluid motion we draw a closed curve, and if from every point of this curve we draw lines of motion, these lines of motion will generate a tubular surface which we may call a tube of fluid motion."[3]
## Flux tube strength
The flux tube's strength, ${\displaystyle F}$, is defined to be the magnetic flux through a surface ${\displaystyle S}$ intersecting the tube, equal to the surface integral of the magnetic field ${\displaystyle \mathbf {B} (\mathbf {x} )}$ over ${\displaystyle S}$
${\displaystyle F=\int _{S}\mathbf {B} \cdot \mathbf {\hat {n)) \;dS}$
Since the magnetic field is solenoidal, as defined in Maxwell's equations (specifically Gauss' law for magnetism): ${\displaystyle \nabla \cdot \mathbf {B} =0}$.[4] the strength is constant at any surface along a flux tube. Under the condition that the cross-sectional area, ${\displaystyle A}$, of the flux tube is small enough that the magnetic field is approximately constant, ${\displaystyle F}$ can be approximated as ${\displaystyle F\approx BA}$.[4] Therefore, if the cross sectional area of the tube decreases along the tube from ${\displaystyle A_{1))$ to ${\displaystyle A_{2))$, then the magnetic field strength must increase proportionally from ${\displaystyle B_{1))$ to ${\displaystyle B_{2))$ in order to satisfy the condition of constant flux F.[5]
${\displaystyle {\frac {B_{2)){B_{1))}={\frac {A_{1)){A_{2))))$
## Plasma physics
### Flux conservation
Main article: Alfvén's theorem
In magnetohydrodynamics, Alfvén's theorem states that the magnetic flux through a surface, such as the surface of a flux tube, moving along with a perfectly conducting fluid is conserved. In other words, the magnetic field is constrained to move with the fluid or is "frozen-in" to the fluid.
This can be shown mathematically for a flux tube using the induction equation of a perfectly conducting fluid
${\displaystyle {\frac {\partial \mathbf {B} }{\partial t))={\boldsymbol {\nabla ))\times (\mathbf {v} \times \mathbf {B} )}$
where ${\displaystyle \mathbf {B} }$ is the magnetic field and ${\displaystyle \mathbf {v} }$ is the velocity field of the fluid. The change in magnetic flux over time through any open surface of the flux tube ${\displaystyle \mathbf {S} }$ enclosed by ${\displaystyle C}$ with a differential line element ${\displaystyle d\mathbf {l} }$ can be written as
${\displaystyle {\frac {d\Phi _{B)){dt))=\int _{S}{\partial \mathbf {B} \over \partial t}\cdot d\mathbf {S} +\oint _{C}\mathbf {B} \cdot \mathbf {v} \times d\mathbf {l} }$
. Using the induction equation gives
${\displaystyle {\frac {d\Phi _{B)){dt))=\int _{S}{\boldsymbol {\nabla ))\times (\mathbf {v} \times \mathbf {B} )\cdot d\mathbf {S} +\oint _{C}\mathbf {B} \cdot \mathbf {v} \times d\mathbf {l} }$
which can be rewritten using Stokes' theorem and an elementary vector identity on the first and second term, respectively, to give[6]
${\displaystyle \int _{S}\mathbf {B} \cdot d\mathbf {S} ={\text{const)).}$
### Compression and extension
In ideal magnetohydrodynamics, if a cylindrical flux tube of length ${\displaystyle L_{0))$ is compressed while the length of tube stays the same, the magnetic field and the density of the tube increase with the same proportionality. If a flux tube with a configuration of a magnetic field of ${\displaystyle B_{0))$ and a plasma density of ${\displaystyle \rho _{0))$ confined to the tube is compressed by a scalar value defined as ${\displaystyle \lambda }$, the new magnetic field and density are given by:[4]
${\displaystyle B={\frac {B_{0)){\lambda ^{2))))$
${\displaystyle \rho ={\frac {\rho _{0)){\lambda ^{2))))$
If ${\displaystyle \lambda <1}$, known as transverse compression, ${\displaystyle B}$ and ${\displaystyle \rho }$ increase and are scaled the same while transverse expansion decreases ${\displaystyle B}$ and ${\displaystyle \rho }$ by the same value and proportion where ${\displaystyle B/\rho }$ is constant.[4]
Extending the length of the flux tube by ${\displaystyle \lambda ^{*))$ gives a new length of ${\displaystyle L=\lambda ^{*}L_{0))$ while the density of the tube remains the same, ${\displaystyle \rho _{0))$, which then results in the magnetic field strength increasing by ${\displaystyle \lambda ^{*}B_{0))$. Reducing the length of the tubes results in a decrease of the magnetic field's strength.[4]
### Plasma pressure
In magnetohydrostatic equilibrium, the following condition is met for the equation of motion of the plasma confined to the flux tube:[4]
${\displaystyle 0=-\nabla p+j\times B-\rho g}$
where
• ${\displaystyle p}$ is the plasma pressure
• ${\displaystyle j}$ is the current density of the plasma
• ${\displaystyle \rho g}$ is the gravitational force
With the magnetohydrostatic equilibrium condition met, a cylindrical flux tube's plasma pressure of ${\displaystyle p(R)}$ is given by the following relation written in cylindrical coordinates with ${\displaystyle R}$ as the distance from the axis radially:[4]
${\displaystyle 0={\frac {dp}{dR))+{\frac {d}{dR))\left({\frac {B_{\phi }^{2}+B_{z}^{2)){2\mu ))\right)+{\frac {B_{\phi }^{2)){\mu R))}$
The second term in the above equation gives the magnetic pressure force while the third term represents the magnetic tension force.[4] The field line's twist around the axis from one end of the tube of length ${\displaystyle L}$ to the other end is given by:[4]
${\displaystyle \Phi (R)={\frac {LB_{\phi }(R)}{RB_{z}(R)))}$
## Examples
### Solar
Diagram of coronal loops that consist of plasma confined to magnetic flux tubes.
Examples of solar flux tubes include sunspots and intense magnetic tubes in the photosphere and the field around the solar prominence and coronal loops in the corona.[4]
Sunspots occur when small flux tubes combine into a large flux tube that breaks the surface of the photosphere.[1] The large flux tube of the sunspot has a field intensity of around 3 kG with a diameter of typically 4000 km.[1] There are extreme cases of when the large flux tubes have diameters of ${\displaystyle 6\times 10^{4))$ km with a field strength of 3 kG.[1] Sunspots can continue to grow as long as there is a constant supply of new flux from small flux tubes on the surface of the sun.[1] The magnetic field within the flux tube can be compressed by decreasing the gas pressure inside and therefore the internal temperature of the tube while maintaining a constant pressure outside.[1]
Intense magnetic tubes are isolated flux tubes that have diameters of 100 to 300 km with an overall field strength of 1 to 2 kG and a flux of around ${\displaystyle 3\times 10^{9))$Wb.[4] These flux tubes are concentrated strong magnetic fields that are found between solar granules.[7] The magnetic field causes the plasma pressure in the flux tube to decrease, known as the plasma density depletion region.[7] If there is a significant difference in the temperatures in the flux tube and the surroundings, there is a decrease in plasma pressure as well as a decrease in the plasma density causing some of the magnetic field to escape the plasma.[7]
Plasma that is trapped within magnetic flux tubes that are attached to the photosphere, referred to as footpoints, create a loop-like structure known as a coronal loop.[8] The plasma inside the loop has a higher temperature than the surroundings causing the pressure and density of the plasma to increase.[8] These coronal loops get their characteristic high luminosity and ranges of shapes from the behavior of the magnetic flux tube.[8] These flux tubes confine plasma and are characterized as isolated. The confined magnetic field strength varies from 0.1 to 10 G with diameters ranging from 200 to 300 km.[8][9]
The result of emerging twisted flux tubes from the interior of the sun cause twisted magnetic structures in the corona, which then lead to solar prominences.[10] Solar prominences are modeled using twisted magnetic flux tubes known as flux ropes.[11]
### Planetary
Graphic of the magnetosphere of Jupiter with a flux tube connecting Jupiter and Io shown in yellow.
Magnetized planets have an area above their ionospheres which traps energetic particles and plasma along magnetic fields, referred to as magnetospheres.[12] The extension of the magnetosphere away from the sun known as a magnetotail is modeled as magnetic flux tubes.[12] Mars and Venus both have strong magnetic fields resulting in flux tubes from the solar wind gathering at high altitudes of the ionosphere on the sun side of the planets and causing the flux tubes to distort along the magnetic field lines creating flux ropes.[12] Particles from the solar wind magnetic field lines can transfer to the magnetic field lines of a planet's magnetosphere through the processes of magnetic reconnection that occurs when a flux tube from the solar wind and a flux tube from the magnetosphere in opposite field directions get close to one another.[12]
Flux tubes that occur from magnetic reconnection forms into a dipole-like configuration around the planet where plasma flow occurs.[12] An example of this case is the flux tube between Jupiter and its moon Io approximately 450 km in diameter at the points closest to Jupiter.[13]
## References
1. Parker, E. N. (1979). "Sunspots and the Physics of Magnetic Flux Tubes. I The General Nature of the Sunspot". The Astrophysical Journal. 230: 905–913. Bibcode:1979ApJ...230..905P. doi:10.1086/157150.
2. ^ Roberts, B (1990). "Waves in Magnetic Flux Tubes". Basic Plasma Processes on the Sun: Proceedings of the 142nd Symposium of the International Astronomical Union Held in Bangalore, India, December 1–5, 1989. Edition 1.
3. ^ Maxwell, J. C. (1861). "On Physical Lines of Force". Philosophical Magazine and Journal of Science. 4.
4. Priest, E. (2014). Magnetohydrodynamics of the Sun. Cambridge University Press. pp. 100–103. ISBN 978-0-521-85471-9.
5. ^ Priest, E. R.; Forbes, T. G. (2001). "Magnetohydrodynamics" (PDF). Nature.
6. ^ Parker, E. N. (1979). Cosmic Magnetic Fields Their Origin and Their Activity. Bristol, UK: Oxford University Press. ISBN 0-19-851290-2.
7. ^ a b c Roberts, B. (2001). "Solar Photospheric Magnetic Flux Tubes: Theory" (PDF). Encyclopedia of Astronomy and Astrophysics. doi:10.1888/0333750888/2255. ISBN 0333750888.
8. ^ a b c d Reale, F. (2014). "Coronal Loops: Observations and Modeling of Confined Plasma". Living Reviews in Solar Physics. 11 (1): 4. arXiv:1010.5927. Bibcode:2014LRSP...11....4R. doi:10.12942/lrsp-2014-4. PMC 4841190. PMID 27194957.
9. ^ Peter, H.; et al. (2013). "Structure of Solar Coronal Loops: from Miniature to Large-Scale". Astronomy & Astrophysics. 556: A104. arXiv:1306.4685. Bibcode:2013A&A...556A.104P. doi:10.1051/0004-6361/201321826. S2CID 119237311.
10. ^ Fan, Y. (2015). Solar Prominences. Springer. ISBN 978-3-319-10416-4.
11. ^ Jibben, P.R.; et al. (2016). "Evidence for a Magnetic Flux Rope in Observations of a Solar Prominence-Cavity System". Frontiers in Astronomy and Space Sciences. 3: 10. Bibcode:2016FrASS...3...10J. doi:10.3389/fspas.2016.00010.
12. Kivelson, M. G.; Bagenal, F. (2007). "Planetary Magnetospheres" (PDF). Encyclopedia of the Solar System. pp. 519–540. Bibcode:2007ess..book..519K. doi:10.1016/B978-012088589-3/50032-3. ISBN 9780120885893. ((cite book)): Missing or empty |title= (help)
13. ^ Bhardwaj, A.; Gladstone, G. R.; Zarka, P. (2001). "An Overview of Io Flux Tube Footpoints in Juptier's Auroral Ionosphere". Advances in Space Research. 27 (11): 1915–1922. Bibcode:2001AdSpR..27.1915B. doi:10.1016/s0273-1177(01)00280-0. | 4,085 | 16,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 62, "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} | 3.109375 | 3 | CC-MAIN-2022-33 | latest | en | 0.926942 |
http://www.roseindia.net/tutorial/java/core/findAverage.html | 1,502,950,419,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102967.65/warc/CC-MAIN-20170817053725-20170817073725-00265.warc.gz | 680,412,802 | 14,456 | # Determine Average by removing Largest and Smallest Number
In this section, you will learn how to remove largest and smallest number from the 10 numbers and then determine the average.
### Determine Average by removing Largest and Smallest Number
In this section, you will learn how to remove largest and smallest number from the 10 numbers and then determine the average. For this, an user is allowed to input 10 numbers. These numbers are then stored into an array in order to find the largest and smallest number. Now to remove these highest and lowest numbers from the array, we have created a method deleteFromArray().
```public static int[] deleteFromArray(int index, int[] arr, int value) {
boolean[] deleteNumber = new boolean[arr.length];
int size = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == value) {
deleteNumber[i] = true;
} else {
deleteNumber[i] = false;
size++;
}
}
int[] newArr = new int[size];
int in = 0;
for (int i = 0; i < arr.length; i++) {
if (!deleteNumber[i]) {
newArr[in++] = arr[i];
}
}
return newArr;
}
```
After that, we have got the final array of 8 numbers. The sum of all these numbers are stored into a variable 'sum' which is further used in determining the average.
Here is the code:
```import java.util.*;
class Calculate {
public static int[] deleteFromArray(int index, int[] arr, int value) {
boolean[] deleteNumber = new boolean[arr.length];
int size = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == value) {
deleteNumber[i] = true;
} else {
deleteNumber[i] = false;
size++;
}
}
int[] newArr = new int[size];
int in = 0;
for (int i = 0; i < arr.length; i++) {
if (!deleteNumber[i]) {
newArr[in++] = arr[i];
}
}
return newArr;
}
public static void main(String[] args) {
int sum = 0;
int arr[] = new int[10];
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
int j = 0, k = 0;
Scanner input = new Scanner(System.in);
System.out.println("Enter 10 numbers:");
for (int i = 0; i < arr.length; i++) {
arr[i] = input.nextInt();
if (arr[i] < min) {
min = arr[i];
j = i;
}
if (arr[i] > max) {
max = arr[i];
k = i;
}
}
System.out.println("Minimum=" + min);
System.out.println("Maximun=" + max);
int newArr[] = Calculate.deleteFromArray(j, arr, min);
int finalArr[] = Calculate.deleteFromArray(k, newArr, max);
for (int i = 0; i < finalArr.length; i++) {
sum += finalArr[i];
}
System.out.println("Final Score: " + sum);
int average = sum / 8;
System.out.println("Average Score: " + average);
}
}
```
Output:
```Enter 10 numbers:
2
3
1
4
9
5
6
7
10
8
Minimum=1
Maximum=10
Final Score: 44
Average Score: 5
``` | 720 | 2,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-34 | latest | en | 0.636312 |
https://www.reference.com/world-view/circles-used-real-life-5a9cee1968769425 | 1,656,840,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00366.warc.gz | 999,260,140 | 16,463 | # How Are Circles Used in Real Life?
Circles are present in real life, both in the natural world and in manmade creations. Manicouagan Reservoir in Canada is a ring-shaped lake that formed in the remains of a crater. Mushrooms with domed caps have circular bases. Ferris wheels take the circle to vertical heights at amusement parks and carnivals. Many household items, including cups, candles, and doorknobs have circles in their designs.
What Are Circles?
A circle is a geometric shape defined as a set of points that are equidistant from a single point on the plane. The connected dots form a series of arcs that surround the center point. Although the perimeter of a circle has no straight lines, straight lines do play a part in calculations. A line between any point in the circle and the center point is called a radius. The circumference of the circle is the perimeter of the circle.
Architecture
Circles appear frequently in architecture around the world. Domes, like those that top the United States Capitol in Washington, D.C., the Duomo of the Florence Cathedral, and St. Peter’s Basilica in Vatican City are all examples of circles used in architecture. Architects also use circles as decorative features in their buildings. For example, the library at Phillips Exeter Academy in New Hampshire has towering slabs of concrete with circles cut out to let viewers see the stacks of books on each floor. The Chartres Cathedral in France features a large circular-shaped window above the front door.
Science
One application of circles in science is in the design of particle separators. The Large Hadron Collider in Europe is a tunnel in the shape of a circle. This shape helps force the particles to move. NASA uses pi ― the ratio of the circumference to the diameter ― in several applications. This includes calculation trajectories, determining the size of distant planets, and measuring craters.
Construction
The Roman arch is one of the most famous examples of how circles are used in construction. Roman architects used wedge-shaped blocks to create the arches that supported their massive aqueducts and domed ceilings. These arches were able to support more weight than the vertical posts and horizontal support beams used in other buildings. Today, arches are still common in construction for this reason.
Transportation
The invention of the wheel remains one of the most important inventions of all time. This circle made it possible for people to move and move things greater distances at faster speeds. Circles are still evident in transportation where they appear in vehicle tires, roundabouts in roads, engine crankshafts, and road designs. GPS also relies on circles when determining distance. It identifies points and calculates the distance between the satellite and the point using a circle theory.
Video Games
Video game creators rely on geometric concepts, including circle theorems when developing virtual worlds for their games. This is how they create the pathways characters follow to navigate around objects. They use their knowledge of circles to transfer two-dimensional ideas to a three-dimensional format. | 611 | 3,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | latest | en | 0.931262 |
http://www.filetransit.com/freeware.php?name=Time_Series_Analysis-5-1 | 1,539,662,772,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00321.warc.gz | 474,944,951 | 11,238 | Home | About Us | Link To Us | FAQ | Contact
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# Time Series Analysis
Added: May 04, 2013 | Visits: 271
There is much more information in a stochastic non-Gaussian or deterministic signal than is conveyed by its autocorrelation and powerspectrum. Higher-order spectra which are defined in terms of the higher-order moments or cumulants of a signal, contain this additional information. The...
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Exercises and case studies for a rigorous approach to risk- and portfolio-management. This booklet stems from the review sessions of the six-day ARPM bootcamp.Contents include:Advanced multivariate statistics; copula-marginal decompositionAnnualization/projection (FFT, cumulants,...
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This user interface allows a user to load some Affymetrix data in an Excel file and perform some basic operations on it - log transform, log ratio, normalization, noise reduction, filtering, PCA and clustering, in addition to some visualizations like profile plots, boxplots and matrix...
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GGobi is an open source visualization program for exploring high-dimensional data. It provides highly dynamic and interactive graphics such as tours, as well as familiar graphics such as the scatterplot, barchart and parallel coordinates plots. Plots are interactive and linked with brushing and...
Platforms: Mac
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This is a simple program written so the equations look like classical matrix equations and it solves the voltage out of the standard current driven loop filter. This is intended to be used in some future PLL analysis program, but it outlines the way to set up the equations, and solve them. There...
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EZFFT(T,U) plots the power spectrum of the signal U(T) , where T is a 'time' and U is a real signal (T can be considered as a space coordinate as well). If T is a scalar, then it is interpreted as the 'sampling time' of the signal U. If T is a vector, then it is interpreted as the 'time' itself....
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The current submission illustrates a way to use the popular open source arduino boards with MATLAB illustrating the capabilities of data acquisition and visualization in real time .The Arduino Demilanove based on the ATMEGA 328 is used to send a series of time series through the serial interface...
Platforms: Matlab
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The Strategy Pattern is just one of many patterns defined by the "Gang of Four" that are commonly used in many other object-oriented programming languages.Now with Matlab 2008b we can define Interfaces and Abstract classes that can make use of many of these patterns. Not only is this good coding...
Platforms: Matlab
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Read metastock files (symbols index files: master, emaster, xmaster; data files: .dat and .mwd; no .dop support)SYNTAXES:(1) METASTOCKREAD Read the symbols index file selected with uigetfile and import the data(2) METASTOCKREAD(FULLPATH) Read the file specified by FULLPATHOUT =...
Platforms: Matlab
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The Matlab code read an XLS file that represent EURO evolution per years. After that the evolution is represented by graphics. Also is displayed some statistics information about this financial time series
Platforms: Matlab
Added: September 09, 2013 | Visits: 147
thresholding_demo.m interactively loads an image stack, performs thresholding based on the Laplace operator and extracts time-series information from the image stack based on a binary mask provided by the thresholding operation.mthresh.m performs thresholding based on the Laplace...
Platforms: Matlab
Added: April 06, 2013 | Visits: 242
PHPGraphs is a set of graphing classes for php.The basic graph types are line, point, area, and bar. Also, an advanced Time Series graph is included. PHPGraphs supports large numbers of large sized data sets.
Platforms: PHP
Added: June 05, 2013 | Visits: 271
PHPGraphs is a set of graphing classes for php.The basic graph types are line, point, area, and bar. Also, an advanced Time Series graph is included. PHPGraphs supports large numbers of large sized data sets.
Platforms: PHP
Added: November 01, 2013 | Visits: 197
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Platforms: *nix
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BioImageXD - free open source software for analysis, processing and 3D rendering of multi dimensional, multi data channel, time series image data from microscopy and other sources. Features . 3D rendering of biological images . image processing and analysis . segmentation with ITK ....
Platforms: *nix
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E-Data Now!TM Live Tracking Software System One company. One database. Instant supply chain information. Seamless - Live Tracking Software U.S.A., Canada, Europe E-Data Now!TM uses seamless technology to track and monitor manufacturing in real time. The patented software uses integrated data...
Platforms: iOS
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SafeSquid Presonal is a Free Content Filter Web Proxy Server for Windows. SafeSquid Content Filter Web Proxy has a BROWSER BASED INTERFACE. SafeSquid Content Filter Web Proxy Server offers arguably, worlds biggest set of Content Filtering features. It gives you Total Access Control, Total Content...
Platforms: Windows
Released: December 15, 2009 | Added: December 29, 2009 | Visits: 1.145
The free Pingdom Desktop Notifier application runs in the background and notifies you if your website is down. It uses the free Pingdom uptime monitoring service and lets you easily get started right inside the application. The Pingdom uptime monitoring service uses a global network of...
Platforms: Windows
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PDL::Impatient is a PDL for the impatient. SYNOPSIS A brief summary of the main PDL features and how to use them. Perl is an extremely good and versatile scripting language, well suited to beginners and allows rapid prototyping. However until recently it did not support data structures...
Platforms: *nix | 1,602 | 6,527 | {"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-2018-43 | latest | en | 0.850877 |
http://ncatlab.org/nlab/show/braided+monoidal+category | 1,448,534,106,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2015-48/segments/1448398447043.45/warc/CC-MAIN-20151124205407-00175-ip-10-71-132-137.ec2.internal.warc.gz | 159,654,400 | 9,836 | # nLab braided monoidal category
### Context
#### Monoidal categories
monoidal categories
# Contents
## Idea
Intuitively speaking, a braided monoidal category is a category with a tensor product and an isomorphism called the ‘braiding’ which lets us ‘switch’ two objects in a tensor product like $x \otimes y$.
## Definition
A braided monoidal category, or braided tensor category, is a monoidal category $V$ equipped with a natural isomorphism
$B_{x,y} : x \otimes y \to y \otimes x$
called the braiding, which must satisfy two axioms called hexagon identities encoding the compatibility of the braiding with the associator for the tensor product.
If the braiding squares to the identity, then the braided monoidal category is a symmetric monoidal category.
### The coherence laws
To see the hexagon identities, let us write $a_{x,y,z} : (x \otimes y) \otimes z \to x \otimes (y \otimes z)$ for the components of the associator in $V$. Then the first hexagon identity says that for all $x,y,z \in Obj(V)$ the following diagram commutes:
$\array{ (x \otimes y) \otimes z &\stackrel{a_{x,y,z}}{\to}& x \otimes (y \otimes z) &\stackrel{B_{x,y \otimes z}}{\to}& (y \otimes z) \otimes x \\ \downarrow^{B_{x,y}\otimes Id} &&&& \downarrow^{a_{y,z,x}} \\ (y \otimes x) \otimes z &\stackrel{a_{y,x,z}}{\to}& y \otimes (x \otimes z) &\stackrel{Id \otimes B_{x,z}}{\to}& y \otimes (z \otimes x) }$
The second hexagon identity says that for all $x,y,z \in Obj(V)$ the following diagram commutes:
$\array{ x \otimes (y \otimes z) &\stackrel{a^{-1}_{x,y,z}}{\to}& (x \otimes y) \otimes z &\stackrel{B_{x \otimes y, z}}{\to}& z \otimes (x \otimes y) \\ \downarrow^{Id \otimes B_{y,z}} &&&& \downarrow^{a^{-1}_{z,x,y}} \\ x \otimes (z \otimes y) &\stackrel{a^{-1}_{x,z,y}}{\to}& (x \otimes z) \otimes y &\stackrel{B_{x,z} \otimes Id}{\to}& (z \otimes x) \otimes y }$
Intuitively speaking, the first hexagon identity says we can braid $x \otimes y$ past $z$ all at once or in two steps. The second hexagon identity says we can braid $x$ past $y \otimes z$ ‘all at once’ or in two steps.
From these axioms, it follows that the braiding is compatible with the left and right unitors $l_x : I \otimes x \to x$ and $r_x : x \otimes I \to x$. That is to say, for all objects $x$ the diagram
$\array{ I \otimes x &&\stackrel{B_{I,x}}{\to}&& x \otimes I \\ & {}_{l_x}\searrow && \swarrow_{r_x} \\ && x }$
commutes.
### In terms of higher monoidal structure
In terms of the language of k-tuply monoidal n-categories a braided monoidal category is a doubly monoidal 1-category .
Accordingly, by delooping twice, it may be identified with a tricategory with a single object and a single 1-morphism.
However, unlike the definition of a monoidal category as a bicategory with one object, this identification is not trivial; a doubly-degenerate tricategory is literally a category with two monoidal structures that interchange up to isomorphism. It requires the Eckmann-Hilton argument to deduce an equivalence with braided monoidal categories.
A commutative monoid is the same as a monoid in the category of monoids. Similarly, a braided monoidal category is equivalent to a monoidal-category object (that is, a pseudomonoid) in the monoidal 2-category of monoidal categories. This result goes back to the 1986 paper by Joyal and Street.
A braided monoidal category is equivalently a category that is equipped with the structure of an algebra over the little 2-cubes operad.
Details are in example 1.2.4 of
### The 2-category of braided monoidal categories
There is a strict 2-category BrMonCat with:
## Properties
### Tannaka duality
Tannaka duality for categories of modules over monoids/associative algebras
monoid/associative algebracategory of modules
$A$$Mod_A$
$R$-algebra$Mod_R$-2-module
sesquialgebra2-ring = monoidal presentable category with colimit-preserving tensor product
bialgebrastrict 2-ring: monoidal category with fiber functor
Hopf algebrarigid monoidal category with fiber functor
hopfish algebra (correct version)rigid monoidal category (without fiber functor)
weak Hopf algebrafusion category with generalized fiber functor
quasitriangular bialgebrabraided monoidal category with fiber functor
triangular bialgebrasymmetric monoidal category with fiber functor
quasitriangular Hopf algebra (quantum group)rigid braided monoidal category with fiber functor
triangular Hopf algebrarigid symmetric monoidal category with fiber functor
supercommutative Hopf algebra (supergroup)rigid symmetric monoidal category with fiber functor and Schur smallness
form Drinfeld doubleform Drinfeld center
trialgebraHopf monoidal category
2-Tannaka duality for module categories over monoidal categories
monoidal category2-category of module categories
$A$$Mod_A$
$R$-2-algebra$Mod_R$-3-module
Hopf monoidal categorymonoidal 2-category (with some duality and strictness structure)
3-Tannaka duality for module 2-categories over monoidal 2-categories
monoidal 2-category3-category of module 2-categories
$A$$Mod_A$
$R$-3-algebra$Mod_R$-4-module
## References
The original papers on braided monoidal categories are by Joyal and Street. The published version does not completely supersede the Macquarie Math Reports version, which has some nice extra results:
Around the same time the same definition was also proposed independently by Lawrence Breen in a letter to Pierre Deligne:
• Lawrence Breen, Une lettre à P. Deligne au sujet des $2$-catégories tressées (1988) (pdf)
For a review of definitions of braided monoidal categories, braided monoidal functors and braided monoidal natural transformations, see:
For an elementary introduction to braided monoidal categories using string diagrams, see:
Eventually we should include all these diagrams here, along with the definition of braided monoidal functor and braided monoidal natural transformation! Can anyone help out?
Revised on June 19, 2014 10:43:32 by Urs Schreiber (145.116.130.115) | 1,646 | 5,963 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 31, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.605403 |
http://slideplayer.com/slide/4167994/ | 1,527,365,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00082.warc.gz | 262,674,977 | 28,908 | # Ch. 10: What is a number?. MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space)
## Presentation on theme: "Ch. 10: What is a number?. MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space)"— Presentation transcript:
Ch. 10: What is a number?
MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space) that leaves the object apparently unchanged.
What is a rigid motion of the plane? Of space? This doesn’t make sense until we decide:
MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space) that leaves the object apparently unchanged. What is a rigid motion of the plane? Of space? This doesn’t make sense until we decide: What is the plane? What is space? This doesn’t make sense until we decide:
MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space) that leaves the object apparently unchanged. What is a rigid motion of the plane? Of space? This doesn’t make sense until we decide: What is the plane? What is space? This doesn’t make sense until we decide: Since the plane is made of pairs (x,y) of numbers and space is made of triples (x,y,z) of numbers, this doesn’t make sense until we decide: What is a number?
MAIN DEFINITION OF THE COURSE: A symmetry of an object (in the plane or space) means a rigid motion (of the plane or space) that leaves the object apparently unchanged. What is a rigid motion of the plane? Of space? This doesn’t make sense until we decide: What is the plane? What is space? This doesn’t make sense until we decide: What is a number? Since the plane is made of pairs (x,y) of numbers and space is made of triples (x,y,z) of numbers, this doesn’t make sense until we decide: Let’s start our backtrack here!
What is a number?
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers”
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Good for counting sheep
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…}
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25 = 3×2×2×5×5.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25 = 3×2×2×5×5. …or like this: 300 = 10×30
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25 = 3×2×2×5×5. …or like this: 300 = 10×30 = 2×5×3×10
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. For example, 20 is not prime because 20 = 5×4. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25 = 3×2×2×5×5. …or like this: 300 = 10×30 = 2×5×3×10 = 2×5×3×2×5.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Including the primes {2, 3, 5, 7, 11, 13, 17, 19…} DEFINITION: A prime number is a natural number greater than 1 that cannot be expressed as a product of two smaller natural numbers. THEOREM: Every natural number greater than 1 is either prime or can be expressed in a unique way as a product of primes. For example, the prime factorization of 300 is: 300 = 2×2×3×5×5. We figured this out by breaking 300 down step-by-step until the pieces could not be further broken down… …like this: 300 = 3×100 = 3×4×25 = 3×2×2×5×5. …or like this: 300 = 10×30 = 2×5×3×10 = 2×5×3×2×5. Same answer after re-ordering! Same answer after re-ordering! This is the hard part to prove!
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000?
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000? 90000 = 300×300
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000? 90000 = 300×300 = 2×2×3×5×5×2×2×3×5×5
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000? 90000 = 300×300 = 2×2×3×5×5×2×2×3×5×5 (reorder) = 2×2×2×2×3×3×5×5×5×5.
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000? 90000 = 300×300 = 2×2×3×5×5×2×2×3×5×5 (reorder) = 2×2×2×2×3×3×5×5×5×5. NOTE: 300 2 has twice as many of each prime as 300 has.
The prime factorization of a square The prime factorization of 300 is: 300 = 2×2×3×5×5. What is the prime factorization of 300 2 = 300×300 = 90000? 90000 = 300×300 = 2×2×3×5×5×2×2×3×5×5 (reorder) = 2×2×2×2×3×3×5×5×5×5. NOTE: 300 2 has twice as many of each prime as 300 has. FACT: The square of any natural number greater than 1 has an even number of occurrences of each prime in its prime factorization.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Good for counting your sheep But has limits…
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers”
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Good for counting positive and negative sheep… But still has limits…
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} “the rational numbers” Examples: 1/2, -17/4, 7/1, -15/38
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} “the rational numbers” synonyms: quotient, fraction, ratio, rational number
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} “the rational numbers” synonyms: quotient, fraction, ratio, rational number Some fractions are the same as other, like 2/3 = 4/6.
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} “the rational numbers” synonyms: quotient, fraction, ratio, rational number Some fractions are the same as other, like 2/3 = 4/6. General rule: a/b = c/d whenever ad = bc. This rule is an addendum to the definition of Q
What is a number? N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers”
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers” What is a number? Good for counting parts of sheep… But are the rational numbers enough?
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers” What is a number? “All numbers are rational.” Greek Mathematician, 500 BC
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers” What is a number? “All numbers are rational.” Greek Mathematician, 500 BC SOME REASONS FOR THIS BELIEF: (1)Finite divisibility of all matter (2)Rational numbers are gifts from the gods (3)They can measure arbitrarily small lengths.
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers” What is a number? “All numbers are rational.” Greek Mathematician, 500 BC
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = { all quotients “a/b” of integers with b≠0} with the understanding that a/b = c/d whenever ad = bc. “the rational numbers” What is a number? “All numbers are rational.” Greek Mathematician, 500 BC “Which rational number squares to 2?”
“All numbers are rational.” Greek Mathematician, 500 BC “Which rational number squares to 2?” THEOREM: There is no rational number that squares to 2.
“All numbers are rational.” Greek Mathematician, 500 BC “Which rational number squares to 2?” THEOREM: There is no rational number that squares to 2. “ὦ περίσσωμα”
“All numbers are rational.” Greek Mathematician, 500 BC “Which rational number squares to 2?” THEOREM: There is no rational number that squares to 2. Greek for “Oh shit.” “ὦ περίσσωμα”
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. Today, we call this length “the square root of 2,” And we refer to it as an “irrational number.” But to the Greek mathematicians in 500 BC, “number” meant “rational number”.
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF:
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2.
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2 This has an even number of 2s in its prime factorization. (twice as many as p has)
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2 This has an even number of 2s in its prime factorization. This has an odd number of 2s in its prime factorization. (twice as many as p has) (one more than twice as many as q has)
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2 This has an even number of 2s in its prime factorization. This has an odd number of 2s in its prime factorization. But the prime factorization of a number is unique. If the left and right side of this equation really equaled each other, there would not be a difference between their prime factorizations.
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. PROOF: Suppose that there are two integers, p and q, such that the fraction p/q squares exactly to 2. That is: (p/q) 2 = 2 ↔ p 2 /q 2 = 2 ↔ p 2 = 2×q 2 This has an even number of 2s in its prime factorization. This has an odd number of 2s in its prime factorization. But the prime factorization of a number is unique. If the left and right side of this equation really equaled each other, there would not be a difference between their prime factorizations. SUMMARY: We assumed that there is a rational number that squares to 2, but this assumption lead to a contradiction; therefore, this assumption must be wrong!
Greek Mathematician, 500 BC THEOREM: There is no rational number that squares to 2. “I need a new kind of number.”
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237…
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… indicates a continuation of the pattern (an unending string of 6s) indicates some unending string of digits, which might not have a pattern that you understand.
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R.
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2?
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = ???
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = ??? The answer is less than 12.75 – 12.74 = 0.01
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = ??? The answer is less than 12.75 – 12.74 = 0.01 and less than 12.750 – 12.749 = 0.001
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = ??? The answer is less than 12.75 – 12.74 = 0.01 and less than 12.750 – 12.749 = 0.001 and less than 12.7500 – 12.7499 = 0.0001
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = ??? The answer is less than 12.75 – 12.74 = 0.01 and less than 12.750 – 12.749 = 0.001 and less than 12.7500 – 12.7499 = 0.0001 and less than 12.75000 – 12.74999 = 0.00001 and so on…
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = 0 The answer is less than 12.75 – 12.74 = 0.01 and less than 12.750 – 12.749 = 0.001 and less than 12.7500 – 12.7499 = 0.0001 and less than 12.75000 – 12.74999 = 0.00001 and so on… THE ANSWER MUST BE ZERO!
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. COMMENT: It takes work to precisely define +,-,×,/ of real numbers. For example, what is 2/3 + √2? GUESS: 12.75000000000 … – 12.74999999999 … = 0 CONCLUSION: 12.75000000000 … = 12.74999999999 … If they differ by zero, they must be the same real number! (This is similar to how 2/3 and 2/6 are two different ways of writing the same rational number)
Today, we put fractions like 2/3 and lengths like √2 on equal footing by them as decimal expressions. 2/3 = 0.66666666… and √2 = 1.41421356237… DEFINITION: A real number means a “decimal expression”; that is, an expression formed from an integer followed by a decimal point followed by infinitely many digits. The set of all real numbers is denoted R. CONCLUSION: 12.75000000000 … = 12.74999999999 … REAL REDUNDANCY RULE: A digit (other than 9) followed by an unending string of 9s can be replaced by the next larger digit followed by an unending string of 0s. There are no other redundancies among real numbers.
N = {1, 2, 3, 4, 5, 6, …} “the natural numbers” Z = {…, -3, -2, -1, 0, 1, 2, 3, …} “the integers” Q = {all quotients “a/b” of integers with b≠0} “the rational numbers” with the understanding that a/b = c/d whenever ad = bc. R = {all decimal expressions} “the real numbers” with the understanding that the real redundancy rule determines which are the same. What is a number?
Picture each real number as a point on the real number line. 0123456-2-3-4-5-6
Picture each real number as a point on the real number line. 0123456-2-3-4-5-6 -13/5π13/3
Picture each real number as a point on the real number line. 0123456-2-3-4-5-6 -13/5π13/3 The number line is like a yard stick, which matches the ancient Greek idea that numbers should represent all possible lengths… …like this one:
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. 0123456-2-3-4-5-6
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. 0123456-2-3-4-5-6 √2 = 1.41421356237 … How do these digits locate the number?
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. 0123456-2-3-4-5-6 √2 = 1.41421356237 …
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. 12 √2 = 1.41421356237 …
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 12
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.02.0
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.02.0 0123456789
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.02.0 0123456789
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 4
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 4
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.41.5
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.41.5 0123456789
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.41.5 0123456789
Picture each real number as a point on the real number line. The digits of a decimal expression locate the number on the real number line. √2 = 1.41421356237 … 1.41.5 0123456789 … and so on. Each next digit provides a ten fold increase in the accuracy with which we know the number’s location on the real number line.
Which real numbers are rational?
Q: Q: Use long division to convert the fraction 3/7 into a decimal expression.
Which real numbers are rational? Q: Q: Use long division to convert the fraction 3/7 into a decimal expression. A: A: 3/7 = 0.428571428571428517….
Which real numbers are rational? Q: Q: Use long division to convert the fraction 3/7 into a decimal expression. A: A: 3/7 = 0.428571428571428517…. With long division, as soon as the remainder repeats, the digits always begin repeating. How are you guaranteed that the remainder will repeat?
Which real numbers are rational? Q: Q: Use long division to convert the fraction 3/7 into a decimal expression. A: A: 3/7 = 0.428571428571428517…. With long division, as soon as the remainder repeats, the digits always begin repeating. How are you guaranteed that the remainder will repeat? FACT: Long division converts any fraction into an eventually repeating decimal expression. FACT: Long division converts any fraction into an eventually repeating decimal expression.
Which real numbers are rational? Q: Q: Use long division to convert the fraction 3/7 into a decimal expression. A: A: 3/7 = 0.428571428571428517…. With long division, as soon as the remainder repeats, the digits always begin repeating. How are you guaranteed that the remainder will repeat? FACT: Long division converts any fraction into an eventually repeating decimal expression. FACT: Long division converts any fraction into an eventually repeating decimal expression. Like this: 103.45873065306530653065… or like this: 4/5 = 0.80000000000…
Which real numbers are rational? Q: Q: Convert N = 0.428571428571428517…. into a fraction.
Which real numbers are rational? Q: Q: Convert N = 0.428571428571428517…. into a fraction. TRICK: Since N has a 6-digit repeating string, we multiply it by 1,000,000 (which has 6 zeros), and then subtract: 1000000×N = 428571. 428571428571 … N = 0. 428571428571 … subtract 999999×N = 428571.000000000000 … We learn that N = 428571/999999 (which reduces to N = 3/7).
Which real numbers are rational? Q: Q: Convert N = 0.428571428571428517…. into a fraction. TRICK: Since N has a 6-digit repeating string, we multiply it by 1,000,000 (which has 6 zeros), and then subtract: 1000000×N = 428571. 428571428571 … N = 0. 428571428571 … subtract 999999×N = 428571.000000000000 … We learn that N = 428571/999999 (which reduces to N = 3/7). FACT: This trick works to convert any eventually repeating decimal expression into a fraction.
Which real numbers are rational? FRACTION Eventually repeating decimal expression Long division Subtraction trick
Which real numbers are rational? FRACTION Eventually repeating decimal expression Long division Subtraction trick THEOREM: A real number is rational precisely when its decimal expression is eventually repeating.
Which real numbers are rational? FRACTION Eventually repeating decimal expression Long division Subtraction trick THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating.
Which real numbers are rational? FRACTION Eventually repeating decimal expression Long division Subtraction trick THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating. COROLLARY: √2 = 1.41421356237 … is NOT eventually repeating! (It’s surprising that we could prove this without understanding the digits)
Which real numbers are rational? THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating. COROLLARY: √2 = 1.41421356237 … is NOT eventually repeating! (It’s surprising that we could prove this without understanding the digits) Q: Q: Is this rational or irrational: 0.01001000100001000001… (pattern continues)
Which real numbers are rational? THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating. COROLLARY: √2 = 1.41421356237 … is NOT eventually repeating! (It’s surprising that we could prove this without understanding the digits) Q: Q: Is this rational or irrational: 0.01001000100001000001… (pattern continues) A: A: Irrational. The pattern is more complicated than “eventually repeating”.
Which real numbers are rational? THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating. COROLLARY: √2 = 1.41421356237 … is NOT eventually repeating! (It’s surprising that we could prove this without understanding the digits) Q: Q: Is this rational or irrational: 0.01001000100001000001… (pattern continues) A: A: Irrational. The pattern is more complicated than “eventually repeating”. Make up your own irrational numbers.
Which real numbers are rational? THEOREM: A real number is rational precisely when its decimal expression is eventually repeating. That is: Every rational number has a decimal expression that’s eventually repeating. Every irrational number has a decimal expression that’s not eventually repeating. COROLLARY: √2 = 1.41421356237 … is NOT eventually repeating! (It’s surprising that we could prove this without understanding the digits) Q: Q: Is this rational or irrational: 0.01001000100001000001… (pattern continues) A: A: Irrational. The pattern is more complicated than “eventually repeating”. The famous numbers π and e are irrational, but this is hard to prove. Make up your own irrational numbers.
How many primes are there?
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … How long until we reach the largest prime? Is there a largest prime?
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers.
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. We know: (1) There are infinitely many natural numbers. (2) Each natural number is a product of primes. But this does not prove Euclid’s Theorem. After all, infinitely many things can be built from unlimited supplies of finitely many different building blocks.
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes.
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. In Andy’s list, the next prime, 17, is missing. But “choosing the next prime” is not a good general procedure, since it begs the question of whether there always is a next prime. In Andy’s list, the next prime, 17, is missing. But “choosing the next prime” is not a good general procedure, since it begs the question of whether there always is a next prime.
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. First multiply them and add 1: L = 2×3×5×7×11×13 + 1 = 30031
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. First multiply them and add 1: L = 2×3×5×7×11×13 + 1 = 30031 This is bigger than anything on Andy’s list It would be nice if numbers formed this way were always prime. This is bigger than anything on Andy’s list It would be nice if numbers formed this way were always prime.
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. First multiply them and add 1: L = 2×3×5×7×11×13 + 1 = 30031 = 59×509 But it’s not prime. Here is its prime factorization
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. First multiply them and add 1: L = 2×3×5×7×11×13 + 1 = 30031 = 59×509 But here are two primes that are missing from Andy’s list!
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. I think that these are ALL of the primes: 2, 3, 5, 7, 11, 13. Andy How do we know Andy is wrong? Euclid found a concrete procedure for identifying a prime that is missing from Andy’s (or any) finite list of primes. First multiply them and add 1: L = 2×3×5×7×11×13 + 1 = 30031 = 59×509 But here are two primes that are missing from Andy’s list! 59 and 509 divide evenly into 30031. Andy’s primes all leave a remainder 1 when divided into 30031. Thus, 59 and 509 are different from Andy’s primes!
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. PROOF: Given any finite list of primes p 1, p 2, p 3, p 4, …, p n we can find a prime that is missing from the list as follows:
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. PROOF: Given any finite list of primes p 1, p 2, p 3, p 4, …, p n we can find a prime that is missing from the list as follows: First multiply them and add 1: N = p 1 ×p 2 ×p 3 ×p 4 ×…×p n + 1
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. PROOF: Given any finite list of primes p 1, p 2, p 3, p 4, …, p n we can find a prime that is missing from the list as follows: First multiply them and add 1: N = p 1 ×p 2 ×p 3 ×p 4 ×…×p n + 1 Each prime factor of N is a prime that’s missing from the list!
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. PROOF: Given any finite list of primes p 1, p 2, p 3, p 4, …, p n we can find a prime that is missing from the list as follows: First multiply them and add 1: N = p 1 ×p 2 ×p 3 ×p 4 ×…×p n + 1 Each prime factor of N is a prime that’s missing from the list! (because they divide evenly into N, while everything on the list leaves a remainder 1 when divided into N.)
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. PROOF: Given any finite list of primes p 1, p 2, p 3, p 4, …, p n we can find a prime that is missing from the list as follows: First multiply them and add 1: N = p 1 ×p 2 ×p 3 ×p 4 ×…×p n + 1 Each prime factor of N is a prime that’s missing from the list! (because they divide evenly into N, while everything on the list leaves a remainder 1 when divided into N.) Thus, no finite list of primes could ever be complete!
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? Are primes in abundance, or are they a rare breed? Do most United States citizens have a prime social security number, or very few?
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 10 100 1,000 10,000 100,000 1,000,000
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 100 1,000 10,000 100,000 1,000,000 Up to N=10:1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 4/10 = 0.40 (40% are prime)
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 100 0.25 1,000 10,000 100,000 1,000,000 Up to N=100: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100. 25/100 = 0.25 (25% are prime)
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 1000.25 1,0000.168 10,0000.1229 100,0000.09592 1,000,0000.078498
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 1000.25 1,0000.168 10,0000.1229 100,0000.09592 1,000,0000.078498 What is the pattern? How does the pattern continue for larger and larger N? What is the pattern? How does the pattern continue for larger and larger N?
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 1000.25 1,0000.168 10,0000.1229 100,0000.09592 1,000,0000.078498
How many primes are there? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … EUCLID’S THEOREM: There are infinitely many prime numbers. After learning Euclid’s Theorem, we still might wonder: how frequently occurring are the prime numbers are among the natural numbers? NFraction of numbers up to N that are prime 100.40 1000.25 1,0000.168 10,0000.1229 100,0000.09592 1,000,0000.078498 1/(twice 7) = 1/14 = 0.0714… 7 digits
Famous unsolved questions about primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … Q: Q: Are there infinitely many “twin primes”? consecutive odd numbers that are both prime, like: 11 & 13, 29 & 31, 41 & 43 …
Famous unsolved questions about primes 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, … Q: Q: Are there infinitely many “twin primes”? consecutive odd numbers that are both prime, like: 11 & 13, 29 & 31, 41 & 43 Q: Q: (Goldbach) Can every positive even number larger than 2 be written as a sum of two primes? 4 = 2 + 2 6 = 3 + 3 8 = 3 + 5 10 = 5 + 5 12 = 7 + 5 14 = 7 + 7 …
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https://www.techwhiff.com/issue/find-arccos-cos-7pi-2-please-show-work-i-will-mark--166776 | 1,669,505,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00787.warc.gz | 1,093,261,025 | 12,502 | # Find arccos[cos(7pi/2)] please show work. I will mark brainliest
###### Question:
Find arccos[cos(7pi/2)] please show work. I will mark brainliest
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What is the molarity of a solution in which 58g of nacl are dissolved in 1.0l of solution?... | 1,040 | 4,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.941737 |
https://puzzlefry.com/puzzles/find-6-hidden-word-newly-painted-room-puzzle/ | 1,656,407,973,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00547.warc.gz | 504,691,483 | 39,073 | # Find 6 hidden word in the newly painted room puzzle
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Find 6 hidden word in the newly painted room
SherlockHolmes Expert Asked on 11th October 2016 in
Viji_Pinarayi Expert Answered on 11th October 2016.
1.Rush
2.paint
3.bucket
4.tap
5.Roller
Sasikumar Starter Answered on 6th April 2020.
Color
Tap
Brush
Bucket
Paint
Roller
Nikhilesh Expert Answered on 15th April 2020.
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#### Published on 22 Apr 2015 | over 3 years ago
Nancy is on a new channel, NancyPi! New videos with Nancy will be at: youtube.com/NancyPi Follow: instagram.com/mathbff facebook.com/mathbff twitter.com/mathbff
What is a limit? How do you read the notation? What does it mean on a graph? How do you find the limit on a graph? To skip ahead: 1) For how to understand limit NOTATION and the CONCEPT of the limit, skip to time 0:34. 2) For WHICH WAY TO LOOK AT THE GRAPH to find the limit, including when to use the X and when to use the Y, skip to time 1:52. 3) For ONE-SIDED LIMITS notation, including the LEFT-SIDED LIMIT and RIGHT-SIDED LIMIT, skip to time 7:54. 4) For how to understand limits where X APPROACHES INFINITY or negative infinity, skip to time 10:24.
For HOW TO FIND THE LIMIT (at a finite value), jump to youtu.be/hewJikMkYFc .
For HOW TO FIND THE LIMIT AT INFINITY, jump to youtu.be/kae8X6aplf0 .
1) LIMIT NOTATION and WHAT A LIMIT MEANS: You can read the limit notation as "the limit, as x approaches 1, of f(x)". This means "when x gets very close to 1, what number is y getting very close to?" The limit is always equal to a y-value. It is a way of predicting what y-value we would expect to have, if we tend toward a specific x-value. Why do we need the limit? One reason is that there are sometimes "blindspots" such as gaps (holes) in a function in which we cannot see what the function is doing exactly at a point, but we can see what it is doing as we head toward that point.
2) HOW TO LOOK AT THE GRAPH to find the limit: a) For a removable discontinuity (hole), b) For a removable discontinuity with a point defined above, and c) For a normal line. When you're finding an overall limit, the hidden, implied meaning is that YOU MUST CHECK BOTH SIDES OF THE X-VALUE, from the left and from the right. If both sides give you the same limit value, then that value is your overall limit. In our example, to find the limit from the left side, TRACE X VALUES from the left of 1 but headed toward 1 (the actual motion is to the right), and check to SEE WHAT Y-VALUE the function is tending toward. That y-value is the left-hand limit. To find the limit from the right side, trace x values from the right of 1 but headed toward 1 (the actual motion is to the left), and again check to see what Y-VALUE the function is heading toward. That y-value is the right-hand limit. Since the left limit (2) and the right limit (2) are the same in our example, the overall limit answer is 2. If they were not the same, we could not give a limit value (see #3). IMPORTANT TAKEAWAY: For the limit, we DO NOT CARE what is happening EXACTLY AT THE X-VALUE and ONLY CARE what y-values the function is hitting NEAR the x-value, as we get closer and closer to that x. In other words, the limit, as x approaches 1, of f(x) can equal 2, even if (1) = 3 or some other number different from 2, or even if f(1) is not defined or indeterminate.
3) ONE-SIDED LIMITS (RIGHT-SIDED LIMIT and LEFT-SIDED LIMIT) for a jump discontinuity: as you saw in #2, to find the overall limit, you have to check both the left and right limits. Sometimes the left limit and right limit are not the same. If you get a limit question with notation in which the x is approaching a number but with a plus sign or minus sign as a superscript, that is notation for a one-sided limit. The minus sign means the limit from the left, and the plus sign means the limit from the right. IF THE LEFT limit AND RIGHT limit are NOT THE SAME, then the overall limit DOES NOT EXIST (sometimes written as "DNE"). Even if the left and right limits are different, you can still write the left-sided limit and right-sided limit values separately.
4) LIMITS in which X APPROACHES INFINITY (or negative infinity): Another "blindspot" is when x goes toward infinity or negative infinity. Since we can never "see" exactly at infinity (or negative infinity), we can use the idea of the limit to say what y-value it looks like the function is headed toward when our x value approaches infinity. If x is approaching INFINITY, TRACE x values TOWARD THE RIGHT (the large positive direction) on the graph, and see what y-value the function is approaching. That y-value is the limit. If x is approaching NEGATIVE INFINITY, trace x values TOWARD THE LEFT (the large negative direction), and check what y-value the function is getting closer and closer to on the graph. That y-value is the limit.
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https://www.coursehero.com/tutors-problems/Statistics-and-Probability/11705619-You-have-disaggregated-the-length-of-stay-values-and-separated-them-in/ | 1,542,618,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745522.86/warc/CC-MAIN-20181119084944-20181119110944-00466.warc.gz | 823,463,520 | 21,997 | View the step-by-step solution to:
# You have disaggregated the length of stay values and separated them into their appropriate MS-DRG category, but you still haven't answered the...
You have disaggregated the length of stay values and separated them into their appropriate MS-DRG category, but you still haven't answered the question that prompted the initial analysis: Is your facility's ALOS for heart failure consistent with that of all Medicare heart failure patients? You want to complete the analysis by testing the hypothesis that your facility's ALOS is greater than the Medicare benchmark. More formally, you have specified the following null and alternative hypotheses:
• Ho: OUR ALOS FOR DRG = BENCHMARK ALOS FOR DRG
• H1: OUR ALOS FOR DRG > BENCHMARK ALOS FOR DRG
Instead of the benchmark data used in Assignment 3.1 that are derived from state estimates using AHRQ H-CUP data, you choose to use the ALOS figures found in the Federal Register for 2014:
CMS Data
The CMS Data PDF contains each MS-DRG with charge and patient days (Average Patient Days = ALOS) for 2014. These figures are what CMS reported for 2014. Use the z-test function to test the hypothesis for each of the three MS-DRGs: 291, 292, 293.
You have some familiarity with the Excel software and you have read about finding probabilities using standardized scores. You decide to answer the probability question posed above using Excel's z.test function given as:
=z.test(range:range,pdf ALOS), where range:range represents the cells where the facility data are located, and pdf ALOS is the national average against which to compare.
For a demonstration of how to use Excel's z test function, view the video tutorial Calculating a z test in Excel (Transcript of Calculating a z test in Excel Tutorial)
The results of this test will tell you the probability that you will reject the null hypothesis when it is actually true (i.e., that you will say your ALOS is greater than the benchmark when in reality your ALOS is not statistically different than the benchmark). This is known as a "Type I" error, and you want the probability of making this type of error to be as small as possible. The conventional threshold (or comfort level) for making this type of mistake is five percent (i.e., you are willing to accept that in five samples out of 100 you will mistakenly reject the null hypothesis when it is true). Naturally, a probability that is smaller is preferred since the likelihood of having made a mistake is smaller. This is essentially what a "p-value" represents: the probability of making a Type I error (of rejecting the null hypothesis when it is true, or in practical terms, concluding that a difference exists when in reality none exists).
Run your calculations using the data specified above to determine if your facility's ALOS for heart failure is consistent with that of all Medicare heart failure patients. Then, make a brief so that you can replicate your analysis at some future date. Your brief should include the following:
1. Discuss whether, in future analyses, you will collect data from a sample of heart failure patients at your facility or look at all heart failure patients at your facility. Be sure to support this decision.
2. Distinguish between the statistics and parameters that you are using (i.e., what are the statistics and what are the parameters?).
3. Once you have estimated the probabilities for each MS-DRG, explain whether you believe your facility has ALOS that are within an acceptable range (i.e. are statistically the same) compared to the national average.
4. Paste a screenshot of your Excel calculations into the last page of your assignment submission.
Here is data for my facility
MS-DRG ALOS Std dev ALOS Ave charges std dev charges
291 6.9 1.97 \$40,067 15,428
292 5.05 1.32 \$22,564 5,088
293 3.6 0.75 \$16,324 8,426
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Browse Documents | 925 | 4,249 | {"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} | 3.109375 | 3 | CC-MAIN-2018-47 | latest | en | 0.8973 |
https://www.scribd.com/document/109254880/Caryl-Pabas-Cooling-of-Hot-Material-2 | 1,544,943,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827281.64/warc/CC-MAIN-20181216051636-20181216073636-00192.warc.gz | 1,024,530,279 | 38,673 | You are on page 1of 6
# Experiment #6 COOLING OF A HOT MATERIAL
Group 2
ALCISO, Hannah-lyn CASADIA, Ariel DUENAS, Princess Marie GABRAL, Genikka NAQUITA, Marjorie PABAS, Caryl L. * III-31 BSE General Science August 24, 2012
Abstract:
The experiment was conducted to determine the factors that affect the rate of cooling of a hot material. The material that was tested is the hot water. The container that was used is 250 ml and 100 ml of beakers. Also different volume of water was added in the same beakers and different beakers. These three variations will tell us if these three trials were able to affect the rate of the cooling. Every 3 minutes the temperature was monitored after getting the initial thermal equilibrium. The amount of water and the size of the beaker didn’t affect the rate of cooling. The result of the experiment shows that the higher the initial temperature of the material the faster the rate of cooling down of it. The graph of this is inversely proportional. The temperature curve has an upward direction when the initial temperature is less than the surrounding temperature.
This experiment is conducted to determine the factors affect the rate of cooling of water. THEORY: Temperature difference in any situation results from energy flow into a system or energy flow from a system to surroundings. . it is cooling down and rate of change of temperature is negative.Introduction: The important concept here that needs to be considered here is how fast the water cool off or how it’s the temperature change. Newton’s law of cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium. And also to prove the Newton’s Law of cooling to answer the question that we encounter in everyday life. since the temperature of the body is higher than the temperature of the surroundings then T-T2 is positive. (1) Where k is a positive proportionality constant. Let the temperature of the body be °C at time t. There is a law – Newton’s law of cooling – that treats this observable fact. The former leads to heating whereas latter leads to cooling of an object. including the discharge of a capacitor and the decay in radioactivity.e. Newton’s Law says that the time a substance takes to cool off depends on the temperature difference between the substance and the surroundings. Suppose that a body with initial temperature T 1°C. Newton's Law of Cooling is useful for studying water heating because it can tell us how fast the hot water in pipes cools off. Also the temperature of the body is decreasing i. This statement leads to the classic equation of exponential decline over time which can be applied to many phenomena in science and engineering. Then by Newton’s law of cooling. is allowed to cool in air which is maintained at a constant temperature T 2°C. Need to find out the essential factors that affect the rate of cooling. and also tells us how fast a water heater cools down if you turn off the breaker when you go on vacation.
The constant ‘k’ depends upon the surface properties of the material being cooled. This will be the initial temperatures at the time 0 minutes. Add a legend for each curve too. Where. k = positive constant and t = time Methodology: To make experiment successful these procedures should be follow. TA = Ambient temperature (temp of surroundings). Repeat the same procedure in the first part. The graph drawn between the temperature of the body and time is known as cooling curve. In general. Next filled up the 250 ml and 100 ml beakers with hot water at 80 °C or higher. Or we can say that the temperature of the body approaches that of its surroundings as time goes. Initial condition is given by T=T1 at t=0 This equation represents Newton’s law of cooling. lim t --> ∞. T (t) = Temperature at time t. TH = Temperature of hot object at time 0. Record all the observation in the data or table given. The slope of the tangent to the curve at any point gives the rate of fall of temperature. e-kt = 0 and T= T2 . If k <0. Then make a graph of the temperature of hot water with the time and draw a best fit line. . First filled the 250 ml beaker with 100 ml of hot water at 80 °C or higher and the other 250 ml beaker with the same amount of water at 50 °C or lower. Make sure to monitor the temperature of hot water every 3 minutes for at least 30 minutes or until it reached thermal equilibrium.
867 100 mL of water in 250 mL of beaker at t1 y = -3.867 100 mL of water in 250 mL of beaker at t2 Linear (100 mL of water in 250 mL of beaker at t1) Linear (100 mL of water in 250 mL of beaker at t2) TIME 6 9 12 15 .DATA AND RESULTS: Table 1.0571x + 89. Cooling of a Hot Material Temperature in °C Time in minutes 0 3 6 9 12 15 18 21 24 27 30 100 mL of water in 250 mL of beaker at t1 100 mL of 250 mL of water in 250 water at t mL of beaker at t2 100 mL of water at t 100 mL of water in 250 mL beaker 100 mL of water in 100 mL beaker 85 70 61 59 55 51 48 46 44 42 40 65 60 55 53 50 48 46 44 43 41 40 80 74 70 66 63 61 58 56 54 51 49 80 69 67 62 59 52 46 44 42 40 39 80 70 61 55 53 50 48 46 44 42 41 80 71 60 56 52 50 48 45 42 41 40 Title of graph: Variation of the temperature of 100 ML of water in 250 ml beker with t1 and t2 90 80 70 60 50 40 30 20 10 0 0 3 y = -7.3429x + 66.
393 100 mL of water in 250 mL beaker 100 mL of water in 100 mL beaker Linear (100 mL of water in 250 mL beaker) Linear (100 mL of water in 100 mL beaker) .Title of graph: Variation of the temperature of 100 ML and 250 ml water in a beaker 90 80 70 60 50 40 30 20 10 0 0 3 6 9 y = -5.857 y = -4.6x + 83.5595x + 78.6905x + 78.6x + 84 Title of graph: Variation of the temperature of 100 mL of water in 250 mL beaker and 100 mL of water in 100 mL beaker 90 80 70 60 50 40 30 20 10 0 0 3 6 9 12 15 18 21 y = -4.5 250 mL of water at t 100 mL of water at t Linear (250 mL of water at t) Linear (100 mL of water at t) y = -4.
the water in 250 ml and 100 ml of beaker also cooled down as the time increases every 3 minutes. Many factors may affect the cooling of a hot material. 4. Still has doesn’t affect the rate of cooling of water. 6. safety precaution is very much necessary because we are dealing with the hot water. Physics: Principles with Application/ Douglas C.ANALYSIS AND DISCUSSION OF RESULTS: In this experiment. These three variations almost the same in their initial temperature ranges from 85-80 °C because hot water is still used and it cools down at the rate of 39-40 °C. 100 ml of water in 250 ml beaker when it reached thermal equilibrium at 85 °C. 5. the following graphs just tell us that the rate of cooling of a hot material was the higher the body’s temperature which was the initial temperature. As we can see in the first graph. 2. Lastly. Conceptual Physics/ Paul G. http://en. . 6.math.ca/coursedoc/math100/notes/diffeqs/cool. the temperature curve has an upward direction when the initial temperature is less than the surrounding temperature. The amount of water and the size of the beaker didn’t affect the rate of cooling. 6thed. Douglas C. Paul G.ugrad. the faster the rate of cooling.html Laboratory Manual in Heat and Thermodynamics. Sears and Zemanky’s University Physics 12thed. the initial temperature only affects the cooling of hot material because the higher the initial temperature the faster the rate of cooling. Same with the 2 nd graph. These are the temperature of the surrounding is higher than then initial temperature the cooling of hot material is slow. Hewitt. Reference: 1. 5. Giancoli. the cooling process is faster. Conclusion: Experiment aims to determine the factors affect the rate of cooling of a hot material. The result of the experiment shows that the higher the body’s temperature the faster the rate at which it cools down.org/wiki/Lumped_capacitance_model http://www. The 3rd graph differs in the beakers that was used. equal volume of water in a different beakers and the values obtain has the same with the first and second table. But when the initial temperature is higher than on the temperature of surrounding. Recommendation: In performing the experiment. In addition. the temperature easily cools down as the time increases every 3 minutes. Make sure that you are handling the beaker with care so that the water will not spill on you.ubc.wikipedia. Hewitt. 10thed. The amount of water and the different beakers that we used doesn’t affect the rate of cooling even though 250 ml is larger than 100 ml of water. 3. Giancoli. | 2,180 | 8,533 | {"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} | 3.25 | 3 | CC-MAIN-2018-51 | latest | en | 0.940824 |
https://cs.stackexchange.com/questions/12139/is-the-language-of-words-containing-equal-number-of-001-and-100-regular | 1,718,266,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00222.warc.gz | 182,617,869 | 46,483 | Is the language of words containing equal number of 001 and 100 regular?
I was wondering when languages which contained the same number of instances of two substrings would be regular. I know that the language containing equal number of 1s and 0s is not regular, but is a language such as $L$, where $L$ = $\{ w \mid$ number of instances of the substring "001" equals the number of instances of the substring "100" $\}$ regular? Note that the string "00100" would be accepted.
My intuition tells me it isn't, but I am unable to prove that; I can't transform it into a form which could be pumped via the pumping lemma, so how can I prove that? On the other hand, I have tried building a DFA or an NFA or a regular expression and failed on those fronts also, so how should I proceed? I would like to understand this in general, not just for the proposed language.
• Did you see How to prove that a language is not regular?
– Juho
Commented May 19, 2013 at 21:25
• Yes. I tried the pumping lemma, without success. I also cannot see any applicable closure properties. And I couldn't get my head round the Myhill–Nerode theorem. Also thanks for the LaTeX. I tried applying it myself, but it split the description akwardly across multiple lines. Commented May 19, 2013 at 21:29
• Probably there should be an additional $0$ loop at $q5$? Commented May 20, 2013 at 9:41
• A similar example of this phenomenon, but for the substrings "01" and "10" was discussed at our sister site Proving a language is regular or irregular. The answer has a similar remark as wece made in his comment: "That is, a 01 transition cannot be followed by another $01$ transition without an intervening $10$ transition.". Commented May 20, 2013 at 9:44
• @Juho: "How to prove that a language is not regular"? Step 1: Pick a language that is not regular. That's where such a proof fails in this case. Commented Nov 12, 2017 at 15:30
An answer extracted from the question.
Yes, it is regular; below is an automaton that accepts it.
As pointed out by Hendrik Jan, there should be an additional 0 self-loop at q5.
• in CS classes for simple problems sometimes just DFAs are given, but it doesnt prove that it exactly accepts the language. you have to [somehow] show for every input string it functions correctly. "how does it work?"
– vzn
Commented Jun 18, 2013 at 15:41
• I think you could merge $q_5$ and $q_2$. Is that right? Commented Aug 13, 2013 at 16:28
It's a trick question. Try constructing a string that contains two 001 and doesn't contain a 100, and see why you can't do it. If X = "number of 001", and Y = "number of 100", then X = Y or X = Y ± 1.
Once you realise the trick, it becomes highly unlikely that the language is irregular, and then constructing a DFA is quite simple. There are only 8 states with their transitions if the next symbol is 0/1:
State S0: Input is empty. -> S1/C0
State S1: Input is 0. -> C2/C0
State A: Y = X + 1, input ends in 00. -> A/C0
State B0: X = Y + 1, input ends in 1. -> B1/B0
State B1: X = Y + 1, input ends in 10. -> C2/B0
State C0: X = Y, input ends in 1. -> C1/C0
State C1: X = Y, input ends in 10. -> A/C0
State C2: X = Y, input ends in 00. -> C2/B0
The initial state is S0, and S0, S1, C0, C1, C2 are accepting states.
We can write every string in $$\{0,1\}^*$$ in the form $$0^{i_0} 1 0^{i_1} 1 0^{i_2} \cdots 0^{i_{m-1}} 1 0^{i_m}$$ Here $$i_j \geq 0$$, and $$m$$ is the number of $$1$$s.
The number of copies of $$001$$ is the number of indices $$i_0,\ldots,i_{m-1}$$ which are at least $$2$$.
The number of copies of $$100$$ is the number of indices $$i_1,\ldots,i_m$$ which are at least $$2$$.
We conclude that the number of copies of $$001$$ is the same as the number of copies of $$002$$ iff $$i_0 \geq 2 \Leftrightarrow i_m \geq 2.$$ This leads to the following regular expression: $$0^* + (\epsilon+0)(10^*)^*1(\epsilon+0) + 000^*(10^*)^*1000^*.$$
$$L=\{\epsilon, 0, 1, 01, 10, 010, 101, 00, 000, 0000,..... , 1, 11, 11111,......, 01110, 1001, 00100,.........\}$$ The pattern I can observe here is whenever we see a '001' as a substring then it has to be followed by 00 to make $$n(001)=n(100)$$ and whenever we see '100' as a substring then it has to be followed by 1 to make it '1001' to make $$n(100)=n(001)$$
• Please tell me if I'm going wrong somewhere Commented May 10, 2020 at 12:45
• I'm having a hard time seeing how this answers the question. The question was, "is L regular?" A good answer should presumably say either "yes" or "no" and then provide justification for its answer. I can't tell whether you are trying to say that L is regular or L is not. If you just wanted to describe a pattern you observed, that does not count as an answer to the question and should not be posted in the 'Your Answer' box. (We want you to provide useful answers to other questions before you can leave incomplete observations in the comments.)
– D.W.
Commented May 14, 2020 at 17:58
• Also there is a picture of an automaton but with no explanation of what that is supposed to mean. We seek answers that come with explanation and/or justification. If you can edit your question to expand on these points, please do so.
– D.W.
Commented May 14, 2020 at 17:59
• isn't it obvious that if a language is regular then a FA exists for it? Commented May 15, 2020 at 3:55
• Please don't answer in the comments. Instead, edit your question to improve it based on the feedback here. Your question was flagged for possible deletion, so I'm offering some suggestions for how it can be improved. Personally, I prefer to see answers improved instead of deleted. It might help to make explicit the things that you find obvious. Also, it might help to explain why that automaton correctly recognizes the language, if there's a reasonable way to do that.
– D.W.
Commented May 15, 2020 at 5:04
Several years ago I and several colleagues generalized when the language of all strings with equal number of $$x$$ and $$y$$ substrings over an alphabet $$\Sigma$$ is regular: https://arxiv.org/abs/1804.11175. The condition depends on whether $$x$$ is "interlaced" by $$y$$ or vice versa. For $$x$$ to be "interlaced" by $$y$$, it must be the case that $$x$$ is a substring of every string in $$\Sigma^\star$$ that starts and ends with $$y$$. We also give a construction whenever the condition is satisfied.
In the case of $$x=001$$ and $$y=100$$ over $$\Sigma = \{0,1\}$$, every string that starts and ends with $$y=100$$ must have the form $$100\cdots100$$. No matter what, $$x=001$$ is a substring of it. Therefore, this language is regular. | 1,891 | 6,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.961238 |
http://www.poly-ed.com/2017/09/ | 1,701,380,288,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00515.warc.gz | 83,883,159 | 9,257 | # Prime Sieve in Java
A very concise prime sieve implementation in Java.
```/******************************************************************************
* Compilation: javac PrimeSieve.java
* Execution: java -Xmx1100m PrimeSieve n
*
* Computes the number of primes less than or equal to n using
* the Sieve of Eratosthenes.
*
* % java PrimeSieve 25
* The number of primes <= 25 is 9
*
* % java PrimeSieve 100
* The number of primes <= 100 is 25
*
* % java -Xmx100m PrimeSieve 100000000
* The number of primes <= 100000000 is 5761455
*
* % java PrimeSieve -Xmx1100m 1000000000
* The number of primes <= 1000000000 is 50847534
*
*
* The 110MB and 1100MB is the amount of memory you want to allocate
* to the program. If your computer has less, make this number smaller,
* but it may prevent you from solving the problem for very large
* values of n.
*
*
* n Primes <= n
* ---------------------------------
* 10 4
* 100 25
* 1,000 168
* 10,000 1,229
* 100,000 9,592
* 1,000,000 78,498
* 10,000,000 664,579
* 100,000,000 5,761,455
* 1,000,000,000 50,847,534
*
******************************************************************************/
public class PrimeSieve {
public static void main(String[] args) {
int n = Integer.parseInt(args[0]);
// initially assume all integers are prime
boolean[] isPrime = new boolean[n+1];
for (int i = 2; i <= n; i++) {
isPrime[i] = true;
}
// mark non-primes <= n using Sieve of Eratosthenes
for (int factor = 2; factor*factor <= n; factor++) {
// if factor is prime, then mark multiples of factor as nonprime
// suffices to consider mutiples factor, factor+1, ..., n/factor
if (isPrime[factor]) {
for (int j = factor; factor*j <= n; j++) {
isPrime[factor*j] = false;
}
}
}
// count primes
int primes = 0;
for (int i = 2; i <= n; i++) {
if (isPrime[i]) primes++;
}
System.out.println("The number of primes <= " + n + " is " + primes);
}
}
```
# Merge Sort in Python
```def merge(a,b):
""" Function to merge two arrays """
c = []
while len(a) != 0 and len(b) != 0:
if a[0] < b[0]:
c.append(a[0])
a.remove(a[0])
else:
c.append(b[0])
b.remove(b[0])
if len(a) == 0:
c += b
else:
c += a
return c
# Code for merge sort
def mergesort(x):
""" Function to sort an array using merge sort algorithm """
if len(x) == 0 or len(x) == 1:
return x
else:
middle = round(len(x)/2)
a = mergesort(x[:middle])
b = mergesort(x[middle:])
return merge(a,b)
```
Code by by Anirudh Jayaraman. | 794 | 2,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.57544 |
http://7puzzleblog.com/36/ | 1,519,444,254,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815318.53/warc/CC-MAIN-20180224033332-20180224053332-00758.warc.gz | 4,763,540 | 8,267 | # DAY 36:
Today’s Challenge
You’ve rolled the numbers 2, 3 and 4 with your three dice. Using these once each, with + – × ÷ available, what is the lowest positive whole number it is NOT possible to make?
Visit Roll3Dice.com for further details.
The 7puzzle Challenge
The playing board of the 7puzzle game is a 7-by-7 grid containing 49 different numbers, ranging from up to 84.
The 1st & 4th rows of the playing board contain the following fourteen numbers:
2 3 9 10 14 15 22 32 35 40 44 54 60 72
Which three different numbers have a sum of 100?
Make 36 Challenge
Can you arrive at 36 by inserting 2, 3, 4 and 6 into the gaps on each line?
• (◯+◯+◯)× = 36
• (◯²+◯)×◯÷ = 36
• (◯–◯)×◯× = 36
• ◯²×(◯+◯–◯) = 36 | 293 | 747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-09 | longest | en | 0.605072 |
https://infinitylearn.com/surge/maths/class-7/comparing-quantities/percentage/different-forms-of-percentages/ | 1,675,369,988,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00734.warc.gz | 333,653,221 | 12,777 | Different Forms of Percentages
# Different Forms of Percentages
• Different Forms of Expressing Percentage
• Summary
• What’s Next?
In the previous segment, we had looked at a few examples of percentage. In this segment
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of the chapter ‘Comparing Quantities’, we will look at different forms of percentage.
## What are the Different forms of expressing percentage?
Percentage can be expressed in 3 ways:
1. A ratio or a fraction
2. A fraction with denominator 100
3. Decimals
Let us understand this with the help of a table by drawing a square of four parts.
Parts of a square Ratio/Fraction Denominator 100 Decimal Percentage As one part of the square is coloured out of4, this is 14 To get the denominator as 100 both the numerator and denominator of1 will be4multiplied by 25.1 × 25 25=4 × 25 100 If we divide the fraction 25 we100get the decimalform as 0.25. 25 is 25%.100
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Verify OTP Code (required) | 256 | 1,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.854454 |
https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Alternative_algebra.html | 1,720,952,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00247.warc.gz | 277,032,805 | 6,968 | # Alternative algebra
"Artin's theorem" redirects here. For Artin's theorem on primitive elements, see Primitive element theorem.
In abstract algebra, an alternative algebra is an algebra in which multiplication need not be associative, only alternative. That is, one must have
for all x and y in the algebra.
Every associative algebra is obviously alternative, but so too are some strictly non-associative algebras such as the octonions. The sedenions, on the other hand, are not alternative.
## The associator
Alternative algebras are so named because they are precisely the algebras for which the associator is alternating. The associator is a trilinear map given by
By definition a multilinear map is alternating if it vanishes whenever two of its arguments are equal. The left and right alternative identities for an algebra are equivalent to[1]
Both of these identities together imply that the associator is totally skew-symmetric. That is,
for any permutation σ. It follows that
for all x and y. This is equivalent to the flexible identity[2]
The associator of an alternative algebra is therefore alternating. Conversely, any algebra whose associator is alternating is clearly alternative. By symmetry, any algebra which satisfies any two of:
• left alternative identity:
• right alternative identity:
• flexible identity:
is alternative and therefore satisfies all three identities.
An alternating associator is always totally skew-symmetric. The converse holds so long as the characteristic of the base field is not 2.
## Properties
Artin's theorem states that in an alternative algebra the subalgebra generated by any two elements is associative.[4] Conversely, any algebra for which this is true is clearly alternative. It follows that expressions involving only two variables can be written unambiguously without parentheses in an alternative algebra. A generalization of Artin's theorem states that whenever three elements in an alternative algebra associate (i.e., ), the subalgebra generated by those elements is associative.
A corollary of Artin's theorem is that alternative algebras are power-associative, that is, the subalgebra generated by a single element is associative.[5] The converse need not hold: the sedenions are power-associative but not alternative.
hold in any alternative algebra.[2]
In a unital alternative algebra, multiplicative inverses are unique whenever they exist. Moreover, for any invertible element and all one has
This is equivalent to saying the associator vanishes for all such and . If and are invertible then is also invertible with inverse . The set of all invertible elements is therefore closed under multiplication and forms a Moufang loop. This loop of units in an alternative ring or algebra is analogous to the group of units in an associative ring or algebra.
Zorn's theorem states that any finite-dimensional non-associative alternative algebra is a generalised octonion algebra.[6]
## Applications
The projective plane over any alternative division ring is a Moufang plane.
The close relationship of alternative algebras and composition algebras was given by Guy Roos in 2008: He shows (page 162) the relation for an algebra A with unit element e and an involutive anti-automorphism such that a + a* and aa* are on the line spanned by e for all a in A. Use the notation n(a) = aa*. Then if n is a non-singular mapping into the field of A, and A is alternative, then (A,n) is a composition algebra.
## References
1. Schafer (1995) p.27
2. Schafer (1995) p.28
3. Conway, John Horton; Smith, Derek A. (2003). On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry. A. K. Peters. ISBN 1-56881-134-9. Zbl 1098.17001.
4. Schafer (1995) p.29
5. Schafer (1995) p.30
6. Schafer (1995) p.56
• Guy Roos (2008) "Exceptional symmetric domains", §1: Cayley algebras, in Symmetries in Complex Analysis by Bruce Gilligan & Guy Roos, volume 468 of Contemporary Mathematics, American Mathematical Society.
• Schafer, Richard D. (1995). An Introduction to Nonassociative Algebras. New York: Dover Publications. ISBN 0-486-68813-5. | 937 | 4,113 | {"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-2024-30 | latest | en | 0.911643 |
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### Converting Celsius to Fahrenheit and Why It Works
```Date: 04/03/2004 at 20:54:04
From: Kittra
Subject: Temperature
What are the Fahrenheit and Celsius patterns that will help me
predict Fahrenheit temperature based on the Celsius temperature?
```
```
Date: 04/03/2004 at 22:20:47
From: Doctor Peterson
Subject: Re: Temperature
Hi, Kittra.
I don't know exactly what "pattern" you are expected to see,
especially at your age, but I can describe how the two are related.
The Fahrenheit and Calsius scales are just like using two different
rulers to measure the same lengths, and also starting at two different
places. If I wanted to describe where I live, along Interstate 90, I
might choose to say it in terms of the number of miles from Boston,
and someone else might give me the number of kilometers from Albany.
They will use different numbers to identify the same place. To change
one measurement to the other, you have to know how many miles there
are in a kilometer, and how many kilometers or miles it is from Boston
to Albany.
Here is a "road map" that shows all the information you need to figure
out one kind of temperature measurement when you know the other:
0 32 212 F
<---+-----+-------------------------------------------+--->
0 100 C
freezing boiling
Looking at this, you can see how many degrees F there are in 100
degrees C; and how much you have to add because F starts at a
different place. It takes 180 degrees F (212-32) to cover the same
"distance" as only 100 degrees C; so every time you go up 10 degrees
C, you go up 18 degrees F. Each degree Celsius is the same as 1.8
degrees Fahrenheit. So to convert a Celsius temperature to
Fahrenheit, you have to multiply by 1.8 (or 9/5) to get the right size
of a degree, and then add 32 to change the starting point.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Temperature
Middle School Temperature
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# Understand the problem
Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs? $\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
# 2010 AMC 8 Problem 25
##### Topic
Recursion – Combinatorics
Easy
Easy
##### Suggested Book
Excursion in Mathematics
Do you really need a hint? Try it first!
Try to recall the concept of recursion . Then proceed to apply in this question . https://en.wikipedia.org/wiki/Recursion#In_mathematics
The number of ways to climb one stair is $1$. $$\\$$ There are $2$ ways to climb two stairs: $1$,$1$ or $2$. $$\\$$ For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$) . $$\\$$ Try to aply this for 4 stairs in recursive manner .
For 4 stairs : After climbing 1 stair in one step , Jo can climb the rest 3 stairs in (1 , 1 , 1) ,(1 , 2), (2, 1) ,(3) i.e. in 4 ways . $$\\$$ After climbing 2 stairs in one step , Jo can climb the rest 2 stairs in (1 , 1 ) and ( 2) i.e. in 2 ways . $$\\$$ After climbing 3 stairs in one step , Jo can climb the rest 1 stair in a single step. $$\\$$ So to climb 4 stairs Jo can have (1+ 2 + 4) = 7 ways . Now time arrives to apply the recursion so proceed to apply .
Why will we apply recursion ? $$\\$$ Notice that to climb 5 stairs , $$\\$$ After climbing 1 stair in one step , we again have to find how many ways Jo can follow to climb the rest 4 stairs . $$\\$$ After climbing 2 stairs in one step , we again have to find how many ways Jo can follow to climb the rest 3 stairs . $$\\$$ After climbing 3 stairs in one step ,we again have to find how many ways Jo can follow to climb the rest 2 stairs . $$\\$$ So the no ways Jo can have for 5 stairs is ( 7 + 4 + 2) = 13 . $$\\$$ So for 6 stairs , the no ways Jo has (13 + 7 +4 ) = 24 .
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# Which of the following most logically completes the
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Which of the following most logically completes the argument?
The irradiation of food kills bacteria and thus retards spoilage. However, it also lowers the nutritional value of many foods. For example, irradiation destroys a significant percentage of whatever vitamin B1 a food may contain. Proponents of irradiation point out that irradiation is no worse in this respect than cooking. However, this fact is either beside the point, since much irradiated food is eaten raw, or else misleading, since _______.
A. many of the proponents of irradiation are food distributors who gain from food’s having a longer shelf life
B. it is clear that killing bacteria that may be present on food is not the only effect that irradiation has
C. cooking is usually the final step in preparing food for consumption, whereas irradiation serves to ensure a longer shelf life for perishable foods
D. certain kinds of cooking are, in fact, even more destructive of vitamin B1 than carefully controlled irradiation is
E. for food that is both irradiated and cooked, the reduction of vitamin B1 associated with either process individually is compounded
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14 Jun 2007, 14:58
E.
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Betweeb A and C .I would go for A
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14 Jun 2007, 19:04
I would go for E. It is miselading if each effect is compounded resulting in even less vitamin B1.
14 Jun 2007, 19:04
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#### AMSDOS
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##### Dodging Obstacles in BASIC and Assembly
« on: 12:23, 18 April 18 »
The other week, I constructed a simple BASIC game of Dodging the Rocket around the Grumpy faces, which I thought worked quite nicely.
Code: [Select]
`10 MODE 0:DEFINT a-z:INK 0,11:INK 1,26:BORDER 11:PEN 120 RANDOMISE TIME:RANDOMIZE RND30 FOR a=1 TO 2540 x=INT(RND*20)+150 LOCATE x,160 PRINT CHR\$(225)61 GOSUB 100070 NEXT a80 x=10:y=12:PEN 3:LOCATE x,y:PRINT CHR\$(239)90 d=1100 WHILE d=1101 IF INKEY(1)=0 THEN IF x<20 THEN LOCATE x,y:PRINT" ":x=x+1:LOCATE x,y:PEN 3:PRINT CHR\$(239)102 IF INKEY(=0 THEN IF x>1 THEN LOCATE x,y:PRINT" ":x=x-1:LOCATE x,y:PEN 3:PRINT CHR\$(239)110 ax=INT(RND*20)+1120 LOCATE ax,1:PEN 1130 PRINT CHR\$(225)135 IF TEST((x-1)*32+4,238)=1 THEN d=0140 LOCATE x,y:PRINT" "150 GOSUB 1000160 LOCATE x,y:PEN 3:PRINT CHR\$(239)170 WEND180 END1000 LOCATE 1,1:PRINT CHR\$(11)1010 RETURN`
The aim of this simple game was to write something similar in Assembly and thought it would make a good demonstration of coding in Assembly and make use of some Collision Detection code I recently wrote.
Though when I tried to do the same thing with my Rocket in Assembly, it looked as if the Rocket didn't have time to display itself, so I ended up with this, so it looks like there's 2 Rockets onscreen. It appears the Scroll routine I'm using here is very slow too, which makes use of SCR SW ROLL. My BASIC version seems to be kicking the Assembly versions butt.
Code: [Select]
` org &8000 ld a,0 call &bc0e call genseed ld a,1 call &bb90 ld b,24.drawobstacles push bc call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,225 call &bb5a call scroll pop bc djnz drawobstacles call printrocket.maingame call scroll ld hl,(ypos2) call &bb75 ld a,32 call &bb5a call printrocket call updateobstacle call collision ld a,(dead) cp 0 jr z,skip ld a,1 call &bb1e jr z,checkleft ld a,(xpos) cp 16 jr z,checkleft inc a ld (xpos),a ld (xpos2),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 add hl,de ld (ex),hl.checkleft ld a,8 call &bb1e jr z,skip ld a,(xpos) cp 1 jr z,skip dec a ld (xpos),a ld (xpos2),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 and a sbc hl,de ld (ex),hl.skip ld a,(dead) and a jr nz,maingame ld a,(dead) inc a ld (dead),a ret ;; Return to BASIC.removerocket ld hl,(oy) call &bb75 ld a,32 call &bb5a ret.printrocket ld hl,(ypos) call &bb75 ld a,3 call &bb90 ld a,239 call &bb5a ret .updaterocket ld hl,(ypos) call &bb75 ld a,32 call &bb5a call scroll ld hl,(ypos) call &bb75 ld a,3 call &bb90 ld a,239 call &bb5a call updateobstacle ret.updateobstacle call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,1 call &bb90 ld a,225 call &bb5a ret.collision ld hl,(ex) ex hl,de ld hl,(ey) call &bbf0 cp 1 jr nz,endcoll ld a,(dead) dec a ld (dead),a.endcoll ret.genseed ld a,r ld (seed),a ret.rand ld a,(seed) ld b,a add a,a add a,a add a,b inc a ld (seed),a ret.scroll ld b,0 ld a,0 ld h,0 ld d,19 ld l,0 ld e,24 call &bc50 ret.scalenum ld a,(seed) srl a srl a srl a srl a inc a ld (result),a ret .ypos defb 12.xpos defb 10.oy defb 12.ox defb 10.ypos2 defb 13.xpos2 defb 10.dead defb 1.seed defb 0.result defb 0.ex defw 292.ey defw 238`
« Last Edit: 07:54, 29 April 18 by AMSDOS »
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#### ervin
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #1 on: 16:05, 18 April 18 »
I notice that .updaterocket is never called, but that doesn't really matter.
SCR_SOFTWARE_ROLL is indeed the culprit, regarding the speed.
Try SCR_HARDWARE_ROLL &BC4D instead. The speed difference is phenomenal.
#### AMSDOS
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #2 on: 09:57, 19 April 18 »
I notice that .updaterocket is never called, but that doesn't really matter.
I was using it initially to generate a PRINT rocket/Enter MainLoop/Delete Rocket/Scroll/PRINT Rocket scenario. At that stage, I hadn't added any controls and it was just checking for a collision and to exit the Loop if one had occurred. Unfortunately at that stage I couldn't see where the Rocket was until the Obstacle was either Deleted by it (which told me the collision failed), or the Program stopped and Rocket finally displayed.
Quote
SCR_SOFTWARE_ROLL is indeed the culprit, regarding the speed.
I'm guessing then it's not really meant for games, I used it in my introductory animated sequence for the game I coded a couple of years ago, which worked fine, but not strictly meant for during gameplay. Still, coding this simple assembly game and experimenting with different aspects is an interesting experience.
Last night I updated the scroll routine with a LDI scroll routine I made up a couple of years ago, though I thought the result was similar to the SCR_SW_ROLL, obviously substituting that code to the ".scroll" routine within my Dodging Obstacles code:
Code: [Select]
`;; LDI Scroll - This shifts the screen down in rows at a time.;; Top Row needs to be blank so contents don't push it all the way down the screen org &4000 xor a ;; Position for Loop equals 0..loop ld hl,(adr_hl) ;; This holds the address pointer and is place into HL ld e,(hl) ;; Though the Contents from that address goes into DE inc hl ;; Cause it's 16bit data this is the only way I know ld d,(hl) ;; How to do it, there maybe other ways. I dunno. ex de,hl ;; The data above needs to go into HL and not DE push hl ;; I need to protect my data in HL ld hl,(adr_de) ;; This holds the address pointer and is place into HL ld e,(hl) ;; Though the Contents from that address goes into DE inc hl ;; Cause it's 16bit data this is the only way I know ld d,(hl) ;; How to do it, there maybe other ways. I dunno. pop hl ;; But restore it again for this routine repeat 80 ldi ;; Moves things along. rend ld hl,(adr_hl) ;; Increment Start Position to the next spot inc hl inc hl ld (adr_hl),hl ;; And store it in that pointer variable. ld hl,(adr_de) ;; Increment Destination to the next position inc hl inc hl ld (adr_de),hl ;; And store it in that pointer variable inc a ;; Increment Loop cp 192 ;; Check if this equals 192. jp nz,loop ;; returns if not equal to 192, otherwise exit ld hl,begin ;; Restores pointer to the beginning ld (adr_hl),hl ;; should the user want to continue calling routine ld hl,dest ;; Restores pointer to the Destination ld (adr_de),hl ;; should the user want to continue calling routine ret ;; exit routine.adr_hl defw begin ;; pointer position to where the start position is for HL register.adr_de defw dest ;; pointer position to where the start position is for DE register;; Below these are Screen co-ordinate positions for DE, move from bottom to top of the screen.dest defw &C780,&CF80,&D780,&DF80,&E780,&EF80,&F780,&FF80.begin defw &C730,&CF30,&D730,&DF30,&E730,&EF30,&F730,&FF30 defw &C6E0,&CEE0,&D6E0,&DEE0,&E6E0,&EEE0,&F6E0,&FEE0 defw &C690,&CE90,&D690,&DE90,&E690,&EE90,&F690,&FE90 defw &C640,&CE40,&D640,&DE40,&E640,&EE40,&F640,&FE40 defw &C5F0,&CDF0,&D5F0,&DDF0,&E5F0,&EDF0,&F5F0,&FDF0 defw &C5A0,&CDA0,&D5A0,&DDA0,&E5A0,&EDA0,&F5A0,&FDA0 defw &C550,&CD50,&D550,&DD50,&E550,&ED50,&F550,&FD50 defw &C500,&CD00,&D500,&DD00,&E500,&ED00,&F500,&FD00 defw &C4B0,&CCB0,&D4B0,&DCB0,&E4B0,&ECB0,&F4B0,&FCB0 defw &C460,&CC60,&D460,&DC60,&E460,&EC60,&F460,&FC60 defw &C410,&CC10,&D410,&DC10,&E410,&EC10,&F410,&FC10 defw &C3C0,&CBC0,&D3C0,&DBC0,&E3C0,&EBC0,&F3C0,&FBC0 defw &C370,&CB70,&D370,&DB70,&E370,&EB70,&F370,&FB70 defw &C320,&CB20,&D320,&DB20,&E320,&EB20,&F320,&FB20 defw &C2D0,&CAD0,&D2D0,&DAD0,&E2D0,&EAD0,&F2D0,&FAD0 defw &C280,&CA80,&D280,&DA80,&E280,&EA80,&F280,&FA80 defw &C230,&CA30,&D230,&DA30,&E230,&EA30,&F230,&FA30 defw &C1E0,&C9E0,&D1E0,&D9E0,&E1E0,&E9E0,&F1E0,&F9E0 defw &C190,&C990,&D190,&D990,&E190,&E990,&F190,&F990 defw &C140,&C940,&D140,&D940,&E140,&E940,&F140,&F940 defw &C0F0,&C8F0,&D0F0,&D8F0,&E0F0,&E8F0,&F0F0,&F8F0 defw &C0A0,&C8A0,&D0A0,&D8A0,&E0A0,&E8A0,&F0A0,&F8A0 defw &C050,&C850,&D050,&D850,&E050,&E850,&F050,&F850 defw &c000,&c800,&d000,&d800,&e000,&e800,&f000,&f800`
Quote
Try SCR_HARDWARE_ROLL &BC4D instead. The speed difference is phenomenal.
Yep, so I replaced the SCR_SW_ROLL with SCR_HW_ROLL, and braced for impact. To give my rocket a chance I placed it lower down the screen (line 24). I've brought ".updaterocket" back into play too (I can now see a blinking Red Rocket), and the HW ROLL Scrolls twice, to prevent the vast number of obstacles which were onscreen. I still had to put a couple of MC_FRAME_WAIT (&BD19s) after that, I had some after the Obstacles were drawn too, but I've since removed. So this is what I've currently done:
Code: [Select]
` org &8000 ld a,0 call &bc0e call genseed ld a,1 call &bb90 ld b,24.drawobstacles push bc call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,225 call &bb5a call scroll call scroll pop bc djnz drawobstacles call printrocket.maingame call updaterocket call collision ld a,(dead) cp 0 jr z,skip ld a,1 call &bb1e jr z,checkleft ld a,(xpos) cp 16 jr z,checkleft inc a ld (xpos),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 add hl,de ld (ex),hl.checkleft ld a,8 call &bb1e jr z,skip ld a,(xpos) cp 1 jr z,skip dec a ld (xpos),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 and a sbc hl,de ld (ex),hl.skip ld a,(dead) and a jr nz,maingame ld a,(dead) inc a ld (dead),a ret ;; Return to BASIC.removerocket ld hl,(oy) call &bb75 ld a,32 call &bb5a ret.printrocket ld hl,(ypos) call &bb75 ld a,3 call &bb90 ld a,239 call &bb5a ret .updaterocket call removerocket call scroll call scroll call printrocket call &bd19 call &bd19 call updateobstacle ret.updateobstacle call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,1 call &bb90 ld a,225 call &bb5a ret.collision ld hl,(ex) ex hl,de ld hl,(ey) call &bbf0 cp 1 jr nz,endcoll ld a,(dead) dec a ld (dead),a.endcoll ret.genseed ld a,r ld (seed),a ret.rand ld a,(seed) ld b,a add a,a add a,a add a,b inc a ld (seed),a ret.scroll ld b,0 ld a,0 call &bc4d ret.scalenum ld a,(seed) srl a srl a srl a srl a inc a ld (result),a ret .ypos defb 24.xpos defb 10.oy defb 24.ox defb 10.dead defb 1.seed defb 0.result defb 0.ex defw 292.ey defw 46`
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#### ervin
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #3 on: 15:23, 19 April 18 »
Nice one.
Much more playable that way!
#### SRS
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #4 on: 22:31, 19 April 18 »
Nice
roundabout 8 bytes smaller and lets say ... a few ... T-states faster:
Code: [Select]
`org &8000 xor a call &bc0e call genseed ld a,1 call &bb90 ld b,24.drawobstacles push bc call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,225 call &bb5a call scroll call scroll pop bc djnz drawobstacles call printrocket.maingame call updaterocket call collision ld a,(dead) or a jr z,skip ld a,1 call &bb1e jr z,checkleft ld a,(xpos) cp 16 jr z,checkleft inc a ld (xpos),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 add hl,de ld (ex),hl.checkleft ld a,8 call &bb1e jr z,skip ld a,(xpos) dec a jr z,skip ld (xpos),a call removerocket call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 and a sbc hl,de ld (ex),hl.skip ld a,(dead) and a jr nz,maingame ld a,(dead) inc a ld (dead),a ret ;; Return to BASIC.removerocket ld hl,(oy) call &bb75 ld a,32 jp &bb5a .printrocket ld hl,(ypos) call &bb75 ld a,3 call &bb90 ld a,239 jp &bb5a.updaterocket call removerocket call scroll call scroll call printrocket call &bd19 call &bd19 jp updateobstacle.updateobstacle call rand call scalenum ld a,(result) ld h,a ld l,1 call &bb75 ld a,1 call &bb90 ld a,225 jp &bb5a.collision ld hl,(ex) ex hl,de ld hl,(ey) call &bbf0 dec a jr nz,endcoll ld a,(dead) dec a ld (dead),a.endcoll ret.genseed ld a,r ld (seed),a ret.rand ld a,(seed) ld b,a add a,a add a,a add a,b inc a ld (seed),a ret.scroll xor a ld b,a jp &bc4d.scalenum ld a,(seed) srl a srl a srl a srl a inc a ld (result),a ret .ypos defb 24.xpos defb 10.oy defb 24.ox defb 10.dead defb 1.seed defb 0.result defb 0.ex defw 292.ey defw 46`
#### AMSDOS
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #5 on: 12:57, 20 April 18 »
Nice
roundabout 8 bytes smaller and lets say ... a few ... T-states faster:
Appreciate the improvements. Was keeping the update & collision routines separate, so I could work on adding additional code to it. The collision for example isn't perfect and thought I could make the game check if a collision with an obstacle has occurred when you move your rocket left or right.
I can also remove some code to remove the Rocket from the screen by simply moving the rocket to the bottom of the screen too, which will reduce the flicker I hope.
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#### AMSDOS
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##### Re: Dodging Obstacles in BASIC and Assembly
« Reply #6 on: 07:50, 29 April 18 »
I've made some more updates to my assembly code using a modified version David Hall's Ariom Sprites to work in with my assembly code. Unfortunately I encountered a problem when I tried to use it with SCR HW ROLL. Not only does SCR HW ROLL roll the screen, it shifts the screens offset, so the Sprite Driver didn't know where to draw the sprites, I tried to rectify this with a variable to store a new base address, so I could do the SCR HW ROLL, followed by SCR CHAR POSITION (&BC1A), to get the new offset position, subtract &54 to it to get my new base address, which initially worked, but then the game crashed (obviously because of some sensitive area was being poked). Aaarrrrruggghhh!
So I replaced that Scroll routine with my own, which appears to be running at the same pace as the SCR SW ROLL routine I was using earlier. So it looks like for this game to work I'll need to go back to SCR HW ROLL and use some of my other firmware routines to Draw the characters.
I'll be posting an updated DSK in my top post, the program is in Binary Format, so
Code: [Select]
`MEMORY &7FFFLOAD"obacles3.bin"CALL &8000`
or
Code: [Select]
`MEMORY &7FFFLOAD"obacles5.bin"CALL &8000`
To load either of those programs.
Code: [Select]
` org &8000 xor a call &bc0e ;; Mode 0 ld hl,colours call setinks call genseed ;; Randomize Seed ld a,1 call &bb90 ;; Pen 1 ld b,24.drawobstacles push bc call rand call scalenum ld a,(result) ld (xpos1),a ld a,2 ld (ypos1),a ld a,3 ld (spr_num),a call ariom_spr call scroll pop bc djnz drawobstacles call printrocket.maingame ld a,(dead) or a jr z,skip ld a,1 call &bb1e ;; KM Test Key jr z,checkleft ld a,(xpos) cp 16 jr z,checkleft call clearrocket ld a,(xpos) inc a ld (xpos),a call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 add hl,de ld (ex),hl.checkleft ld a,8 call &bb1e ;; KM Test Key jr z,skip ld a,(xpos) cp 1 jr z,skip call clearrocket ld a,(xpos) dec a ld (xpos),a call printrocket ld a,(xpos) ld (ox),a ld hl,(ex) ld de,32 and a sbc hl,de ld (ex),hl.skip call collision call updaterocket call updateobstacle ld a,(dead) and a jr nz,maingame call printexplosion ld a,(dead) inc a ld (dead),a ret ;; Return to BASIC.printrocket ld hl,(ypos) ld (ypos1),hl ld a,1 ld (spr_num),a call ariom_spr ret .updaterocket call clearrocket call scroll call printrocket call &bd19 ;; FRAME call &bd19 call &bd19 call &bd19 ret.clearrocket ld hl,(ypos) ld (ypos1),hl ld a,0 ld (spr_num),a call ariom_spr ret.printexplosion ld hl,(ypos) ld (ypos1),hl ld a,2 ld (spr_num),a call ariom_spr ret.updateobstacle call rand call scalenum ld a,(result) ld (xpos1),a ld a,2 ld (ypos1),a ld a,3 ld (spr_num),a call ariom_spr ret.collision ld hl,(ex) ex hl,de ld hl,(ey) call &bbf0 ;; TEST(ex,ey) cp 0 ;; PEN:=0 jr z,endcoll ;; IF yes endcoll ld a,(dead) ;; dec a ;; You're Dead ld (dead),a ;;.endcoll ret.genseed ld a,r ld (seed),a ret.rand ld a,(seed) ld b,a add a,a add a,a add a,b inc a ld (seed),a ret.setinks ld c,(hl) ld b,c push af push hl call &bc32 pop hl pop af inc hl inc a cp 16 jr c,setinks ret.scroll ld b,192 ;; This is how many lines to scroll down..loop push bc ;; This protects the Loop Counter (not required at this stage) ld hl,(adr_hl) ;; This holds the address pointer and is place into HL ld e,(hl) ;; Though the Contents from that address goes into DE inc hl ;; Cause it's 16bit data this is the only way I know ld d,(hl) ;; How to do it, there maybe other ways. I dunno. ex de,hl ;; The data above needs to go into HL and not DE push hl ;; I need to protect my data in HL ld hl,(adr_de) ;; This holds the address pointer and is place into HL ld e,(hl) ;; Though the Contents from that address goes into DE inc hl ;; Cause it's 16bit data this is the only way I know ld d,(hl) ;; How to do it, there maybe other ways. I dunno. pop hl ;; But restore it again for this routine ld bc,&40 ;; BC is used here to address the length of the data on screen upto &40. ldir ;; This instruction shifts the screen. pop bc ;; Restores Loop Counter value so loop can address it. ld hl,(adr_hl) ;; Increment Start Position to the next spot inc hl inc hl ld (adr_hl),hl ;; And store it in that pointer variable. ld hl,(adr_de) ;; Increment Destination to the next position inc hl inc hl ld (adr_de),hl ;; And store it in that pointer variable djnz loop ;; returns to loop if any remaining lines, otherwise proceed to exit ld hl,begin ;; Restores pointer to the beginning ld (adr_hl),hl ;; should the user want to continue calling routine ld hl,dest ;; Restores pointer to the Destination ld (adr_de),hl ;; should the user want to continue calling routine ret ;; exit routine.scalenum ld a,(seed) srl a srl a srl a srl a inc a ld (result),a ret;; Ariom Sprites by David Hall.ariom_spr ld a,(spr_num) ld b,a ;; Sprite Number ld de,&0020 ;; Incrementer for 8x8 sprite (4 x 8 = 32 or 20h) ld hl,Rocket-&20 ;; address of sprites - &20 = &7d18.spr_table add hl,de ;; add &20 to &7d18 to obtain start of sprite table djnz spr_table ;; loop until spr_num = 0, each loop adds &20 to obtain the correct sprite data push hl ;; preserve sprite detail ld a,(xpos1) ld b,a ;; xpos details ld hl,&bfac ;; this address serves as a base ld e,&04 ;; 4 is incremented as each byte holds 4 colours.x_scr_pos add hl,de ;; djnz x_scr_pos ;; loop until b reaches zero ld e,&50 ;; &50 is added to place sprite in screen memory ld a,(ypos1) ld b,a ;; ypos details.y_scr_pos add hl,de ;; djnz y_scr_pos ;; loop until b reaches zero, by this stage hl holds an address somewhere on-screen pop de ;; restore sprite detail in de ld c,&08 ;; c = height of image.nxt_line push hl ;; preserve screen address in hl ld b,&04 ;; b = width of image.drw_spr ld a,(de) ;; a = byte contents ld (hl),a ;; draw byte contents to screen inc hl ;; increase screen address inc de ;; increase to next position of sprite djnz drw_spr ;; loop until b = 0 pop hl ;; restore original screen address ld a,&08 ;; with a = 8 add h ;; add to h to position to a new line ld h,a ;; and placed into h dec c ;; decrement the height of the image jr nz,nxt_line ;; if this doesn't equal 0 proceed to next line ret ;; return to BASIC or M/C routine .ypos defb 10.xpos defb 10.ypos1 defb 2.xpos1 defb 5.ox defb 10.dead defb 1.seed defb 0.result defb 0.ex defw 296.ey defw 270.spr_num defb 3.colours defb 0,24,20,6,26,0,2,8,10,12,14,16,18,22,13,3.adr_hl defw begin ;; pointer position to where the start position is for HL register.adr_de defw dest ;; pointer position to where the start position is for DE register;; Sprites.Rocketdefb &00,&15,&00,&00; line 0defb &00,&3A,&2A,&00; line 1defb &15,&64,&35,&00; line 2defb &15,&30,&35,&00; line 3defb &00,&3A,&2A,&00; line 4defb &15,&3A,&3F,&00; line 5defb &44,&44,&44,&00; line 6defb &00,&00,&00,&00; line 7.Explosiondefb &00,&FF,&00,&00; line 0defb &55,&CC,&AA,&00; line 1defb &EE,&60,&DD,&00; line 2defb &EE,&C0,&DD,&00; line 3defb &EE,&C0,&DD,&00; line 4defb &55,&CC,&AA,&00; line 5defb &00,&FF,&00,&00; line 6defb &00,&00,&00,&00; line 7.Bolderdefb &00,&FF,&00,&00; line 0defb &00,&FF,&AA,&00; line 1defb &55,&FF,&FF,&AA; line 2defb &FF,&FF,&FF,&FF; line 3defb &55,&FF,&FF,&FF; line 4defb &7F,&FF,&FF,&BF; line 5defb &15,&7F,&7F,&2A; line 6defb &00,&15,&15,&00; line 7.dest defw &C780,&CF80,&D780,&DF80,&E780,&EF80,&F780,&FF80.begin defw &C730,&CF30,&D730,&DF30,&E730,&EF30,&F730,&FF30 defw &C6E0,&CEE0,&D6E0,&DEE0,&E6E0,&EEE0,&F6E0,&FEE0 defw &C690,&CE90,&D690,&DE90,&E690,&EE90,&F690,&FE90 defw &C640,&CE40,&D640,&DE40,&E640,&EE40,&F640,&FE40 defw &C5F0,&CDF0,&D5F0,&DDF0,&E5F0,&EDF0,&F5F0,&FDF0 defw &C5A0,&CDA0,&D5A0,&DDA0,&E5A0,&EDA0,&F5A0,&FDA0 defw &C550,&CD50,&D550,&DD50,&E550,&ED50,&F550,&FD50 defw &C500,&CD00,&D500,&DD00,&E500,&ED00,&F500,&FD00 defw &C4B0,&CCB0,&D4B0,&DCB0,&E4B0,&ECB0,&F4B0,&FCB0 defw &C460,&CC60,&D460,&DC60,&E460,&EC60,&F460,&FC60 defw &C410,&CC10,&D410,&DC10,&E410,&EC10,&F410,&FC10 defw &C3C0,&CBC0,&D3C0,&DBC0,&E3C0,&EBC0,&F3C0,&FBC0 defw &C370,&CB70,&D370,&DB70,&E370,&EB70,&F370,&FB70 defw &C320,&CB20,&D320,&DB20,&E320,&EB20,&F320,&FB20 defw &C2D0,&CAD0,&D2D0,&DAD0,&E2D0,&EAD0,&F2D0,&FAD0 defw &C280,&CA80,&D280,&DA80,&E280,&EA80,&F280,&FA80 defw &C230,&CA30,&D230,&DA30,&E230,&EA30,&F230,&FA30 defw &C1E0,&C9E0,&D1E0,&D9E0,&E1E0,&E9E0,&F1E0,&F9E0 defw &C190,&C990,&D190,&D990,&E190,&E990,&F190,&F990 defw &C140,&C940,&D140,&D940,&E140,&E940,&F140,&F940 defw &C0F0,&C8F0,&D0F0,&D8F0,&E0F0,&E8F0,&F0F0,&F8F0 defw &C0A0,&C8A0,&D0A0,&D8A0,&E0A0,&E8A0,&F0A0,&F8A0 defw &C050,&C850,&D050,&D850,&E050,&E850,&F050,&F850 defw &c000,&c800,&d000,&d800,&e000,&e800,&f000,&f800.theend defb 0`
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https://mathoverflow.net/questions/317598/does-f-have-the-same-minimiser-as-nabla-f-for-f-strictly-convex | 1,611,506,625,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00029.warc.gz | 456,797,184 | 28,927 | # Does $f$ have the same minimiser as $\|\nabla f \|$ for $f$ strictly convex?
This question is migrated from MathStackExchange where it seemed to be too hard. I wonder does anyone here have any ideas?
Suppose $$f: K \to \mathbb R$$ is $$\mathcal C^2$$ and strictly convex on some compact convex $$K \subset \mathbb R^n$$. That means $$f(a x + by) < a f(x) + bf(y)$$ for all $$x,y \in K$$ and $$a,b \in (0,1)$$ with $$a+b=1$$.
Strict convexity implies $$f$$ has a unique minimum over $$K$$. In all examples I have found, this minimum also coincides with the minimum of $$\|\nabla f\|$$. I wonder if it is always true but am unable to find or invent a proof unless $$n=1$$.
For $$n = 1$$ the gradient $$\nabla f(x)=f'(x)$$ is just the derivative and strict convexity is equivalent to $$f'$$ being strictly increasing. In that case we can show $$\min f$$ occurs at the minimiser of $$|f'|$$.
Consider first the case that $$f'(a) = 0$$ for some $$a \in K$$. By strict convexity there is only one such $$a$$. Clearly $$a$$ is the unique minimiser of $$|f'|$$. But $$f'(a) = 0$$ also means $$a$$ is a local minimum, and then convexity implies $$a$$ is the global minimum.
Otherwise $$f'$$ is strictly positive or negative negative. First assume the former. Without loss of generality we have $$K =[0,1]$$. Then since $$f'$$ is strictly increasing $$|f'(x)| = f'(x)$$ has unique minimum at $$0$$. Also by writing $$f(x) = f(0) + \int_0^x f'(y) \, dy$$ as the integral of a strictly positive function we see $$f$$ is also strictly increasing hence also has unique minimum at $$0$$. A symmetric argument shows for $$f'$$ strictly negative both functions have unique minimum at $$1$$.
Of course for $$n>1$$ there is no notion of the gradient vector being positive and the proof fails to generalise. Also the higher-dimensional analogue of $$f(x) = \int_0^x f'(y) \, dy$$ is Stoke's theorem which does not recover values of $$f$$ at a point. Rather the left-hand-side becomes the integral over the boundary of whatever region we're integrating over on the right.
One fact I imagine is useful is the function $$F(x) = \|\nabla f(x)\|^2$$ is differentiable with gradient $$\nabla F(x) = H(x) \nabla f(x)$$ where $$H(x)$$ is the Hessian of $$f$$ at $$x$$. By strict convexity the Hessian is positive definite.
Now $$F(x)$$ achieves its minimum at some $$a \in K$$. Without loss of generality $$a=0$$. Unless $$F(a) = 0$$ we know the gradient vector $$\nabla F(a)$$ is normal to $$K$$. That means $$K$$ is contained in the halfspace $$\{x \in \mathbb R^n: H(x) \nabla f(x) \cdot x\ge 0\}$$. That tells us moving in the direction $$\nabla f$$ a small amount will increase $$F$$ since the directional derivative is $$H(x) \nabla f(x) \cdot \nabla f(x) = \nabla f(x) ^T H(x) \nabla f(x) > 0$$ by positive definiteness.
If $$H(x) \nabla f(x)$$ was parallel to $$\nabla f(x)$$ we would be done but in general this doesn't occur. For example consider the function $$f(x) = x^2 + 2y^2$$ over $$K=[1,2]^2$$. Clearly $$f$$ is minimised at $$(1,1)$$ where the gradient is $$\nabla f(1) = (2,4)$$. We can compute $$F(x) = (2x)^2 + (4y)^2 = 4x^2 + 16y^2$$. This is also minimised at $$(1,1)$$ where the gradient is $$\nabla F(1) =(8,32)$$. In this case the gradients are not parallel but they are both normal to $$K$$.
Note: There is something perverse about the statement that $$f$$ has the same minimiser as $$\|\nabla f \|$$ because $$f$$ is coordinate-independent (obviously) but $$\|\nabla f \|$$ is not -- it depends on some choice of inner-product to define. However I don't think this should be a problem since changing the inner product should only amount to skewing the domain. That transform takes straight lines to straight lines so it should preserve convexity and take the points we're interested to onto each other.
Consider $$f(x,y)=x^3+ay^3$$ ($$a>0$$) on $$K=\{x,y\ge 0,x+y=1\}$$ (or in a thin convex set around this interval). The minimum of $$f$$ is at the point where $$x^2=ay^2$$ or $$x=a^{1/2}y$$. The square of the gradient is (up to a constant) $$x^4+a^2y^4$$ and its minimum is attained when $$x^3=a^2y^3$$ or $$x=a^{2/3}y$$, which is a different point for $$a\ne 1$$.
• Note also that the same example works on $K = \{ 0 \leq x,y,\ x+y \leq 1 \}$, so requiring that $K$ be full dimensional doesn't help. – David E Speyer Dec 13 '18 at 14:42
• @DavidESpeyer You meant $1\le x+y \le 2$, right? – fedja Dec 13 '18 at 15:11 | 1,382 | 4,432 | {"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": 78, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-04 | latest | en | 0.900296 |
https://developpaper.com/buckle-questions-169-171-189-198-and-202/ | 1,632,756,870,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00128.warc.gz | 263,330,622 | 12,948 | Time：2021-2-22
# Buckle questions 169, 171, 189, 198 and 202
## 169. Most elements
### code:
``````class Solution {
public int majorityElement(int[] nums) {
Arrays.sort(nums);
return nums[nums.length/2];
}
}``````
## 171. Serial number of Excel table
### How to solve the problem:
• Tag: String traversal, decimal conversion
• The initialization result is ans = 0. During traversal, each letter is subtracted from A. because a represents 1, after subtraction, each number needs to be added with 1 to calculate its representation
• Num = letter -‘a ‘+ 1
• Because there are 26 letters, it is equivalent to 26 base system, and every 26 numbers advance one bit
• So for every bit traversed, ans = ans * 26 + num
• Take ZY as an example, if Z is 26 and Y is 25, the result is 26 * 26 + 25 = 701
• Time complexity: O (n)
### code:
``````class Solution {
public int titleToNumber(String s) {
int ans = 0;
for(int i=0;i<s.length();i++) {
int num = s.charAt(i) - 'A' + 1;
ans = ans * 26 + num;
}
return ans;
}
}``````
## 189. Turn the array
### The first idea is to solve the problem
The simplest way is to rotate the array K times, one element at a time.
Complexity analysis
Time complexity: O (n * k). Each element is moved one step o (n) k times o (k).
Space complexity: O (1). No extra space is used.
### Code 1:
``````public class Solution {
public void rotate(int[] nums, int k) {
int temp, previous;
for (int i = 0; i < k; i++) {
previous = nums[nums.length - 1];
for (int j = 0; j < nums.length; j++) {
temp = nums[j];
nums[j] = previous;
previous = temp;
}
}
}
}``````
### The second idea is to solve the problem
We can use an extra array to put each element in the correct position, that is, the original array with the subscript i, we put it in the position of (I + k)% of the length of the array. Then copy the new array to the original one.
### Code 2:
``````class Solution {
public void rotate(int[] nums, int k) {
int[] a=new int[nums.length];
for(int i=0;i<nums.length;i++){
a[(i+k)%nums.length]=nums[i];
}
for(int i=0;i<nums.length;i++){
nums[i]=a[i];
}
}
}``````
## 198. Robbing homes
### Question:
You are a professional thief, planning to steal houses along the street. There is a certain amount of cash hidden in each room. The only restricting factor for you to steal is that the adjacent houses are equipped with interconnected anti-theft systems. If two adjacent houses are intruded by thieves at the same night, the system will automatically alarm.
Given an array of non negative integers representing the amount of money stored in each house, calculate the maximum amount you can steal in one night without touching the alarm device.
Example 1:
Input: [1,2,3,1] output: 4 explanation: steal house 1 (amount = 1), then steal house 3 (amount = 3). The maximum amount of money stolen = 1 + 3 = 4.
Example 2:
Input: [2,7,9,3,1] output: 12 explanation: steal house 1 (amount = 2), steal house 3 (amount = 9), and then steal house 5 (amount = 1). The maximum amount of money stolen = 2 + 9 + 1 = 12.
Tips:
0 <= nums.length <= 100 0 <= nums[i] <= 400
### How to solve the problem:
Consider the simplest case first. If there is only one house, the maximum total amount can be obtained by stealing the house. If there are only two houses, because the two houses are adjacent to each other, they can not steal at the same time, only one of them can be stolen. Therefore, choosing the house with higher amount of money to steal can steal the maximum total amount.
1.If the number of houses is more than two, how to calculate the maximum total amount that can be stolen? For the K ~ (k > 2) house, there are two options
2.If you steal the kth house, you can’t steal the Kth-1 house. The total amount of theft is the sum of the maximum total amount of the first K-2 houses and the amount of the kth house.
The total amount of theft is the highest total amount of the first k-1 houses.
In the two options, choose the option with larger total amount of theft, and the total amount of theft corresponding to this option is the highest total amount that the top k houses can steal.
If DP [i] is used to represent the maximum total amount that can be stolen from the first I houses, then there is the following state transition equation:
`dp[i]=Math.max(dp[i-2]+nums[i],dp[i-1]);`
### code:
``````class Solution {
public int rob(int[] nums) {
if(nums==null||nums.length==0){
return 0;
}
int length=nums.length;
if(length==1){
return nums[0];
}
int[] dp=new int[length];
dp[0]=nums[0];
dp[1]=Math.max(nums[0],nums[1]);
for(int i=2;i<length;i++){
dp[i]=Math.max(dp[i-2]+nums[i],dp[i-1]);
}
return dp[length-1];
}
}``````
## 202. Happy number
### Question:
Write an algorithm to determine whether a number n is a happy number.
“Happy number” is defined as: for a positive integer, replace the number with the sum of squares of the numbers in each position each time, and then repeat the process until the number becomes 1, or it may be an infinite cycle but never becomes 1. If it can be changed to 1, then this number is the happy number.
If n is a happy number, it returns true; if n is not, it returns false.
Example:
Input: 19 output: true explanation: 12 + 92 = 82 + 22 = 68 62 + 82 = 100 12 + 02 + 02 = 1
### The first idea is to solve the problem
The algorithm is divided into two parts, we need to design and write code.
1.Given a number n, what is the next number?
2.According to a series of numbers to determine whether we have entered a cycle.
In the first part, we separate the digits and find the sum of squares.
Part 2 can be done with HashSet. Every time we generate the next number in the chain, we check whether it is already in the HashSet.
If it’s not in the HashSet, we should add it.
If it’s in a HashSet, that means we’re in a loop, so we should return false.
The reason we use hashsets instead of vectors, lists, or arrays is because we repeatedly check for the presence of a number. It takes O (1) to check whether the number is in the hash set, and O (n) for other data structures. Choosing the right data structure is the key to solve these problems.
The core idea is to use a HashSet and recursion. Through recursion, we can judge whether result 1 will appear in the end. Through HashSet, we can know whether it is a loop.
### Code 1:
``````class Solution {
private int getNext(int n) {
int totalSum = 0;
while (n > 0) {
int d = n % 10;
n = n / 10;
totalSum += d * d;
}
}
public boolean isHappy(int n) {
Set<Integer> seen = new HashSet<>();
while (n != 1 && !seen.contains(n)) {
n = getNext(n);
}
return n == 1;
}
}``````
### The second idea is to solve the problem
The principle is that we use two pointers, one is fast and the other is slow. If it is a happy number, then the fast pointer will reach the number 1 first, and it will end. If it is not a happy number, then the fast and slow pointers will finally meet on a certain number. In coding, we can make the fast pointer execute one more step each time, which is FAS tRunner = getNext(getNext(fastRunner));
### Code 2:
``````class Solution {
public int getNext(int n){
int totalSum=0;
while(n>0){
int d=n%10;
n=n/10;
totalSum+=d*d;
}
}
public boolean isHappy(int n) {
int slowRunner=n;
int fastRunner=getNext(n);
while(fastRunner!=1&&slowRunner!=fastRunner){
slowRunner=getNext(slowRunner);
fastRunner=getNext(getNext(fastRunner));
}
return fastRunner==1;
}
}``````
## Hot! Front and rear learning routes of GitHub target 144K
Hello, Sifu’s little friend. I’m silent Wang Er. Last week, while appreciating teacher Ruan Yifeng’s science and technology weekly, I found a powerful learning route, which has been marked with 144K on GitHub. It’s very popular. It covers not only the front-end and back-end learning routes, but also the operation and maintenance learning routes. As […] | 2,093 | 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} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.658384 |
https://programming.vip/docs/introduction-to-dynamic-programming-knapsack-problems-such-as-0-1-knapsack.html | 1,643,252,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00145.warc.gz | 518,078,532 | 6,661 | # Introduction to dynamic programming (knapsack problems such as 0-1 knapsack)
## Basic idea of dynamic programming
Understand the basic idea of dynamic programming
1. Determine state variables (functions)
2. Determine the state transition equation (recursive relationship)
3. Determine boundary
## Various knapsack problem contents, ideas, examples and code implementation
### Basic 0-1 knapsack problem
The basic idea of 0-1 knapsack problem
0-1 knapsack detailed problem solution
0-1 knapsack simplification / dimensionality reduction idea – it's a little complicated, but it's very detailed, mainly: 1 Reverse order: because the information obtained in this layer is recursively obtained by the previous layer, the reverse order protects the information of the previous layer to ensure the correctness of the information obtained in this layer; 2. The essence of dimensionality reduction is that the information can be covered in the recursive process without affecting the results, because the information of layer i is only related to layer i-1 and has nothing to do with layer i-2.
Title: The most basic 0-1 knapsack problem
```#include<bits/stdc++.h>
using namespace std;
const int N=1e4+10;
int wei[N],val[N];
//int f[N][N];//f[i][j] represents the maximum value of I items in the backpack with capacity j
int f[N];
int n,wemax;
int main()
{
cin>>n>>wemax;//4 5
for(int i=1;i<=n;i++)
{
cin>>wei[i]>>val[i];
}
/*Not simplified:
for(int i=1;i<=n;i++)
{
for(int j=0;j<=wemax;j++)//j It is the [0,wemax] of capacity to capacity. In each case, consider whether to put it or not, and keep the larger value
{
f[i][j]=f[i-1][j];//Don't put the value of the ith item
if(j>=wei[i]) f[i][j]=max(f[i][j],f[i-1][j-wei[i]]+val[i]);
/* Explanation:
j>=wei[i] The backpack itself has enough capacity to put the i-th item
Do you prefer not to put the i-th item? Is it more valuable or more valuable (if you put it, some of the previous items may have to be taken out)
f[i][j]，That is, f[i-1][j] is the value when it is not released
f[i-1][j-wei[i]]+val[i]Is the maximum value when the capacity is j-wei[i] + the value of the current item
That is, at this time, the value is the largest from the previous situation and the ith item can be put down
The problem with abstraction is why f[i-1] is OK
Because according to the definition, f[i-1] is the optimal solution in its case, the optimal solution + v must be the optimal solution in f[i]
}
}
for(int i=1;i<=n;i++)
{
for(int j=0;j<=wemax;j++)
{
cout<<f[i][j]<<" ";
}
cout<<endl;
}
cout<<f[n][wemax]<<endl;
*/
//simplify:
for(int i=1;i<=n;i++)
{
//Reverse order
//Because the determination of f[j] of layer i must use the f[j] of layer i-1, and the obtained f[j] of this layer cannot be used, the reverse order must be used.
//One dimensional storage is to cover the previous information and reduce the space
for(int j=wemax;j>=wei[i];j--)
{
f[j]=max(f[j],f[j-wei[i]]+val[i]);
//cout<<f[j]<<" ";
}
//cout<<endl;
}
/*for(int i=1;i<=n;i++)
{
cout<<f[i]<<" ";
}*/
cout<<f[wemax]<<endl;
return 0;
}
/*
Not simplified:
f[i][j]:
j:1 2 3 4 5
0 2 2 2 2 2
0 2 4 6 6 6
0 2 4 6 6 8
0 2 4 6 6 8
simplify:
f[i]: 2 4 6 6 8
Change in cycle:
5 4 3 2 1 <-order
2 2 2 2 2
6 6 6 4
8 6 6
8 6
*/
```
### 0-1 knapsack problem with two constraints
Title: NASA's Food Program
```#include<bits/stdc++.h>
using namespace std;
const int N=1e3;
int wei[N],v[N],ca[N];
int f[N][N];
struct node{
int v;
int w;
int ca;
}food[N];
int main()
{
int maxv,maxw;
cin>>maxv>>maxw;
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>food[i].v>>food[i].w>>food[i].ca;
}
for(int i=1;i<=n;i++)
{
for(int j=maxv;j>=food[i].v;j--)//First limit
{
for(int k=maxw;k>=food[i].w;k--)//Second limit
{
f[j][k]=max(f[j][k],f[j-food[i].v][k-food[i].w]+food[i].ca);
}
}
}
cout<<f[maxv][maxw];
return 0;
}
```
### Complete knapsack problem in which items can be selected infinitely
Title: Complete knapsack problem , that is, one more layer of circulation on the basis of 0-1 backpack, Detailed text explanation
```#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int f[N];
int wei[N],val[N];
/*
Combine K selected items with wei[i], f[i][j] = max(f[i][j],f[i-1][j-k*v[i]]+k*w[i])
Optimization:
f[i , j ] = max( f[i-1,j] , f[i-1,j-wei]+val , f[i-1,j-2*wei]+2*val , f[i-1,j-3*wei]+3*val , .....)
f[i , j-v]= max( f[i-1,j-wei] , f[i-1,j-2*wei] + val , f[i-1,j-3*wei]+2*val , .....)
-------->Then f [i] [J] = max (f [I-1] [J], f [I, j-wei] + VAL);
Compare 0-1 knapsack f [i] [J] = max (f [I-1] [J], f [I-1, j-wei] + VAL);
Before further optimization, f[i][j]=f[i-1][j];
After further optimization: if the 0-1 knapsack is f[i-1,j-wei]+val, it needs to be in reverse order, because it is related to the i-1 layer, and the range is [m, wei[i]]
However, the complete knapsack is f[i,j-wei]+val), which does not need to be in reverse order. It can be from small to large, and the range is [wei[i], m]
-------->State transition equation: f[j] = max(f[j],f[j-wei[i]]+val[i])
*/
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>wei[i]>>val[i];
}
for(int i=1;i<=n;i++)
{
for(int j=wei[i];j<=m;j++)//Enumerate from small to large because layer i is independent of layer i-1
{
f[j]=max(f[j],f[j-wei[i]]+val[i]);
}
}
cout<<f[m];
}
```
### Multiple knapsack problem in which objects can be selected finite times 1
Title: Multiple knapsack problem 1
Based on the complete knapsack problem, a limit of K < = num [i] & & K * Wei [i] < = J is added to limit the number of items selected.
```#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int f[N][N];
int wei[N],val[N],num[N];
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>wei[i]>>val[i]>>num[i];
}
for(int i=1;i<=n;i++)
{
for(int j=0;j<=m;j++)
{
for(int k=0;k<=num[i]&&k*wei[i]<=j;k++)
f[i][j]=max(f[i][j],f[i-1][j-wei[i]*k]+val[i]*k);
}
}
cout<<f[n][m];
return 0;
}
```
### Multiple knapsack problem 2, binary optimization method
Title: Multiple knapsack problem 2 ,Topic Video Explanation
I didn't particularly understand. I'll listen to the video carefully when I'm free
```#include<bits/stdc++.h>
using namespace std;
const int N=2010;
int f[N];
struct Good{
int wei,val;
};
int main()
{
int n,m;
cin>>n>>m;
vector<Good> goods;
for(int i=1;i<=n;i++)
{
int w,v,s;
cin>>w>>v>>s;
for(int k=1;k<=s;k*=2)
{
s-=k;
goods.push_back({w*k,v*k});
}
if(s>0) goods.push_back({w*s,v*s});
}
for(auto good:goods)
{
for(int j=m;j>=good.wei;j--)
{
f[j]=max(f[j],f[j-good.wei]+good.val);
}
}
cout<<f[m];
return 0;
}
```
### Grouping knapsack problem
Title: Grouping knapsack problem
Solution: Detailed text explanation,Video problem solving
It is said to be the general case of multiple knapsack problem
```#include<bits/stdc++.h>
using namespace std;
const int N=110;
int f[N];
int wei[N][N],val[N][N],s[N];//s represents the number of items in group i
/*
0-1: Maximum value of the first i items with volume less than j
Group backpack: the maximum value of the first group i items whose volume is less than j
*/
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>s[i];
for(int j=0;j<s[i];j++)
{
cin>>wei[i][j]>>val[i][j];
}
}
for(int i=1;i<=n;i++)//stage
{
for(int j=m;j>=0;j--)//i. j. common composition state
{
for(int k=0;k<s[i];k++)//k is decision
{
if(wei[i][k]<=j)
{
f[j]=max(f[j],f[j-wei[i][k]]+val[i][k]);
}
}
}
}
cout<<f[m]<<endl;
return 0;
}
```
### 0-1 knapsack extension question
Title: Flowers
Resolution: Detailed explanation of Luogu
Idea:
I don't particularly understand this question. I'll consider it later
UN optimized code:
```#include<bits/stdc++.h>
using namespace std;
const int maxn=105, mod = 1000007;
int n, m, a[maxn], f[maxn][maxn];
int main()
{
cin>>n>>m;
for(int i=1; i<=n; i++) cin>>a[i];
f[0][0] = 1;
for(int i=1; i<=n; i++)
for(int j=0; j<=m; j++)
for(int k=0; k<=min(j, a[i]); k++)
f[i][j] = (f[i][j] + f[i-1][j-k])%mod;
cout<<f[n][m]<<endl;
return 0;
}
```
The optimized version code is as follows:
```#include<bits/stdc++.h>
using namespace std;
const int N=110;
int f[N];
int a[N];
int mod=1000007;
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++) cin>>a[i];
f[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=m;j>=0;j--)
{
for(int k=1;k<=min(a[i],j);k++)
{
f[j]=(f[j]+f[j-k])%mod;
}
}
}
cout<<f[m]<<endl;
}
```
In Article 2, each model and state transition formula are analyzed in detail
## summary
In the case of introduction to dynamic planning, I wrote several topics related to backpack, but I still know a little. I don't have the ability to independently analyze and solve new dynamic planning problems. I can only recall the problems and use methods that have been done according to my memory, and then consider how to solve new problems. I hope more and more questions can make the situation not so cramped
Keywords: Algorithm Dynamic Programming
Added by Haggen on Thu, 13 Jan 2022 18:23:28 +0200 | 2,930 | 8,862 | {"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} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.77817 |
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Fraction Calculator. Below are multiple fraction calculators capable of addition, subtraction, multiplication, division, simplification, and conversion between fractions and decimals. …
### Solve | Microsoft Math Solver
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Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.
### Fraction Calculator & Problem Solver - Chegg
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Understand the how and why See how to tackle your equations and why to use a particular method to solve it — making it easier for you to learn.; Learn from detailed step-by-step …
### Fraction calculator - calculation: 2/5+1/5 - hackmath.net
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Add: 2 / 5 + 1 / 5 = 2 + 1 / 5 = 3 / 5 It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common …
### Fraction calculator - calculation: 5 1/2-2 1/5
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Whole number 5 equally 5 * 2. b) Add the answer from the previous step 10 to the numerator 1. New numerator is 10 + 1 = 11. c) Write a previous answer (new numerator 11) over the …
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### How can we make 2+2=5? - Quora
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Answer (1 of 590): It's from the book 1984 by George Orwell. In the novel, it is used as an example of an obviously false dogma that one may be required to believe, similar to …
## 2 2 1 5 & other calculators
Online calculators are a convenient and versatile tool for performing complex mathematical calculations without the need for physical calculators or specialized software. With just a few clicks, users can access a wide range of online calculators that can perform calculations in a variety of fields, including finance, physics, chemistry, and engineering. These calculators are often designed with user-friendly interfaces that are easy to use and provide clear and concise results. | 937 | 3,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.794897 |
https://community.airtable.com/t/add-a-space-to-a-field-before-a-number/24663 | 1,566,307,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00529.warc.gz | 421,987,051 | 5,030 | # Add a space to a field before a number
I regularly pull reports from multiple systems that include product IDs. These IDs are always a series of letters followed by numbers (e.g., ABC 123, WXYZ 9876). The number of letters can vary between 3 and 6 characters and the same goes for the numbers.
The problem I’m having is that in one system, there is no space within the ID (e.g., ABC123) while in the other system, there is always a space between the letters and numbers (e.g., ABC 123). Since the number of characters can vary, I’m having trouble finding a formula that basically looks at the cell and adds a space when it finds a number. Any help is appreciated!
Update:
If it helps, this formula is what does what I’m looking for in Excel
=TRIM(REPLACE(A2,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A2&“1234567890”)),0," "))
TRIM, REPLACE, and FIND all exist in Airtable but I’m not sure how to use FIND to search for any combination of items (e.g., 1 OR 2 OR 3…) like you can in the above Excel formula.
Hi @Jonathan_Gunnell1, I’m sure you’ve thought of this, but just in case…
Generally, spaces in an ID field are not great. If you just have two systems and one provides IDs with a space and one without, could you simply remove the space from the system that gives a space, rather than trying to add a space to the system ID that doesn’t? So much easier to achieve with an AT formula:
`SUBSTITUTE(string, old_text, new_text, [index])`
JB
This is a great suggestion. Unfortunately, I had initially tried this approach but then I ran into issues when I had to pull data out of Airtable and upload .csv files into another system that uses the spaces. When that happens, I just pull it into Excel and use the formula mentioned above first, but I would love to find a way to continue using the spaces within Airtable to eliminate that step.
Here’s something that I got to work:
``````LEFT(ID, (LEN(ID) - LEN("" & VALUE(ID)))) & " " & RIGHT(ID, LEN("" & VALUE(ID)))
``````
Running this on non-spaced IDs will add a space before the number portion. If you need to mix and match spaced and non-spaced IDs and have them all come out with spaces, this slight tweak will work:
``````IF(
ID,
TRIM(
LEFT(
ID,
LEN(ID) - LEN("" & VALUE(ID))
)
) & " " &
RIGHT(
ID,
LEN("" & VALUE(ID))
)
)
``````
1 Like | 593 | 2,295 | {"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-2019-35 | latest | en | 0.92652 |
https://www.physicsforums.com/threads/mathematica-ndsolve-periodic-boundary-conditions.797408/ | 1,606,171,872,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00384.warc.gz | 851,699,551 | 14,924 | # Mathematica Ndsolve periodic boundary conditions
• Mathematica
Hello,
n=1,2,3...,10 and stepsize between n(2)-n(1)=0.1
It is periodic. I mean n(11)=n(1). i have a initial function which depends on n
and i want to solve this equation by NDsolve like that
u[n, t = 0] == 1/(2*n + 1)
Do [ u[n, 0], {n, 0, 10, 0.1}] %
u[n + 1, t] == u[1, t]
u [n,t]== u[0,t]
s == NDsolve [
D [u[n, t]] == -[u[n + 1, t] + u[n - 1, t]] - [[u[n, t]]^3], ,
u[n, t == 0], u, {n, 0, 10}, {t, 0, 5}]
Plot3D [s, {n, 0, 10}, {t, 0, 5}]
but it doesn't work.
Thank you so much | 251 | 550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-50 | latest | en | 0.813591 |
https://www.maplesoft.com/support/help/Maple/view.aspx?path=ImageTools/GetLayer | 1,510,980,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00210.warc.gz | 840,204,476 | 24,529 | ImageTools - Maple Programming Help
Home : Support : Online Help : Graphics : Image Processing : ImageTools Package : ImageTools/GetLayer
ImageTools
GetLayer
extract a layer from a multi-layer image
Calling Sequence GetLayer( img, layer, opts )
Parameters
img - ColorImage or ColorAImage; input image layer - 1,2,3,4; layer index opts - (optional) equation(s) of the form option = value; specify options for the GetLayer command
Options
• output = Image
Specifies a data structure into which the output is written. This can be used to avoid allocating memory. The size and number of layers must match that of the input. The dimensions of the output image are adjusted so that the row and column indices match the input. The default is NULL.
Description
• The GetLayer command extracts and returns a single layer from a multi-layer image.
• The img parameter is the image from which the layer is extracted; it must be of type ColorImage or ColorAImage.
• The layer parameter is the index of the layer to extract. For a ColorImage it is an integer from 1 to 3, for a ColorAImage it is an integer from 1 to 4.
Examples
> $\mathrm{with}\left(\mathrm{ImageTools}\right):$
> $\mathrm{img}≔\mathrm{Create}\left(100,200,\left[\left(r,c\right)→0.5+0.5\mathrm{sin}\left(\left(0.005+0.003r\right)r\right)\left(0.5+0.5\mathrm{sin}\left(0.15c\right)\right),\left(r,c\right)→\mathrm{evalf}\left(0.5+{ⅇ}^{-\frac{r}{50}}\cdot 0.5\mathrm{sin}\left(\frac{c}{20}\right)\right),\left(r,c\right)→\mathrm{evalf}\left(\frac{c}{100}\right)\right]\right):$
> $\mathrm{Embed}\left(\mathrm{img}\right)$
> $\mathrm{layers}≔\mathrm{map2}\left(\mathrm{GetLayer},\mathrm{img},\left[1,2,3\right]\right):$
> $\mathrm{Embed}\left(\left[\mathrm{layers}\right]\right)$
> | 529 | 1,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-47 | latest | en | 0.409871 |
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forum discussions will move to the Trailblazer Community.
ShowAll Questionssorted byDate Posted
james2000
# Apex Math Bug?
I've run into a disconcerting issue while writing a test case: there seems to be at least one case of decimal division that doesn't produce an accurate result. Since we deal with a lot financial/currency calculations, this is a little scary to me. Has anyone seen anything like this before or can anyone tell me if I'm doing something wrong here:
Decimal num = 2.10;Decimal denom = 2.0;Decimal result = num / denom;System.debug('Result=' + result.setScale(2).toPlainString());if (result != 1.05) System.debug('Not equal!');
If I run this code in an anonymous block, it displays the following:
20090626192509.249:AnonymousBlock.appirio_sm1: line 4, column 1: Result=1.00
20090626192509.249:AnonymousBlock.appirio_sm1: line 6, column 2: Not equal!
Thoughts?
JimRae
If you use the decimal divide method, with a scale set, the result is correct:
Decimal num = 2.10;
Decimal denom = 2.0;
Decimal result = num / denom;
system.debug('\n\nresult='+result);
Decimal resultdiv =num.divide(denom,2);
system.debug('\n\ndivtest'+resultdiv);
System.debug('Result2=' + result.setScale(2).toPlainString());
system.debug('\n\n');
if (result != 1.05){
System.debug('Not equal!');
}
if (resultdiv != 1.05){
System.debug('Divide Method Not equal!');
}
Additionally, the documention indicates the "/" symbol is used on an integer or a double. If you do your calculation with a double, the answer also comes out correct.
/ Division operator. Divides x, an Integer or Double, by y, another Integer or Double. Note that if a double is used, the result is a Double.
Message Edited by JimRae on 06-26-2009 04:51 PM
JimRae
If you use the decimal divide method, with a scale set, the result is correct:
Decimal num = 2.10;
Decimal denom = 2.0;
Decimal result = num / denom;
system.debug('\n\nresult='+result);
Decimal resultdiv =num.divide(denom,2);
system.debug('\n\ndivtest'+resultdiv);
System.debug('Result2=' + result.setScale(2).toPlainString());
system.debug('\n\n');
if (result != 1.05){
System.debug('Not equal!');
}
if (resultdiv != 1.05){
System.debug('Divide Method Not equal!');
}
Additionally, the documention indicates the "/" symbol is used on an integer or a double. If you do your calculation with a double, the answer also comes out correct.
/ Division operator. Divides x, an Integer or Double, by y, another Integer or Double. Note that if a double is used, the result is a Double.
Message Edited by JimRae on 06-26-2009 04:51 PM
This was selected as the best answer
james2000
Thanks, the divide method does indeed work. So does using a Double. I'm curious why the / operator is allowed for Decimal if it can potentially return faulty or unexpected results though.
Message Edited by james2000 on 06-26-2009 05:22 PM | 777 | 2,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.680822 |
https://www.sarkarirush.com/sd-yadav-math-book-pdf/ | 1,721,561,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00158.warc.gz | 818,483,526 | 25,694 | SD yadav math book pdf: Hello friends, in this article we are going to share with you the one of the very popular books on arithmetic by SD yadav sir that is sharda publication math book sd yadav pdf download in Hindi. This book is specially designed for the aspirants preparing for various competition examination like SSC, Income tax inspector, Bank PO and Clerk, CDS, NDA, CAT, MAT, Railway and various other competition exams.
This book is specially based on the latest SSC exam pattern and the revised and updated edition is well designed with the latest content. There are millions of students preparing for various competition exam as mentioned above and looking for the best resources and study material that will help him to succeed in the examination and this book is the one that will help you to command over the arithmetic part and help you to cover it in a very easy and simple manner.
Contents
The best part about this book is it is available in Hindi, generally those aspirants who have difficulty with full English book then this book is the solution for those, it will help you to easily understand the concept and learn the way to solve the problem.
This entire book comprises of 44 chapter that cover each topic required to cover the arithmetic portion. Below we have provided the chapter wise pdf of each chapter, you can download it one by one and begin your preparations.
Book Name sd yadav math book Author Name sd yadav Format PDF Size MB Pages 167 Language English and Hindi
Contents of sd yadav math book pdf in hindi
Chapter- 01. संख्या पद्धति (Number System Question)
Chapter- 02. महत्तम समापवर्तक तथा लघुत्तम समापवर्त्य (HCF and LCM Question)
Chapter- 03. दशमलव भिन्न (Decimal Fraction Question)
Chapter- 04. सरलीकरण (Simplification Question)
Chapter- 05. वर्गमूल तथा घनमूल (Square Root and Cube Root Question)
Chapter- 06. औसत (Average Question)
Chapter- 07. संख्याओं पर आधारित प्रश्न (Problems on Number Question)
Chapter- 08. आयु संबंधित प्रश्न (Problems on Age Question)
Chapter- 09. घातांक तथा करणी (Surds and Incides Question)
Chapter- 10. प्रतिशतता (Percentage Question)
Chapter- 11. लाभ तथा हानि (Profit and Loss Question)
Chapter- 12. अनुपात व समानुपात (Ratio and Proportion Question)
Chapter- 13. साझा (Partnership Question)
Chapter- 14. मिश्र समानुपात (Compound Proportion Question)
Chapter- 15. समय तथा कार्य (Time and Work Question)
Chapter- 16. पाईप तथा टंकी के प्रश्न (Pipes and Cistems Question)
Chapter- 17. समय तथा दूरी (Time and Distance Question)
Chapter- 18. रेल संबंधित प्रश्न (Problems on Trains Question)
Chapter- 19. धारा तथा नाव संबधित प्रश्न (Boats and Streams Question)
Chapter- 20. मिश्रण (Alligation and Mixture Question)
Chapter- 21. साधारण ब्याज (Simple Interest Question)
Chapter- 22. चक्रवृद्धि ब्याज (Compound Interest PDF)
Chapter- 23. क्षेत्रफल (Area Based Question)
Chapter- 24. ठोस वस्तुओं के आयतन (Volume of Solid Based Question)
Chapter- 25. दौड (Races related Questions)
Chapter- 26. कैलेण्डर (Calendar related Question)
Chapter- 27. घडी (Clock related Question)
Chapter- 28. स्टॉक तथा शेयर (Stock and Shares Question)
Chapter- 29. मिती काटा (True Discount Question)
Chapter- 30. महाजनी बट्टा (Banker’s Discount Question)
Chapter- 31. बीजगणित (Algebra Question)
Chapter- 32. दो चरों में रैखिक समीकरण (Linear Equation in Two Variable Question)
Chapter- 33. द्विघात समीकरण (Quadratic Equation Question)
Chapter- 34. त्रिकोणमिति (Trigonometry Question)
Chapter- 35. रेखायें तथा कोण (Lines and Angles Question)
Chapter- 36. त्रिभुज (Triangles Question)
Chapter- 38. वृत्त (Circles Question)
Chapter- 39. बहुभुज (Polygons Question)
Chapter- 40. सारणीयन (Tabulation Question)
Chapter- 41. दण्ड आलेख (Bar Graphs Question)
Chapter- 42. रेखाचित्र (Line Graphs Question)
Chapter- 43. पाई चार्ट (Pie Chart Question)
Chapter- 44. संख्या श्रेणी (Number Series Question)
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This post may contain Amazon affiliate links. If you make a purchase through these links, we may earn a commission at no additional cost to you. We only recommend products and services we genuinely believe in. | 1,206 | 4,448 | {"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} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.886634 |
https://www.financialmodellinghandbook.org/calculation-block-components/ | 1,723,094,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00069.warc.gz | 619,256,024 | 6,840 | # Calculation blocks: Part 2
Calculation block components
There can only be four different kinds of line items in a calculation block.
1. A calculation
3. A placeholder
4. An input
In this section, we are only going to look at what these things are, and how and why we use them. In the next section, we will go step by step through how we build them.
Let's look at the types of line items we have in the block above:
• Row 23: an input - we’ll move this later.
• Row 24: an offsheet link
• Row 25: a local link -
• Row 26: a calculation. Always the last item in the block.
What we are aiming for here is the ability to look at the model and understand immediately what you’re looking at, without wasting your time interrogating formulas or trying to figure stuff out.
💡
The structure of the model tells you exactly what each item is visually, and therefore instantly.
## 1 A calculation
The calculation is always the last item in the block. You can immediately recognise a calculation block from the structure of the model; and, therefore, immediately know that the last item is a calculation.
By extension, if something is not the last item in the block, it must not be a calculation and will instead be an ingredient to that calculation.
Take a minute to internalise this as it's core to all the modelling we will do in the rest of this book:
💡
The calculation is always the last item in the block. If it's not the last item in the block, it's not a calculation.
There will be times that you need to break this rule. Those times should be the exceptions. We will talk about some exceptions later in the book.
If it's not the last item in the block, and therefore not a calculation, there are only three possible things it can be.
Wherever possible, we present all of a calculation's ingredients in the block, next to the calculation itself. This helps to reduce thinking time. If everything driving the calculation is visible next to it, we reduce the time required to understand and decode a formula.
## 3 A placeholder
Sometimes, when building up a calculation block, we don't have all the ingredients we need. We, therefore, need a structured and consistent plan for creating placeholders. These placeholders are temporary. Later in the process, we will replace them with links.
## 4 An input
This one might make you uncomfortable. Most modellers know that separating inputs, calculations, and outputs is essential. And ultimately, all inputs will end up on a dedicated input sheet.
However, while building the model, it can be more productive to put the input next to the calculation. We have an automated process for relocating these inputs. Like placeholders, this is temporary. When the model is complete, we will replace these inputs with links.
See Modelling Skill 10 to learn about relocating inputs. | 614 | 2,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-33 | latest | en | 0.922717 |
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11:22
What is Correlation in Urdu, What is Correlation in Hindi, What is Correlation Analysis in Urdu, what is Correlation Analysis in Hindi, What is Correlation, Pearson correlation, Correlation formula, Correlation analysis, linear correlation, correlation coefficient formula, pearson correlation interpretation, MBA - MCA - CA - CS - CWA - BBA - BCA - BCom - MCom - GRE - GMAT - Grade 11 - Grade 12 - Class 11 - Class 12 - IAS - CAIIB - FIII - IBPS - BANK PO - UPSC - CPA - CMA - Competitive Exams
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Views: 258076 ECOHOLICS
04:18
Basic Statistical Tests Training session with Dr Helen Brown, Senior Statistician, at The Roslin Institute, December 2015. ************************************************ These training sessions were given to staff and research students at the Roslin Institute. The material is also used for the Animal Biosciences MSc course taught at the Institute. ************************************************ *Recommended YouTube playback settings for the best viewing experience: 1080p HD ************************************************ Content: Measuring association How associated are two measurements? Correlation coefficient calculated from raw data values --The more values deviate from a perfect line, the lower the correlation Most easily calculated within a package Sometimes referred to as ’Pearson’s correlation coefficient’ Correlation between platelets and WBC in new born calves Correlation for non-normal data -Standard correlation coefficient (r) is less appropriate for non-normal data and is particularly sensitive to outlying values -An alternative Spearman rank correlation coefficient is calculated based on the ranks of the data
04:31
Views: 256 Ruth Guthrie
18:02
Learn about correlation analysis in this video For Training & Study packs on Analytics/Data Science/Big Data, Contact us at [email protected] Find all the study packs available with us here: http://analyticuniversity.com/ SUBSCRIBE TO THIS CHANNEL for free tutorials on Analytics/Data Science/Big Data/SAS/R/Hadoop
Views: 10637 Analytics University
14:53
correlation
Views: 319195 Yasser Khan
02:48
Views: 5101 Data Science Dojo
10:25
This playlist/video has been uploaded for Marketing purposes and contains only selective videos. For the entire video course and code, visit [http://bit.ly/2EirTMs]. This video discusses the theory and assumptions of the Pearson correlation coefficient. • Purpose of the Pearson correlation • Hypotheses for the Pearson correlation • Pearson correlation theory and formulas, assumptions For the latest Big Data and Business Intelligence tutorials, please visit http://bit.ly/1HCjJik Find us on Facebook -- http://www.facebook.com/Packtvideo Follow us on Twitter - http://www.twitter.com/packtvideo
Views: 23 Packt Video
09:56
10:16
In this brief presentation, Kelly Clement shows you what correlation analysis is, and how to use it in your market analysis.
Views: 32318 MetaStock
04:51
Views: 22486 Data Science Dojo
02:37
This video is part of an online course, Intro to Statistics. Check out the course here: https://www.udacity.com/course/st101.
Views: 37220 Udacity
09:56
http://www.gdawgenterprises.com This video demonstrates different categories or types of correlation, including positive correlation, negative correlation, and no correlation. Also, strength of correlation is shown. The graphing calculator is used to quantify the strength of correlation by finding the correlation coefficient, sometimes called R.
Views: 44970 gdawgrapper
11:56
Follow me on Twitter @amunategui Check out my new book "Monetizing Machine Learning": https://amzn.to/2CRUOKu A great way to explore new data is to use a pairwise correlation matrix. This will pair every combination of your variables and measure the correlation between them. Code and walkthrough: http://amunategui.github.io/Exploring-Your-Data-Set/ Follow me on Twitter https://twitter.com/amunategui and signup to my newsletter: http://www.viralml.com/signup.html More on http://www.ViralML.com and https://amunategui.github.io Thanks!
Views: 52226 Manuel Amunategui
07:03
Correlation analysis is the process of studying the strength of that relationship with available statistical data.
Views: 500 DARIGA MUSSINA
03:03
TUGAS DATA MINING - CORRELATION Sistem Informasi - Universitas Darma Persada Kelompok 2 : 1. Ryo Gusti N. 2. Osvaldo S. 3. Zulfikar 4. Istiana 5. Karina A. 6. Evan Sandika
Views: 1204 Evan Sandika
34:07
In this video you will learn how to measure the strength of relation between variables by calculating correlation and interpreting it. For Training & Study packs on Analytics/Data Science/Big Data, Contact us at [email protected] Find all free videos & study packs available with us here: http://analyticsuniversityblog.blogspot.in/ SUBSCRIBE TO THIS CHANNEL for free tutorials on Analytics/Data Science/Big Data/SAS/R/Hadoop
Views: 2819 Analytics University
01:53
Mr. Peterson's Stats Class example to find the correlation coefficient using Google SpreadSheets
Views: 25289 Jacob Peterson
01:56
Views: 916 Victor Lavrenko
05:39
Views: 9349 CARAJACLASSES
08:55
Demo covers how you can use the correlation functions in R and uses Rs rich visualisation to see and understand correlation.
Views: 13422 Melvin L
08:20
x2 correlation test (Pearson's chi-squared test) with example - explained. read more at: www.engineeringway.com
Views: 596 Dharmik Thummar
24:58
Course web page: http://web2.slc.qc.ca/pcamire/
Views: 507187 [email protected]
05:56
This video explains what is meant by the covariance and correlation between two random variables, providing some intuition for their respective mathematical formulations. Check out https://ben-lambert.com/econometrics-course-problem-sets-and-data/ for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: https://ben-lambert.com/bayesian/ Accompanying this series, there will be a book: https://www.amazon.co.uk/gp/product/1473916364/ref=pe_3140701_247401851_em_1p_0_ti
Views: 285487 Ben Lambert
06:03
This video will show you how to make a simple scatter plot. Remember to put your independent variable along the x-axis, and you dependent variable along the y-axis. For more videos please visit http://www.mysecretmathtutor.com
Views: 205460 MySecretMathTutor
02:29
Statistica 11 Video Tutorial developed by DLSU PsychLab 2013 Best if viewed on 480p resolution or higher
Views: 10152 DLSUPsychLab
03:58
Cucudata Short Videos: SPSS Data Analysis Tutorial Please Visit：www.cucudata.com For More Videos and Files Download More Data and Information Ask Experts & Get Answers Our Professional Team Provides a Comprehensive Data Analysis & Data Mining Services
Views: 22 spss
23:38
FinTree website link: http://www.fintreeindia.com FB Page link :http://www.facebook.com/Fin... We love what we do, and we make awesome video lectures for CFA and FRM exams. Our Video Lectures are comprehensive, easy to understand and most importantly, fun to study with! This Video was recorded during a one of the CFA Classes in Pune by Mr. Utkarsh Jain.
Views: 22512 FinTree
10:55
Learn how to make predictions using Simple Linear Regression. To do this you need to use the Linear Regression Function (y = a + bx) where "y" is the dependent variable, "a" is the y intercept, "b" is the slope of the regression line, and "x" is the independent variable. This video also shows you how to determine the slope (b) of the regression line, and the y intercept (a). In order to determine the slope of a line you will need to first determine the Pearson Correlation Coefficient - this is described in a separate video (https://www.youtube.com/watch?v=2SCg8Kuh0tE).
Views: 465802 Eugene O'Loughlin
20:53
. Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "FAIR USE" for purposes such as criticism, comment, news reporting, teaching, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, educational or personal use tips the balance in favor of fair use. .
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Example of how to find covariance for a set of data points.
Views: 99106 Stephanie Glen
19:14
Get this full course at http://www.MathTutorDVD.com. In this lesson, you will learn how to identify and construct scatter plots in statistics. In a scatter plot, the x and y variable are plotted as a pair, and use look at the relationship between x and y to determine if there is any relationship between the variables. If the variables are related by positive correlation, the line tends to trend upwards. If the variables are negatively correlated, the line tends to slope downwards. We will learn about scatter plots and how to determine negative and positive correlation in this lesson by looking at the slope of the line connecting the data points.
Views: 14844 mathtutordvd
04:32
This video is suitable for REGRESSION ANALYSIS | REGRESSION STATISTICS | REGRESSION BY CHANDAN PODDAR | REGRESSION EQUATION | REGRESSION CHANGE IN ORIGIN AND SCALE | REGRESSION QUESTION | REGRESSION CONCEPTS | REGRESSION CONCEPTS AND NUMERICAL | REGRESSION NUMERICAL | CORRELATION | CORRELATION IN HINDI | CORRELATION BY CHANDAN PODDAR | TYPES OF CORRELATION | MEANING OF CORRELATION | MEANING OF CORRELATION IN HINDI | CORRELATION FOR CPT . To watch complete course click here :- https://www.vidyakul.com/super-saver/super-saver-by-chandan-sir For Videos related call at :- 9818434684 For Books related enquiry :- 8010201786 For any other Enquiry :- 9953633448 Mail ID :- [email protected]
12:47
This video is related to finding the similarity between the users. It tells us that how much two or more user are similar in terms of liking and disliking the things. How netflix suggest the video to other user if two users have same similarity index.
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11:48
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This video is suitable for REGRESSION ANALYSIS | REGRESSION STATISTICS | REGRESSION BY CHANDAN PODDAR | REGRESSION EQUATION | REGRESSION CHANGE IN ORIGIN AND SCALE | REGRESSION QUESTION | REGRESSION CONCEPTS | REGRESSION CONCEPTS AND NUMERICAL | REGRESSION NUMERICAL | CORRELATION | CORRELATION IN HINDI | CORRELATION BY CHANDAN PODDAR | TYPES OF CORRELATION | MEANING OF CORRELATION | MEANING OF CORRELATION IN HINDI | CORRELATION FOR CPT . To watch complete course click here :- https://www.vidyakul.com/super-saver/super-saver-by-chandan-sir For Videos related call at :- 9818434684 For Books related enquiry :- 8010201786 For any other Enquiry :- 9953633448 Mail ID :- [email protected]
02:06
Learn what is correlation. You will also what is regression in the next video Follow us https://www.facebook.com/AnalyticsUniversity/ For Training & Study packs on Analytics/Data Science/Big Data, Contact us at [email protected] Find all the study packs available with us here: http://analyticsuniversityblog.blogspot.in/ SUBSCRIBE TO THIS CHANNEL for free tutorials on Analytics/Data Science/Big Data/SAS/R/Hadoop
Views: 6936 Analytics University
20:01
Basic introduction to correlation - how to interpret correlation coefficient, and how to chose the right type of correlation measure for your situation. 0:00 Introduction to bivariate correlation 2:20 Why does SPSS provide more than one measure for correlation? 3:26 Example 1: Pearson correlation 7:54 Example 2: Spearman (rhp), Kendall's tau-b 15:26 Example 3: correlation matrix I could make this video real quick and just show you Pearson's correlation coefficient, which is commonly taught in a introductory stats course. However, the Pearson's correlation IS NOT always applicable as it depends on whether your data satisfies certain conditions. So to do correlation analysis, it's better I bring together all the types of measures of correlation given in SPSS in one presentation. Watch correlation and regression: https://youtu.be/tDxeR6JT6nM ------------------------- Correlation of 2 rodinal variables, non monotonic This question has been asked a few times, so I will make a video on it. But to answer your question, monotonic means in one direction. I suggest you plot the 2 variables and you'll see whether or not there is a monotonic relationship there. If there is a little non-monotonic relationship then Spearman is still fine. Remember we are measuring the TENDENCY for the 2 variables to move up-up/down-down/up-down together. If you have strong non-monotonic shape in the plot ie. a curve then you could abandon correlation and do a chi-square test of association - this is the "correlation" for qualitative variables. And since your 2 variables are ordinal, they are qualitative. Good luck
Views: 513214 Phil Chan
07:37
Views: 176956 Two-Point-Four
12:19
Data Mining | Chi-Square Test for Nominal Data | Correlation Test for Nominal Data ******************************************************* categorical data analysis spss, Chi-Square Test for Nominal Data, chi square test nominal data, Correlation Test for Nominal Data, statistical analysis for nominal data, t test categorical data, when to use chi square test in research, chi square test spss, f distribution, test of independence, anova test, chi square test table, chi square test calculator, chi square test genetics, fisher's exact test, chi square test p value, t distribution, test interpretation, correlation analysis spss, chi square test excel, spss regression analysis, Please Subscribe My Channel
Views: 1435 Learning With Mahamud
04:37
This video discusses numerical and graphical methods for exploring relationships between two categorical variables, using contingency tables, segmented bar plots, and mosaic plots.
Views: 27485 Mine Çetinkaya-Rundel | 3,617 | 15,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-13 | longest | en | 0.852108 |
https://hotstarforpc.com/about-golf/how-long-of-a-walk-is-9-holes-of-golf.html | 1,653,040,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00474.warc.gz | 348,518,928 | 19,648 | # How long of a walk is 9 holes of golf?
Contents
## How far do you walk for 9 holes of golf?
Golf can be great exercise. It is said that walking 9-holes on the course is the equivalent of walking over 3 miles! But if you do walk, you still want to be able to play at a decent pace to keep up with the 15- minute a hole standard.
## Is walking 9 holes good exercise?
If a golfer choose to walk while playing golf — either nine holes or 18 holes — they’re going to burn a substantial number of calories. The golf industry likes to say that a golfer will burn as many as 2,000 calories playing golf if they walk all 18 holes while carrying a golf bag — which is usually 20-30 lbs.
## How long does 18 holes of golf take walking?
In golf, 18 holes take 4 hours on average. It takes this time for a standard foursome to get around the course while walking. For 4 players, that gives an average of 13 minutes for each hole.
## How long does it take to play 9 holes of golf with 4 players walking?
Related questions. How long does it take to play nine holes of golf? 9 holes of golf is expected to take no longer than 1 hour 15 minutes for 1 player, 1 hour 30 minutes for a group of 2 players, 1 hour 45 minutes for a group of 3 players and 2 hours for a group of 4 players.
## How many calories do you burn walking 9 holes?
Walking at a slow rate burns 180 calories per hour, so using two hours at this rate as an estimate for the walking component, and subtracting those 360 calories from the previous result, you get about 460 calories for nine holes while riding a golf cart.
## How many steps is 9 holes of golf with a cart?
I clipped a pedometer on Gary’s golf partner, his wife, Karen, to see how many steps she’d get during a round of golf using a cart. She surprised me, clocking 2,880 steps — more than a mile — during nine holes.
## Can you lose weight playing golf?
Golf can absolutely help you lose weight. Walking an 18-hole golf course in four hours can burn up to 800 calories, or even more if the terrain is hilly and if you are carrying your golf clubs in a carry bag.
## Do you play better golf walking or riding?
One of the most obvious advantages of walking during a golf round is the health benefits. A golfer usually burns twice as many calories walking 18 holes as does one who rides. Walking 18 holes is equivalent to a 3.5 – 4 mile run and can exceed 10,000 steps during a typical golf round.
THIS IS EXCITING: Are golf carts allowed in WaterSound Florida?
## Does golf keep you fit?
Golf can be good for your health and your heart. Walking an average course for a round of golf can be between five to seven kilometres. If you walk 18 holes three to five times a week, you’ll get an optimal amount of endurance exercise for your heart.
## How long does a 4 man scramble take?
Playing in a four-person scramble golf tournament may take you an estimate of 5 hours on the course. When the course is packed with people and everyone begins on different holes, the tournament play can get really slow.
## What is an average golfer?
In 2019, according to the National Golf Foundation, golfers played an average of 18.2 rounds. This number varies greatly by age however. Golfers over the age of 65 played an average of 36 rounds a year while those ages 18-34 only play an average of 12 rounds.
## How far do golfers walk at the Masters?
The five-mile or so walk between the Georgia pines at Augusta National is 11,000-plus steps of up and down and up again. It requires hitting shots from uneven lies. Of digging into the pine straw when required.
## What is 9 holes of golf called?
A “9-hole course”, typically the type referred to as an “executive course”, has only 9 holes instead of 18, but with the otherwise normal mix of par-3, par-4 and par-5 holes (typically producing a par score of between 34 and 36), and the course can be played through once for a short game, or twice for a full round.
## What does 9 holes mean in golf?
A 9-hole course is exactly what it says – a golf course with 9 holes. Like 18-hole golf courses, a 9-hole golf course is mostly comprised of par-4 holes plus a combination of par-3 and par-5 holes.
THIS IS EXCITING: What TV channel is showing Olympic golf?
## How long does it take PGA players to play 18 holes?
As you add players to your group, you can expect the time to complete a round to grow by roughly thirty minutes for each additional player. Two players could play in three hours, three players could play in three hours and thirty minutes, and four players, on average, could take four hours to play 18 holes of golf. | 1,109 | 4,610 | {"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} | 3.421875 | 3 | CC-MAIN-2022-21 | latest | en | 0.936822 |
http://work.thaslwanter.at/Stats/html/statsBayes.html | 1,545,219,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832259.90/warc/CC-MAIN-20181219110427-20181219132427-00321.warc.gz | 302,502,848 | 7,310 | # Bayesian vs. Frequentist Interpretation¶
Calculating probabilities is only one part of statistics. Another is the interpretation of them - and the consequences that come with different interpretations.
So far we have restricted ourselves to the frequentist interpretation, which interprets $$p$$ as the frequency of an occurrence: if an outcome of an experiment has the probability $$p$$, it means that if that experiment is repeated $$N$$ times (where $$N$$ is a large number), then we observe this specific outcome $$N*p$$ times.
The Bayesian interpretation of $$p$$ is quite different, and interprets $$p$$ as our believe of the likelihood of a certain outcome. For some events, this makes a lot more sense. For example, in the upcoming semi-final of the soccer worldcup in Brazil, Argentine will play against the Netherlands, with Lionel Messi leading the Argentinian team. Since Messi is (for a soccer player) already fairly old, this may be a one-time event, and we will never have a large number $$N$$ repetitions.
But in addition to this difference in interpretation, the Bayesian approach has another advantage: it lets us bring in prior knowledge into the calculation of the probability $$p$$, through the application of Bayes’ Theorem:
In its most common form, it is:
$\label{eq:BayesTheorem} P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}\cdot$
In the Bayesian interpretation, probability measures a degree of belief. Bayes’s theorem then links the degree of belief in a proposition before and after accounting for evidence. For example, suppose it is believed with 50% certainty that a coin is twice as likely to land heads than tails. If the coin is flipped a number of times and the outcomes observed, that degree of belief may rise, fall or remain the same depending on the results.
John Maynard Keynes, a great economist and thinker, said “When the facts change, I change my mind. What do you do, sir?” This quote reflects the way a Bayesian updates his or her beliefs after seeing evidence.
For proposition A and evidence B,
• P(A), the prior probabiliby, is the initial degree of belief in A.
• P(A|B), the posterior probability, is the degree of belief having accounted for B. It can be read as the probability of A, given that B is the case.
• the quotient P(B|A)/P(B) represents the support B provides for A.
If the number of available data points is large, the difference in interpretation does typically not change the result significantly. If the number of data points is small, however, the possibility to bring in external knowledge may lead to a significantly improved estimation of $$p$$.
# Bayesian Example¶
(This example, and the above definition of Bayes’ Theorem, has been taken from Wikipedia.)
Suppose a man told you he had a nice conversation with someone on the train. Not knowing anything about this conversation, the probability that he was speaking to a woman is 50% (assuming the speaker was as likely to strike up a conversation with a man as with a woman). Now suppose he also told you that his conversational partner had long hair. It is now more likely he was speaking to a woman, since women are more likely to have long hair than men. Bayes’ theorem can be used to calculate the probability that the person was a woman.
To see how this is done, let W represent the event that the conversation was held with a woman, and L denote the event that the conversation was held with a long-haired person. It can be assumed that women constitute half the population for this example. So, not knowing anything else, the probability that W occurs is P(W) = 0.5.
Suppose it is also known that 75% of women have long hair, which we denote as P(L |W) = 0.75 (read: the probability of event L given event W is 0.75, meaning that the probability of a person having long hair (event “L”), given that we already know that the person is a woman (event “W”) is 75%). Likewise, suppose it is known that 15% of men have long hair, or P(L |M) = 0.15, where M is the complementary event of W, i.e., the event that the conversation was held with a man (assuming that every human is either a man or a woman).
Our goal is to calculate the probability that the conversation was held with a woman, given the fact that the person had long hair, or, in our notation, P(W |L). Using the formula for Bayes’ theorem, we have:
$P(W|L) = \frac{P(L|W) P(W)}{P(L)} = \frac{P(L|W) P(W)}{P(L|W) P(W) + P(L|M) P(M)}$
where we have used the law of total probability to expand P(L). The numeric answer can be obtained by substituting the above values into this formula (the algebraic multiplication is annotated using $$\cdot$$, the centered dot). This yields
$P(W|L) = \frac{0.75\cdot0.50}{0.75\cdot0.50 + 0.15\cdot0.50} = \frac56\approx 0.83,$
i.e., the probability that the conversation was held with a woman, given that the person had long hair, is about 83%.
Another way to do this calculation is as follows. Initially, it is equally likely that the conversation is held with a woman as with a man, so the prior odds are 1:1. The respective chances that a man and a woman have long hair are 15% and 75%. It is 5 times more likely that a woman has long hair than that a man has long hair. We say that the likelihood ratio or Bayes factor is 5:1. Bayes’ theorem in odds form, also known as Bayes’ rule, tells us that the posterior odds that the person was a woman is also 5:1 (the prior odds, 1:1, times the likelihood ratio, 5:1). In a formula:
$\frac{P(W|L)}{P(M|L)} = \frac{P(W)}{P(M)} \cdot \frac{P(L|W)}{P(L|M)}.$
# The Bayesian Approach in the Age of Computers¶
Bayes’ theorem was named after the Reverend Thomas Bayes (1701 - 1761), who studied how to compute a distribution for the probability parameter of a binomial distribution. So it has been around for a long time. The reason Bayes’ Theorem has become so popular in statistics in recent years is the cheap availability of massive computational power. This allows the empirical calculation of posterior probabilities, one-by-one, for each new piece of evidence. This, combined with statistical approaches like Markov Chain Monte Carlo (MCMC) simulations, has allowed radically new statistical analysis procedures, and has led to what may be called “statistical trench warfare” between the followers of the different philosophies. If you don’t believe me, check the corresponding discussions on the WWW.
For more information on that topic, check out (in order of rising complexity)
• Wikipedia, which has some nice explanations under Bayes ...
• Bayesian Methods for Hackers, a nice, free ebook, providing a practical introduction to the use of PyMC (see below).
• The PyMC User Guide: PyMC is a very powerful Python package which makes the application of MCMC techniques very simple.
• Pattern Recognition and Machine Learning, a comprehensive, but often quite technical book by Christopher M. Bishop .
# Example: The Challenger Disaster¶
This is an excerpt of the excellent “Bayesian Methods for Hackers”. For the whole book, check out Bayesian Methods for Hackers.
On January 28, 1986, the twenty-fifth flight of the U.S. space shuttle program ended in disaster when one of the rocket boosters of the Shuttle Challenger exploded shortly after lift-off, killing all seven crew members. The presidential commission on the accident concluded that it was caused by the failure of an O-ring in a field joint on the rocket booster, and that this failure was due to a faulty design that made the O-ring unacceptably sensitive to a number of factors including outside temperature. Of the previous 24 flights, data were available on failures of O-rings on 23, (one was lost at sea), and these data were discussed on the evening preceding the Challenger launch, but unfortunately only the data corresponding to the 7 flights on which there was a damage incident were considered important and these were thought to show no obvious trend. The data are shown below:
To simulate the probability of the O-rings failing, we need a function that goes from one to zero. One of the most frequently used functions for that is the logistic function:
$\label{eq:logisticFcn} p(t) = \frac{1}{ 1 + e^{ \;\beta t + \alpha } }$
In this model, the variable $$\beta$$ that describes how quickly the function changes from 1 to 0, and $$\alpha$$ indicates the location of this change.
Using the Python package PyMC, a Monte-Carlo simulation of this model can be done remarkably easily (Note: at the time of writing, PyMC was not yet released for Python 3!):
# --- Perform the MCMC-simulations ---
temperature = challenger_data[:, 0]
D = challenger_data[:, 1] # defect or not?
# Define the prior distributions for alpha and beta
# 'value' sets the start parameter for the simulation
# The second parameter for the normal distributions is the "precision",
# i.e. the inverse of the standard deviation
beta = pm.Normal("beta", 0, 0.001, value=0)
alpha = pm.Normal("alpha", 0, 0.001, value=0)
# Define the model-function for the temperature
@pm.deterministic
def p(t=temperature, alpha=alpha, beta=beta):
return 1.0 / (1. + np.exp(beta * t + alpha))
# connect the probabilities in p with our observations through a
# Bernoulli random variable.
observed = pm.Bernoulli("bernoulli_obs", p, value=D, observed=True)
# Combine the values to a model
model = pm.Model([observed, beta, alpha])
# Perform the simulations
map_ = pm.MAP(model)
map_.fit()
mcmc = pm.MCMC(model)
mcmc.sample(120000, 100000, 2)
From this simulation, we obtain not only our best estimate for $$\alpha$$ and $$\beta$$, but also information about our uncertainty about these values:
The probability curve for an O-ring failure thus looks as follows:
One advantage of the MCMC simulation is, that it also provides confidence intervals for the probability:
On the day of the Challenger disaster, the outside temperature was 31 degrees Fahrenheit. The posterior distribution of a defect occurring, given this temperature, almost guaranteed that the Challenger was going to be subject to defective O-rings.
challenger.py Full implementation of the MCMC simulation. | 2,373 | 10,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-51 | latest | en | 0.933949 |
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# Number 486068751
four hundred eighty six million sixty eight thousand seven hundred fifty one
### Properties of the number 486068751
Factorization 3 * 3 * 7 * 79 * 127 * 769 Divisors 1, 3, 7, 9, 21, 63, 79, 127, 237, 381, 553, 711, 769, 889, 1143, 1659, 2307, 2667, 4977, 5383, 6921, 8001, 10033, 16149, 30099, 48447, 60751, 70231, 90297, 97663, 182253, 210693, 292989, 425257, 546759, 632079, 683641, 878967, 1275771, 2050923, 3827313, 6152769, 7715377, 23146131, 54007639, 69438393, 162022917, 486068751 Count of divisors 48 Sum of divisors 820019200 Previous integer 486068750 Next integer 486068752 Is prime? NO Previous prime 486068749 Next prime 486068753 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 11100111110001101001000001111 Octal 3476151017 Duodecimal 116949953 Hexadecimal 1cf8d20f Square 236262830698700001 Square root 22046.966934252 Natural logarithm 20.001860634823 Decimal logarithm 8.6866977015014 Sine 0.56647366777432 Cosine -0.82407984062123 Tangent -0.68740143836948
Number 486068751 is pronounced four hundred eighty six million sixty eight thousand seven hundred fifty one. Number 486068751 is a composite number. Factors of 486068751 are 3 * 3 * 7 * 79 * 127 * 769. Number 486068751 has 48 divisors: 1, 3, 7, 9, 21, 63, 79, 127, 237, 381, 553, 711, 769, 889, 1143, 1659, 2307, 2667, 4977, 5383, 6921, 8001, 10033, 16149, 30099, 48447, 60751, 70231, 90297, 97663, 182253, 210693, 292989, 425257, 546759, 632079, 683641, 878967, 1275771, 2050923, 3827313, 6152769, 7715377, 23146131, 54007639, 69438393, 162022917, 486068751. Sum of the divisors is 820019200. Number 486068751 is not a Fibonacci number. It is not a Bell number. Number 486068751 is not a Catalan number. Number 486068751 is not a regular number (Hamming number). It is a not factorial of any number. Number 486068751 is a deficient number and therefore is not a perfect number. Binary numeral for number 486068751 is 11100111110001101001000001111. Octal numeral is 3476151017. Duodecimal value is 116949953. Hexadecimal representation is 1cf8d20f. Square of the number 486068751 is 236262830698700001. Square root of the number 486068751 is 22046.966934252. Natural logarithm of 486068751 is 20.001860634823 Decimal logarithm of the number 486068751 is 8.6866977015014 Sine of 486068751 is 0.56647366777432. Cosine of the number 486068751 is -0.82407984062123. Tangent of the number 486068751 is -0.68740143836948
### Number properties
Examples: 3628800, 9876543211, 12586269025
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Примеры - прочитай устно и реши
Classement
Retournez les tuiles est un modèle à composition non limitée. Il ne génère pas de points pour un classement. | 403 | 941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-14 | latest | en | 0.254824 |
https://www.wallstreetoasis.com/forums/accounting-help-will-give-sb-to-fellow-monkeys | 1,566,359,235,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00091.warc.gz | 1,031,634,558 | 43,690 | HI monkeys,
i am in need of your help~~
i attached the picture of my question..
please follow the red lines for question 1, and blue lines for question 2...
image here: http://i47.tinypic.com/nd671f.jpg thanks!!!!!!!!!
will give out many silver bananas
Look at the A/A of "Streaming" between 2010 and 2011. It increased by less than 500K, while the CF shows 700K for 2011 amortization.
With regards to DVDs, take a look at Acquisition of content (85K), Total content lib in 2010 (627K), and Total content lib in 2011 (599K). Now look at the Amortization in 2011 - 97K. Difference between amortization and investment is about 12K, doesn't explain it all, but might give you a start in terms of hunting for clues.
Good luck
• 1
Red:
Right now the assumption you're making is that the depreciation expense incurred in a year = the change in accumulated depreciation between two years on the balance sheet.
Think about what must be true for that to happen. Hint: it's about fixed asset movement: within a year a company can (1) purchase fixed assets, (2) dispose / sell fixed assets, or (3) both.
Notice the logic fallacy yet?
Blue:
Look at your current comparison between the two years on the balance sheet. What's weird about your comparison figures?
Hint: for (1), you're using the cost of the DVDs, but for (2), you're using the cost of the DVDs less amortization, given there's no current content library for both years.
When you understand this, think again about the potential movement of DVDs within a year. To do a mini-test for your understanding, try figuring out how much the company has disposed in DVDs between the two years. You have enough info in your picture to figure this out.
Hopefully this helps more than giving you the answers
• 1
quattroblanc:
Red:
Right now the assumption you're making is that the depreciation expense incurred in a year = the change in accumulated depreciation between two years on the balance sheet.
Think about what must be true for that to happen. Hint: it's about fixed asset movement: within a year a company can (1) purchase fixed assets, (2) dispose / sell fixed assets, or (3) both.
Notice the logic fallacy yet?
Blue:
Look at your current comparison between the two years on the balance sheet. What's weird about your comparison figures?
Hint: for (1), you're using the cost of the DVDs, but for (2), you're using the cost of the DVDs less amortization, given there's no current content library for both years.
When you understand this, think again about the potential movement of DVDs within a year. To do a mini-test for your understanding, try figuring out how much the company has disposed in DVDs between the two years. You have enough info in your picture to figure this out.
Hopefully this helps more than giving you the answers
hi quattro, thanks for your answer.. let me just see if i got this correct
red:
because the cash flow amortization was 795K, and the accumulated amortization between 2011-2010 was 477K. it means the company disposed of some assets during the year such that on the balance sheet the accumulated amortization is smaller than cash flow amortization..
so the logic is they disposed of these assets that were supposed to be amortized such that they can record less accumulated amortization on the balance sheet..
is this correct?
blue:
I see what you mean now..
so the change in gross amount is 27K, and the acquisition of DVD library was 85K... which means the company disposed around ~112K of DVD library assets during the year..
is this correct?
LOL at the SBs on the sheet.
The HBS guys have MAD SWAGGER. They frequently wear their class jackets to boston bars, strutting and acting like they own the joint. They just ooze success, confidence, swagger, basically attributes of alpha males.
+1 to quattro. you're ignoring disposals.
7,548 questions across 469 investment banks. The WSO Investment Banking Interview Prep Course has everything you'll ever need to start your career on Wall Street. Technical, Behavioral and Networking Courses + 2 Bonus Modules. Learn more.
.
Quattro is on track about additional stock movements, other than the linear "easier to spot" you have on the sheet. These may include e.g. sales/disposals and write-offs/downs. Depending on treatment and disclosure they could be a point to consider for red issue.
Also your logic on blue is hard to follow, as you take the FY gross amount less the PY net amount, which is affected by the acc. amortisation. Try PY Total Gross + investments = FY Total gross + (NET other movements, as per above). Same for acc. amortisation (as movements also need to be account for corresponding amortisation). This can help you spot any other movements (perhaps on the notes).
Hope this helps.
• 1
thanks for the clue monkeys, i will start looking at those things and figure out whats going on!! | 1,098 | 4,862 | {"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} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.949051 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=18104 | 1,580,307,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00503.warc.gz | 876,592,410 | 5,795 | # Chapter 5
The flashcards below were created by user ajocson on FreezingBlue Flashcards.
1. Population
The entire set of individuals or other entities to which study finding are to be generalized.
2. Sample
A subset of a population that is used to study the population as a whole.
3. Elements
The individual members of the population whose characteristics are to be measured.
4. Sampling Frame
A list of all elements or other units containing the elements in a population.
5. Enumeration units
Units that contain one or more elements and that are listed ina sampling frame.
6. Sampling Units
Units listed at each stage of a multistage sampling design.
7. Sampling Error
Any difference between the characteristics of a sample and the characteristics of a population. The larger the sampling error, the less representative the sample.
8. Target Population
A set of elements larger than or different from the population sampled and to which the researcher would like to generalize study findings.
9. Representative Sample
A sample that "looks like" the population from which it was selected in all respects that are potentially relevant to the study. The distribution of characteristics among elements of a representative sample is the same as the distibution of those characteristics among the total population. In an unrepresentative sample, some characteristics are overrepresented or underrepresented.
10. Census
Research in which information is obtained through the responses that all available members of an entire population give to questions.
11. Probability Sampling Method
A sampling method that relies on a random, or chance, selection method so that the probability of selection of population elements is known.
12. Nonprobability Sampling Method
Sampling method in which the probability of selection of population elements is unknown.
13. Probability of Selection
The likelihood that an element will be selected from the population for inclusion in the sample. In a census of all elements of a population, the probability that any particular element will be selected is 1.0. If half of the elements in the population are sampled on the basis of chance (say, by tossing a coin), the probability of selection for each element is one half, or .5. As the size of the sample as a proportion of the population decreases, so does the porbability of selection.
14. Random Sampling
A method of sampling that relies on a random, or chance, selection method so that every element of the sampling frame has a known probability of being selected.
15. Nonrespondents
People or other entities who do not participate in a study although they are selected for the sample.
16. Systematic Bias
Overrepresentation of some population characteristics in a sampel due to the method used to select the sample. A sample shaped by systematic sampling error is a biased sample.
17. Simple Random Sampling
A method of sampling in which every sample element is selected only on the basis of chance, through a random process.
18. Random Number Table
A table containing lists of numbers that are ordered solely on the basis of chance; it is used for drawing a random sample.
19. Random Digit Dialing
The random dialing by a machine of numbers withing designated phone prefixes, which creates a random sample for phone surveys.
20. Replacement Sampling
A method of sampling in which sample of elements are reutrned to the sampling frame after being selected, so they may be sampled again. Random samples may be selected with or without replacement.
21. Systematic Random Sampling
A method of sampling in which sample elements are selected from a list or from sequential files.
22. Sampling Interval
The number of cases from one sampled case to another in a systematic random sample.
23. Periodicity
A sequence of elements (in a list to be sampled) that varies in some regular, periodic pattern.
24. Stratified Random Sampling
A method of sampling in which sample elements are selected separately from population strata that are identified in advance by the researcher.
25. Proportionate Stratified Sampling
Sampling method in which elements are selected from strata in exact proportion to their representation in the population.
26. Disproportionate Stratified Sampling
Sampling in whcih elements are selcted from strata in different proportions from those that appear in the population.
27. Cluster Sampling
Sampling in whcih elements are selected in two or more stages, with the first stage being the random selection of naturally occuring clusters and the last stage being the random selection of elements within clusters.
28. Cluster
A naturally occuring, mixed aggregate of elements of the population.
29. Availability Sampling
Sampling in which elements are selected on the basis of convenience.
30. Quota Sampling
A nonprobability sampling method in which elements are selected to ensure that the sample represents certain characteristics in proportion to their prevalence in the population.
31. Purposive Sampling
A nonprobability sampling method in which elements are selected for a purpose, usually because of their unique position.
32. Snowball Sampling
A method of sampling in which sample elements are selected as they are identified by successive informants or interviewees.
33. Inferential Statistics
A mathematical tool for estimating how likely it is that a statistical result based on data from a random sample is representative of the population from which the sample is assumed to have been selected.
34. Random Sampling Error (Chance Sampling Error)
Differences between the population and the sameple that are due only to chance factors (random error), not to systematic sampling error. Random sampling error may or may not result in an unrepresentative sample. The magnitude of sampling error due to chance factors can be estimated statistically.
35. Sample Statistic
The value of a statistic, such as a mean, computed from sample data.
36. Population Parameter
The value of a statistic, such as a mean, computed using data for the entire population; a sample statistic is an estimate of a population parameter.
Author: ajocson ID: 18104 Card Set: Chapter 5 Updated: 2010-05-06 17:45:22 Tags: chapter five schutt research methods Folders: Description: Investigating the Social World - Ch. 5 Key Terms Show Answers: | 1,250 | 6,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-05 | longest | en | 0.937936 |
https://www.physicsforums.com/threads/what-do-you-call-this-homomorphism.264559/ | 1,539,933,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00195.warc.gz | 1,041,741,454 | 12,952 | # What do you call this homomorphism?
1. Oct 15, 2008
### tgt
f:A->A but f(a) does not equal a for all a in A.
2. Oct 15, 2008
### waht
Endomorphism?
3. Oct 15, 2008
### Jang Jin Hong
If set of all homomorphism is denoted as U,
and identity mapping is denoted as I
That is complementary set of identity map.
denoted as U-I
4. Oct 16, 2008
### HallsofIvy
So you have a "set of homomorphisms" minus a single homomorphism? U- {I} would make more sense. But I would interpret "does not equal a for all a in A" as meaning f(a) is NEVER equal to a.
5. Oct 16, 2008
### geor
I am confused, too.
A group homomorphism always
maps 1 to 1, so...
6. Oct 17, 2008
### tgt
fair point. It's probably an endomorphism.
7. Oct 17, 2008
### tgt
that seems right. | 261 | 765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-43 | latest | en | 0.945868 |
https://kidsworksheetfun.com/4th-grade-division-and-multiplication-word-problems/ | 1,709,328,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00550.warc.gz | 336,175,194 | 24,909 | # 4th Grade Division And Multiplication Word Problems
These worksheets require the students to differentiate between the phrasing of a story problem that requires multiplication versus one that requires division to reach the answer. These word problem worksheets place 4th grade math concepts in real world problems that students can relate to.
Word Problems Worksheets Dynamically Created Word Problems Fraction Word Problems Division Word Problems Math Words
### The following collection of free 4th grade maths word problems worksheets cover topics including addition subtraction multiplication division mixed operations fractions and decimals.
4th grade division and multiplication word problems. Math word problem worksheets for grade 4. All numbers are whole numbers with 1 to 4 digits. Division questions may have remainders which need to be interpreted e g.
Worksheets math grade 4 word problems multiplication division. Students need to gain a strong understanding of place value in order to understand the relationship between digits and how these relationships apply to. Fourth grade f 4 addition subtraction multiplication and division word problems qks.
The division problems do not include remainders. We provide math word problems for addition subtraction multiplication division time money fractions and measurement volume mass and length. Below are six versions of our grade 4 math worksheet with mixed multiplication and division word problems.
How many left over. Worksheets math grade 4 word problems. This worksheets combine basic multiplication and division word problems.
Mixed multiplication division.
Monster Math Free Printable World Problems For Halloween Word Problem Worksheets Math Word Problems Math Words
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4th Grade Number Math Word Problems Math Words Multi Step Word Problems | 554 | 3,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.853598 |
https://tildesites.bowdoin.edu/~sbarker/teaching/courses/systems/21spring/schedule.php | 1,720,944,478,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00562.warc.gz | 488,225,535 | 3,565 | CSCI 2330
Foundations of Computer Systems
Bowdoin College
Spring 2021
Instructor: Sean Barker
Schedule
The course schedule is tentative and subject to change. This page will be updated frequently to reflect the most up-to-date schedule. Chapter sections refer to the Computer Systems textbook.
MonFeb 81remoteIntroduction, Lab 0Ch. 1 (skim) Intro Slides
WedFeb 102remoteNumbering SystemsCh. 2.1 Representation Slides
FriFeb 123remoteBitwise OperatorsBinary Exercises
MonFeb 154remoteInteger RepresentationsCh. 2.2
WedFeb 175in-personInteger Representations, Lab 1Representation Exercises
FriFeb 196in-personInteger RepresentationsCh. 2.3 Bitpuzzle Exercises
MonFeb 227in-personFloating-Point RepresentationsCh. 2.4 Logic Exercises
WedFeb 248in-personFloating-Point RepresentationsFloat Exercises
FriFeb 269in-personMemory and Pointers Pointers tutorial Memory Slides
MonMar 110in-personPointers and ArraysPointer Exercises
WedMar 311remoteStrings, Lab 2
FriMar 512in-personMemory Allocation
MonMar 813in-personDebugging, Lab D
WedMar 1014in-personx86-64 ISA, Data MovementCh. 3.1-3.4 x86-basics Slides
FriMar 1215in-personx86 ArithmeticCh. 3.5 Operand Exercises
MonMar 1516in-personCondition Codes & ConditionalsCh. 3.6 x86-control Slides
WedMar 1717in-personReverse Engineering, Lab 3GDB Reference
FriMar 19in-personMidterm Exam
MonMar 22No Class: Spring Break
WedMar 2418in-personLoops & Switches
FriMar 2619in-personJump Tables
MonMar 2920in-personProcedures & StacksCh. 3.7 x86-procedures Slides
WedMar 3121in-personProcedure Memory
FriApr 222in-personArrays & StructsCh. 3.8-3.9 x86-structures Slides Array Exercises
MonApr 523remoteStructs & Buffer OverflowCh. 3.10 Buffers Slides
WedApr 724remoteCode Injection Attacks
FriApr 925remoteBuffer Overflow Defenses
MonApr 1226in-personROP Attacks, Lab 4
WedApr 1427in-personCaching & Cache DesignsCh. 6.1 Caching Slides
FriApr 1628in-personDirect-Mapped CachesCh. 6.4 Caching Exercises
MonApr 1929in-personAssociative CachesAssociative Exercises
WedApr 2130in-personLocality, Lab 5Ch. 6.2-6.3, 6.5
FriApr 2331in-personProcessesCh. 8.2 Process Slides
MonApr 2632in-personExceptional Control FlowCh. 8.1
WedApr 2833in-personProcess Control & ShellsCh. 8.3 Fork Exercises
FriApr 3034in-personLab 5 Discussion
MonMay 335in-personProcess ControlCh. 8.4
WedMay 536in-personSignalsCh. 8.5 | 702 | 2,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.627525 |
http://salon-rouce.com/4phsi27/dca10e-how-to-calculate-capacitor-value-for-12v-power-supply | 1,624,237,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00280.warc.gz | 42,276,613 | 9,835 | The choice of the capacitor value needs to fulfil a number of requirements. The highest output voltage has fallen a bit; but the lowest output voltage has gone from 0V to 11.6. It is too difficult to find the exact power rating resistors that you have calculated. The 0.1uf capacitor reducing output voltage oscillations. Rearranging Vpk-pk ripple = Iload / fC we get C = Iload / 4 * f * Vpk-pk ripple, C = 2A /4* 50Hz * 1V whence C = 2 / 200 Farads = 10,000 uF, (* in fact the voltage ripple also depends on the internal resistance of the transformer and rectifier. Common capacitor value for SMD capacitor is almost same as ceramic and electrolytic capacitors. If you're using 3 motors, and a 12V power supply, your total current should not exceed 0.66A per motor x 3 motors = 1.98A. This post explains how to calculate resistor and capacitor values in transformerless power supply circuits using simple formulas like ohms law. This is shown in the graph below. Example 2: Must calculate the voltage of a 100nF capacitor after being charged a period of 1ms through 10 kilo-ohm resistor with 5V supply: View example: Example 3: Must calculate the time to discharge a 470uF capacitor from 385 volts to 60 volts with 33 kilo-ohm discharge resistor: View example The input capacitor is decided in reference with output power derived from the power supply and ripple voltage allowed in the switch voltage. 10^4 = 10000. C is the value of the capacitor. You have 2 phases, and a current per phase of 0.33A, so your total current shouldn't exceed 0.66A per motor. Vpk-pk ripple= Iload /4 f C (see below) where Your email address will not be published. Although the final ripple content which is the difference between the peak value and the minimum value of the smoothed DC, never seem to eliminate completely, and directly relies on the load current. Because these capacitors have a DC value, they are actually storing a lot of energy that never gets used. In the mentioned formula we can see that the ripple and the capacitance are inversely proportional, meaning if the ripple needs to be minimum, the capacitor value needs to increase and vice versa. Lets see how adding the capacitor changes this. A very good post that I have learnt a lot. The most important formula for calculating the smoothing capacitor is: $$C = I \cdot \frac{\Delta t}{\Delta U}$$ The smoothing capacitor formula, alternatively: $$I = C \cdot \frac{\Delta U}{\Delta t}$$ If you have any circuit related query, you may interact through comments, I'll be most happy to help! Sir I planned to design an inverter for my home which should light up 3,20 watt cfl bulb and also for mobile charging.hence I assumed that my total watt requirement is not more than 100watt.please suggest any circuit based on my requirement and say the information about the battery that I need to use for operating an inverter for 5 hours. More resistance gives better smoothing but worse load regulation). It is a common mistake to calculate the rms current load by adding … We saw that the output from the transformer and rectifier was a DC voltage; but it contains a large unwanted AC component. Thus, the above formula clearly shows how the required filter capacitor may be calculated with respect to the load current and the minimum allowable ripple current in the DC component. This causes heating of the capacitor and can be destructive. Let's try to understand the relation between load current, ripple and the optimal capacitor value from the following evaluation. Sir, I have seen more number of inverter circuits on your site. very good post and site and about calculating filter capacitor voltage what’s your idea? The yellow line shows the output voltage from the previous unsmoothed supply with a 2A load, We saw in the previous page that the rms value of our "dc" wave is roughly 10.6V. The rms value for a sawtooth wave is Vrms = Vpp / 2*sqrt(3) = Vpp / 3.46, Here Vpp ripple is 1.3V so Vrms for the ac wave is 1.3 / 3.46V = 0.375V (unsmoothed value was 5.4V). Calculate the required capacity of Capacitor in both kVAR and Farads. And you need to know how to calculate capacitor values. Vpk-pk ripple= Iload /4 f C (see below) where. Therefore the rectance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.To get current I divide mains Volt by the rectance in kilo ohm.That is 230 / 14.4 = 15.9 mA. On the next page we evaluate the size of this current. For code “104″ The two figures 10 indicate the significant figures and the 4 indicates the multiplier , i.e. A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. Your email address will not be published. Generally, Resistors come in 1/4 watt, 1/2 watt, 1 watt, 2 watt, 5 watt, and so on. As before all calculated figures apply to a 12V RMS voltage from the transformer. I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. The time constant of a resistor-capacitor series combination is defined as the time it takes for the capacitor to deplete 36.8% (for a discharging circuit) of its charge or the time it takes to reach 63.2% (for a charging circuit) of its maximum charge capacity given that it has no initial charge. Power Supplies –Filter Capacitor 1 by Kenneth A. Kuhn July 26, 2009 The energy storage process of a capacitor is such as to oppose a change in voltage. A larger capacitor produces less ripple or a higher resistance load (drawing less current thus less time for the capacitor to discharge) will reduce the level of ripple because the capacitor has less time to discharge. Hence 0.22 microfarad is 0.22 x 1/1,000,000 farads. The capacitor ripple current in a typical power supply is a combination of ripple currents at various frequencies. In the previous article we learned about ripple factor in power supply circuits, here we continue and evaluate the formula for calculating ripple current, and consequently the filter capacitor value for eliminating the ripple content in the DC output. Power supply decoupling capacitors must be selected with care to ensure sufficient effective capacitance for the nRF power system, because insufficient capacitance can cause instability and malfunction in power system operation mode engine. f is the frequency before rectification (here 50Hz) and The 200mA fuse will protect the circuit from mains during shot circuit or component failures. = 0.01 Farads or 10,000uF (1Farad = 1000000 uF) Where did you get the decimal point from? Last Updated on December 1, 2020 by Swagatam 21 Comments. sir, your circuit is great but i have questions to you …how did you do ? = 0.01 Farads or 10,000uF (1Farad = 1000000 uF) Thus, the above formula clearly shows how the required filter capacitor may be calculated with respect to the load current and the minimum allowable ripple current in the DC component. In Capacitor Power Supplies we use a Voltage Dropping Capacitor in series with the phase line. With no load at all, just the capacitor and the rectifier, the capacitor will charge to … You have helped so much. The 1000uf capacitor is not critical, but this is a good value. 3: choose transformer: The nearest suitable transformer is 24V at 8A - that will be fine. In the second circuit diagram, the smoothing capacitor is located behind the bridge rectification. No batteries or anything. Please how do we measure/calculate/obtain the Vpp? Rearranging Vpk-pk ripple = Iload / fC we get C = Iload / 4 * f * Vpk-pk ripple. Vpp = the minimum ripple (the peak to peak voltage after smoothing) that may be allowable or OK for the user, because practically it's never feasible to make this zero, as that would demand an unworkable, non-viable monstrous capacitor value, probably not feasible for anybody to implement. The effective capacitance of chip capacitors may only be a small fraction of the marked nominal value. Vijay, if you are interested t calculate the exact value of the capacitor then you'll need to evaluate charging current first, which can be found by dividing the AH of your battery with 10. In the following section we will try to evaluate the formula for calculating filter capacitor in power supply circuits for ensuring minimum ripple at the output (depending on the connected load current spec). So single phase induction motors, can be made to start running by temporary connecting an “start” winding though a resistor, or capacitor. Thank you so much for your clarification. its "almost" a sawtooth wave. Yes great question. We use Q (charge) = C V = I t , and rearrange to get V = I t / C so V = I T / 4 C. The minimum output voltage is Vout = Vpk - (Vpk-pk ripple), In the above example Vpk = 14.6V and Vpk-pk ripple= 1.3V so Vout (min) = 13.3 V. The storage and release of charge in the capacitor results in an AC current flowing through it. The RMS value of the output waveform is 12.0 V. This is higher than the 10.6V for the unsmoothed supply. The amount of ripple voltage is given (approximately *) by Vijay, you can try the following circuit: https://homemade-circuits.com/2012/09/mini-50-watt-mosfet-inverter-circuit.html, the battery should be rated at at least 12V, 75 AH, the inverter is capable of handling up to 200 watts if the trafo is appropriately rated, nice post sir.really helpful….. thanks sir, Previous: Digital Power Meter for Reading Home Wattage Consumption, Next: What’s Ripple Current in Power Supplies. Why the Capacitor in Your Power Supply Filter is Too Big January 21, 2016 by David Williams The job of the capacitor in the output filter of a DC power supply is to maintain a constant DC value by removing as much power ripple as possible. Calculate smoothing capacitor – formula. Can u suggest the circuit which should produce an exact sinewave as same grid supply. 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Https: //www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials answers! The circuit from mains during shot circuit or component failures note also the addition of switch! Sinewave as same grid supply of power rating resistors that you are happy with it be.... Ensures smooth, no-dropout operation for the unsmoothed supply that never gets used out the of. The marked nominal value the decimal point from value, they are actually storing a lot of energy that gets. 0V to 11.6 now have a DC value, they are actually a... See below ) where did you do by Swagatam 21 Comments the choice of the driver resistor! The load is relatively higher, the smoothing capacitor increases the average output voltage 2 X 100 X 1 =. My comment to ensure that we give you the best experience on our website the value you for! Post explains how to calculate capacitor values that have developed over time determine current. By adding a smoothing capacitor increases the average output voltage capacitor and can power low current devices f is frequency. Post that I have seen more number of requirements I 'll be happy! As a result, this how to calculate capacitor value for 12v power supply ensures smooth, no-dropout operation for entire! Reference with output power derived from the wall the motor power factor has to be improved to 0.9 connecting... | 4,595 | 19,468 | {"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": 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} | 3.5625 | 4 | CC-MAIN-2021-25 | longest | en | 0.882755 |
https://math.stackexchange.com/questions/2538473/evaluate-the-line-integral-trouble-with-setting-up-and-parameterizing-and-integ | 1,566,679,226,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00296.warc.gz | 541,887,124 | 31,231 | # evaluate the line integral, trouble with setting up and parameterizing and integrating
I tried following the example in my book, and I got stuck when trying to take the integral for the second part with $y = x^2$. I'm not completely confident on what to parameterize with here. I tried replacing $x^2$ with $y$, then in the radical I had $4 + y^3$, but when trying to integrate that with u substitution I couldn't finish it. I'm not sure I'm setting everything up right.
I'm also not sure how to tell which integral goes with which segment. Do you just assume the first goes with the first and second with the second?
hint
the first segment can be parametrised as
$$x=0+2t \;\;\;,\;\; y=0+t$$
the integral along this segment is
$$\int_0^1 \Bigl ((2t+2t)(2dt)+4t^2 (dt)\Bigr)=$$
$$4\int_0^1t (2+t)dt=\frac {16}{3}$$
You can do it for the second segment defined by
$$x=2+t \;\;\;, \;\; y=1-t$$
• also can you explain how you parametrized those equations? – 2316354654 Nov 26 '17 at 20:00
• @2316354654 the line AB is parametrized by $x=x_A+t (x_B-x_A)$ and $y=...$ – hamam_Abdallah Nov 26 '17 at 20:14
• why is your integral from 1 to 0? I thought it would be 2 to 0. – 2316354654 Nov 26 '17 at 21:02
• is the integral always from 1 to 0when you parameterize in terms of t? – 2316354654 Nov 26 '17 at 21:15
• @2316354654 Yes. t=0 gives the first point of the segment and t=1 gives the second. – hamam_Abdallah Nov 27 '17 at 11:36 | 465 | 1,440 | {"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} | 3.828125 | 4 | CC-MAIN-2019-35 | latest | en | 0.909044 |
https://betterlesson.com/lesson/633529/frequency-pitch-and-amplitude-developing-models?from=breadcrumb_lesson | 1,638,430,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00354.warc.gz | 219,697,141 | 26,656 | # Frequency, Pitch, and Amplitude: Developing Models
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## Objective
Students will be able to develop models demonstrating the relationships between characteristic properties of waves.
#### Big Idea
Students practice making models of wave properties using sound makers, slinkies, lasers and more!
## Introduction and Connection to the NGSS and Common Core
Students at this point have just been introduced to wave properties. This lesson is designed to develop a deeper understanding of wave properties through the creation of models. I have found that this lesson typically takes two days. As students are building their foundational understandings in models and in wave properties, I find that this lesson in particular is a lesson not to rush through.
Waves can be a challenging topic for a middle school student. With many waves being "invisible", it can be a difficult concept for a student to relate to and explain. As a result, they have many misconceptions about wave behavior. In order to allow students to make connections, this lesson helps make wave properties visible as they create models to demonstrate property relationships (SP2 Developing and Using Models, CCC Patterns). The Next Generation Science Standards have identified an essential question that can drive student learning as, "What are the characteristic properties of waves and how can they be used?" This lesson focuses specifically on the following NGSS standards:
MS-PS4-1 Use mathematical representations that describes a simple model for waves that includes how the amplitude of a wave is related to the energy in the wave.
The disciplinary core idea PS4-A explains that a simple wave has a repeating pattern with a specific frequency, wavelength, and amplitude. In turn, creating relationships between these wave properties is critical for student understanding. Graphs and charts are a great method for representing patterns in data mathematically. With the use of slinkies, students can discover these relationships and gather data for developing their own graphs showing property relationships.
## Connection to the Essential Question: What are you going to be learning today?
5 minutes
Have the students refer to their Unit Plan that includes the Essential Question, "What are the characteristic properties of waves and how are they used?" along with the "I can" statements/skills for the unit. In the previous lesson, the students worked to break down what the essential question means. As this is only the second day of the unit, take time to break the question down again to make sure that all students are connecting to what their focus is.
Here's how to break down the question. First read the EQ as a class. Then examine key words. For example, ask students what "characteristic", "properties", and "how are they used" might mean. Many middle school students have an especially difficult time verbalizing what a property is. Without this understanding, it will be challenging for them to connect to what they are supposed to be learning. The conversation may be very similar to the previous day's lesson; however, hopefully, it may go a little faster! A conversation may unfold something like this:
Teacher: What does the word characteristic mean? Your response does not have to be connected to science. In your everyday life when you hear that word, what does it mean to you?
Student: Something that describes something.
Teacher: Absolutely! One characteristic about me is that I love science. What does the word property mean to you? Again, your response does not have to be connected to science. In your everyday life when you hear the word property, what does it mean to you?
Student: Something you own, something that is yours.
Teacher: Absolutely! Now, I am wondering how a wave can own something. What is it that a wave can own?
Student: Like what type of wave it is or what it looks like.
Teacher: Hmmm....characteristic and property both seem like they mean similar things. They are both words that make us think of descriptions of things. I wonder why both words are in this Essential Question. (Pause....Wait time! Let the students think about why both words are needed.) Let me ask you this. I notice that in this question the word characteristic is used a little differently. It says, "characteristic properties". I wonder how the meaning of the word characteristic changes when it is put in front of another word like this. For example, what if I said, "What are the characteristic properties of a science teacher?" What does the word characteristic mean now?
Student: Typical or common.
Teacher: Ahh! That makes sense. So, what we are looking for here it seems is typical or common ways to describe waves. Maybe it is the things that waves might have in common.
Teacher: I am noticing that there is a second part to this EQ. It says, "...how can they be used". What do you think we should look for when making connections to that?
Student: Maybe...ways that humans might use them in the real world?
Teacher: Today we are going to continue to identify characteristic properties of waves. In addition, we are going to investigate the relationships between these properties.
## Mini Lesson: Creating Models With a Laser Light Show
20 minutes
Say, "Yesterday, we looked at the Skills 2 and 3 "I Can" statements. We identified that we needed to look at properties such as amplitude, frequency, wavelength, and pitch. What is important when looking at an "I Can" statement is to identify what you need to do with each of those properties. Look at these skills again, what is it that we should be able to do with these skills?"
Students should offer things like "find relationships", "provide evidence" and "create models".
Say, "Create models? Hmmm. What does that mean?"
Students may offer many things, but ultimately they come to the conclusion that a model is a visual representation of a concept.
Allow students to brainstorm ideas of what a model could mean. Students often come up with these ideas:
To help guide your discussion it may be helpful to read the following article. In the article, Joseph Krajcik and Joi Merritt explain in the article "Engaging Students in Scientific Practices: What does constructing and revising models look like in the science classroom? Understanding A Framework for K−12 Science Education" that:
"Models provide scientists and engineers with tools for thinking, to visualize and make sense of phenomena and experience, or to develop possible solutions to design problems (NRC 2011). Models are external representations of mental concepts. Models can include diagrams, three-dimensional physical structures, computer simulations, mathematical formulations, and analogies."
Explain that today students will be creating models that represent relationships between some of the characteristic properties of waves. Explain that to practice this, they will be using a "Laser Light Show" to model the relationship between any characteristic properties of waves of their choosing. Explain how important evidence is in this activity. When students create their model, they must demonstrate it for you in order to explain the evidence that their model accurately shows of the relationship.
I provide the students with a "Laser Light Show" "contraption" to experiment with various properties of waves such as amplitude and pitch. Want a Laser Light Show Apparatus? Here is where I got them. If you don't have them, it is fine! Ask students to come up with their own ideas, provide them with slinkies, tuning forks, etc. The great thing about models is that the possibilities are endless! The key here is that when students show you their models that they can do three things:
1. State the relationship.
2. Produce an accurate model.
3. Provide evidence their model demonstrates.
As students are experimenting with the Laser Show to identify relationships between characteristic properties of waves, I stop in and ask students, "Are you noticing any relationships between any characteristic properties of waves?". Without checking in, students can lose focus of their goal.
( I want to emphasize that this is hard to capture on film. I wanted to give you a visual of what the Laser Light Show Apparatus does; but it is just impossible to capture all of the awesomeness of this apparatus on video. They are so cool!)
## Student Activity : Creating Models of Relationships Between Wave Properties
60 minutes
Once each group has presented their models to you, tell them that they will be creating three more models to present to the class. Students have access to whistling tubes, slinkies, tuning forks, paper, pencils, or classroom materials of their choosing that they feel like would help them in their model. There are no specific materials that students have to use. Students create models of the following wave properties:
1. The relationship between energy and amplitude.
2. The relationship between frequency and pitch.
3. The idea that pitch/frequency are independent of amplitude (they show no relationship).
Number 3 is definitely the most challenging. One misconception that students have is that amplitude changes frequency. They think that adding energy speeds up the frequency when in fact they are independent of each other (you can have a high amplitude, high frequency/pitch and a low amplitude, high frequency/pitch).
As students work, encourage them to practice what they are going to share to the class. They should discuss and plan which group members will demonstrate each model and which members will explain each model to the class. Emphasize that they must include evidence of the relationship in their explanation.
After each group presents, ask the students in the audience to pose any follow up questions they still have. I especially encourage students to ask about what evidence was shown of the relationships. I explain to the students that when students ask follow up questions, they are not being critical. We are simply engaging in science discourse. We are going to have a conversation about science. In our quest for understanding, we are all in it together!
Here are a few examples:
Energy vs. Amplitude
Frequency vs. Pitch
Amplitude vs Frequency/Pitch
## Closure
5 minutes
To close, have students use their hands to demonstrate direct and inversely related wave properties. For a direct relationship, students would put two thumbs up indicating that as one variable increases, the other increases. For an inverse relationship, student put one thumb up and one thumb down to show that as one variable increases, the other decreases. I ask them to show the following:
• Frequency and Pitch (two thumbs up)
• Energy and Amplitude (two thumbs up)
• Frequency and Wavelength (one thumb up and the other thumb down)
• Wavelength and Pitch (one thumb up and the other thumb down)
I end with the toughest -- Amplitude and Frequency. (Students should do something to show that they are not proportionally related). | 2,358 | 11,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.918411 |
http://frequentmiler.boardingarea.com/2012/03/14/pie-at-3-14x-for-pi-day-314/ | 1,519,387,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814700.55/warc/CC-MAIN-20180223115053-20180223135053-00252.warc.gz | 138,684,857 | 17,555 | ## Pie at 3.14X for Pi day 3/14
I don’t know how I missed it, but I just found out that today is Pi day! March 14 = 3.14. It’s too late in the day for me to get the free pie at 3:14 pm that some local bakeries offered. Instead, I’ll tackle an important challenge: how can one buy pie for 3.14 points per dollar?
At first blush, this challenge seems impossible. I mean, sure, we can find ways to get 2X, 3X, or even 5X or 10X.. But 3.14X, no way. Right? But then I remembered that with a Sapphire Preferred card all points earned get a 7% end of year dividend. So, if you simply buy a piece of pie at a restaurant you’ll get 2X for dining + .14X end of year dividend. The total comes tantalizingly close: 2.14 points per dollar instead of 3.14. Now we’re getting somewhere! But, how do we get that final point?
How about dining programs like SkyMiles dining or MileagePlus dining? If we buy pie at a restaurant enrolled in one of these programs then we’ll get extra points per dollar over the 2.14! Sadly though, there doesn’t seem to be a way to get just 1 extra point. If you’re a basic member you only get half a point per dollar. If you’re an “online member” you get 3 miles per dollar. VIP members get even more. Rats.
OK, how about a different approach. What if we find a merchant in the Ultimate Rewards Mall that offers 2X and sells pie. The closest I could find was World Market which has this Pie Contest in a Box:
OK, so its not pie, but if you buy it correctly you can get 3.14 points per dollar. Just log into the Ultimate Rewards Mall with your Sapphire Preferred account, click through to World Market and then buy this item. Don’t checkout with your Sapphire Preferred card though! If you do, you’ll get a total of 3.21 points per dollar (because of the annual 7% dividend). Instead, use a card that only gives 1X (your Ink Bold would work fine). Now you’ll get 2.14 points from the Ultimate Rewards Mall and 1 point from your other credit card. There! 3.14 points per dollar. Not for pie, but for a pie contest. Close enough?
#### About Greg The Frequent Miler
Greg is the owner, founder, and primary author of the Frequent Miler. He earns millions of points and miles each year, mostly without flying, and dedicates this blog to teaching others how to do the same.
More articles by Greg The Frequent Miler »
### Pingbacks
1. […] For the final amusing story this week, I couldn’t pick just one, so you get a couple. First, head over to View From the Wing to see how Gary got a little bit too much information from his readers after a recent post. Details HERE. Then go over and read about The Frequent Miler’s quest to purchase Pie and earn 3.14 points per dollar earlier this week on Pi Day. Details HERE. […]
2. […] earn 7.04 points per dollar. This is similar to the Pi day challenge I did in March (see “Pie at 3.14X for Pi day 3/14”), but with an added twist: today we want to use a Chase Freedom card to make the […]
1. Ollie says:
Sounds good, now time to get some real pie!
2. Daniel G says:
I love your posts, by far my favorite blog!
3. Bender says:
Hahaha this is good, original stuff
4. jackie says:
Have to say — you are great! Love your blog!
5. AlwaysTravel says:
Brilliant, and totally funny! Thanks!
6. Mom Miler says:
Brilliant but no longer useful. I read this too late – how sad. Finally you found a use for my Sapphire preferred card. Can’t wait until next year.
7. Mom Miler says:
I just looked. I don’t have sapphire preferred. Sad again
8. James says:
Of course, a more accurate Pi day happens in 4 years… | 950 | 3,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-09 | latest | en | 0.836921 |
http://emathematics.matematicas.us/g7_linear_function.php?def=find_constant_word | 1,498,373,989,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320443.31/warc/CC-MAIN-20170625064745-20170625084745-00651.warc.gz | 120,870,876 | 4,392 | User:
Find the constant of variation: word problems
Peter walked a total of 12 kilometres by making 2 trips to school. Later on, he walked 18 kilometres during 3 trips. How many kilometres does Reid walk per trip?
Create a chart.
To find the constant of variation, divide "Total distance walked (kilometres)" by "Number of trips".
The constant of variation is 6.
The voltage (in volts) measured across a resistor is directly proportional to the current (in amperes) flowing through the resistor. If 8 volts is measured across a resistor carrying a current of 2 amperes, and 36 volts is measured across a resistor carrying a current of 9 amperes, what is the constant of variation (resistance in ohms)?
Solution: | 165 | 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} | 3.421875 | 3 | CC-MAIN-2017-26 | longest | en | 0.951178 |
https://oeis.org/A007459 | 1,511,382,690,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806660.82/warc/CC-MAIN-20171122194844-20171122214844-00057.warc.gz | 659,724,882 | 4,776 | This site is supported by donations to The OEIS Foundation.
Annual appeal: Please make a donation to keep the OEIS running! Over 6000 articles have referenced us, often saying "we discovered this result with the help of the OEIS". Other ways to donate
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A007459 Higgs's primes: a(n+1) = smallest prime > a(n) such that a(n+1)-1 divides the product (a(1)...a(n))^2. (Formerly M0660) 8
2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 43, 47, 53, 59, 61, 67, 71, 79, 101, 107, 127, 131, 139, 149, 151, 157, 173, 181, 191, 197, 199, 211, 223, 229, 263, 269, 277, 283, 311, 317, 331, 347, 349, 367, 373, 383, 397, 419, 421, 431, 461, 463, 491, 509, 523, 547, 557, 571 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 REFERENCES N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 S. Burris and S. Lee, Tarski's high school identities, Amer. Math. Monthly 100 (1993), 231-236. R. G. Wilson, V, Note to N. J. A. Sloane with attachment, (Annotated scanned copy of The Am. Math. Mo. Vol. 100 No. 3 pp. 233, Mar. 1993). R. G. Wilson, V, Letter to N. J. A. Sloane, Oct. 1993 MAPLE a:=[2]; P:=1; j:=1; for n from 2 to 32 do P:=P*a[n-1]^2; for i from j+1 to 250 do if (P mod (ithprime(i)-1)) = 0 then a:=[op(a), ithprime(i)]; j:=i; break; fi; od: od: a; # N. J. A. Sloane, Feb 12 2017 MATHEMATICA f[ n_List ] := (a = n; b = Apply[ Times, a^2 ]; d = NextPrime[ a[ [ -1 ] ] ]; While[ ! IntegerQ[ b/(d - 1) ] || d > b, d = NextPrime[ d ] ]; AppendTo[ a, d ]; Return[ a ]); Nest[ f, {2}, 75 ] PROG (Haskell) a007459 n = a007459_list !! (n-1) a007459_list = f 1 a000040_list where f q (p:ps) = if mod q (p - 1) == 0 then p : f (q * p ^ 2) ps else f q ps -- Reinhard Zumkeller, Apr 14 2013 (PARI) step(v)=my(N=vecprod(v)^2); forprime(p=v[#v]+1, , if(N%(p-1)==0, return(concat(v, p)))) first(n)=my(v=[2]); for(i=2, n, v=step(v)); v \\ Charles R Greathouse IV, Jun 11 2015 CROSSREFS Cf. A057447, A057448, A057459, A282027. Sequence in context: A042987 A089189 A097375 * A129944 A176162 A152900 Adjacent sequences: A007456 A007457 A007458 * A007460 A007461 A007462 KEYWORD nonn,nice,changed AUTHOR EXTENSIONS More terms from David W. Wilson Definition clarified by N. J. A. Sloane, Feb 12 2017 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 1,010 | 2,661 | {"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-2017-47 | longest | en | 0.565438 |
http://chronos.msu.ru/en/component/zoo/knowpublications/knowpublications-16-biblioteka-elektronnykh-publikatsij/a-new-simple-method-for-generation-and-detection-of-elementary-dm-particles-via-collisions-between-elementary-antiparticles-positrons-positrons-antiprotons-antiprotons-positrons-antiprotons-in-colliders | 1,723,056,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00840.warc.gz | 8,061,768 | 5,255 | A new simple method for generation and detection of elementary DM particles via collisions between elementary antiparticles (positrons-positrons), (antiprotons-antiprotons), (positrons-antiprotons) in colliders - Institute for Time Nature Explorations
Copyright © 2024 Institute for Time Nature Explorations. All Rights Reserved.Joomla! is Free Software released under the GNU General Public License.
A new simple method for generation and detection of elementary DM particles via collisions between elementary antiparticles (positrons-positrons), (antiprotons-antiprotons), (positrons-antiprotons) in colliders
Грибов Ю.А. (Gribov Y.A.) A new simple method for generation and detection of elementary DM particles via collisions between elementary antiparticles (positrons-positrons), (antiprotons-antiprotons), (positrons-antiprotons) in colliders // http://vixra.org. 2015. 8 p.
Категории: Исследование, Авторский указатель, Время в математике, Математика
## A new simple method for generation and detection of elementary DM particles via collisions between elementary antiparticles (positrons-positrons), (antiprotons-antiprotons), (positrons-antiprotons) in colliders 0.0/5 rating (0 votes)/*<![CDATA[*/jQuery(function($) {$('#c4aab8cf-bd11-4dce-bfe9-a2fdc567666a-66b3c0344241f').ElementRating({ url: '/en/component/zoo/?task=callelement&format=raw&item_id=7869&element=c4aab8cf-bd11-4dce-bfe9-a2fdc567666a' }); });/*]]>*/
### Summary
The substantially new simple method for generating & detection of elementary particles of Dark Matter (DM) is proposed [1], distinguished by the fact that pairs of elementary (DM /(Ordinary Antimatter (OAM), particles like (dark electron/positron), (dark proton/antiproton) etc. could be easily created in conventional low energy colliders, but only in collision between elementary antiparticles (preferably positrons-positrons, antiprotons-antiprotons or positrons- antiprotons).
This method is predicted by the new physical concept of DM by the author, where DM particles are intrinsically identical to our Ordinary Matter (OM) particles, but are shifted in two the nearest adjacent DM-Universes [2]. The method can be used for calibration the proposed direct-DMdetectors by the author , using captured antiparticles - physical mediators between OM and DM [3]. The dark DM-protons etc. created in the collider via the colliding antimatter will atypically annihilate with the visible OAM antiprotons captured in the DM-detector with outcome of two gamma quanta (one dark- undetectable and one detectable).
You have no rights to post comments | 657 | 2,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.534565 |
https://www.hekedbay.com/how-to-add-or-subtract-fractions/ | 1,723,250,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00866.warc.gz | 624,191,515 | 14,868 | # How To Add Or Subtract Fractions
How To Add Or Subtract Fractions – We use cookies to do better. By using our website, you consent to our cookie policy. Cookie settings
This article was co-authored by David Jia and author Jessica Gibson. David Jia is an educator and founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grade levels in a variety of subjects, as well as advising on college admissions and preparing for the SAT, ACT, ISEE and more. After achieving a perfect math score of 800 on the SAT and an English score of 690, David received a Dickinson Scholarship from the University of Miami where he earned a BA in Business Administration. Additionally, David has worked as an online video instructor for textbook companies such as Larson Texts, Big Idea Learning, and Big Idea Math.
## How To Add Or Subtract Fractions
Subtracting fractions can seem a little confusing at first, but after some basic multiplication and division, you’ll be ready for some simple subtraction. If the fraction is correct, make sure the denominator is the same before subtracting the numerator. If the fractions are messy and you have whole numbers, convert them to improper fractions. You also need to make sure the denominators are the same before subtracting the numerators.
## Awesome Activities For Adding And Subtracting Fractions With Like Denominators
To subtract fractions, first make sure the denominators are equal. If not, find the least common multiple of the two denominators. Then multiply each denominator so that it is equal to the least common multiple. Then multiply the numerator by the same number you multiplied the denominator by. Finally, subtract the numerator and leave the denominator as the lowest common multiple. To learn how to subtract mixed numbers, scroll down! This week we started one of the most stressful topics. We will be adding and subtracting fractions. We came back from winter break and these kids looked like they were going to cry when I said we were fighting fractions. The truth is – if you know your multiplication facts, you’ll be fine! Hope that gives you some ideas
Often, the fractions are completely next to each other. Instead of lying next to each other, place fractions on top of each other. I teach my students how to “blackboard”. This is the first thing I tell them. Yes – they have to rewrite fractions, but let me tell you… when they draw correctly, they change the game. The image below shows how I set up the array.
### Subtracting Whole Number With Fraction Worksheet
I start by showing the fractions on top of each other. I don’t let them adjust it or change the order, because when they have to subtract they will get negative numbers. Then we put the multiplication sign, the fraction bar, the equal sign and then the last fraction bar. Usually I like to put a line under the last pieces so I can keep track of what I’ve added or subtracted.
Let them use the multiplication table. I know… it sounds so simple. It probably is. I know many teachers hate having their kids use multiplication tables … however, are you testing your multiplication skills or your ability to add and subtract fractions? Of course, the unit will be adding and subtracting fractions, so it will be easier for you and the student if you let them use the multiplication table. Just remember to ask yourself – what strategies are you teaching / using?
When you need to add or subtract fractions – keep it clean! Seriously. If students are not neat writers, ask them to use lined paper. I have kids who can write a number in 4 lines. They are very confused. I got a revised paper for them. I made sure that the students had papers according to their needs. I gave one of my students ten blank “boards” of paper (see tip 1). Another student was only allowed to use lined paper. Another student was using grid paper. Watch your students write and make a simple change for them.
## Ways To Add And Subtract Fractions
If you’re looking for an amazing Google Classroom class that fits fractions – check out my store here! We use cookies to do better. By using our website, you consent to our cookie policy. Cookie settings
This article was co-authored by David Jia. David Jia is an educator and founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grade levels in a variety of subjects, as well as advising on college admissions and preparing for the SAT, ACT, ISEE and more. After achieving a perfect math score of 800 on the SAT and an English score of 690, David received a Dickinson Scholarship from the University of Miami where he earned a BA in Business Administration. Additionally, David has worked as an online video instructor for textbook companies such as Larson Texts, Big Idea Learning, and Big Idea Math.
This article provides 10 references which can be found at the bottom of the page.
### Addition And Subtraction Of Similar Fractions
Adding and subtracting fractions is a basic skill. From elementary school to college, fractions appear all the time in everyday life, especially in math lessons. Follow these steps to learn how to add and subtract fractions, whether they’re like fractions, mixed fractions, or opposites of improper fractions. Once you know one way, the rest is pretty easy!
To add and subtract fractions with the same denominator or base, place 2 fractions next to each other. Add or subtract digits or higher numbers and write the result in a new fraction at the top. The lower number of responses will be the same as the denominator of the original fractions. To learn how to add and subtract fractions with different denominators, read on! Now that we’re done with geometry, we turn to fractions. Oh, fractions! Children come with preconceived notions that groups are big and bad, scary, harsh and mean. I think it’s my job to prove them wrong. I’m posting a bit inappropriately but want to post our fraction addition and subtraction anchor chart and collapsible.
## Subtracting Fractions With Same Denominator Interactive Worksheet
The kids copied a fantastic freebie from 4mulaFun, the anchor chart for their awesome addition and subtraction of compound fractions. Click the link to follow her on Pinterest. He is a great pinner to follow.
Soon I will be adding our anchor charts and note pages for all of our fraction concepts. Meanwhile, if you’re looking for faction resources, they’re in my TpT store. My favorite right now? All the fractions task cards you could ever want (256 cards to be exact), all in one package.
#### Free Printable Adding And Subtracting Fractions Worksheets
Another amazing resource I use when teaching fractions is fun and awesome fraction stories. Their kids can’t get enough of these stories and they are perfect homework because they provide an overview of the concepts we cover throughout the day!
*Note: I get nothing for mentioning Funny and Awesome Fraction Stories and am not paid to mention it. This is a book I have had for years and love it!
If you are looking for | 1,483 | 7,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.92902 |
https://www.freemathhelp.com/forum/threads/please-anyone-help-me-understand-dy-dx-2-dz-2-y-3n-2x-4-ay-n-3-az-m-2-y-3-y-3-dz.137998/#post-590209 | 1,718,467,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00180.warc.gz | 688,955,190 | 9,861 | # Please anyone help me understand: dy/dx = (2 * dz^2 * y^(3n) - 2x^4 * ay^(-n^3) - az * (m^2 - y^3)) / (y^3 * dz)
#### kaif78
##### New member
Please anyone help me understand how to select these multiplier while solving this kind questions.
dy/dx = (2 * dz^2 * y^(3n) - 2x^4 * ay^(-n^3) - az * (m^2 - y^3)) / (y^3 * dz) | 133 | 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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.740303 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1503&CurriculumID=37&Num=4.20 | 1,540,211,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515041.68/warc/CC-MAIN-20181022113301-20181022134801-00510.warc.gz | 491,167,359 | 3,848 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 6 Mathematics4.20 Regular Polygons
A polygon in which all the sides are equal and all the angles are equal is called a regular polygon. Directions: Answer the following. Also draw at least five examples of regular polygon.
Q 1: A regular quadrilateral is also called asquareparallelogramrectangle Q 2: The figure shown is a regularpentagonoctagonhexagon Q 3: The figure shown is a regularoctagonsquarepentagon Q 4: The figure shows regular triangle which is also calledright angle triangleequilateral triangleisosceles triangle Q 5: A regular polygon hasangles of equal measuresides of equal lengthsides of equal length and angles of equal measure Q 6: The figure shown is a regularoctagonpentagonhexagon Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 238 | 1,022 | {"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-2018-43 | longest | en | 0.888222 |
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module1introductiontokinematics-181227112050.pptx
1. 1. Department of Mechanical Engineering JSS Academy of Technical Education, Bangalore-560060 Kinematics of Machines (Course Code:17ME42)
2. 2. TEXT BOOKS Theory of Machines, Rattan S.S, Tata McGraw-Hill Publishing Company. Theory of Machines, Sadhu Singh Theory of Machines, by R S Khurmi REFERENCE BOOKS: “Theory of Machines & Mechanisms", J.J. Uicker, , G.R. Pennock, J.E. Shigley. OXFORD 3rd Ed. Further Reference: National Programme on Technology Enhanced Learning (NPTEL) http://nptel.ac.in/courses/112104121/1
3. 3. At the end of the course, students will be able to, CO1: Describe the concepts of machines, mechanisms and relatedterminologies. CO2: Identify the mechanisms and predict their motions in mechanical components. CO3: Analyze planar mechanism for displacement, velocity and acceleration analytically and graphically . CO4: Analyze various motion transmission elements like gears, gear trains and cams. CO5: Utilize analytical, mathematical and graphical aspects of kinematics of machines for effective design. Course Outcomes
4. 4. Kinematics of Machines CHAPTER 1: Introduction to Kinematics of Machines
5. 5. Module 1 Introduction: Definitions Link or element, kinematic pairs, Degrees of freedom, Grubler's criterion (without derivation), Kinematic chain, Mechanism, Structure, Mobility of Mechanism, Inversion, Machine. Kinematic Chains and Inversions: Inversions of Four bar chain; Single slider crank chain and Double slider crank chain
6. 6. Terminologies Machine • Device for transferring and/or transforming motion and force (power) from source (Input) to the load (output). or • A machine is a device which receives energy in some available form and utilizes to do some work.
7. 7. Classification of Motion • Continuous rotation motion • Linear motion / Rectilinear Motion / translatory motion • Intermittent motion • Angular Oscillations
8. 8. Basic Definitions • Link or element • Kinematic pair • Kinematic chain • Mechanism, Machine & Structure • Degrees of freedom (DOF) • Grubler’s criterion • Inversion Kinematic Chains and Inversions • Inversions of Four bar chain • Single slider crank chain and • Double slider crank chain
9. 9. Basic Definitions Link/Eleme nt Kinematic Pair/Joint Kinematic Chain Mechanism Machine
10. 10. Kinematic Link or Element Each part of a machine, which moves relative to some other part, is known as a kinematic link (or simply link) or element.
11. 11. Types of Links • Rigid link : Link which does not undergo any deformation while transmitting motion. • Flexible link: Link which is partly deformed transmission of motion. e.g.: belts, ropes, chains and wires in a manner not to affect the • Fluid link: A link formed by having a fluid in a receptacle and the motion is transmitted through the fluid by pressure. e.g.: hydraulic presses, jacks and brakes.
12. 12. Classification of Links • Unary Link : Link with one Node (bucket of an excavator) • Binary link : Link connected to other links at two points • Ternary link: Link connected to other links at three points • Quaternary link: Link connected to other links at four points
13. 13. Structure An assembly of a no. of resistant bodies (members) having no relative motion between them and meant for carrying loads having straining action. Examples: A railway bridge, a roof truss, machine frames. etc.
14. 14. Difference Between a Machine and a Structure • The parts of a machine move relative to one another, whereas the members of a structure do not move relative to one another • A machine transforms the available energy into some useful work, whereas in a structure no energy is transformed into useful work • The links of a machine may transmit both power and motion, while the members of a structure transmit forces only (load bearing member).
15. 15. Kinematic Pair The two links or elements of a machine, when in contact with each other, are said to form a pair, If the relative motion between them is completely or successfully constrained (i.e. in a definite direction) Successfully constrained Completely constrained
16. 16. Constrained Motions 1. Completely constrained motion When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion. Example: The motion of a square bar in a square hole The motion of a shaft with collars at each end in a circular hole
17. 17. 2. Incompletely constrained motion When the motion between a pair can take place in more than one direction, then it can be an incompletely constrained motion. E.g.: A circular bar or shaft in a circular hole It may either rotate or slide in a hole. These both motions have no relationship with the other (automobile wheel).
18. 18. 3. Successfully constrained motion When the motion between the elements, forming a pair, is such that the constrained motion is not completed by itself, but by some other means E.g.: Shaft in a foot-step bearing, I C engine valve, Piston inside an cylinder.
19. 19. Classification of Kinematic Pairs According to the type of relative motion between the elements 1. Sliding pair / Prismatic pair (P) When the two elements of a pair are connected in such a way that one can only slide relative to the other, has a completely constrained motion. E.g.: The piston and cylinder, Cross-head and guides of a reciprocating steam engine, ram and its guides in shaper, tail stock on the lathe bed DOF = 1
20. 20. 2. Turning Pair / Revolute Pair (R) When the two elements of a pair are connected in such a way that one can only turn or revolve about a fixed axis of another link. Examples Lathe spindle supported in head stock. cycle wheels turning over their axles. DOF = 1
21. 21. 3. Spherical pair (S) When the two elements of a pair are connected in such a way that one element turns or swivels about the other fixed element. Eg: The ball and socket joint. Attachment of a car mirror. Pen stand. DOF = 3
22. 22. 4. Screw pair When the two elements of a pair are connected in such a way that one element can turn about the other by screw threads. Examples: • The lead screw of a lathe with nut • Bolt with a nut DOF = 1
23. 23. 5. Cylindrical pair If the relative motion between the pairing elements is the combination of turning and sliding, then it is called as cylindrical pair. DOF = 2
24. 24. 6. Rolling Pair When the two elements of a pair are connected in such a way that one rolls over another fixed link. E.g.: Ball and roller bearings, railway wheel rolling over a fixed rail Belt and pulley DOF = 1
25. 25. Classification of Kinematic Pairs According to the type / nature of contact between the elements / links. 1. Lower pair When the two elements of a pair have a surface contact, and the surface of one element slides or rolls over the surface of the other. E.g. sliding pairs, turning pairs and screw pairs form lower pairs.
26. 26. Classification of Kinematic Pairs According to the type / nature of contact between the elements / links. 2. Higher pair When the two elements of a pair have a line or point contact, and the motion between the two elements is partly turning and partly sliding. E.g.: toothed gearing, belt and rope drives, ball and roller bearings and cam and follower.
27. 27. Classification of Kinematic Pairs According to the Mechanical arrangement. 1. Self closed pair When the two elements of a pair are connected together mechanically in such a way that only required relative motion occurs. E.g. Lower pairs are self closed pair. 2. Force - closed pair When the two elements of a pair are not connected mechanically but are kept in contact by the action of external forces, the pair is said to be a force-closed pair. E.g.: Cam and follower
28. 28. Lower pair Higher pair Self closed pair Force - closed pair • Sliding pairs • Turning pairs • Screw pairs Rolling pair Kinematic Pairs
29. 29. • Based on the possible motions (Few Important Types only) Name of Pair Letter Symbol D.O.F 1. Revolute / Turning Pair R 1 2. Prismatic / Sliding Pair P 1 3. Helical / Screw Pair H 1 4. Cylindrical Pair C 2 5. Spherical / Globular Pair S (or) G 3 6. Flat / Planar Pair E 3
30. 30. Kinematic Chain When the kinematic pairs are coupled in such a way that the last link is joined to the first link to transmit definite motion (i.e. completely or successfully constrained motion), it is called a kinematic chain. Or Assembly of links (Kinematic link / element) and Kinematic pairs to transmit required / specified output motion(s) for given input motion(s)
31. 31. Mechanism • When one of the links of a kinematic chain is fixed, the chain is known as mechanism. • It may be used for transmitting or transforming motion e.g. printing machine, windshield wiper, robot arms • A mechanism may be regarded as a machine in which each part is reduced to the simplest form to transmit the required motion.
32. 32. Degrees of Freedom (DOF) / Mobility of a Mechanism Number of independent coordinates, required to describe / specify the configuration or position of all the links of the mechanism, with respect to the fixed link at any given instant.
33. 33. Degrees of freedom (DOF) In a kinematic pair, depending on the constraints imposed on the motion, the links may loose some of the six degrees of freedom.
34. 34. Grubler’s Criterion / Equation Grubler’s mobility equation M = 3 (n-1) - j1 - j2 M = Mobility or Total no. of DOF n = Total no. of links in a mechanism J1 = No. of joints having 1 DOF J2 = No. of joints having 2 DOF If, M>0, It gives mechanism with M DOF M=0, it gives a statically determinate structure M<0, it gives statically indeterminate structure
35. 35. Find the Degrees of Freedom
36. 36. Find the Degrees of Freedom
37. 37. Inversion of Mechanism • A mechanism is one in which one of the links of a kinematic chain is fixed. • Different mechanisms can be obtained by fixing different links of the same kinematic chain.
38. 38. Kinematic Chain When the kinematic pairs are coupled in such a way that the last link is joined to the first link to transmit definite motion (i.e. completely or successfully constrained motion).
39. 39. Types of Kinematic Chains 1. Four bar chain or quadric cyclic chain 2. Single slider crank chain 3. Double slider crank chain
40. 40. 1. Four bar chain or quadric chain • Four bar chain (mechanism) is the simplest and the basic kinematic chain and consists of four rigid links • Each of them forms a turning pair at A, B, C and D. The link that makes a complete revolution is called a crank • The four links may be of different lengths. • According to Grashof ’s law for a four bar mechanism, “the sum of the shortest and longest link lengths should not be greater than the sum of the remaining two link lengths” if there is to be continuous relative motion between the two links.
41. 41. 1. Four bar chain or quadric chain • The shortest link, will make a complete revolution relative to the other three In Fig., AD (link 4 ) is a crank. links crank or driver. • link BC (link 2) which makes a partial rotation or oscillates is known as lever or rocker or follower • link CD (link 3) which connects the crank and lever is called connecting rod or coupler. • The fixed link AB (link 1) is known as frame of the mechanism. The mechanism transforms rotary motion into oscillating motion.
42. 42. • In this mechanism, the links AD and BC (having equal length) act as cranks and are connected to the respective wheels. • The link CD acts as a coupling rod. • The link AB is fixed in order to maintain a constant centre to centre distance betweenthem. Purpose: Transmitting rotary motion from one wheel to the otherwheel.
43. 43. Purpose of this mechanism is to convert rotary motion into reciprocating motion. When the crank AB rotates about the fixed pointA. The lever oscillates about another fixed point D. The end E of lever is connected to a piston rod which reciprocates in a cylinder.`
44. 44. The four links are : fixed link at A, link AC, link CE and link BFD. Links CE and BFD act as levers. On displacement of the mechanism, the tracing point E at the end of the link CE traces out approximately a straight line.
45. 45. Single Slider Crank Chain & Inversion
46. 46. 2. Single Slider Crank Chain
47. 47. Pendulum Pump or Bull Engine When the crank (link 2) is given a rotary motion, the connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at A. The piston attached to the piston rod (link 1) reciprocates.
48. 48. Link 3 forming the turning pair is fixed and it corresponds to the connecting rod of a reciprocating steam engine mechanism. When the crank (link 2) rotates, the piston attached to piston rod (link 1) reciprocates and the cylinder (link 4) oscillates about a pin pivoted to the fixed link at A A
49. 49. It consists of seven cylinders in one plane all revolve about fixed centre D. Cylinders form link 1, Crank (link 2) is fixed. When the connecting rod (link4) rotates, the piston (link 3) reciprocates inside the cylinder.
50. 50. Radial Engine
51. 51. A mechanism used in shaping and slotting machines, where the metal is cut intermittently. • Link AC (i.e. link 3) is fixed. • Crank CB revolves with uniform angular speed about the fixed center C. • Sliding block attached to the crank pin at B slides along the slotted bar AP, thus causing AP to oscillate about the pivoted pointA. • Short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2
52. 52. • The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. • The return stroke occurs when the crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction.
53. 53. Double Slider Crank Chain
54. 54. • This inversion is obtained by fixing the slotted plate (link 4). • fixed plate or link 4 has two straight grooves cut in it, at right angles to eachother. • link 1 and link 3, are known as sliders and form sliding pairs with link 4. • link AB (link 2) is a bar which forms turning pair with links 1 and3. • When the links 1 and 3 slide along their respective grooves, any point on the link 2 such as P traces out an ellipse on the surface of link 4
55. 55. Show that AP and BP are the semi-major axis and semi-minor axis of the ellipse. OX and OY as horizontal and vertical axes let the link BA is inclined at an angle θ with the horizontal, Now the co-ordinates of the point P on the link BA will be This is the equation of an ellipse
56. 56. • This mechanism is used for converting rotary motion into a reciprocating motion. • The inversion is obtained by fixing either the link 1 or link 3. link 1 is fixed. • In this mechanism, when the link 2 (crank) rotates about B as centre, the link 4 (frame) reciprocates. • The fixed link 1 guides the frame.
57. 57. End of Module | 3,721 | 15,260 | {"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-2023-14 | latest | en | 0.758723 |
https://coalcountrychamber.com/how-is-it-used/how-is-coal-loaded-onto-trains.html | 1,632,538,738,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00330.warc.gz | 231,893,152 | 19,171 | # How is coal loaded onto trains?
Contents
At the largest surface mine operations, coal is loaded directly from overhead silos with storage capacity as high as 48,000 tons into gondola cars that move along a track loop that can accommodate multiple unit trains at once.
## How is coal used in trains?
Coal is carried in the tender of the locomotive and is hand-shoveled by the fireman into the firebox. … Water surrounds the outside of the firebox. Heat from the burning coal turns water to steam, which rises to the top of the boiler. The area surrounding the firebox and tubes is the “steam generator” of the locomotive.
## How long does it take to load a coal train?
Coal loading capacity is to be at a minimum rate of 4,000 tons per hour and permit loading a train of 10,000 tons within 4 hours. 5,000 tons per hour loading capability should be considered by the origin should 15,000 ton trains be the maximum train size desired.
## Where do coal trains go?
The coal train corridor extends from mines in Montana and Wyoming through Sandpoint, Idaho to Spokane, down through the Columbia River Gorge, then up along the Puget Sound coast, passing through Longview, Tacoma, Seattle, Edmonds, Everett, Mt. Vernon, Bellingham, Ferndale and all points in between.
## What are coal trains called?
Steam locomotives
## Do trains still run on coal?
Initially, both coal and wood were utilized to power locomotives, however, electric and diesel power grew to prominence in the 20th century. What do trains use for fuel? Trains use diesel, electric, and steam power for fuel.
fireman
## How much is a train car full of coal?
And a coal car is about 143 tons loaded, that would be \$2059.20 per car. Then taking the \$2059.20 per car cost and times it by a 135 car train and it should be about \$277,992.00 per train.
120 tons
## How much does a loaded train weigh?
Short Answer: A train weight can range anywhere between 4,000 tons (8,818,490 lbs) and 20,000 tons (44,092,452 lbs) or even more under some particular instances.
## How much money is coal worth?
In 2019, the national average sales price of bituminous, subbituminous, and lignite coal at coal mines was \$30.93 per short ton, and the average delivered coal price to the electric power sector was \$38.53 per short ton.
## How fast do freight trains go?
That would be quite a feat. Trains carrying freight are currently allowed to travel at speeds of up to 70 mph or 80 mph, but unloaded many trains generally only travel from 40-50 mph, according to FRA researchers.
## How long is a railroad coal car?
Some recently built equipment is capable of 143 tons. ** Required length for rotary gondolas but only required for bottom discharge equipment if equipped with rotary couplers.
Steel Manual Hopper.Specifications TableDiagramCoupled Length**53.1 ftNet Carrying Capacity*102 tonsGross Weight on Rail*131.5 tonsЕщё 1 строка
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## Why are coal trains not covered?
The US Department of Transportation classifies coal dust as a “pernicious ballast foulant” that can weaken and destabilize rail tracks. … If shippers wished to reduce or prevent coal dust from escaping, they could do so by filling cars less full or covering them with tarps or chemical sprays.4 мая 2016 г.
## Why is coal transported by rail?
Railroads accounted for over 70% of coal shipments to power plants in 2005. The balance moved by truck, barge, and conveyor. Most coal moved by rail because coal mines are often distant from power plants, and rail is usually the most economical means for moving bulk commodities long distances.
## What does coal train mean?
Noun. coal train (plural coal trains) (rail transport) a freight train which carries a single commodity: coal. | 881 | 3,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-39 | latest | en | 0.957082 |
http://mathhelpforum.com/calculus/177839-antiderivatives-print.html | 1,527,451,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00351.warc.gz | 187,524,787 | 2,624 | # Antiderivatives
• Apr 16th 2011, 08:45 AM
rowdy3
Antiderivatives
Find the following.
∫ (v^2 - e^(3v)) dv.
I did
∫(V^2-e^(3v)) dv
∫(v^2)dv - I (e^(3v) )dv
∫(v^3)/3- (e^(3v))/3
∫(v^3-e^(3v))/3
Did I do it right?
• Apr 16th 2011, 09:02 AM
e^(i*pi)
Quote:
Originally Posted by rowdy3
Find the following.
∫ (v^2 - e^(3v)) dv.
I did
∫(V^2-e^(3v)) dv
∫(v^2)dv - I (e^(3v) )dv
∫(v^3)/3- (e^(3v))/3
∫(v^3-e^(3v))/3
Did I do it right?
Lose the integral sign on your last two lines and add +C
• Apr 16th 2011, 09:10 AM
rowdy3
∫(v^2-e^(3v)) dv
= ∫(v^2)dv - ∫ (e^(3v) )dv
= (v^3)/3- (e^(3v))/3 + C
This could also be written as
(1/3)v3 - (1/3)e3v + C
or as (1/3)(v3 - e3v) + C | 364 | 669 | {"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-2018-22 | latest | en | 0.840564 |
https://www.calculatorology.com/70-degrees-fahrenheit-to-celsius/ | 1,702,319,815,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00225.warc.gz | 761,257,386 | 30,860 | код 70 degrees Fahrenheit to Celsius - Calculatorology
# 70 degrees Fahrenheit to Celsius
Conversion of 75 degree Fahrenheit to degrees Celsius is determined by the unit size of these two scales.
## 21.111 degrees Celsius(°C)
Conversion of 75 degree Fahrenheit to degrees Celsius is determined by the unit size of these two scales. Celsius is an SI derived unit that measures temperature expressed as (°C) while the Fahrenheit scale also measures temperature but is mostly used in the United States. The only difference between the two scales is that they start with different numbers and rise gradually at different rates.
The Fahrenheit and degrees Celsius are defined by two fixed points. When the Fahrenheit scale shows a temperature reading of 70 °F, it will be 21.111°C on the Celsius scale. This gives a difference of 100 and 180 degrees apart in the unit size of Celsius and Fahrenheit scales respectively.
To determine the calculation of 70 degrees Fahrenheit to degrees Celsius;
180/100 can be simplified as 9/5 whereas 100/180 can be simplified as 5/9.
If you want to get the temperature in degrees Celsius, we subtract 32 from the temperature reading in degrees Fahrenheit. Multiply by 5/9 to determine the reading.
### Fahrenheit to Celsius Converter
#### `70 degrees Fahrenheit(°F) = 21.111 degrees Celsius(°C)`
##### Formula
70°F = 21.111°C is the standard principle used in the conversion. Therefore, it implies that the 70 degrees Fahrenheit is equivalent to 21.111 degrees Celsius.
T (°C) = [T (°F) – 32] x 5/9 is the method for converting degrees Fahrenheit to degrees Celsius.
##### For example;
Convert 70 degrees Fahrenheit to degrees Celsius.
##### Solution
T (°C) = [70°F– 32] x 5/9
= 21.111° Celsius
#### Other methods that work
We can also use a different procedure to determine the results. First, you will add 40 to 70 degrees Fahrenheit and then multiply by 5/9. Add 40 to the result to get the temperature in degrees Celsius.
##### For example;
Convert 70 degrees Fahrenheit to degrees Celsius
##### Solution
70°F + 40 = 110
= 110 X 5/9 = 61.111
= 61.111 - 40 = 21.111° Celsius | 506 | 2,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-50 | longest | en | 0.744238 |
https://br.msx.org/ar/node/5427 | 1,696,426,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00252.warc.gz | 155,377,190 | 6,351 | # Build that game :)
صفحة 1/2
| 2
Here is a concept for a game I quickly made:
```100 SCREEN 7
110 G=9.8/10:'gravitation
120 W=1.5:'wind
130 F=2:'force/power behind shot
140 Y=200:'start coords Y-pos
150 PSET (10,200)
160 A\$=INPUT\$(1)
170 FOR I=0 TO512 STEP W
180 LINE-(I,Y/500),15
181 F=F-G
190 Y=Y-F
200 'G=(G*G)
201 X=X+10
210 NEXT
100 SCREEN 7
110 G=9.6/10
120 W=-.4
130 F=2
140 Y=100
150 PSET (0,Y)
160 A\$=INPUT\$(1)
170 FOR I=0 TO512 STEP W
180 LINE-(I,Y),15
181 F=F-G
190 Y=Y-F
191 IF Y>210 THEN F=F*-1
200 'G=(G*G)
201 X=X+10
210 NEXT```
The ultimate goal is to have 2 players shooting at eachothers tanks. Who first destroys the other wins
It roughly based on this amiga game
Aaaah, Scorched Earth.. a Classic.
Would be great to see this one ported to MSX
shouldn't be too hard in basic
would be nice on MSX and seems quite possible... great game
I go for worms
ARTRAG: yeah! Just guess what I played on my Amiga, 10 minutes ago
"First blood!"
Hi !
It's my first post to MSX.org !
Concerning Scortched Tank, if I remember correctly, it was programmed in Basic on Amiga (Amos). It is maybe possible to find the source code since I believe this game is a DP game. Anyway it could be cool to port it to the MSX because it is a very entertaining game !
Welcome, AmiMSX!
Thank you !
a DP game eh? (PD you mean, as in Public Domain)
Well having the tank doing a DP (deep or double penetration) would've been cool tho
a DP game eh? (PD you mean, as in Public Domain)
Yes, indeed PD, here in France we call it "Domaine Public" so it's DP is the french spelling,
sorry there has been a bug somewhere between my brain and my hands while I was
typing !
Well having the tank doing a DP (deep or double penetration) would've been cool tho
صفحة 1/2
| 2 | 557 | 1,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.899143 |
https://www.oliveboard.in/blog/accounting-ratios-rbi-grade-b-notes/ | 1,709,598,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00303.warc.gz | 935,057,651 | 51,776 | # Financial / Accounting Ratios – Free eBook for RBI Grade B & SEBI Grade A
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c. Rate or time
d. None of the above
Q.2. Which of the following is not included in current assets?
a. Debtors
b. Stock
c. Cash at bank
d. Cash in hand
Q. 3. Quick ratio is 1.8:1, the current ratio is 2.7:1 and current liabilities are Rs 60,000. Determine the value of the stock.
a. Rs 54,000
b. Rs 60,000
c. Rs 1,62,000
d. None of the above
Solution:
Quick Ratio =Quick assets /Current liabilities
1.8 = Quick assets /Current liabilities
1.8 = Quick assets/60000
Quick assets = 60000*1.8
Current Ratio = Current Assets / Current Liabilities.
2.7= Current assets/60000
Current assets =60000*2.7
Quick Assets are current assets without inventories as inventories are least liquid in current assets.
Hence quick assets =Current assets – stock
Stock= Current assets-quick assets
Stock = 60000*2.7 – 60000*1.8
=60000*0.9= 54000
These ratios form the most important part of the syllabus of the Numericals in the phase 2 Exam and hence these need to be studied in totality and not each ratio as a separate entity. Try to find the relation between ratios as problems are often asked in a combination of ratios.
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## Study material for RBI Grade B 2020 by Oliveboard
Oliveboard’s will be your one-stop destination for all your preparation needs
### 1. Course Details
RBI Grade B Cracker is designed to cover the complete syllabus for the 3 most important subjects: GA for Phase 1 and ESI + F&M for Phase 2 exam. Not just that, it also includes Mock Tests & Live Strategy Sessions for English, Quant & Reasoning for Phase 1. The course aims to complete your preparation in time for the release of the official notification.
#### 1.1. Features:
Phase I Phase II Weekly Live Classes for GK Special Live Classes on Exam Pattern & Strategy for Quant, Reasoning & English Sectional + Topic + GK Tests 20 Full-Length Mock Tests Weekly Live Classes & 50+ Video Lectures for ESI, and F&M Live Practice Sessions Class and Tips for Descriptive Writing Complete Notes for ESI, F&M ESI, F&M Mock Tests | 1,221 | 4,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.843138 |
http://www.thebigger.com/chemistry/thermodynamics/first-law-of-thermodynamics/calculate-q-w-%E2%88%86u-and-%E2%88%86h-for-the-reversible-isothermal-expansion-of-one-mole-of-an-ideal-gas-at-37oc-from-a-volume-of-20dm3-to-a-volume-of-30-dm3/ | 1,519,098,626,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00552.warc.gz | 566,958,232 | 4,555 | # Calculate q, w, ∆U and ∆H for the reversible isothermal expansion of one mole of an ideal gas at 37oC from a volume of 20dm3 to a volume of 30 dm3
Temperature T = 37oC = 37 + 273 = 310 K
Since the process is Isothermal,
Therefore, ∆U=0 and ∆H = 0 (as for an isothermal expansion of an ideal gas ∆U=0 and ∆H = 0)
As work done in reversible isothermal expansion is given by:
w= -nRT ln (V2/ V1)
= – (1 mol) (8.314 J K-1 mol-1) (310 K) ln (30 dm3 / 20dm3)
= – 1045.02 J
From first law, ∆U= q + w
Since ∆U=0, q = – w = 1045.02 J
Category: First Law of Thermodynamics | 217 | 570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-09 | latest | en | 0.83743 |
https://www.jiskha.com/display.cgi?id=1415172375 | 1,518,902,657,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807825.38/warc/CC-MAIN-20180217204928-20180217224928-00387.warc.gz | 884,546,679 | 3,910 | # statistics
posted by .
Waiting times at a doctor office are normally distributed with a mean of 35 minutes, and a standard deviation of 10 minutes. What is the chance a patient would have to wait over 50 minutes?
• statistics -
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
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https://www.chiefdelphi.com/t/motion-magic-position-puzzler/432319 | 1,726,478,072,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00051.warc.gz | 655,191,026 | 7,740 | Motion Magic position puzzler
We have code to turn a turntable using a Falcon. We control it both with percent output and with motionmagic. The code is here:
The logic is that we run the turntable at a constant speed. We use an IR sensor to look for the cone; if we find the two edges, we stop and command a ‘softer’ move to an offset set point. Softer because an abrupt stop often leaves the cone in variable positions. The ‘softer’ move is a motion magic move at a reduced velocity and acceleration. (Side note, this is not actually the ideal solution, because the motion profile starts slow and ramps up, so we get abrupt movement anyway, but it is what we had time for).
And this works. It works in the lab. It works on the practice field. It’s slick. But sometimes, just sometimes, usually after a relatively long running session, we get a novel behavior. After the cone stops in the correct position, the turntable will start spinning slowly clockwise.
Critically, that is not at any of the speeds we normally command. It is perhaps at the intended velocity of the motion magic command. Now, our position control is a not especially well tuned PID, so I had the theory that if we exited motion magic with a last I driven command, maybe that would go on forever. But my understanding is that motion magic will seek to hold the destination point after the move, so that doesn’t add up. My only other hypothesis is that sometimes GetSelectedSensorPosition() will return 0, or will ‘wrap’ such that our commanded move is actually taking us back to 0, instead of just using an offset. That is, if we spin a motor for a long time, can we get into a state where commanding a move to GetSelectedSensorPosition() + 1500 will go to an unexpected point?
It is the case that once the problem starts happening, it is quite consistent and happens every time.
At any rate, I find it to be an interesting puzzler so I thought I’d share and see if anyone else has any clue bats.
I would be really curious to see a self-test of that motor when its in that failure state… If you need a quick solution you could add some logic to set the motor to 0 velocity once it’s at its setpoint just to ensure it doesn’t try and go somewhere.
Well, the issue is that we can’t easily create the failure state. I think we’ve got a number of paths to ‘fixing’ it (including getting rid of the motion magic in favor of a soft slowdown). But I’m a curious type; I really want to know what’s happening.
I hate stuff that’s hard to reproduce. When you get in that failure state using tuner to self test the motor and plot relevant fields should shine light on it I hope.
While I’m not quite sure of what’s going on, I do know that MotionMagic can cause some funky results when stuff changes outside of it (typically while robot is disabled) that can cause it to ignore its motion profiling.
Is there a reason you are using a non-zero kI term on the turntable PID (i.e. was it failing to reach its target without it)? Most of the time, a simple P or PD controller is more than enough to produce consistent motion, especially with motion magic, so that may be the culprit.
Another thing that could happen is that the offset is getting repeatedly applied (i.e. motor is always targeting 1500 ticks ahead of its current position) because the command is not being properly terminated. This lines up with your symptoms of a.) constant rotation in the given direction at a speed consistent with your motion magic limits and b.) inability to terminate the behavior once it starts. You could use a timeout on the command to see if this is the source.
Thanks for listening :-).
The mechanism came together late and had some binding spots; the kI helps us overcome the flaws in the mechanism. In an ideal world, we wouldn’t need it (and we’d also use velocity control instead of % output control, but that’s a different issue).
That’s a good theory. But the method that sets the motor is guarded by a boolean, and I really can’t see how it would get called more than once. But I’d also expect it to consistently fail in that fashion if that was the bug, rather than only every blue moon.
This topic was automatically closed 365 days after the last reply. New replies are no longer allowed. | 929 | 4,262 | {"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-2024-38 | latest | en | 0.943094 |
https://www.coursehero.com/file/6142844/Graphs-of-Functions-Part-2/ | 1,513,385,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948580416.55/warc/CC-MAIN-20171215231248-20171216013248-00116.warc.gz | 710,572,270 | 149,517 | Graphs of Functions (Part 2)
# Graphs of Functions (Part 2) - Domain Recall that the...
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Domain Recall that the domain is the set of all input values to which the rule applies. These are called your independent variables. These are the values that correspond to the first components of the ordered pairs it is associated with. If you need a review on the domain, feel free to go to Tutorial 30: Introduction to Functions. On a graph, the domain corresponds to the horizontal axis. Since that is the case, we need to look to the left and right to see if there are any end points to help us find our domain. If the graph keeps going on and on to the right then the domain is infinity on the right side of the interval. If the graph keeps going on and on to the left then the domain is negative infinity on the left side of the interval. If you need a review on finding the domain given a graph, feel free to go to Tutorial 31: Graphs of Functions, Part I. Range Recall that the range is the set of all output values. These are called your dependent variables. These are the values that correspond to the second components of the ordered pairs it is associated with. If you need a review on the range, feel free to go to Tutorial 30: Introduction to Functions. On a graph, the range corresponds to the vertical axis. Since that is the case, we need to look up and down to see if there are any end points to help us find our range. If the graph keeps going up with no endpoint then the range is infinity on the right side of the interval. If the graph keeps going down then the range goes to negative infinity on the left side of the interval. If you need a review on finding the domain given a graph, feel free to go to Tutorial 31: Graphs of Functions, Part I x- intercept No matter what type of graph that you have, recall that the x -intercept is where
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the graph crosses the x axis. The word 'intercept' looks like the word 'intersect'. Think of it as where the graph intersects the x -axis . If you need more review on intercepts, feel free to go to Tutorial 26: Equations of Lines. y -intercept No matter what type of graph that you have, recall that the y -intercept is where the graph crosses the y axis. The word 'intercept' looks like the word 'intersect'. Think of it as where the graph intersects the y -axis . If you need more review on intercepts, feel free to go to Tutorial 26: Equations of Lines. Functional Value Recall that the functional value correlates with the second or y value of an ordered pair. If you need a review on functional values, feel free to go to Tutorial 30: Introduction to Functions . Example 1 : Use the graph to determine a) the domain, b) the range, c) the x -intercepts, if any d) the y -intercept, if any, and e) the functional value indicated.
a) Domain We need to find the set of all input values. In terms of ordered pairs, that correlates with the first component of each one. In terms of this two dimensional graph, that corresponds with the x values (horizontal axis). Since that is the case, we need to look to the left and right and see if there are any end points. In
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Graphs of Functions (Part 2) - Domain Recall that the...
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Ask a homework question - tutors are online | 840 | 3,705 | {"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} | 4.46875 | 4 | CC-MAIN-2017-51 | latest | en | 0.889675 |
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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:
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[Solved]: How to store factorials?
, ,
Problem Detail:
Can someone help me to store the factorial of large numbers such as 100! efficiently?
UPDATE: obviously, storing the argument rather than the factorial digits themselves achieves a significant saving. The true challenge is to find a data compression scheme that achieves a significant compression ratio, but with a lighter computational complexity than recomputing the factorial(s) from the argument.
More precisely, can you design an algorithm to produce all decimal digits of \$n!\$, for \$n\le N\$, using \$o(N\log N!)\$ storage and small computational cost, such as the lower bound \$O(\log n!)\$.
We can calculate 100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000. You can store this in any way you want: as `100!`, as the ASCII string `93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000`, as a Unicode string, or in binary in any number of formats. It all depends on what you want to do with the value. If you want to display the value on screen as a decimal, you might as well store the decimal string. If you want to do arithmetic, then you should be using a library for dealing with "big numbers" such as GMP, and in this case the library will provide its own means for storing numbers. If you're just interested in storing the expression 100!, and intend to use it in a computer algebra software (CAS), you might as well store it as just `100!`. | 415 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-16 | longest | en | 0.859795 |
https://brilliant.org/discussions/thread/zeroth-law-of-thermodynamics/ | 1,532,243,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593051.79/warc/CC-MAIN-20180722061341-20180722081341-00463.warc.gz | 613,921,483 | 14,494 | # **Zeroth law of thermodynamics**
I decided to make a note on this.
What does thermodynamics mean? it means heat change.
Zeroth law states that: if two objects $$A$$ and $$B$$ are not in thermal contact but are in thermal equilibrum with object $$C$$. Then, object $$A$$ and $$B$$ are in thermal equilibrum with each other.
If you have additional comment. you can post them....
Note by Samuel Ayinde
4 years, 1 month ago
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So basically the Transitive property applied to thermodynamics. Cool!
- 4 years, 1 month ago
Yes @Finn Hulse
- 4 years, 1 month ago
well,in most of zeroth law statements,many of us think that only of thermal equilibrium, but tje systems A B and C must be in mechanical equilibrium too.
- 3 years, 3 months ago
yeap sure......
- 3 years, 3 months ago | 532 | 1,658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-30 | latest | en | 0.792258 |
https://edurev.in/v/92685/Types-of-RelationsReflexive-Symmetric-Transitive-a | 1,721,623,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00672.warc.gz | 189,011,374 | 61,204 | Types of Relations: Reflexive Symmetric Transitive & Equivalence
# Types of Relations: Reflexive Symmetric Transitive & Equivalence Video Lecture | Mathematics (Maths) Class 12 - JEE
## Mathematics (Maths) Class 12
204 videos|288 docs|139 tests
## FAQs on Types of Relations: Reflexive Symmetric Transitive & Equivalence Video Lecture - Mathematics (Maths) Class 12 - JEE
1. What is a reflexive relation?
Ans. A reflexive relation is a relation where every element is related to itself. In other words, for every element 'a' in the set, the relation contains the pair (a, a). For example, the relation "is equal to" is reflexive because every element is equal to itself.
2. Can a relation be both reflexive and symmetric?
Ans. Yes, a relation can be both reflexive and symmetric. Reflexivity means that every element is related to itself, while symmetry means that if 'a' is related to 'b', then 'b' is also related to 'a'. So, a relation can satisfy both properties by containing all the pairs (a, a) for reflexivity and (a, b) for symmetry.
3. What does it mean for a relation to be transitive?
Ans. Transitivity is a property of relations where if 'a' is related to 'b' and 'b' is related to 'c', then 'a' is also related to 'c'. In other words, if there is a chain of relationships, the transitive relation implies that the end elements are also related. For example, if 'x' is a parent of 'y' and 'y' is a parent of 'z', then the transitive relation implies that 'x' is also a parent of 'z'.
4. How can we determine if a relation is an equivalence relation?
Ans. To determine if a relation is an equivalence relation, we need to check three properties: reflexivity, symmetry, and transitivity. If the relation satisfies all three properties, it is an equivalence relation. Reflexivity ensures that each element is related to itself, symmetry guarantees that if 'a' is related to 'b', then 'b' is also related to 'a', and transitivity ensures that the relation follows a chain of relationships.
5. Give an example of an equivalence relation.
Ans. An example of an equivalence relation is the relation "is congruent to" in geometry. If two geometric figures have the same shape and size, they are considered congruent. This relation is reflexive because any figure is congruent to itself, symmetric because if figure 'A' is congruent to figure 'B', then 'B' is also congruent to 'A', and transitive because if figure 'A' is congruent to 'B' and 'B' is congruent to 'C', then 'A' is also congruent to 'C'.
## Mathematics (Maths) Class 12
204 videos|288 docs|139 tests
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https://codedump.io/share/NIFGqIrXUckE/1/how-to-aggregate-data-frame-and-add-0-for-categories-not-found | 1,527,152,468,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866107.79/warc/CC-MAIN-20180524073324-20180524093324-00361.warc.gz | 538,230,638 | 9,130 | user299791 - 1 year ago 43
R Question
I have a data frame like:
``````> prova
sent weeknumber processed
1 100 1 1
2 23 1 0
3 254 1 1
4 321 2 0
5 1241 2 0
6 323 2 1
7 1221 3 1
structure(list(sent = c(100, 23, 254, 321, 1241, 323, 1221),
weeknumber = c(1, 1, 1, 2, 2, 2, 3), processed = c(1, 0,
1, 0, 0, 1, 1)), .Names = c("sent", "weeknumber", "processed"
), row.names = c(NA, -7L), class = "data.frame")
``````
If I want to extract the number of Sent by week number for rows with processed = 0, I can do:
``````aggregate(prova[prova\$processed==0,]\$sent, by=list(prova[prova\$processed==0,]\$weeknumber), FUN = sum)
Group.1 x
1 1 23
2 2 1562
``````
And if I want to extract the sum of Sent by week number when processed = 1 I do:
``````aggregate(prova[prova\$processed==1,]\$sent, by=list(prova[prova\$processed==1,]\$weeknumber), FUN = sum)
Group.1 x
1 1 354
2 2 323
3 3 1221
``````
However, I'd like to find a way to have always the same result length, i.e. in case of processed=0, something like this:
`````` Group.1 x
1 1 23
2 2 1562
3 3 0 // this is the new row I'd like to add
``````
If I simply pass the whole list of possible week numbers, I get:
``````aggregate(prova[prova\$processed==0,]\$sent, by=list(prova\$weeknumber), FUN = sum)
Error in aggregate.data.frame(as.data.frame(x), ...) :
arguments must have same length
``````
We can use an `if/else` condition with `data.table`. Convert the 'data.frame' to 'data.table' (`setDT(prova)`), grouped by 'weeknumber', `if` there are not `any` 0 values in 'processed', return 0 or `else` get the `sum` of 'sent' where 'processed' is 0.
``````library(data.table)
setDT(prova)[, .(sent = if(!any(processed==0)) 0
else sum(sent[processed==0])), by = weeknumber]
# weeknumber sent
#1: 1 23
#2: 2 1562
#3: 3 0
``````
But, if we need the `sum` of 'sent' for each value of 'processed' grouped by 'weeknumber', a convenient option is `dcast`
``````dcast(setDT(prova), weeknumber~processed, value.var="sent", sum)
# weeknumber 0 1
#1: 1 23 354
#2: 2 1562 323
#3: 3 0 1221
``````
Or with `xtabs` from `base R` which also does the `sum` of 'sent' for each of the combinations of 'weeknumber' with 'processed'.
``````xtabs(sent~weeknumber + processed, prova)
``````
If we are using `aggregate`, one option is `merge` the output of `aggregate` with the `unique` set of 'weeknumber' and then replace the `NA` elements in 'sent' to 0.
``````res <- merge(data.frame(weeknumber = unique(prova\$weeknumber)),
aggregate(sent~weeknumber, prova, subset = processed ==0, FUN = sum),
all.x=TRUE)
res\$sent[is.na(res\$sent)] <- 0
res
# weeknumber sent
#1 1 23
#2 2 1562
#3 3 0
``````
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https://www.physicsforums.com/threads/does-this-differential-equation-have-a-solution.585156/ | 1,542,810,940,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748901.87/warc/CC-MAIN-20181121133036-20181121155036-00113.warc.gz | 936,635,144 | 13,677 | Homework Help: Does this differential equation have a solution?
1. Mar 8, 2012
xvudi
1. The problem statement, all variables and given/known data
I'm examining the equation Ohm_Max + dOhm/dr = Ohm_Max - dOhm/dr and can't find any solutions other than the trivial one, Ohm(r) = 0 for all r.
It's meant to determine if it is possible to build a length of conductor such that, upon dividing it at any arbitrary point, you'll find that the resistance behind is the same as the resistance ahead.
3. The attempt at a solution
Ohm(r) = 0. You can't have one because if Ohm(r) is even then dOhm/dr is odd.
Ohm(r) = Ohm_Max at r = 0 for bounds -L/2 to L/2, captured in the use of Ohm_Max the constant.
So the integral from -L/2 to 0 must equal the integral from 0 to L/2 meaning that dOhm/dr has to be even. This can't be as if Ohm(r) is odd then Ohm(0) must be 0 and not Ohm_Max. Odd functions cannot be valued at 0.
Now, mathematically why can't this work? I apologize for the absence of TeX. I'm still getting used to the forum interface.
2. Mar 8, 2012
Staff: Mentor
Instead of using Ohm as a variable, let's switch to R, for resistance. Your equation simplifies to 2 dR/dr = 0.
This is a very simple differential equation to solve, and the solution is not necessarily the trivial solution, although that is one solution.
3. Mar 8, 2012
xvudi
Ohm is resistance. I apologize for the confusion.
So R(r) = C1 * exp(0 * r) = C1
Particular solution for R_Max is 0 as all of its derivatives are zero, putting in our constraint at R(0) we solve for our constant C1.
R(0) = R_Max => C1 = R_Max except...
R(-L/2) = 0
R(L/2) = 0
So C1 must be a function of r except C1 cannot be a function of r.
This means that R does not depend on r. Meaning that dR/dr has to be zero (easily verified) and I just realized where my last post went off the rails.
I meant to add that for bounds -L/2 and L/2 that R(r) must be zero.
So the only way you can do it is if you stick a slider on some rails next to a circuit containing two perfectly matched resistors and call the resistors the system because I forgot the above constraint and I apologize.
So, with this added constraint, what is the grand mathgalactic reason why it isn't a solvable problem? It's something fundamental. It has to do with the problem type and I can't put my finger on it.
4. Mar 9, 2012
Staff: Mentor
This is really the long way around.
R'(r) = 0 ==> R(r) = C
The idea is that if the derivative of something is zero, the something must be a constant.
Now, using the initial condition R(0) = Rmax, then C = Rmax
If you also know that R(L/2) = R(-L/2) = 0, then C = Rmax = 0.
Am I missing something? | 762 | 2,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-47 | latest | en | 0.94532 |
https://datascience.stackexchange.com/questions/42517/what-affects-the-magnitude-of-lasso-penalty-of-a-feature | 1,660,704,875,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.78/warc/CC-MAIN-20220817001643-20220817031643-00281.warc.gz | 206,679,150 | 65,313 | # What affects the magnitude of lasso penalty of a feature?
Is there a way to intuitively tell if the lasso penalty for a particular feature will be small or large? Consider the following scenario:
Imagine we use Lasso regression on a dataset of 100 features $$[X_1, X_2, X_3, ..., X_{100}]$$. Now, if we take one feature and scale up its values by a factor of 10, so for example $$X_1 = X_1 * 10$$, and then we fit lasso regression on the new dataset, which of the following might be true?
1. X1 probably would be excluded from the fitted model
2. X1 probably would be included in the fitted model
Someone told me that 2 is true. Somehow the operation on $$X_1$$ would result in less lasso penalty for $$X_1$$, therefore the feature is likely to be kept.
I cannot understand why. Can anyone tell if this whole statement is not correct to begin with or if there is some truth in it?
When you increase the scale of $$X_1$$ by 10, the scale of the corresponding weight in you linear regression sees its scale divided by 10 (it is exactly 10 for standard linear regression and the same order of magnitude holds for lasso loss). As a consequence, the $$L_1$$ penalty for this coefficient is also divided by a factor 10 and it is less likely to be set to 0. | 313 | 1,258 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-33 | latest | en | 0.936905 |
https://studysoup.com/guide/2352982/10x9x8x7x6x5x4 | 1,618,804,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00015.warc.gz | 653,291,997 | 14,070 | ×
×
# 10x9x8x7x6x5x4 Description
##### Description: These are examples of the material that is on the exam. Some are re-worked examples we got in class
6 Pages 160 Views 1 Unlocks
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Exam 2 Wednesday, October 19, 2016 11:53 AMContents: - Counting and tree diagrams - Rearrangements - probability Counting and tree diagrams: Suppose that you have the set S= {1,2,3,4} How many 3 element subsets can one create? A: 4 Use a tree diagram Levels represent a selection round Leaves represent an outcome Since we don’t care about the order you don't count all the leaves (the definition of a subset is that the elements are the same no matter the order. Ex. 1234 and 2314) With bigger alphabets you can figure out the answer by taking the number of total rearrangements and dividing it by the number of nce we on care aou e orer you on coun a the leaves (the definition of a subset is that the elements are the same no matter the order. Ex. 1234 and 2314) With bigger alphabets you can figure out the answer by taking the number of total rearrangements and dividing it by the number of subsets. Ex. If you have the alphabet {0,1,2,3,4,5,6,7,8,9} and you want to find the number of 4 element subsets that can be created with that set: First: there are 10x9x8x7 possible 4 element arrangements Then figure out that there are 24 possible subsets because 1 subset contains 4 elements that can be arranged 4! Ways Now let's try a harder example:How many codes of length 7 can you create using the alphabet {0,1,2,3,4,5,6,7,8,9} (repeated digits allowed) First Step: Determine the size of your alphabet You have a 10 digit alphabet
Step two: Determine how many levels of the tree you will have You have 7 choices for each digit in the code. Raise 10 to the 7th because you have 5 choices
## How many 3 element subsets can one create?
We also discuss several other topics like combining small pieces of information into larger clusters is known as
If you want to learn more check out How should you use eye-contact while giving a speech?
We also discuss several other topics like math 241 oregon state | 600 | 2,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-17 | latest | en | 0.890934 |
https://www.scribd.com/document/23350383/2004-Amaths-Mock1-Paper-Marking-Scheme | 1,547,715,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00450.warc.gz | 936,369,898 | 47,225 | You are on page 1of 20
# KING’S GLORY EDUCATIONAL CENTRE
CE MOCK EXAMINATION 2004
A. LEUNG
Prepared by A. Leung
King’s Glory Educational Centre
-1-
Section A
1. Let P(n) be the proposition.
n(n + 1)(n + 5)
P(n) : T1 + T2 + T3 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +Tn =
3
1(1 + 1)(1 + 5)
When n = 1, L.H.S. = T1 = 1(1 + 3) = 4 , R.H.S. = =4
3
∴ P (1) is true.
k (k + 1)(k + 5)
Assume P (k ) is true, i.e. T1 + T2 + T3 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +Tk =
3
When n = k + 1, L.H.S. = (T1 + T2 + T3 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +Tk ) + Tk +1
k (k + 1)(k + 5)
= + (k + 1)(k + 4)
3
k +1 2
=
3
(
k + 5k + 3k + 12 )
(k + 1)(k + 2)(k + 6)
=
3
∴ P (k + 1) is also true.
By the principle of mathematical induction, the proposition is true for all positive integers n.
r
th
2. The (r+1) term in descending powers of x = C 18
r (x )
2 18 − r ⎛2⎞
⎜ ⎟
⎝ x⎠
( )
= 2 r C r18 x 36−3r
For x 3 , power of x = 3, 36 − 3r = 3 , r = 11
Coefficient of x 3 = 211 C1118 = 65175552.
-2-
3. f ( x) = x 2 + (3 + m) x + (2m − 2) . Let A(α , 0) , B( β , 0) be the coordinates of intersection.
α , β are the roots of the equation x 2 + (3 + m) x + (2m − 2) = 0 .
α + β = −(3 + m) αβ = 2m − 2
α−β =2 5
(α − β )2 = 20
(α + β )2 − 4αβ = 20
[− (3 + m)]2 − 4αβ = 20
(m − 3)(m + 1) = 0
m=3 or m = −1
4. (2 x + 1)(y 3 − 2 y + 1) + (x 2 + x + 5)(3 y 2 − 2) dy = 0
dx
dy (2 x + 1) (y 3 − 2 y + 1)
=− 2
dx (x + x + 5)(3 y 2 − 2)
dy 3
=
dx (1, 0 ) 14
Equation of tangent is y =
3
(x − 1)
14
i.e. 3x − 14 y − 3 = 0
-3-
5. (a) I = ∫ 15 cos 4 θ (cosθ dθ )
Let t = sin θ , dt = cosθ dθ
I = ∫ 15(1 − t 2 ) 2 dt
∫ 15(1 − 2t + t 4 ) dt
2
=
= 15t − 10t 3 + 3t 5 + c
= 15 sin θ − 10 sin 3 θ + 3 sin 5 θ + c
dy
(b) y=∫ dx
dx
∫ 15 cos
5
= x dx
= 15 sin x − 10 sin 3 x + 3 sin 5 x + c
π
When x = ,y=6
2
15 − 10 + 3 + c = 6
c = −2
Equation of C is y = 3 sin 5 x − 10 sin 3 x + 15 sin x − 2
-4-
3
6. (a) tan 2θ =
4
2 tan θ 3
=
1 − tan θ 4
2
8 tan θ = 3 − 3 tan 2 θ
3 tan 2 θ + 8 tan θ − 3 = 0
(3 tan θ − 1)(tan θ + 3) = 0
1
tan θ = (tan θ > 0)
3
r
(b) tan θ =
h
1
r= h
3
1
V = πr 2 h
3
2
1 ⎛1 ⎞
= π ⎜ h⎟ h
3 ⎝3 ⎠
π
= h3
27
Differentiate both sides with respect to time t,
dV π ⎛ 2 dh ⎞
= ⎜ 3h ⎟
dt 27 ⎝ dt ⎠
-5-
π dh
= h2
9 dt
dV
Now, h = 8 , = 40π
dt
π
40π =
9
( 8 )
2 dh
dt
dh 45
=
dt 8
45
water level is increasing at cm / min at that instant.
8
7. 2 cos 5 x cos 2 x = 3 cos 5 x
cos 5 x(2 cos 2 x − 3 ) = 0
3
cos 5 x = 0 or cos 2 x =
2
5 x = (2m + 1) 90 o or 2 x = 360 o n ± 30 o where m, n are integers.
x = (2m + 1)18 o or x = 180 o n + 15 o
-6-
m−4 2
8. (a) = where m is the slope of the required line.
1 + 4m 9
9m − 36 = 2 + 8m or 9m − 36 = −2 − 8m
m = 38 or m=2
⎛ 1+ k ⎞
(b) − ⎜ ⎟=2 (m < 3)
⎝ 7 − 2k ⎠
1 + k = −14 + 4k
k =5
Equation of required line is 6 x − 3 y − 17 = 0
10k + 3 v 4k + 11 v
9. (a) OP = i+ j
1+ k 1+ k
v v
(b) AB = 7i − 7 j
Q OP ⊥ AB , OP ⋅ AB = 0
⎛ 10k + 3 v 4k + 11 v ⎞ v v
⎜ i+ j ⎟ ⋅ (7i − 7 j ) = 0
⎝ 1+ k 1+ k ⎠
-7-
7
[(10k + 3)(1) − (4k + 11)(1)] = 0
1+ k
6k − 8 = 0
4
k=
3
v v v v
(c) OP = 7i + 7 j , OA = 3 i + 11 j
OP ⋅ OA = OP OA cos ∠AOP
v v v v
(7i + 7 j ) ⋅ (3 i + 11 j ) = 7 2 + 7 2 3 2 + 112 cos ∠AOP
(7)(3) + (7)(11) = 98 130 cos ∠AOP
∠AOP = 29.7 o
-8-
10. (a) Maximum point : (−1, 3) Minimum point : (1, − 3)
(b)
11. Q ( x + 3) 2 = x + 3
2
∴ x+3 − 7 x + 3 − 18 ≤ 0
2
( x + 3 − 9 )( x + 3 − 2 ) ≤ 0
x+3 ≤9 (Q x + 3 + 2 ≥ 2 )
−9 ≤ x+3≤ 9
− 12 ≤ x ≤ 6
-9-
Section B
12. (a) (i) Equation of S1 is x( x + 2) + ( y − 2)( y + 2) = 0
i.e. x 2 + y 2 + 2 x − 4 = 0
2 − (− 2)
(ii) Equation of BC is y + 2 = (x − 0)
−2
i.e. 2x + y + 2 = 0
(iii) Equation of family of circle through B and C is
x 2 + y 2 + 2 x − 4 + k (2 x + y + 2) = 0 for all values of k.
the circle passes through A(3, 7)
∴ 9 + 6 + 49 − 4 + k (6 + 7 + 2 ) = 0
k = −4
∴ Equation of S 2 is x 2 + y 2 − 6 x − 4 y − 12 = 0
(b) (i) Equation of tangent is y = mx + 8
⎧ x 2 + y 2 − 6 x − 4 y − 12 = 0
⎨ have equal roots
⎩ y = mx + 8
On substituting y, we have
(1 + m )x
2 2
+ 2(6m − 3)x + 20 = 0
∴ [2(6m − 3)]2 − 4(1 + m 2 )(20) = 0
-10-
On simplifying, giving 16m 2 − 36m − 11 = 0
9 11
∴ m1 + m2 = , m1 m2 = −
4 16
m1 − m2
(ii) tan θ =
1 + m1 m2
(m1 + m2 )2 − 4m1m2
=
1 + m1 m2
⎛ − 11 ⎞
2
⎛9⎞
⎜ ⎟ − 4⎜ ⎟
⎝4⎠ ⎝ 16 ⎠
=
11
1−
16
= 4 5
⎧y = x2
13. (a) For intersection of P and L, ⎨
⎩ y = ax
On solving, we have x = 0 or x = a
a
Hence ∫ 0
(ax − x 2 ) dx = 288
a
⎡a 2 1 3 ⎤
⎢ 2 x − 3 x ⎥ = 288
⎣ ⎦0
a3 a3
− = 288
2 3
a = 12
-11-
(b) Required volume
= π∫
12
0
[(12 x + 1) 2
− ( x 2 + 1) 2 dx ]
12
= π∫ (− x 4 + 142 x 2 + 24 x) dx
0
12
⎡ 1 142 3 ⎤
= π ⎢− x 5 + x + 12 x 2 ⎥
⎣ 5 3 ⎦0
⎡ 1 142 ⎤
= π ⎢− (12) 5 + (12) 3 + 12(12) 2 ⎥
⎣ 5 3 ⎦
= 33753.6 π
h
(c) Capacity of bowl = π ∫ x 2 dy
0
h
= π ∫ y dy
0
h
⎡ y2 ⎤
=π ⎢ ⎥
⎣ 2 ⎦0
πh 2
=
2
πh 2
(d) = 33753.6 π
2
h = 260
-12-
π
14. (a) V = (l sin θ ) 2 (l cosθ )
3
πl 3 ⎛ π⎞
= sin 2 θ cosθ ⎜0 < θ < ⎟
3 ⎝ 2⎠
dV πl 3
(b)
=
3
[
sin 2 θ (− sin θ ) + cosθ (2 sin θ cosθ ) ]
πl 3
=
3
(
sin θ 2 − 3 sin 2 θ )
πl 3
When V is increasing,
dV
> 0,
3
( )
sin θ 2 − 3 sin 2 θ > 0
2 ⎛ π⎞
sin 2 θ < ⎜0 < θ < ⎟
3 ⎝ 2⎠
6
sin θ <
3
πl 3
When V is decreasing,
dV
< 0,
3
( )
sin θ 2 − 3 sin 2 θ < 0
2 ⎛ π⎞
sin 2 θ > ⎜0 < θ < ⎟
3 ⎝ 2⎠
6
sin θ >
3
6
Since V changes from increasing to decreasing via sin θ = , hence when V is
3
6
maximum, sin θ = .
3
1
S= (2l sin θ )(l cosθ )
2
1 2
= l sin 2θ
2
Clearly, S is maximum when sin 2θ = 1 ,
-13-
π π
2θ = , i.e. θ =
2 4
Therefore, when V is a maximum, S is not a maximum.
π dθ
(d) When l = 6, θ = , = 0.1
6 dt
dV π ⎡ π π⎤
= (6) 3 ⎢2 sin − 3 sin 3 ⎥
dθ 3 ⎣ 6 6⎦
= 45π
dV dV dθ
Q = ⋅
dt dθ dt
dV
= (45π )(0.1) = 4.5π .
dt
Hence the capacity is increasing at 4.5π unit 3 per second
-14-
v v v 2
15. (a) a ⋅ a = a =1,
v v v v ⎛1⎞
a ⋅ b = a b cosθ = (1)(3)⎜ ⎟ = 1
⎝ 3⎠
v v v 2
b ⋅b = b =9
v v
a + 2b
(b) OC =
3
v v
(k + 1)a + (2k + 2)b
(c) OD =
3
BD = OD − OB
v v
(k + 1)a + (2k + 2)b v
= −b
3
v v
(k + 1)a + (2k − 1)b
=
3
Q OD⊥BD , ∴ OD ⋅ BD = 0
( )⎤⎥ ⋅ 13 [(k + 1)av + (2k − 1)b ] = 0
⎡k +1 v v v
⎢⎣ 3 a + 2b
k +1
9
[ v v v v v v v v
]
(k + 1)a ⋅ a + (2k − 1)a ⋅ b + 2(k + 1)b ⋅ a + 2(2k − 1)b ⋅ b = 0
(k + 1)(1) + (2k − 1)(1) + 2(k + 1)(1) + 2(2k − 1)(9) = 0
41k = 16
16
k=
41
-15-
( )
1 v v
BD = 19a − 3b
41
( )⎤⎥ ⋅ ⎡⎢ 411 (19av − 3b )⎤⎥ ⋅
⎡1 v v v
BD ⋅ BD = ⎢ 19a − 3b
⎣ 41 ⎦ ⎣ ⎦
(
v v
)
⎛1⎞
2
2
v v v v
BD = ⎜ ⎟ 361a ⋅ a − 114a ⋅ b + 9b ⋅ b
⎝ 41 ⎠
2
⎛1⎞
= ⎜ ⎟ [361(1) − 114(1) + 9(9)]
⎝ 41 ⎠
2 2 32
BD = 82 , ∴ BD = 82
41 41
(e)
v v v
(
AO ⋅ AB = − a ⋅ b − a )
v v v v
= a ⋅a − a ⋅b
= 1−1 = 0
∠OAB = 90 o , hence ∠OAB = ∠ODB = 90 o
-16-
16. (a) (i) Let f ( x) = ax 3 + bx 2 + cx + d .
Since α is a root of f ( x) = 0 , ∴ x − α is a factor of f ( x) .
Similarly , x − β and x − γ is a factor of f ( x)
Hence f ( x) = a( x − α )( x − β )( x − γ )
ax 3 + bx 2 + cx + d ≡ a( x − α )( x − β )( x − γ )
≡ ax 3 − a(α + β + γ ) x + a(αβ + αγ + βγ ) − aαβγ
Comparing coefficients of appropriate terms, we have
c
a(αβ + αγ + βγ ) = c , αβ + αγ + βγ =
a
−d
− a(αβγ ) = d , αβγ
a
(b) cos 3θ = cos(θ + 2θ )
= cosθ cos 2θ − sin θ sin 2θ
= cosθ (2 cos 2 θ − 1) − sin θ (2 sin θ cosθ )
= cosθ (2 cos 2 θ − 1) − 2 cosθ (1 − cos 2 θ )
= 4 cos 3 θ − 3 cosθ
Let x = cosθ , 8 cos 3 θ − 6 cosθ = 3
3
4 cos 3 θ − 3 cosθ =
2
3
cos 3θ =
2
3θ = 30 o , 330 o , 390 o
-17-
θ = 10 o , 110 o , 130 o
(c) cos10 o , cos110 o and cos130 o are the roots of 8 x 3 − 6 x − 3 = 0
⎛− 3⎞
(i) cos10 o cos110 o cos130 o = − ⎜⎜ ⎟
⎝ 8 ⎠
( )(
cos10 o − cos 70 o − cos 50 o ) =
8
3
3
cos10 o cos 50 o cos 70 o =
8
−6
(ii) cos10 o cos130 o + cos10 o cos110 o + cos110 o cos130 o =
8
−3
=
4
-18-
17. (i) α + β = −4 , αβ = 2
S1 = α + β = −4
S2 = α 2 + β 2
= (α + β ) − 2αβ
2
= 12
(ii) Q α , β are the roots of the equation x 2 + 4 x + 2 = 0
∴ S n + 2 + 4S n +1 + 2 S n
( ) ( ) (
= α n + 2 + β n + 2 + 4 α n +1 + β n +1 + 2 α n + β n )
( )
= α n α 2 + 4α + 2 + β n ( β 2 + 4 β + 2)
=0
(iii) Put n = 1, S 3 + 4 S 2 + 2 S1 = 0
S 3 + 4(12) − 2(− 4 ) = 0
S 3 = −40
Put n = 2, S 4 + 4S 3 + 2S 2 = 0
S 4 + 4(− 40 ) + 2(12 ) = 0
S 4 = 136
-19-
(iv) Put n = 3, S 5 + 4S 4 + 2S 3 = 0
S 5 + 4(136) + 2(− 40) = 0
S 5 = −464
Since x 2 + 4 x + 2 = 0
− 4 ± 4 2 − 4(1)(2)
x=
2(1)
x = −2 + 2 or x = −2 − 2
Q S 5 = −464
(
∴ −2+ 2 + −2− 2 ) (
5
)
5
= −464
(
− 2− 2 − 2+ 2) (
5
)
5
= −464
(2 − 2 ) + (2 + 2 )
5 5
= 464
-20- | 5,070 | 8,338 | {"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} | 3.90625 | 4 | CC-MAIN-2019-04 | latest | en | 0.721513 |
https://www.kiasuparents.com/kiasu/question/18133-2/ | 1,495,571,700,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607702.58/warc/CC-MAIN-20170523202350-20170523222350-00638.warc.gz | 911,692,871 | 18,393 | # Question
Can anyone help me with this question. Thank you
42 metal strips will make 21 of the designs, since he needs 2 metal strip to make one design.
Length wise,
1st piece is 11cm in length and subsequent 20 pieces adds 3 cm each to the length.
1st piece + 20 X 3 = 11 + 60 = 71 cm | 83 | 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} | 3.625 | 4 | CC-MAIN-2017-22 | latest | en | 0.947154 |
http://www.nelson.com/mathfocus/grade9/quizzes/ch07/mf9_ch07_02.htm | 1,542,463,156,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743521.59/warc/CC-MAIN-20181117123417-20181117145417-00489.warc.gz | 468,105,639 | 8,653 | Name: Lesson 7.2: Using Probabilities to Make Decisions
Matching
Two players spun the spinner below twice and then calculated the product of the two numbers. They did this 20 times and recorded the numbers in the table below. Player 1 won when the product was odd and Player 2 won when the product was even.
Game Number 1 2 3 4 5 6 7 8 9 10 Spin 1 5 2 3 1 2 3 4 3 5 5 Spin 2 1 4 5 3 5 1 3 5 5 2 Product 5 8 15 3 10 3 12 15 25 10
Game Number 11 12 13 14 15 16 17 18 19 20 Spin 1 3 4 2 1 1 2 2 3 5 5 Spin 2 3 2 3 3 3 2 1 3 3 2 Product 9 8 6 3 3 4 2 9 15 10
Outcome Table for the Game
´ 1 2 3 4 5 1 1 2 3 4 5 2 2 4 6 8 10 3 3 6 9 12 15 4 4 8 12 16 20 5 5 10 15 20 25
Match each question below with its answer.
a. b. c. No, because the outcome table shows that there are more even products than odd products. This means that Player 2 has more chances to win. d. e. f. Player 2, because the outcome table shows that there are more even products. g. Player 1 h. Yes, because each number on the spinner is equally likely to be landed on.
1.
What percentage of times did Player 1 win?
2.
What percentage of times did Player 2 win?
3.
Which player won the most games?
4.
What is the theoretical probability that Player 1 will win?
5.
What is the theoretical probability that Player 2 will win?
6.
Which player is more likely to win if they played another 20 games?
7.
Is the game fair? How do you know?
8.
Is the spinner fair? | 515 | 1,446 | {"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} | 4.375 | 4 | CC-MAIN-2018-47 | latest | en | 0.923337 |
https://customwritings.pro/demand-estimation-paper-4-6-pages/ | 1,618,635,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00436.warc.gz | 313,266,090 | 20,240 | April 1, 2021
###### Demonstrate synthesis of the knowledge and skills acquired in preceding graduate nursing core and specialty curriculum content.
April 1, 2021
Econ Paper
Imagine that you work for the maker of a leading brand of low-calorie, frozen microwavable food that estimates the following demand equation for its product using data from 26 supermarkets around the country for the month of April.
Option 1
Note: The following is a regression equation. Standard errors are in parentheses for the demand for widgets.
QD = – 5200 – 42P + 20PX + 5.2I + .20A + .25M
(2.002) (17.5) (6.2) (2.5) (0.09) (0.21)
R2 = 0.55 n = 26 F = 4.88
Your supervisor has asked you to compute the elasticities for each independent variable. Assume the following values for the independent variables:
Q = Quantity demanded of 3-pack units
P (in cents) = Price of the product = \$500 per 3-pack unit
PX (in cents) = Price of leading competitor’s product = \$600 per 3-pack unit
I (in dollars) = Per capita income of the standard metropolitan statistical area
(SMSA) in which the supermarkets are located = \$5,500
A (in dollars) = Monthly advertising expenditures = \$10,000
M = Number of microwave ovens sold in the SMSA in which the
supermarkets are located = 5,000
Write a four to six (4-6) page paper in which you:
1. Compute the elasticities for each independent variable. Note: Write down all of your calculations.
2. Determine the implications for each of the computed elasticities for the business in terms of short-term and long-term pricing strategies. Provide a rationale in which you cite your results.
3. Recommend whether you believe that this firm should or should not cut its price to increase its market share. Provide support for your recommendation.
4. Assume that all the factors affecting demand in this model remain the same, but that the price has changed. Further assume that the price changes are 100, 200, 300, 400, 500, 600 dollars.
a) Plot the demand curve for the firm.
b) Plot the corresponding supply curve on the same graph using the following MC / supply function Q = -7909.89 + 79.0989P with the same prices.
c) Determine the equilibrium price and quantity.
d) Outline the significant factors that could cause changes in supply and demand for the product. Determine the primary manner in which both the short-term and the long-term changes in market conditions could impact the demand for, and the supply, of the product.
5. Indicate the crucial factors that could cause rightward shifts and leftward shifts of the demand and supply curves.
6. Use at least three (3) quality academic resources in this assignment. Note: Wikipedia does not qualify as an academic resource.
· Be typed, double spaced, using Times New Roman font (size 12), with one-inch margins on all sides; citations and references must follow APA or school-specific format. Check with your professor for any additional instructions.
· Include a cover page containing the title of the assignment, the student’s name, the professor’s name, the course title, and the date. The cover page and the reference page are not included in the required assignment page length.
The specific course learning outcomes associated with this assignment are:
· Analyze how production and cost functions in the short run and long run affect the strategy of individual firms.
· Apply the concepts of supply and demand to determine the impact of changes in market conditions in the short run and long run, and the economic impact on a company’s operations.
· Use technology and information resources to research issues in managerial economics and globalization.
· Write clearly and concisely about managerial economics and globalization using proper writing mechanics.
##### Do you need a similar assignment done for you from scratch? We have qualified writers to help you. We assure you an A+ quality paper that is free from plagiarism. Order now for an Amazing Discount! Use Discount Code “Newclient” for a 15% Discount!NB: We do not resell papers. Upon ordering, we do an original paper exclusively for you.
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Answer the following questions using the information below:
The Atheist Hospital is considering the purchase of a new x-ray machine. The existing machine
is operable for five more years and will have a zero disposal price. If the machine is disposed
now, it can be sold for $90,000. 16 12 The new machine will cost$669,000. It is expected to reduce the amount of time to take x-rays
and result in an estimated $59,000 in additional cash inflows during the year of acquisition and$246,000 during each additional year of operation.
The new machine has the same five-year life as the existing one, and zero disposal value. These
cash flows will generally occur throughout the year and are recognized at the end of each year.
The hospital is a not-for-profit organization and does not pay income taxes.
Questions:
1. What is the net present value of the investment, assuming the required rate of return is 16%?
NPV:
650
20
2. Would the hospital want to purchase the new machine?
Comment if desired:
60
3. What is the internal rate of return of the investment to nearest 10th percent?
.101 is 10.1%
4. Based on the information given above, does the IRR indicate that the hospital would favor the investment?
5. Explain why or why not:
6. What is the net present value of the investment, assuming the required rate of return is 20%?
NPV:
7 Would the hospital want to purchase the new machine?
8. Explain any difference from the
response at 12%:
Space below is available for supporting calculations:
60
0.1%
Oval Tire Company needs to overhaul and keep its auto lift system or buy a new one. Either option would have a 5-year
life, as the shop's lease expires then. The facts have been gathered, and they are as follows:
Current
Machine
Purchase Price, New
Current book value
Overhaul needed now
Annual cash operating costs
Current salvage value
Salvage value in five years
New
Machine
Relevant
was
2
151,000 $148,000 3 25,500 27,500 4 61,500 63,000 5 45,000 48,000 6 38,500 40,000 7 30,000 35,000 8 16 15$112,500 $151,000 33,500 25,500 61,500 45,000 38,500 8,000 35,000 Required: Which alternative would you advise Oval to choose, given a current required rate of return of 16%? Show computations below, and assume no taxes. How much better off, in terms of present value, will Oval Tire be if they follow your advice? [Be sure this is clearly calculated and labeled below.] Area below is available for your calculations: RETURN TO WELCOME PAGE AND ACKNOWLEDGE TERMS Network Service Center is considering purchasing a new computer network for$90,000. It will require additional working capital
of $11,000. Its anticipated eight-year life will generate additional client revenue of$37,000 annually with operating costs,
excluding depreciation, of $15,000. At the end of eight years, it is expected to have a salvage value of$9,500 and return $7,000 in working capital. Taxes are not considered. Required: a. If the company has a required rate of return of 14%, what is the net present value of the proposed investment? b. c. d. What is the internal rate of return, to nearest 1/10% percentage point? .101 is 10.1% years in decimal terms to nearest 10th year. One year and 6 months = 1.5, etc. What is the payback period? Based on this limited information, what can you recommend to management? Area below is available for calculations: was 2 3 4 5 6 7 90 11 37 15 500 7 82 13 33 15 500 5 RETURN TO WELCOME PAGE AND ACKNOWLEDGE TERMS Custer Construction Company is considering four proposals for the construction of new loading facilities that will include the latest in ship loading/unloading equipment. After careful analysis, the company's accountant has developed the following information about the four proposals: Proposal 1 Proposal 2 Proposal 3 Proposal 4 Payback period 4 years 4.5 years 6 years 7 years$80,000
$178,000$166,000
\$308,000
Internal rate of return
12%
14%
11%
13%
Accrual accounting rate of return
8%
6%
4%
7%
Net present value
Required:
How can this information be used in the decision-making process for the new loading facilities? Does it cause any confusion?
Explain carefully. | 1,053 | 4,135 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.928836 |
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Posts
## Green function for linearized Navier-Stokes around boundary layers: away from critical layers
I’ve just submitted this new paper with E. Grenier (ENS de Lyon) on arxiv (scheduled to announce next Tuesday 1:00GMT), in which we construct the Green function for the classical Orr-Sommerfeld equations and derive sharp semigroup bounds for linearized Navier-Stokes equations around a boundary layer profile. This is part of the long program to understand the stability of classical Prandtl’s layers appearing in the inviscid limit of incompressible Navier-Stokes flows.
More precisely, let ${\nu}$ be the small viscosity and let ${L_\nu}$ be the linearized Navier-Stokes operator around a stationary boundary layer ${U = [U(y),0]}$ on the half-line ${y\ge 0}$, together with the classical no-slip boundary condition. Naturally, there are two cases: either ${U(y)}$ is spectrally stable or spectrally unstable to the corresponding Euler operator ${L_0}$. In this paper, we consider the unstable case, giving the existence of the maximal unstable eigenvalue ${\lambda_0}$, with ${\Re \lambda_0>0}$.
Our first main result in this paper is roughly as follows:
Theorem 1 There holds the sharp semigroup bound:
$\displaystyle \| e^{L_\nu t} \|_{\eta} \le C_\epsilon e^{( \Re \lambda_0 + \epsilon) t} , \qquad \forall~\epsilon>0$
uniformly in time ${t\ge 0}$ and uniformly in the inviscid limit ${\nu \rightarrow0}$. Here, ${\|\cdot \|_\eta}$ denotes some exponentially weighted ${L^\infty}$ norm.
Certainly, the key difficulty in such a theorem is to derive sharp bounds in term of time growth and uniform estimates in the inviscid limit. Standard energy estimates yield precisely a semigroup bound of order ${e^{\| \nabla U\|_{L^\infty} t}}$, which is far from being sharp.
In fact, such a sharp semigroup bound is a byproduct of our much delicate study on the Orr-Sommerfeld problem, the resolvent equations of the linearized Navier-Stokes. In order to accurately capture the behavior of (unbounded) vorticity on the boundary, we are obliged to derive pointwise estimates on the corresponding Green function. We follow the seminal approach of Zumbrun-Howard developed in their study of stability of viscous shocks in the system of conservation laws.
That is to say, our second main result of this paper is to provide uniform bounds on the Green function of the classical Orr-Sommerfeld problem and derive pointwise bounds on the corresponding Green function of Navier-Stokes. This paper is the first in our program of deriving sharp semigroup bounds for Navier-Stokes around a boundary layer profile. | 630 | 2,619 | {"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": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.846658 |
https://www.unitpedia.com/capsule-volume-and-area-calculator/ | 1,638,178,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00406.warc.gz | 1,172,926,631 | 10,338 | # Capsule Volume and Area Calculator
Height
Select Unit
### How To Calculate Capsule Volume?
Capsule Volume =π * radius2 * ( 4/3 * radius + height) .
ex : Rectangle have radius = 15, height = 20 .
Volume = π * 152 * ( 4/3 * 15 + 20) = 28274.33 m3
.
π = 3.14159265359
The capsule is a basic three-dimensional geometric shape consisting of a cylinder with hemispherical ends.
### How To Calculate Capsule area?
The surface area formula =2 * π * radius * ( 2 * radius + height) .
### How To Calculate Capsule lenght ?
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https://research.stlouisfed.org/fred2/series/EVLXSSNQ | 1,448,931,442,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464386.98/warc/CC-MAIN-20151124205424-00171-ip-10-71-132-137.ec2.internal.warc.gz | 842,939,361 | 18,957 | # Total Value of Loans for More than 365 Days, Small Domestic Banks
2015:Q3: 784 Million of Dollars
Quarterly, 1st Full Wk. in 2nd Mo. Of Qtr., Not Seasonally Adjusted, EVLXSSNQ, Updated: 2015-10-07 2:16 PM CDT
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For further information, please refer to the Board of Governors of the Federal Reserve System's E.2 release, online at http://www.federalreserve.gov/releases/e2/about.htm or in the footnotes of the E.2, Survey of Terms of Business Lending Release. These data are collected during the middle month of each quarter and are released in the middle of the succeeding month.
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# Classical decomposition
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• The following presentation is meant to familiarize individuals with classical decomposition. It does not contain an entirely comprehensive study of this statistical tool; however, it should make individuals aware of the benefits that classical decomposition can provide. Individuals who will benefit the most from this learning tool will have a basic background in introductory business statistics and knowledge of simple linear regression. Sources that are used throughout this presentation are cited in the notes below each slide and in the bibliography and readings list. Please contact the creator of this presentation with any questions or comments: Kurt Folke E-mail: kfolke75@hotmail.com Boise State University College of Business & Economics
• Also included in this presentation are solutions to the exercise. These solutions are available in hidden slide form in Appendix B.
• Definitions taken from … StatSoft Inc. (2003). Time Series Analysis. Retrieved April 21, 2003, from http://www.statsoft.com/textbook/sttimser.html Time series models are functions that relate time to previous values of the model. Accordingly, these models presume what has happened in the past will reoccur in the future. Further examples of possible time series data include earnings, market share, and cash flows.
• The multiplicative model Y = TCSe is the product of the trend, cyclical, seasonal, and error (or random) components at some time t . Likewise, the additive model Y = T+C+S+e is the sum of trend, cyclical, seasonal, and error (or random) components at some time t. The trend component is the gradual upward or downward movement found in a time series as a result of many possible factors (such as demand). Cyclical influences are recurrent up or down movements that last for long periods of time (longer than a year). Examples of events that might trigger a cyclical influence include recessions or booms. The seasonal component is an upward and downward movement that is repeated periodically as a result of holidays, seasons, etc. These seasonal influences may be observed as being weekly, monthly, quarterly, yearly, or some other periodic term. The error or random component of a time series is the usually small, erratic movement that does not follow a pattern and can be the result of the weather, strikes, and other unpredictable events.
• Definitions taken from … Shim, Jae K. Strategic Business Forecasting. New York: St Lucie, 2000. 269. Classical decomposition is a powerful tool for decomposing the elements of a time series model and studying each component’s sub-patterns. By analyzing each component, management can make educated decisions concerning trend and demand for future periods.
• A possible example is … classical decomposition can help a company learn when it is experiencing an abnormally high/low demand for its products.
• Steps summarized from … DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management . Homewood: Business One Irwin, 1991. 297-298. The basic steps outlined above encompass the major tasks in classical decomposition. Many sub-steps of these general tasks are shown in the explanation and illustration that follows.
• Although both multiplicative and additive time series models can be used in classical decomposition, this presentation will only include the multiplicative model as it is most commonly used. Methods for choosing between using the additive or multiplicative model can be found in … DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management . Homewood: Business One Irwin, 1991. 289-290.
• The classical decomposition demonstrated here models trend and cyclical effects together for simplicity and lack of a simple modeling technique for cyclical influences. Note: This assumption is appropriate for short-term forecasts, but forecasts for periods longer than one year should include an adjustment for cyclical influences. Just as the four-quarter moving average is used when dealing with quarterly data, likewise the 12-month moving average should be used when working with monthly data. Regardless of the periodic term used in the moving average, the outcome is the same: the seasonal influences are averaged, and therefore are neither seasonally high, nor seasonally low. Note: The hyperlinks provided in these slides will navigate the operator between the classical decomposition explanation and the classical decomposition illustration . This provides the learner with a conceptual explanation followed by actual application of the process.
• Since the simple moving average is centered at the end of one period and the beginning of the next, computing the centered moving average is necessary to ensure that the average is centered at the middle of the period. Through this process, the centered moving average is created to contain no seasonality, and therefore is the trend-cyclical component of the model.
• Using the identity Se = (Y/TC) , the seasonal-error component is derived by dividing the original data ( Y ) by the trend-cyclical data ( TC ). The trend-cyclical data is the centered moving average that was developed earlier.
• By taking the seasonal-error components and averaging them across the available periods, the unadjusted seasonal index is computed. This computation is demonstrated on slide 22. The adjusting factor is created by dividing the number of periods per year (four since the data is quarterly) by the sum of the unadjusted seasonal indexes. This ensures that the average seasonal index is one since all of the seasonal indexes must equal the number of periods in the year. If this were not done, error would be introduced into the final forecast.
• The adjusted seasonal index is the product of the unadjusted seasonal index and the adjusting factor. A quarterly adjusted seasonal index of 0.942 suggests the data was 94.2 percent of the typical trend-cyclical value during the quarter. Accordingly, adjusted seasonal indexes greater than one indicate that the data was higher than the typical trend-cyclical values for that quarter.
• To create the deseasonalized data, the original data values ( Y ) must be divided by their appropriate seasonal indexes ( S ). Once the deseasonalized data is computed, it can be analyzed to identify true fluctuations in the time series. These fluctuations can help management in strategic planning.
• By using simple linear regression, the trend of the time series can be estimated. This process is done by using the deseasonalized values to create a trend-cyclical regression equation of the form Tt = a + bt…where t is equal to the period. Hence, t = 1 refers to year 1, quarter 1. Although actual simple linear regression computations are not shown, recommendations are to use Excel or Minitab when creating the trend-cyclical regression equations from the deseasonalized data.
• Using the trend-cyclical regression equation, the trend data can be created by imputing each period’s assigned number into the equation. Note: At this point, it would be suitable to check the model to see how closely it fits the data. This process is omitted from this demonstration; however, in practice it should be performed before continuing. A demonstration of this is presented in … DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management . Homewood: Business One Irwin, 1991. 296-297. The final forecast is developed by multiplying the trend values by their appropriate seasonal indexes. This produces a more accurate forecast for management. Note: As previously mentioned, this method is appropriate for short-term forecasts, but forecasts for periods longer than one year should include an adjustment for cyclical influences.
• Steps summarized from … DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management . Homewood: Business One Irwin, 1991. 297-298.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Use the hyperlink to navigate to the explanation slide that includes conceptual details in the notes.
• Graphing the trend, original, and deseasonalized data can be very helpful for identifying fluctuations in trend. Deviations from the norm can be invaluable knowledge for management to analyze and use for planning future capacity, production, and allocations of resources.
• This example provides individuals with the opportunity to apply the new skills they have learned through this presentation. It is highly recommended that a spreadsheet program such as Excel or Minitab be used for computations and for building the trend-cyclical regression equation. In Excel, simple linear regression can be performed by going to Tools , Data Analysis , and using the Regression tool. Preformatted Excel templates have been created for this exercise and are available in Appendix A. Solutions for all steps are presented in hidden slides in Appendix B.
• Definitions taken from … StatSoft Inc. (2003). Time Series Analysis. Retrieved April 21, 2003, from http://www.statsoft.com/textbook/sttimser.html Time series models are based on the assumption that what has happened in the past will reoccur in the future. Classical decomposition can be used to segregate the elements of a time series model; after studying each component’s sub-patterns, management can apply the new learned knowledge when making decisions regarding strategic planning.
• The sources provided in the bibliography and readings list are highly recommended to individuals wishing to expand their knowledge in classical decomposition and similar statistical tools.
• Directions for use: Double-click on the desired table Highlight the cells of the table Select “copy” from the right-click pop-up menu or the Edit pull-down menu Open a spreadsheet program Paste the table into the spreadsheet program
• Directions for use: Double-click on the desired table Highlight the cells of the table Select “copy” from the right-click pop-up menu or the Edit pull-down menu Open a spreadsheet program Paste the table into the spreadsheet program
• Directions for use: Double-click on the desired table Highlight the cells of the table Select “copy” from the right-click pop-up menu or the Edit pull-down menu Open a spreadsheet program Paste the table into the spreadsheet program
• Directions for use: Double-click on the desired table Highlight the cells of the table Select “copy” from the right-click pop-up menu or the Edit pull-down menu Open a spreadsheet program Paste the table into the spreadsheet program
• Directions for use: Double-click on the desired table Highlight the cells of the table Select “copy” from the right-click pop-up menu or the Edit pull-down menu Open a spreadsheet program Paste the table into the spreadsheet program
• To show these slides in the presentation: Select the Normal View tab In the left-hand screen, select slide From the Slide Show pull down menu, press Hide Slide
• To show these slides in the presentation: Select the Normal View tab In the left-hand screen, select slide From the Slide Show pull down menu, press Hide Slide
• To show these slides in the presentation: Select the Normal View tab In the left-hand screen, select slide From the Slide Show pull down menu, press Hide Slide
• To show these slides in the presentation: Select the Normal View tab In the left-hand screen, select slide From the Slide Show pull down menu, press Hide Slide
• To show these slides in the presentation: Select the Normal View tab In the left-hand screen, select slide From the Slide Show pull down menu, press Hide Slide
• ### Classical decomposition
1. 1. Classical Decomposition Boise State University By: Kurt Folke Spring 2003
2. 2. Overview: <ul><li>Time series models & classical decomposition </li></ul><ul><li>Brainstorming exercise </li></ul><ul><li>Classical decomposition explained </li></ul><ul><li>Classical decomposition illustration </li></ul><ul><li>Exercise </li></ul><ul><li>Summary </li></ul><ul><li>Bibliography & readings list </li></ul><ul><li>Appendix A: exercise templates </li></ul>
3. 3. Time Series Models & Classical Decomposition <ul><li>Time series models are sequences of data that follow non-random orders </li></ul><ul><li>Examples of time series data: </li></ul><ul><ul><li>Sales </li></ul></ul><ul><ul><li>Costs </li></ul></ul><ul><li>Time series models are composed of trend, seasonal, cyclical, and random influences </li></ul>
4. 4. Time Series Models & Classical Decomposition <ul><li>Decomposition time series models: </li></ul><ul><li>Multiplicative: Y = T x C x S x e </li></ul><ul><li>Additive: Y = T + C + S + e </li></ul><ul><li>T = Trend component </li></ul><ul><li>C = Cyclical component </li></ul><ul><li>S = Seasonal component </li></ul><ul><li>e = Error or random component </li></ul>
5. 5. Time Series Models & Classical Decomposition <ul><li>Classical decomposition is used to isolate trend, seasonal, and other variability components from a time series model </li></ul><ul><li>Benefits: </li></ul><ul><ul><li>Shows fluctuations in trend </li></ul></ul><ul><ul><li>Provides insight to underlying factors affecting the time series </li></ul></ul>
6. 6. Brainstorming Exercise <ul><li>Identify how this tool can be used in your organization… </li></ul>
7. 7. Classical Decomposition Explained <ul><li>Basic Steps: </li></ul><ul><li>Determine seasonal indexes using the ratio to moving average method </li></ul><ul><li>Deseasonalize the data </li></ul><ul><li>Develop the trend-cyclical regression equation using deseasonalized data </li></ul><ul><li>Multiply the forecasted trend values by their seasonal indexes to create a more accurate forecast </li></ul>
8. 8. Classical Decomposition Explained: Step 1 <ul><li>Determine seasonal indexes </li></ul><ul><li>Start with multiplicative model… </li></ul><ul><li>Y = TCSe </li></ul><ul><li>Equate… </li></ul><ul><li>Se = (Y/TC) </li></ul>
9. 9. Classical Decomposition Explained: Step 1 <ul><li>To find seasonal indexes, first estimate trend-cyclical components </li></ul><ul><li>Se = (Y/ TC ) </li></ul><ul><li>Use centered moving average </li></ul><ul><ul><li>Called ratio to moving average method </li></ul></ul><ul><li>For quarterly data, use four-quarter moving average </li></ul><ul><ul><li>Averages seasonal influences </li></ul></ul>Example
10. 10. Classical Decomposition Explained: Step 1 <ul><li>Four-quarter moving average will position average at… </li></ul><ul><ul><ul><li>end of second period and </li></ul></ul></ul><ul><ul><ul><li>beginning of third period </li></ul></ul></ul><ul><li>Use centered moving average to position data in middle of the period </li></ul>Example
11. 11. Classical Decomposition Explained: Step 1 <ul><li>Find seasonal-error components by dividing original data by trend-cyclical components </li></ul><ul><li>Se = ( Y/TC ) </li></ul><ul><li>Se = Seasonal-error components </li></ul><ul><li>Y = Original data value </li></ul><ul><li>TC = Trend-cyclical components </li></ul><ul><li>(centered moving average value) </li></ul>Example
12. 12. Classical Decomposition Explained: Step 1 <ul><li>Unadjusted seasonal indexes (USI) are found by averaging seasonal-error components by period </li></ul><ul><li>Develop adjusting factor (AF) so USIs are adjusted so their sum equals the number of quarters (4) </li></ul><ul><ul><li>Reduces error </li></ul></ul>Example Example
13. 13. Classical Decomposition Explained: Step 1 <ul><li>Adjusted seasonal indexes (ASI) are derived by multiplying the unadjusted seasonal index by the adjusting factor </li></ul><ul><li>ASI = USI x AF </li></ul><ul><li>ASI = Adjusted seasonal index </li></ul><ul><li>USI = Unadjusted seasonal index </li></ul><ul><li>AF = Adjusting factor </li></ul>Example
14. 14. Classical Decomposition Explained: Step 2 <ul><li>Deseasonalized data is produced by dividing the original data values by their seasonal indexes </li></ul><ul><li>( Y/S ) = TCe </li></ul><ul><li>Y/S = Deseasonalized data </li></ul><ul><li>TCe = Trend-cyclical-error component </li></ul>Example
15. 15. Classical Decomposition Explained: Step 3 <ul><li>Develop the trend-cyclical regression equation using deseasonalized data </li></ul><ul><li> T t = a + bt </li></ul><ul><li>T t = Trend value at period t </li></ul><ul><li>a = Intercept value </li></ul><ul><li>b = Slope of trend line </li></ul>Example
16. 16. Classical Decomposition Explained: Step 4 <ul><li>Use trend-cyclical regression equation to develop trend data </li></ul><ul><li>Create forecasted data by multiplying the trend data values by their seasonal indexes </li></ul><ul><ul><li>More accurate forecast </li></ul></ul>Example Example
17. 17. Classical Decomposition Explained: Step Summary <ul><li>Summarized Steps: </li></ul><ul><li>Determine seasonal indexes </li></ul><ul><li>Deseasonalize the data </li></ul><ul><li>Develop the trend-cyclical regression equation </li></ul><ul><li>Create forecast using trend data and seasonal indexes </li></ul>
18. 18. Classical Decomposition: Illustration <ul><li>Gem Company’s operations department has been asked to deseasonalize and forecast sales for the next four quarters of the coming year </li></ul><ul><li>The Company has compiled its past sales data in Table 1 </li></ul><ul><li>An illustration using classical decomposition will follow </li></ul>
19. 19. Classical Decomposition Illustration: Step 1 <ul><li>(a) Compute the four-quarter simple moving average </li></ul><ul><li>Ex: simple MA at end of Qtr 2 and beginning of Qtr 3 </li></ul><ul><li>(55+47+65+70)/4 = 59.25 </li></ul>Explain
20. 20. Classical Decomposition Illustration: Step 1 <ul><li>(b) Compute the two-quarter centered moving average </li></ul><ul><li>Ex: centered MA at middle of Qtr 3 </li></ul><ul><li>(59.25+61.25)/2 </li></ul><ul><li>= 60.500 </li></ul>Explain
21. 21. Classical Decomposition Illustration: Step 1 <ul><li>(c) Compute the seasonal-error component (percent MA) </li></ul><ul><li>Ex: percent MA at Qtr 3 </li></ul><ul><li>(65/60.500) </li></ul><ul><li>= 1.074 </li></ul>Explain
22. 22. Classical Decomposition Illustration: Step 1 <ul><li>(d) Compute the unadjusted seasonal index using the seasonal-error components from Table 2 </li></ul><ul><li>Ex (Qtr 1): [(Yr 2, Qtr 1) + (Yr 3, Qtr 1) + (Yr 4, Qtr 1)]/3 </li></ul><ul><li>= [0.989+0.914+0.926]/3 = 0.943 </li></ul>Explain
23. 23. Classical Decomposition Illustration: Step 1 <ul><li>(e) Compute the adjusting factor by dividing the number of quarters (4) by the sum of all calculated unadjusted seasonal indexes </li></ul><ul><li>= 4.000/(0.943+0.851+1.080+1.130) = (4.000/4.004) </li></ul>Explain
24. 24. Classical Decomposition Illustration: Step 1 <ul><li>(f) Compute the adjusted seasonal index by multiplying the unadjusted seasonal index by the adjusting factor </li></ul><ul><li>Ex (Qtr 1): 0.943 x (4.000/4.004) = 0.942 </li></ul>Explain
25. 25. Classical Decomposition Illustration: Step 2 <ul><li>Compute the deseasonalized sales by dividing original sales by the adjusted seasonal index </li></ul><ul><li>Ex (Yr 1, Qtr 1): </li></ul><ul><li>(55 / 0.942) </li></ul><ul><li>= 58.386 </li></ul>Explain
26. 26. Classical Decomposition Illustration: Step 3 <ul><li>Compute the trend-cyclical regression equation using simple linear regression </li></ul><ul><li>T t = a + bt </li></ul><ul><li>t-bar = 8.5 </li></ul><ul><li>T-bar = 69.6 </li></ul><ul><li>b = 1.465 </li></ul><ul><li>a = 57.180 </li></ul><ul><li>T t = 57.180 + 1.465t </li></ul>Explain
27. 27. Classical Decomposition Illustration: Step 4 <ul><li>(a) Develop trend sales </li></ul><ul><li>T t = 57.180 + 1.465t </li></ul><ul><li>Ex (Yr 1, Qtr 1): </li></ul><ul><li>T 1 = 57.180 + 1.465(1) = 58.645 </li></ul>Explain
28. 28. Classical Decomposition Illustration: Step 4 <ul><li>(b) Forecast sales for each of the four quarters of the coming year </li></ul><ul><li>Ex (Yr 5, Qtr 1): </li></ul><ul><li>0.942 x 82.085 </li></ul><ul><li>= 77.324 </li></ul>Explain
29. 29. Classical Decomposition Illustration: Graphical Look
30. 30. Classical Decomposition: Exercise <ul><li>Assume you have been asked by your boss to deseasonalize and forecast for the next four quarters of the coming year (Yr 5) this data pertaining to your company’s sales </li></ul><ul><li>Use the steps and examples shown in the explanation and illustration as a reference </li></ul><ul><li>Basic Steps </li></ul><ul><li>Explanation </li></ul><ul><li>Illustration </li></ul><ul><li>Templates </li></ul>
31. 31. Summary <ul><li>Time series models are sequences of data that follow non-arbitrary orders </li></ul><ul><li>Classical decomposition isolates the components of a time series model </li></ul><ul><li>Benefits: </li></ul><ul><ul><li>Insight to fluctuations in trend </li></ul></ul><ul><ul><li>Decomposes the underlying factors affecting the time series </li></ul></ul>
32. 32. Bibliography & Readings List <ul><li>DeLurgio, Stephen, and Bhame, Carl. Forecasting Systems for Operations Management . Homewood: Business One Irwin, 1991. </li></ul><ul><li>Shim, Jae K. Strategic Business Forecasting . New York: St Lucie, 2000. </li></ul><ul><li>StatSoft Inc. (2003). Time Series Analysis. Retrieved April 21, 2003, from http://www.statsoft.com/textbook/sttimser.html </li></ul>
33. 33. Appendix A: Exercise Templates
34. 34. Appendix A: Exercise Templates
35. 35. Appendix A: Exercise Templates
36. 36. Appendix A: Exercise Templates
37. 37. Appendix A: Exercise Templates
38. 38. Appendix B: Exercise Solutions
39. 39. Appendix B: Exercise Solutions
40. 40. Appendix B: Exercise Solutions
41. 41. Appendix B: Exercise Solutions Trend-cyclical Regression Equation T t = 5.402 + 0.514t
42. 42. Appendix B: Exercise Solutions | 5,666 | 22,936 | {"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} | 3.15625 | 3 | CC-MAIN-2016-22 | latest | en | 0.878527 |
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Lecture 16
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Stat 1400 4.20.2017 9:30 am Homework due Tuesday (425) Student Question: How do you set up homework question 9.2.4? ~ ~ ~ Answer: CI : p z* p(1 p) (n+4) The leastsquares regression line The leastsquares regression line is the unique line such that the sum of the vertical distances between the data points and the line is zero, and tsquaredof the vertical distances is the smallest possible. o Square the value to account for negative values Residualso The vertical distances from each point to the leastsquares regression line are called residuals. The sum of all the residuals is by definition 0. Outliers have unusually large residuals (in absolute value). Points above the line have a positive Points below the line have a residual negative residual (over estimation). (under estimation). ^ Predicted y Observed y dist. (y y)residual 1
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So we can recommend you notes for your school. | 363 | 1,432 | {"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-2018-13 | latest | en | 0.836323 |
https://1hotelsouthbeachforsale.com/qa/quick-answer-how-do-you-show-correlation.html | 1,627,338,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00249.warc.gz | 91,758,430 | 8,125 | # Quick Answer: How Do You Show Correlation?
## How do you tell if a correlation is strong or weak?
r > 0 indicates a positive association.
r < 0 indicates a negative association.
Values of r near 0 indicate a very weak linear relationship.
The strength of the linear relationship increases as r moves away from 0 toward -1 or 1..
## How do you represent a correlation?
Pearson’s correlation coefficient is represented by the Greek letter rho (ρ) for the population parameter and r for a sample statistic. This correlation coefficient is a single number that measures both the strength and direction of the linear relationship between two continuous variables.
## How would you visually represent a correlation coefficient?
A plot of paired data points on an x- and a y-axis, used to visually represent a correlation. Who is responsible for the invention of the Pearson product-moment correlation coefficient?
## What does an R squared value of 0.9 mean?
r is always between -1 and 1 inclusive. The R-squared value, denoted by R 2, is the square of the correlation. It measures the proportion of variation in the dependent variable that can be attributed to the independent variable. … Correlation r = 0.9; R=squared = 0.81. Small positive linear association.
## What are the 5 types of correlation?
CorrelationPearson Correlation Coefficient.Linear Correlation Coefficient.Sample Correlation Coefficient.Population Correlation Coefficient.
## What makes a weak correlation?
A weak correlation means that as one variable increases or decreases, there is a lower likelihood of there being a relationship with the second variable. … Earthquake magnitude and the depth at which it was measured is therefore weakly correlated, as you can see the scatter plot is nearly flat.
## What is a good r2?
R-squared should accurately reflect the percentage of the dependent variable variation that the linear model explains. Your R2 should not be any higher or lower than this value. … However, if you analyze a physical process and have very good measurements, you might expect R-squared values over 90%.
## What does an R squared value of 0.3 mean?
– if R-squared value < 0.3 this value is generally considered a None or Very weak effect size, - if R-squared value 0.3 < r < 0.5 this value is generally considered a weak or low effect size, ... - if R-squared value r > 0.7 this value is generally considered strong effect size, Ref: Source: Moore, D. S., Notz, W.
## How do you explain a weak negative correlation?
Negative correlation or inverse correlation is a relationship between two variables whereby they move in opposite directions. If variables X and Y have a negative correlation (or are negatively correlated), as X increases in value, Y will decrease; similarly, if X decreases in value, Y will increase.
## How do you show correlation on a graph?
How to plot a correlation graph in ExcelSelect two columns with numeric data, including column headers. … On the Inset tab, in the Chats group, click the Scatter chart icon. … Right click any data point in the chart and choose Add Trendline… from the context menu.
## What is correlation on a graph?
Graphs can either have positive correlation, negative correlation or no correlation. Positive correlation means as one variable increases, so does the other variable. They have a positive connection. Negative correlation means as one variable increases, the other variable decreases.
## What are the 3 types of scatter plots?
With scatter plots we often talk about how the variables relate to each other. This is called correlation. There are three types of correlation: positive, negative, and none (no correlation). Positive Correlation: as one variable increases so does the other.
## What does R 2 tell you?
R-squared is a statistical measure of how close the data are to the fitted regression line. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. … 100% indicates that the model explains all the variability of the response data around its mean.
## How do you find the correlation between two variables?
We often see patterns or relationships in scatterplots. When the y variable tends to increase as the x variable increases, we say there is a positive correlation between the variables. When the y variable tends to decrease as the x variable increases, we say there is a negative correlation between the variables. | 924 | 4,479 | {"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-2021-31 | latest | en | 0.931321 |
http://fer3.com/arc/m2.aspx/18-june-2013-lunar-distance-Hirose-jul-2013-g24555 | 1,638,155,760,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00573.warc.gz | 28,115,022 | 5,932 | # NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
### Compose Your Message
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Re: 18 june 2013 lunar distance
From: Paul Hirose
Date: 2013 Jul 02, 16:46 -0700
```Dave Walden wrote:
> The additional places are essentially 'free' in modern computers, and are correct, so why not?
>
> They amuse some of us!
When people explore the limits of their software and compare results,
there can be benefits beyond amusement. In December 2010 we had a long
discussion in the HASTRO-L group about whether or not a lunar
occultation of Spica was visible from Alexandria in November 283 BC. One
person using the Swiss Ephemeris software had a .0005° discrepancy in
Spica's position. It was due to a previously undetected bug in the
processing of star catalog data. Right ascension proper motion in the
Hipparcos catalog is expressed in great circle units, but the Swiss
Ephemeris was coded as if they were coordinate units. That was quickly
fixed.
> Using USNO NOVA's, DE405, SIMBAD data for Altair, and Paul's "Delta T
= 32.184 + 35 - 0.0661627", I get:
>
> true xld= in deg
> + 106 deg 34' 28.466124" OR 106 deg 34.474435' OR 106.574574deg
Note that each angle above has 60 times less precision that the one to
its left. A good rule of thumb is to use the same total number of
digits. For instance, DDDMMSS.s and DDD.ddddd have the same number of
digits and about the same precision.
In this case, Lunar3 says 106°34′28.453″. That's 13 mas different from
Dave. (2013-06-18 0330 UT1, 67.1178 delta T, DE406 ephemeris, Hipparcos
second reduction)
I think 1 mas is close enough, but 13 is too much. My use of DE406
should not matter. It's the same integration as DE405 but stored in a
more compact format, "only" accurate to 1 meter for the Moon. (That's
how well DE406 duplicates the original integration, not its accuracy
with respect to the true position of the body.)
What about SIMBAD vs. Lunar3's internal star catalog? With the SIMBAD data
I get 106°34′28.454″, only 1 mas more.
Is the discrepancy in the Moon position, or Altair? I would like to
compare the geocentric apparent places of both bodies. And to eliminate
any question about the precession / nutation model, I wonder if Dave can
put the coordinates in the GCRS. From Lunar3 I get Moon coordinates
-2.44969548e-003 -5.24187596e-004 -3.58294080e-004 au, or
192° 04′ 40.973″ -08° 08′ 21.886″
and Altair coordinates
4.59325899e-001 -8.74784757e-001 1.54179599e-001 au, or
297° 42′ 09.913″ +08° 52′ 09.188″
--
I filter out messages with attachments or HTML.
```
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Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd) | 904 | 3,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.913956 |
https://en.m.wikibooks.org/wiki/Differential_Geometry/Vector_of_Darboux | 1,637,978,037,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00470.warc.gz | 314,846,250 | 6,900 | # Differential Geometry/Vector of Darboux
We consider the rotation of the body formed by the three vectors t, p, and b. We define the rotation vector d as follows:
1. d is in the direction of the axis of rotation.
2. d points in the direction such that the rotation is clockwise in that direction.
3. |d| has the value of the angular velocity of the rigid body, or the velocity when at a distance of 1 away from the axis of rotation. | 103 | 435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.917307 |
https://www.vedantu.com/question-answer/solve-the-differential-equation-left-ex+1-class-11-maths-icse-5ee7101e47f3231af26a22d9 | 1,721,056,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00442.warc.gz | 908,026,774 | 30,255 | Courses
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# Solve the differential equation $\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0$.
Last updated date: 15th Jul 2024
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Hint: Use a variable separable method to solve the given differential equation. Integrate with respect to ‘dx’ and ‘dy’ to both sides and simplify it to get the solution of the given differential equation.
We have
$\left( {{e}^{x}}+1 \right)ydy+\left( y+1 \right)dx=0.............\left( i \right)$
Dividing the whole equation by dx, we get,
$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)\dfrac{dx}{dx}=0$
Or
$\left( {{e}^{x}}+1 \right)y\dfrac{dy}{dx}+\left( y+1 \right)=0..........(ii)$
Now, as we know, we have three types of differential equations i.e., separable, linear and homogeneous.
Now, by observation, we get that if we divide equation (ii), by ‘y’ then we can separate variables ‘x’ and ‘y’ easily. Hence, the given differential equation belongs to a separable type.
So, on dividing equation (ii), we get
$\left( {{e}^{x}}+1 \right)\dfrac{y}{y}\dfrac{dy}{dx}+\left( \dfrac{y+1}{y} \right)=0$
Or
$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}+\dfrac{y+1}{y}=0$
Now transferring $\dfrac{y+1}{y}$ to other side, we get
$\left( {{e}^{x}}+1 \right)\dfrac{dy}{dx}=-\dfrac{\left( y+1 \right)}{y}..........(iii)$
Now, we can transfer functions of variable ‘x’ to one side and functions of variable ‘y’ to another side to integrate the equation with respect to ‘dx’ and ‘dy’.
So, equation (iii) can be written as
$\left( \dfrac{y}{y+1} \right)dy=\dfrac{-1}{{{e}^{x}}+1}dx$
Now, we observe that variable are easily separated, so we can integrate them with respect to
‘x’ and ‘y’ hence, we get
$\int{\dfrac{y}{y+1}dy=-\int{\dfrac{1}{{{e}^{x}}+1}dx............(iv)}}$
Let ${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$ and ${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$
Let us solve both the integration individually.
So, we have ${{I}_{1}}$ as,
${{I}_{1}}=\int{\dfrac{y}{y+1}dy}$
Adding and subtracting ‘1’ in numerator, we get,
${{I}_{1}}=\int{\dfrac{\left( y+1 \right)-1}{\left( y+1 \right)}dy}$
Now, we can separate (y+1) as
${{I}_{1}}=\int{\dfrac{y+1}{y+1}dy}-\int{\dfrac{1}{y+1}dy}$
Or
${{I}_{1}}=\int{1dy}-\int{\dfrac{1}{y+1}dy}$
As we know,
\begin{align} & \int{\dfrac{1}{x}dx}=\ln x \\ & \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\ \end{align}
So, ${{I}_{1}}$ can be simplified as,
${{I}_{1}}=\int{{{y}^{0}}dy-\ln \left( y+1 \right)+{{C}_{1}}}$
${{I}_{1}}=y-\ln \left( y+1 \right)+{{C}_{1}}............\left( v \right)$
Now, we have ${{I}_{2}}$ as
${{I}_{2}}=\int{\dfrac{1}{{{e}^{x}}+1}dx}$
Multiplying by ${{e}^{-x}}$ in numerator and denominator we get,
${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}.{{e}^{-x}}+{{e}^{-x}}}dx}$
As we have property of surds as, ${{m}^{a}}.{{m}^{b}}={{m}^{a+b}}$
So, we can write ${{I}_{2}}$ as
${{I}_{2}}=\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}..............\left( vi \right)$
Let us suppose $1+{{e}^{-x}}=t$.
Differentiating both sides w.r.t. x, we get
$-{{e}^{-x}}=\dfrac{dt}{dx}$
Where $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$, so, we have
${{e}^{-x}}dx=-dt$
Substituting these values in equation (vi), we get
${{I}_{2}}=\int{\dfrac{-dt}{t}=-\int{\dfrac{dt}{t}}}$
Now, we know that $\int{\dfrac{1}{t}dt=\ln t}$ hence,
${{I}_{2}}=-\ln t+{{C}_{2}}$
Since, we have value of t as $1+{{e}^{-x}}$, hence ${{I}_{2}}$ in terms of x can be given as
${{I}_{2}}=-\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}}.............\left( vii \right)$
Hence, from equation (iv), (v) and (vii) we get,
\begin{align} & y-\ln \left( y+1 \right)+{{C}_{1}}=-\left( -\ln \left( 1+{{e}^{-x}} \right)+{{C}_{2}} \right) \\ & y-\ln \left( y+1 \right)+{{C}_{1}}=\ln \left( 1+{{e}^{-x}} \right)-{{C}_{2}} \\ & y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+{{C}_{3}} \\ \end{align}
Where $-{{C}_{2}}-{{C}_{1}}={{C}_{3}}$
Let us replace ${{C}_{3}}$ by ‘ln C’, so we get above equation as
$y-\ln \left( y+1 \right)=\ln \left( 1+{{e}^{-x}} \right)+\ln C........\left( viii \right)$
Now, we know that $\text{ln }a+\ln b=\ln ab$, so, equation (viii) can be given as
$y=\ln \left( y+1 \right)+\ln \left( C\left( 1+{{e}^{-x}} \right) \right)$
$y=\ln \left( C\left( y+1 \right)\left( 1+{{e}^{-x}} \right) \right)$
As we know that if ${{a}^{x}}=N$ then $x={{\log }_{a}}N$ or vice versa.
Hence, above equation can be written as
$C\left( y+1 \right)\left( 1+{{e}^{-x}} \right)={{e}^{y}}$
This is the required solution.
Note: One can go wrong if trying to solve the given differential equation by a homogenous method, as the given equation is not a homogeneous differential equation so we cannot apply this method to a given variable separable differential equation.
Observing the given differential equation as a variable separable equation is the key point of the question. | 1,900 | 4,893 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.767627 |
https://dsearls.org/courses/M131FiniteMath/Samples/Test1.htm | 1,702,173,015,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00696.warc.gz | 217,540,463 | 3,622 | ## MAT 131 Sample Test 1
The questions below are representative of the kinds of questions that will be on the test. The questions do not represent an exhaustive list of the specific types of questions that might be asked.
1. What is the present value of an investment worth \$500 eight years from now if it pays 6.2% interest compounded monthly? How much interest would you earn over the term of this investment? What is the annual percentage yield for this investment? Generate a table showing how the balance changes over the first two periods.
2. If you invested \$120 a month into a savings account paying 3.4% compounded monthly, how much would you have in ten years and how much interest would you have earned? What is the annual percentage yield for this investment? Generate a table showing how the balance changes over the first two periods.
3. How much could you borrow on a four year loan at 6.9% compounded monthly if you could afford to pay \$250 a month? How much interest would you pay? What is the annual percentage rate for this loan? Generate a table showing how the balance changes over the first two periods.
4. If you deposited \$1,000 in a savings account paying 5% compounded semiannually, how much would you have in 60 years and how much interest would you have earned? What is the annual percentage yield? Generate a table showing how the balance changes over the first two periods.
5. How much would you have to invest each quarter at 4.1% compounded quarterly if your goal was to have \$50,000 in ten years? How much interest would you earn? What is your annual percentage yield? Generate a table showing how the balance changes over the first two periods.
6. What would be your monthly payments on a \$200,000 25-year mortgage at 7.2% compounded monthly? How much interest would you pay? What is the annual percentage rate for this loan? Generate a table showing how the balance due changes over the first two periods.
7. Supposed you saved \$400 a month for 30 years in a retirement fund earning 5% compounded monthly. When you retire, you elect to take your retirement as a guaranteed 20-year annuity. That is, you (or your heir, in the event of your death) will receive a monthly payment every month for 20 years at which time the account will be empty. What will be the size of the monthly payment? How much interest will this account have earned over its entire 50-year span?
8. A couple finances a new house with a 25-year, \$150,000 mortgage at 5.67% compounded monthly. How much is the monthly payment on this mortgage? After four years, the interest rate on the mortgage goes up to 8.05%. How much is the new monthly payment?
9. Suppose a man invested \$5,000 ten years ago. The value of that investment today is \$10,000. If interest is compounded quarterly, what nominal annual rate of interest is this investment earning?
10. If a couple puts \$450 a month into an account paying 1.3% compounded monthly, how long would it take to save up \$180,000? | 666 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-50 | latest | en | 0.969875 |
https://aprove.informatik.rwth-aachen.de/eval/JBC-Recursion/Julia_10_Recursive/HanR.jar.AProVE%202011.html | 1,726,383,317,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00257.warc.gz | 81,316,141 | 6,676 | ### (0) Obligation:
JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_20 (Apple Inc.) Main-Class: HanoiR
`public class HanoiR { private void solve(int h, int from, int to, int support) { if (h < 1) return; else if (h == 1) { //System.out.println("from " + from + " to " + to + "\n"); } else { solve(h - 1, from, support, to); //System.out.println("from " + from + " to " + to + "\n"); solve(h - 1, support, to, from); } } public static void main(String[] args) { Random.args = args; new HanoiR().solve(Random.random(),1,2,3); }}public class Random { static String[] args; static int index = 0; public static int random() { if (index >= args.length) return 0; String string = args[index]; index++; return string.length(); }}`
### (1) JBC2FIG (SOUND transformation)
Constructed FIGraph.
### (2) Obligation:
FIGraph based on JBC Program:
HanoiR.main([Ljava/lang/String;)V: Graph of 97 nodes with 0 SCCs.
HanoiR.solve(IIII)V: Graph of 45 nodes with 0 SCCs.
### (3) FIGtoITRSProof (SOUND transformation)
Transformed FIGraph SCCs to IDPs. Logs:
Log for SCC 0:
Generated 33 rules for P and 12 rules for R.
Combined rules. Obtained 3 rules for P and 2 rules for R.
Filtered ground terms:
690_0_solve_ConstantStackPush(x1, x2, x3) → 690_0_solve_ConstantStackPush(x2, x3)
Cond_746_1_solve_InvokeMethod1(x1, x2, x3, x4) → Cond_746_1_solve_InvokeMethod1(x1, x3, x4)
886_0_solve_Return(x1) → 886_0_solve_Return
Cond_746_1_solve_InvokeMethod(x1, x2, x3, x4) → Cond_746_1_solve_InvokeMethod(x1, x3)
723_0_solve_Return(x1) → 723_0_solve_Return
Cond_690_0_solve_ConstantStackPush(x1, x2, x3, x4) → Cond_690_0_solve_ConstantStackPush(x1, x3, x4)
Filtered duplicate args:
690_0_solve_ConstantStackPush(x1, x2) → 690_0_solve_ConstantStackPush(x2)
Cond_690_0_solve_ConstantStackPush(x1, x2, x3) → Cond_690_0_solve_ConstantStackPush(x1, x3)
Filtered unneeded arguments:
Cond_746_1_solve_InvokeMethod1(x1, x2, x3) → Cond_746_1_solve_InvokeMethod1(x1, x2)
Combined rules. Obtained 3 rules for P and 2 rules for R.
Finished conversion. Obtained 3 rules for P and 2 rules for R. System has predefined symbols.
### (4) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
836_1_solve_InvokeMethod(723_0_solve_Return, 1) → 886_0_solve_Return
836_1_solve_InvokeMethod(886_0_solve_Return, x0) → 886_0_solve_Return
The integer pair graph contains the following rules and edges:
(0): 690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(x0[0] > 1, x0[0])
(1): COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(x0[1] - 1), x0[1], x0[1] - 1)
(2): COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2]) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[2] - 1)
(3): 746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1) → COND_746_1_SOLVE_INVOKEMETHOD(x0[3] > 0, 723_0_solve_Return, x0[3], 1)
(4): COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[4] - 1)
(5): 746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5]) → COND_746_1_SOLVE_INVOKEMETHOD1(x0[5] > 0, 886_0_solve_Return, x0[5], x1[5])
(6): COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6]) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[6] - 1)
(0) -> (1), if ((x0[0] > 1* TRUE)∧(x0[0]* x0[1]))
(0) -> (2), if ((x0[0] > 1* TRUE)∧(x0[0]* x0[2]))
(1) -> (3), if ((690_0_solve_ConstantStackPush(x0[1] - 1) →* 723_0_solve_Return)∧(x0[1]* x0[3])∧(x0[1] - 1* 1))
(1) -> (5), if ((690_0_solve_ConstantStackPush(x0[1] - 1) →* 886_0_solve_Return)∧(x0[1]* x0[5])∧(x0[1] - 1* x1[5]))
(2) -> (0), if ((x0[2] - 1* x0[0]))
(3) -> (4), if ((x0[3] > 0* TRUE)∧(x0[3]* x0[4]))
(4) -> (0), if ((x0[4] - 1* x0[0]))
(5) -> (6), if ((x0[5] > 0* TRUE)∧(x0[5]* x0[6])∧(x1[5]* x1[6]))
(6) -> (0), if ((x0[6] - 1* x0[0]))
The set Q consists of the following terms:
836_1_solve_InvokeMethod(723_0_solve_Return, 1)
836_1_solve_InvokeMethod(886_0_solve_Return, x0)
### (5) IDPNonInfProof (SOUND transformation)
The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.
For Pair 690_0_SOLVE_CONSTANTSTACKPUSH(x0) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0, 1), x0) the following chains were created:
• We consider the chain 690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0]), COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1)) which results in the following constraint:
(1) (>(x0[0], 1)=TRUEx0[0]=x0[1]690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])∧(UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥))
We simplified constraint (1) using rule (IV) which results in the following new constraint:
(2) (>(x0[0], 1)=TRUE690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])∧(UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥))
We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6) (x0[0] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[(3)bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
• We consider the chain 690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0]), COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1)) which results in the following constraint:
(7) (>(x0[0], 1)=TRUEx0[0]=x0[2]690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])∧(UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥))
We simplified constraint (7) using rule (IV) which results in the following new constraint:
(8) (>(x0[0], 1)=TRUE690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧690_0_SOLVE_CONSTANTSTACKPUSH(x0[0])≥COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])∧(UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥))
We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11) (x0[0] + [-2] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(12) (x0[0] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[(3)bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
For Pair COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0, 1)), x0, -(x0, 1)) the following chains were created:
• We consider the chain COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1)) which results in the following constraint:
(13) (COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1])≥NonInfC∧COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1])≥746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))∧(UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥))
We simplified constraint (13) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(14) ((UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥)∧[(-1)bso_20] ≥ 0)
We simplified constraint (14) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(15) ((UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥)∧[(-1)bso_20] ≥ 0)
We simplified constraint (15) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(16) ((UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥)∧[(-1)bso_20] ≥ 0)
We simplified constraint (16) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(17) ((UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥)∧0 = 0∧[(-1)bso_20] ≥ 0)
For Pair COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1)) the following chains were created:
• We consider the chain COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1)) which results in the following constraint:
(18) (COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2])≥NonInfC∧COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2])≥690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))∧(UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥))
We simplified constraint (18) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(19) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥)∧[1 + (-1)bso_22] ≥ 0)
We simplified constraint (19) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(20) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥)∧[1 + (-1)bso_22] ≥ 0)
We simplified constraint (20) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(21) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥)∧[1 + (-1)bso_22] ≥ 0)
We simplified constraint (21) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(22) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥)∧0 = 0∧[1 + (-1)bso_22] ≥ 0)
For Pair 746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0, 1) → COND_746_1_SOLVE_INVOKEMETHOD(>(x0, 0), 723_0_solve_Return, x0, 1) the following chains were created:
• We consider the chain 746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1) → COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1), COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1)) which results in the following constraint:
(23) (>(x0[3], 0)=TRUEx0[3]=x0[4]746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1)≥NonInfC∧746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1)≥COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)∧(UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥))
We simplified constraint (23) using rule (IV) which results in the following new constraint:
(24) (>(x0[3], 0)=TRUE746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1)≥NonInfC∧746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1)≥COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)∧(UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥))
We simplified constraint (24) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(25) (x0[3] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥)∧[bni_23 + (-1)Bound*bni_23] + [bni_23]x0[3] ≥ 0∧[1 + (-1)bso_24] ≥ 0)
We simplified constraint (25) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(26) (x0[3] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥)∧[bni_23 + (-1)Bound*bni_23] + [bni_23]x0[3] ≥ 0∧[1 + (-1)bso_24] ≥ 0)
We simplified constraint (26) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(27) (x0[3] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥)∧[bni_23 + (-1)Bound*bni_23] + [bni_23]x0[3] ≥ 0∧[1 + (-1)bso_24] ≥ 0)
We simplified constraint (27) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(28) (x0[3] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥)∧[(2)bni_23 + (-1)Bound*bni_23] + [bni_23]x0[3] ≥ 0∧[1 + (-1)bso_24] ≥ 0)
For Pair COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0, 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1)) the following chains were created:
• We consider the chain COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1)) which results in the following constraint:
(29) (COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1)≥NonInfC∧COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1)≥690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))∧(UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥))
We simplified constraint (29) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(30) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥)∧[(-1)bso_26] ≥ 0)
We simplified constraint (30) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(31) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥)∧[(-1)bso_26] ≥ 0)
We simplified constraint (31) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(32) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥)∧[(-1)bso_26] ≥ 0)
We simplified constraint (32) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(33) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥)∧0 = 0∧[(-1)bso_26] ≥ 0)
For Pair 746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0, x1) → COND_746_1_SOLVE_INVOKEMETHOD1(>(x0, 0), 886_0_solve_Return, x0, x1) the following chains were created:
• We consider the chain 746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5]) → COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5]), COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1)) which results in the following constraint:
(34) (>(x0[5], 0)=TRUEx0[5]=x0[6]x1[5]=x1[6]746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5])≥NonInfC∧746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5])≥COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])∧(UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥))
We simplified constraint (34) using rule (IV) which results in the following new constraint:
(35) (>(x0[5], 0)=TRUE746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5])≥NonInfC∧746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5])≥COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])∧(UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥))
We simplified constraint (35) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(36) (x0[5] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧[bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧[(-1)bso_28] ≥ 0)
We simplified constraint (36) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(37) (x0[5] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧[bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧[(-1)bso_28] ≥ 0)
We simplified constraint (37) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(38) (x0[5] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧[bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧[(-1)bso_28] ≥ 0)
We simplified constraint (38) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(39) (x0[5] + [-1] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧0 = 0∧[bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧0 = 0∧[(-1)bso_28] ≥ 0)
We simplified constraint (39) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(40) (x0[5] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧0 = 0∧[(2)bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧0 = 0∧[(-1)bso_28] ≥ 0)
For Pair COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0, x1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1)) the following chains were created:
• We consider the chain COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1)) which results in the following constraint:
(41) (COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6])≥NonInfC∧COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6])≥690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))∧(UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥))
We simplified constraint (41) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(42) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥)∧[1 + (-1)bso_30] ≥ 0)
We simplified constraint (42) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(43) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥)∧[1 + (-1)bso_30] ≥ 0)
We simplified constraint (43) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(44) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥)∧[1 + (-1)bso_30] ≥ 0)
We simplified constraint (44) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(45) ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_30] ≥ 0)
To summarize, we get the following constraints P for the following pairs.
• 690_0_SOLVE_CONSTANTSTACKPUSH(x0) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0, 1), x0)
• (x0[0] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[(3)bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
• (x0[0] ≥ 0 ⇒ (UIncreasing(COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])), ≥)∧[(3)bni_17 + (-1)Bound*bni_17] + [bni_17]x0[0] ≥ 0∧[(-1)bso_18] ≥ 0)
• COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0, 1)), x0, -(x0, 1))
• ((UIncreasing(746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))), ≥)∧0 = 0∧[(-1)bso_20] ≥ 0)
• COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1))
• ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))), ≥)∧0 = 0∧[1 + (-1)bso_22] ≥ 0)
• 746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0, 1) → COND_746_1_SOLVE_INVOKEMETHOD(>(x0, 0), 723_0_solve_Return, x0, 1)
• (x0[3] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)), ≥)∧[(2)bni_23 + (-1)Bound*bni_23] + [bni_23]x0[3] ≥ 0∧[1 + (-1)bso_24] ≥ 0)
• COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0, 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1))
• ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))), ≥)∧0 = 0∧[(-1)bso_26] ≥ 0)
• 746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0, x1) → COND_746_1_SOLVE_INVOKEMETHOD1(>(x0, 0), 886_0_solve_Return, x0, x1)
• (x0[5] ≥ 0 ⇒ (UIncreasing(COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])), ≥)∧0 = 0∧[(2)bni_27 + (-1)Bound*bni_27] + [bni_27]x0[5] ≥ 0∧0 = 0∧[(-1)bso_28] ≥ 0)
• COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0, x1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0, 1))
• ((UIncreasing(690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_30] ≥ 0)
The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:
POL(TRUE) = 0
POL(FALSE) = 0
POL(836_1_solve_InvokeMethod(x1, x2)) = [-1]
POL(723_0_solve_Return) = [-1]
POL(1) = [1]
POL(886_0_solve_Return) = [1]
POL(690_0_SOLVE_CONSTANTSTACKPUSH(x1)) = [1] + x1
POL(COND_690_0_SOLVE_CONSTANTSTACKPUSH(x1, x2)) = [1] + x2
POL(>(x1, x2)) = [-1]
POL(746_1_SOLVE_INVOKEMETHOD(x1, x2, x3)) = [1] + x2
POL(690_0_solve_ConstantStackPush(x1)) = x1
POL(-(x1, x2)) = x1 + [-1]x2
POL(COND_746_1_SOLVE_INVOKEMETHOD(x1, x2, x3, x4)) = x3
POL(0) = 0
POL(COND_746_1_SOLVE_INVOKEMETHOD1(x1, x2, x3, x4)) = [1] + x3
The following pairs are in P>:
COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[2], 1))
746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1) → COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)
COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6]) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[6], 1))
The following pairs are in Pbound:
690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])
746_1_SOLVE_INVOKEMETHOD(723_0_solve_Return, x0[3], 1) → COND_746_1_SOLVE_INVOKEMETHOD(>(x0[3], 0), 723_0_solve_Return, x0[3], 1)
746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5]) → COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])
The following pairs are in P:
690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(>(x0[0], 1), x0[0])
COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(-(x0[1], 1)), x0[1], -(x0[1], 1))
COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(-(x0[4], 1))
746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5]) → COND_746_1_SOLVE_INVOKEMETHOD1(>(x0[5], 0), 886_0_solve_Return, x0[5], x1[5])
There are no usable rules.
### (7) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
836_1_solve_InvokeMethod(723_0_solve_Return, 1) → 886_0_solve_Return
836_1_solve_InvokeMethod(886_0_solve_Return, x0) → 886_0_solve_Return
The integer pair graph contains the following rules and edges:
(0): 690_0_SOLVE_CONSTANTSTACKPUSH(x0[0]) → COND_690_0_SOLVE_CONSTANTSTACKPUSH(x0[0] > 1, x0[0])
(1): COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(x0[1] - 1), x0[1], x0[1] - 1)
(4): COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[4] - 1)
(5): 746_1_SOLVE_INVOKEMETHOD(886_0_solve_Return, x0[5], x1[5]) → COND_746_1_SOLVE_INVOKEMETHOD1(x0[5] > 0, 886_0_solve_Return, x0[5], x1[5])
(4) -> (0), if ((x0[4] - 1* x0[0]))
(0) -> (1), if ((x0[0] > 1* TRUE)∧(x0[0]* x0[1]))
(1) -> (5), if ((690_0_solve_ConstantStackPush(x0[1] - 1) →* 886_0_solve_Return)∧(x0[1]* x0[5])∧(x0[1] - 1* x1[5]))
The set Q consists of the following terms:
836_1_solve_InvokeMethod(723_0_solve_Return, 1)
836_1_solve_InvokeMethod(886_0_solve_Return, x0)
### (8) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
### (10) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
836_1_solve_InvokeMethod(723_0_solve_Return, 1) → 886_0_solve_Return
836_1_solve_InvokeMethod(886_0_solve_Return, x0) → 886_0_solve_Return
The integer pair graph contains the following rules and edges:
(1): COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 746_1_SOLVE_INVOKEMETHOD(690_0_solve_ConstantStackPush(x0[1] - 1), x0[1], x0[1] - 1)
(2): COND_690_0_SOLVE_CONSTANTSTACKPUSH(TRUE, x0[2]) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[2] - 1)
(4): COND_746_1_SOLVE_INVOKEMETHOD(TRUE, 723_0_solve_Return, x0[4], 1) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[4] - 1)
(6): COND_746_1_SOLVE_INVOKEMETHOD1(TRUE, 886_0_solve_Return, x0[6], x1[6]) → 690_0_SOLVE_CONSTANTSTACKPUSH(x0[6] - 1)
The set Q consists of the following terms:
836_1_solve_InvokeMethod(723_0_solve_Return, 1)
836_1_solve_InvokeMethod(886_0_solve_Return, x0)
### (11) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes. | 11,096 | 28,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.613921 |
https://jp.mathworks.com/matlabcentral/fileexchange/25674-automotive-electrical-system-simulation-and-control | 1,713,131,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00766.warc.gz | 317,769,732 | 24,385 | ## Automotive Electrical System Simulation and Control
バージョン 1.12 (293 KB) 作成者:
A conventional vehicle electrical system model with alternator, battery, loads, and idle control.
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# Automotive Electrical System Simulation and Control
A conventional vehicle electrical system model with alternator, battery, loads, and idle control.
This model shows an example of a conventional vehicle electrical system model, which was shown in the webinar "Optimizing Vehicle Electrical Design through System-Level Simulation". The model is intended to study component sizing, selection, and control. http://www.mathworks.com/videos/optimizing-vehicle-electrical-design-through-system-level-simulation-81919.html
The electrical system model contains a Simscape lead-acid battery model as described in SAE Paper 2007-01-0778. There are two choices for battery size. https://www.mathworks.com/content/dam/mathworks/tag-team/Objects/s/40542_SAE-2007-01-0778-Battery-Modeling-Process.pdf
The model also contains data-driven alternator model options using either look-up tables or Model-Based Calibration Toolbox blocks. The alternator model is described in SAE Paper 2007-01-3471.
Control may be added to monitor the system condition, and increase idle speed to compensate for bad conditions in the vehicle charging system. Vehicle loads are modeled as variable resistive elements.
Simulink Design Optimization can be used to estimate battery parameters from experimental battery data. To see how, please watch this video (5 min): http://www.mathworks.com/videos/estimating-parameters-of-a-battery-68957.html
Parameter estimation for these models from measured data sets at different temperature and current data is a complex task. To learn how MathWorks Consulting can teach you this technique, please contact us or see:
MathWorks Consulting: Battery Simulation and Controls http://www.mathworks.com/services/consulting/proven-solutions/battery-simulation-and-controls.html
### 引用
Robyn Jackey (2024). Automotive Electrical System Simulation and Control (https://github.com/mathworks/automotive-electrical-system), GitHub. 取得済み .
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GitHub の既定のブランチを使用するバージョンはダウンロードできません
バージョン 公開済み リリース ノート
1.12
Fix the name after moving repo to mathworks github
1.11
Moved to mathworks Github
1.10
Connect to GitHub, add R2019b compatibility model
1.9.0.1
1.9.0.0
Updated the description and acknowledgement
1.8.0.0
Updated description only. No file changes.
1.7.0.0
Added an updated model for R2012b and later.
1.6.0.0
Updated the description with new webinar link. Referenced alternator paper available from SAE.
1.5.0.0
Minor update to directions in Readme.txt.
Changed solver setting to "nonadaptive" to avoid warnings.
1.3.0.0
updated broken webinar link in description
1.2.0.0
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1.0.0.0
この GitHub アドオンでの問題を表示または報告するには、GitHub リポジトリにアクセスしてください。
この GitHub アドオンでの問題を表示または報告するには、GitHub リポジトリにアクセスしてください。 | 849 | 3,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.802788 |
https://danburf.wordpress.com/2015/06/02/day-174-guess-my-rule/ | 1,498,207,007,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320040.36/warc/CC-MAIN-20170623082050-20170623102050-00640.warc.gz | 733,973,691 | 19,885 | # Day 174: Guess My Rule
I put this up as the opener for algebra today:
A large chunk of students quickly jumped to “the previous number times two”.
That wasn’t my rule.
Then they shifted to +2, +4 +6…
That wasn’t my rule either.
Silence for about 2 minutes… Then a student asked “Is the next number 10?”
“That fits my rule” I replied.
This threw ’em all off… They dabbled around with different numbers for a bit, I had fun with it, especially when they pitched crazy numbers like 54, 81, and 1092, which fit my rule.
Eventually, they drifted into trying to find numbers that didn’t fit my rule… which is awesome problem solving.
This really is the heart of what I want to get at through all this factoring, solving, graphing and so on. To me math isn’t so much about memorizing or reproducing a certain skill, but more about taking what you know and tweaking it to solve some crazy problem you have never seen before.
That moment where you have abandoned all hope and try some crazy technique in a problem that you learned months or years ago, which ends up working is what I love about math. The struggle leading up to that moment; following hundreds of self-imposed rules and just sheer grit isn’t easy. It is even more difficult to learn and teaching it takes someone really special.
Everyday I try and very delicately move students towards that direction. With the hopes that maybe, at some point in their lives, in a situation that isn’t even close to math related, they will be able to use their critical thinking abilities, which took years to develop, to solve a difficult problem and experience that felling of having everything fall perfectly into place.
For me, that hope makes everyday worth it.
By the way… my rule was each number had to be larger than the previous. They went crazy over this, some nasty reverse psychology on my end: They automatically see something math related and dive into testing different equations and rules, when really the rule doesn’t require anything fancy. Credit.
## 3 thoughts on “Day 174: Guess My Rule”
1. Thom H Gibson
I heard this on a podcast recently with Dan Pink I believe. Can’t remember who he was chatting with. Pretty great. Also, just found your blog. Are you blogging every day of the year?
Like
2. Maya Quinn
Prior to seeing 10, I had wondered if it was number names of increasing evens with letter count: 3, 4, 5, …
So: TWO, FOUR, EIGHT, and the next would be TWELVE, since 12 is the next even that has exactly 6 letters in its name.
Like | 583 | 2,522 | {"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-2017-26 | longest | en | 0.967659 |
http://blog.sleptons.com/2014/12/lambda-calculus-lambda-abstraction.html | 1,511,126,375,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00383.warc.gz | 40,326,325 | 14,858 | ## Tuesday, 30 December 2014
### Lambda Calculus, Lambda Abstraction
Lambda Abstraction is the way we define an anonymous function in λ calculus. Suppose we have a function like f(x) = 2x which in fact is a function that maps any x to 2x, now from the previous posts we already knew that in λ calculus the expression 2x (which is a term) can be shown as * 2 x, but the way we show this as an anonymous function is like:
λx. * 2 x
See there is no equality sign, here λx is something anonymously like f(x) and the dot after it shows the start of the evaluation formula. So a familiar C or Java-like function like the following:
int doubleIt(int x) {
return 2 * x
}
Will be something like the following in Lisp-like language for example (which you can use it inline of course):
(lambda (x) (∗ 2 x))
The above is nothing but what we already talked about it: λx. * 2 x
Just one parameter
Any lambda abstraction can contain only one input parameter, so in general, an abstraction is something like:
λx. t
In which t is any valid lambda expression or simply a term. Here are some more examples:
1. λx. + x 5
2. λx. * 2 * x x
3. λx. + x y
Bound and free variables
Note in the third example y is just an unknown variable for the function, this variable is called free variable while the x is called bound variable. The function gets x and adds y to it. Since we can have abstractions in some part of an anonymous function itself so the third example can be written in the following form to have the meaning we usually like to give it to:
λy.t , t = + x y => λy.λx. + x y
Or simply as nested abstraction like:
λyx. + x y
In Lisp we write it like:
(lambda (x y) (+ x y)) | 445 | 1,675 | {"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} | 3.4375 | 3 | CC-MAIN-2017-47 | longest | en | 0.865211 |
https://r.789695.n4.nabble.com/Failure-in-predicting-parameters-td4768059.html | 1,618,581,796,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00174.warc.gz | 573,737,513 | 17,931 | # Failure in predicting parameters
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## Failure in predicting parameters
Hello, I would like to use the Rutledge equation (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The equation is: Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb I defined the equation and another that subtracts the values from the expectations. I used minpack.lm to get the parameters, but I got an error: ``` > library("minpack.lm") > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, + 12133.57, 12148.89, 12137.09) > high <- h[1:45] > MaxFluo <- max(high) > halfFluo <- MaxFluo/2 > halfCycle = 27 > find_slope <- function(X, Y) { + Slope <- c(0) + for (i in 2:length(X)) { + delta_x <- X[i] - X[i-1] + delta_y <- Y[i] - Y[i-1] + Slope[i] <- delta_y/delta_x + } + return(Slope) + } > slopes <- find_slope(1:45, high) > > rutledge <- function(m, s, M, B, x) { + divisor = 1 + exp(-1* ((x-m)/s) ) + y = (M/divisor) + B + return(y) + } > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > points(1:45, init, type="l", col="red") > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), + fn = rutledge_param, x = 1:45, y = high) Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), : evaluation of fn function returns non-sensible value! ``` Where could the error be? -- Best regards, Luigi ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Failure in predicting parameters
Do the negative values in your data make any sense? Note that if Fb must be >0, Fc must be also. But I have *not* examined your code/equations in detail, so feel free to ignore if this is irrelevant. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <[hidden email]> wrote: > Hello, > I would like to use the Rutledge equation > (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The > equation is: > Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb > I defined the equation and another that subtracts the values from the > expectations. I used minpack.lm to get the parameters, but I got an > error: > ``` > > > library("minpack.lm") > > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, > + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, > 15.01, > + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, > + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, > + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, > 11684.96, > + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, > 12100.63, > + 12133.57, 12148.89, 12137.09) > > high <- h[1:45] > > MaxFluo <- max(high) > > halfFluo <- MaxFluo/2 > > halfCycle = 27 > > find_slope <- function(X, Y) { > + Slope <- c(0) > + for (i in 2:length(X)) { > + delta_x <- X[i] - X[i-1] > + delta_y <- Y[i] - Y[i-1] > + Slope[i] <- delta_y/delta_x > + } > + return(Slope) > + } > > slopes <- find_slope(1:45, high) > > > > rutledge <- function(m, s, M, B, x) { > + divisor = 1 + exp(-1* ((x-m)/s) ) > + y = (M/divisor) + B > + return(y) > + } > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) > + p\$B) - y > > > > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > > points(1:45, init, type="l", col="red") > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > high[1]), > + fn = rutledge_param, x = 1:45, y = high) > Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > high[1]), : > evaluation of fn function returns non-sensible value! > ``` > > Where could the error be? > > > -- > Best regards, > Luigi > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Failure in predicting parameters
In reply to this post by Luigi > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y Did you mean that p\$x to be just x? As is, this returns numeric(0) for the p that nls.lm gives it because p\$x is NULL and NULL-aNumber is numeric(). -Bill On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <[hidden email]> wrote: > > Hello, > I would like to use the Rutledge equation > (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The > equation is: > Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb > I defined the equation and another that subtracts the values from the > expectations. I used minpack.lm to get the parameters, but I got an > error: > ``` > > > library("minpack.lm") > > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, > + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, > + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, > + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, > + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, > + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, > + 12133.57, 12148.89, 12137.09) > > high <- h[1:45] > > MaxFluo <- max(high) > > halfFluo <- MaxFluo/2 > > halfCycle = 27 > > find_slope <- function(X, Y) { > + Slope <- c(0) > + for (i in 2:length(X)) { > + delta_x <- X[i] - X[i-1] > + delta_y <- Y[i] - Y[i-1] > + Slope[i] <- delta_y/delta_x > + } > + return(Slope) > + } > > slopes <- find_slope(1:45, high) > > > > rutledge <- function(m, s, M, B, x) { > + divisor = 1 + exp(-1* ((x-m)/s) ) > + y = (M/divisor) + B > + return(y) > + } > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > > points(1:45, init, type="l", col="red") > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > + fn = rutledge_param, x = 1:45, y = high) > Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > high[1]), : > evaluation of fn function returns non-sensible value! > ``` > > Where could the error be? > > > -- > Best regards, > Luigi > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Failure in predicting parameters
Hello, the negative data comes from the machine. Probably I should use raw data directly, although in the paper this requirement is not reported. The p\$x was a typo. Now I corrected it and I got this error: ``` > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(x-p\$m)/p\$s))) + p\$B) - y > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), + fn = rutledge_param, x = 1:45, y = high) Error in dimnames(x) <- dn : length of 'dimnames' [2] not equal to array extent ``` Probably because 'slopes' is a vector instead of a scalar. Since the slope is changing, I don't think is right to use a scalar, but I tried and I got: ``` > estim <- nls.lm(par = list(m = halfFluo, s = 1, M = MaxFluo, B = high[1]), + fn = rutledge_param, x = 1:45, y = high) > estim Nonlinear regression via the Levenberg-Marquardt algorithm parameter estimates: 6010.94, 1, 12021.88, 4700.49288888889 residual sum-of-squares: 1.14e+09 reason terminated: Relative error in the sum of squares is at most `ftol'. ``` The values reported are the same I used at the beginning apart from the last (the background parameter) which is 4700 instead of zero. If I plug it, I get an L shaped plot that is worse than that at the beginning: ``` after = init = rutledge(halfFluo, 1, MaxFluo, 4700.49288888889, high) points(1:45, after, type="l", col="blue") ``` What did I get wrong here? Thanks On Sun, Mar 14, 2021 at 8:05 PM Bill Dunlap <[hidden email]> wrote: > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > Did you mean that p\$x to be just x? As is, this returns numeric(0) > for the p that nls.lm gives it because p\$x is NULL and NULL-aNumber is > numeric(). > > -Bill > > On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <[hidden email]> wrote: > > > > Hello, > > I would like to use the Rutledge equation > > (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The > > equation is: > > Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb > > I defined the equation and another that subtracts the values from the > > expectations. I used minpack.lm to get the parameters, but I got an > > error: > > ``` > > > > > library("minpack.lm") > > > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, > > + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, > > + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, > > + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, > > + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, > > + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, > > + 12133.57, 12148.89, 12137.09) > > > high <- h[1:45] > > > MaxFluo <- max(high) > > > halfFluo <- MaxFluo/2 > > > halfCycle = 27 > > > find_slope <- function(X, Y) { > > + Slope <- c(0) > > + for (i in 2:length(X)) { > > + delta_x <- X[i] - X[i-1] > > + delta_y <- Y[i] - Y[i-1] > > + Slope[i] <- delta_y/delta_x > > + } > > + return(Slope) > > + } > > > slopes <- find_slope(1:45, high) > > > > > > rutledge <- function(m, s, M, B, x) { > > + divisor = 1 + exp(-1* ((x-m)/s) ) > > + y = (M/divisor) + B > > + return(y) > > + } > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > > > > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > > > points(1:45, init, type="l", col="red") > > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > > + fn = rutledge_param, x = 1:45, y = high) > > Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > > high[1]), : > > evaluation of fn function returns non-sensible value! > > ``` > > > > Where could the error be? > > > > > > -- > > Best regards, > > Luigi > > > > ______________________________________________ > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > and provide commented, minimal, self-contained, reproducible code. -- Best regards, Luigi ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Failure in predicting parameters
Just an update: I tried with desmos and the fitting looks good. Desmos calculated the parameters as: Fmax = 11839.8 Chalf = 27.1102 (with matches with my estimate of 27 cycles) k = 2.76798 Fb = -138.864 I forced R to accept the right parameters using a single named list and re-written the formula (it was a bit unclear in the paper): ``` rutledge <- function(p, x) { m = p\$half_fluorescence s = p\$slope M = p\$max_fluorescence B = p\$back_fluorescence y = (M / (1+exp( -((x-m)/s) )) ) + B return(y) } ``` but when I apply it I get a funny graph: ``` desmos <- rutledge(list(half_fluorescence = 27.1102, slope = 2.76798, max_fluorescence = 11839.8, back_fluorescence = -138.864) , high) ``` On Mon, Mar 15, 2021 at 7:39 AM Luigi Marongiu <[hidden email]> wrote: > > Hello, > the negative data comes from the machine. Probably I should use raw > data directly, although in the paper this requirement is not reported. > The p\$x was a typo. Now I corrected it and I got this error: > ``` > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(x-p\$m)/p\$s))) + p\$B) - y > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > + fn = rutledge_param, x = 1:45, y = high) > Error in dimnames(x) <- dn : > length of 'dimnames' [2] not equal to array extent > ``` > Probably because 'slopes' is a vector instead of a scalar. Since the > slope is changing, I don't think is right to use a scalar, but I tried > and I got: > ``` > > estim <- nls.lm(par = list(m = halfFluo, s = 1, M = MaxFluo, B = high[1]), > + fn = rutledge_param, x = 1:45, y = high) > > estim > Nonlinear regression via the Levenberg-Marquardt algorithm > parameter estimates: 6010.94, 1, 12021.88, 4700.49288888889 > residual sum-of-squares: 1.14e+09 > reason terminated: Relative error in the sum of squares is at most `ftol'. > ``` > The values reported are the same I used at the beginning apart from > the last (the background parameter) which is 4700 instead of zero. If > I plug it, I get an L shaped plot that is worse than that at the > beginning: > ``` > after = init = rutledge(halfFluo, 1, MaxFluo, 4700.49288888889, high) > points(1:45, after, type="l", col="blue") > ``` > What did I get wrong here? > Thanks > > On Sun, Mar 14, 2021 at 8:05 PM Bill Dunlap <[hidden email]> wrote: > > > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > Did you mean that p\$x to be just x? As is, this returns numeric(0) > > for the p that nls.lm gives it because p\$x is NULL and NULL-aNumber is > > numeric(). > > > > -Bill > > > > On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <[hidden email]> wrote: > > > > > > Hello, > > > I would like to use the Rutledge equation > > > (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The > > > equation is: > > > Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb > > > I defined the equation and another that subtracts the values from the > > > expectations. I used minpack.lm to get the parameters, but I got an > > > error: > > > ``` > > > > > > > library("minpack.lm") > > > > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, > > > + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, > > > + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, > > > + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, > > > + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, > > > + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, > > > + 12133.57, 12148.89, 12137.09) > > > > high <- h[1:45] > > > > MaxFluo <- max(high) > > > > halfFluo <- MaxFluo/2 > > > > halfCycle = 27 > > > > find_slope <- function(X, Y) { > > > + Slope <- c(0) > > > + for (i in 2:length(X)) { > > > + delta_x <- X[i] - X[i-1] > > > + delta_y <- Y[i] - Y[i-1] > > > + Slope[i] <- delta_y/delta_x > > > + } > > > + return(Slope) > > > + } > > > > slopes <- find_slope(1:45, high) > > > > > > > > rutledge <- function(m, s, M, B, x) { > > > + divisor = 1 + exp(-1* ((x-m)/s) ) > > > + y = (M/divisor) + B > > > + return(y) > > > + } > > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > > > > > > > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > > > > points(1:45, init, type="l", col="red") > > > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > > > + fn = rutledge_param, x = 1:45, y = high) > > > Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > > > high[1]), : > > > evaluation of fn function returns non-sensible value! > > > ``` > > > > > > Where could the error be? > > > > > > > > > -- > > > Best regards, > > > Luigi > > > > > > ______________________________________________ > > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > > https://stat.ethz.ch/mailman/listinfo/r-help> > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > > and provide commented, minimal, self-contained, reproducible code. > > > > -- > Best regards, > Luigi -- Best regards, Luigi ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Failure in predicting parameters
It worked. I re-written the equation as: ``` rutledge_param <- function(p, x, y) ( (p\$M / ( 1 + exp(-(x-p\$m)/p\$s)) ) + p\$B ) - y ``` and used Desmos to estimate the slope, so: ``` estim <- nls.lm(par = list(m = halfCycle, s = 2.77, M = MaxFluo, B = high[1]), fn = rutledge_param, x = 1:45, y = high) summary(estim) R <- rutledge(list(half_fluorescence = 27.1102, slope = 2.7680, max_fluorescence = 11839.7745, back_fluorescence = -138.8615) , 1:45) points(1:45, R, type="l", col="red") ``` Thanks On Tue, Mar 16, 2021 at 8:29 AM Luigi Marongiu <[hidden email]> wrote: > > Just an update: > I tried with desmos and the fitting looks good. Desmos calculated the > parameters as: > Fmax = 11839.8 > Chalf = 27.1102 (with matches with my estimate of 27 cycles) > k = 2.76798 > Fb = -138.864 > I forced R to accept the right parameters using a single named list > and re-written the formula (it was a bit unclear in the paper): > ``` > rutledge <- function(p, x) { > m = p\$half_fluorescence > s = p\$slope > M = p\$max_fluorescence > B = p\$back_fluorescence > y = (M / (1+exp( -((x-m)/s) )) ) + B > return(y) > } > ``` > but when I apply it I get a funny graph: > ``` > desmos <- rutledge(list(half_fluorescence = 27.1102, slope = 2.76798, > max_fluorescence = 11839.8, back_fluorescence > = -138.864) , high) > ``` > > On Mon, Mar 15, 2021 at 7:39 AM Luigi Marongiu <[hidden email]> wrote: > > > > Hello, > > the negative data comes from the machine. Probably I should use raw > > data directly, although in the paper this requirement is not reported. > > The p\$x was a typo. Now I corrected it and I got this error: > > ``` > > > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(x-p\$m)/p\$s))) + p\$B) - y > > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > > + fn = rutledge_param, x = 1:45, y = high) > > Error in dimnames(x) <- dn : > > length of 'dimnames' [2] not equal to array extent > > ``` > > Probably because 'slopes' is a vector instead of a scalar. Since the > > slope is changing, I don't think is right to use a scalar, but I tried > > and I got: > > ``` > > > estim <- nls.lm(par = list(m = halfFluo, s = 1, M = MaxFluo, B = high[1]), > > + fn = rutledge_param, x = 1:45, y = high) > > > estim > > Nonlinear regression via the Levenberg-Marquardt algorithm > > parameter estimates: 6010.94, 1, 12021.88, 4700.49288888889 > > residual sum-of-squares: 1.14e+09 > > reason terminated: Relative error in the sum of squares is at most `ftol'. > > ``` > > The values reported are the same I used at the beginning apart from > > the last (the background parameter) which is 4700 instead of zero. If > > I plug it, I get an L shaped plot that is worse than that at the > > beginning: > > ``` > > after = init = rutledge(halfFluo, 1, MaxFluo, 4700.49288888889, high) > > points(1:45, after, type="l", col="blue") > > ``` > > What did I get wrong here? > > Thanks > > > > On Sun, Mar 14, 2021 at 8:05 PM Bill Dunlap <[hidden email]> wrote: > > > > > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > > > Did you mean that p\$x to be just x? As is, this returns numeric(0) > > > for the p that nls.lm gives it because p\$x is NULL and NULL-aNumber is > > > numeric(). > > > > > > -Bill > > > > > > On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <[hidden email]> wrote: > > > > > > > > Hello, > > > > I would like to use the Rutledge equation > > > > (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The > > > > equation is: > > > > Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb > > > > I defined the equation and another that subtracts the values from the > > > > expectations. I used minpack.lm to get the parameters, but I got an > > > > error: > > > > ``` > > > > > > > > > library("minpack.lm") > > > > > h <- c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, > > > > + -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, > > > > + 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, > > > > + 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, > > > > + 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, > > > > + 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, > > > > + 12133.57, 12148.89, 12137.09) > > > > > high <- h[1:45] > > > > > MaxFluo <- max(high) > > > > > halfFluo <- MaxFluo/2 > > > > > halfCycle = 27 > > > > > find_slope <- function(X, Y) { > > > > + Slope <- c(0) > > > > + for (i in 2:length(X)) { > > > > + delta_x <- X[i] - X[i-1] > > > > + delta_y <- Y[i] - Y[i-1] > > > > + Slope[i] <- delta_y/delta_x > > > > + } > > > > + return(Slope) > > > > + } > > > > > slopes <- find_slope(1:45, high) > > > > > > > > > > rutledge <- function(m, s, M, B, x) { > > > > + divisor = 1 + exp(-1* ((x-m)/s) ) > > > > + y = (M/divisor) + B > > > > + return(y) > > > > + } > > > > > rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y > > > > > > > > > > > > > > > init = rutledge(halfFluo, slopes, MaxFluo, 0, high) > > > > > points(1:45, init, type="l", col="red") > > > > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]), > > > > + fn = rutledge_param, x = 1:45, y = high) > > > > Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = > > > > high[1]), : > > > > evaluation of fn function returns non-sensible value! > > > > ``` > > > > > > > > Where could the error be? > > > > > > > > > > > > -- > > > > Best regards, > > > > Luigi > > > > > > > > ______________________________________________ > > > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > > > https://stat.ethz.ch/mailman/listinfo/r-help> > > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > > > and provide commented, minimal, self-contained, reproducible code. > > > > > > > > -- > > Best regards, > > Luigi > > > > -- > Best regards, > Luigi -- Best regards, Luigi ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code. | 9,010 | 25,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-17 | latest | en | 0.485102 |
https://ta-east.com/betting/understanding-sports-betting-odds-before-you-make-a-killing/ | 1,620,697,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00431.warc.gz | 546,876,839 | 9,139 | # Understanding sports betting odds before you make a killing
In case you have entered the betting arena without a clear strategy in mind then you could emerge out as a very unhappy bettor, not to mention that you could also lose a ton of money at the same time. Just as a doctor would require the right skills before operating on a patient, you too will have to work tirelessly to understand sports betting odds before you make a killing inside the betting arena.
Each fascinating sport such as boxing, tennis, golf, baseball, football, basketball, horse racing, car racing, and many such more are eligible for betting. Even if the sport is college football handled by way of the ncaa or pro football handled by way of the nfl, bets may be placed, even though the odds will certainly differ in each sport and every game too. Odds are also displayed differently in different countries. By way of example, USA displays odds as American odds, i.e. -100, -230, etc, while Canada, Australia and European countries display odds as Decimal odds, i.e. 2.50, 1.30, etc, and also the UK displays odds as Fractional odds, i.e. 8/8, 8/4, etc.
Sports betting odds also display the underdog team along with the top-dog, the over-under and the spread by which you must win if you decide to bet for or against the spread. A simple example will explain betting odds easily. For example in an nfl match involving the Buffalo Bills and the New York Giants, the odds may be displayed as under
Buffalo Bills -11.2 -120 -160
New York Giants +11.2 -120 +270
The very first sign next to the teams name indicates the underdog and also the top-dog. Hence, Buffalo Bills is the top-dog because of the “-” sign whereas the Giants would be the underdogs because of the “+” sign beside it. The figure 11.2 signifies the spread and the favored team has to win with the points mentioned on the spread. The figure 120 implies that you will need to bet \$120 to win \$100 on that spread. The last figure of 160 next to the Bills indicates that you need to stake \$160 to win \$100 together with your stake on that bet since the Bills are the favored team to win. On the other hand, you only have to bet \$100 to win \$270 should you bet on the Giants since it is the underdog team. arbitrage betting strategy
The above was an example of betting odds from the American odds format. Each bookie will display odds in numerous formats however, you would be able to quickly convert it in your format when using the odds converter available in most websites. The world wide web has today definitely made betting easier although you will definitely have to work on understanding and calculating the betting odds prior to placing your bet.
You might certainly be able to bet efficiently and convert most of your placed bets into huge wins if you read and understand sports betting odds. Each sports book displays odds and you’ll have to choose sites that have the most effective odds for you to bet on. Understanding sports betting odds is essential if you wish to beat the odds and the bookies and make a killing by winning the maximum number of bets. | 678 | 3,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.954173 |
https://www.physicsforums.com/threads/change-in-electrons-direction-changes-the-lorentz-force-direction.756938/ | 1,521,696,365,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00030.warc.gz | 866,751,629 | 18,480 | # Change in electrons direction, changes the Lorentz force direction?
1. Jun 6, 2014
### Dash-IQ
When the Lorentz force is applied on a wire that has current flowing in a magnetic field, by changing the direction of the electron(not it's KE) would this possibly change the direction of the Lorentz force? Can the wire be controlled in a way that the force's direction stays the same path?
Last edited: Jun 6, 2014
2. Jun 6, 2014
### Staff: Mentor
What do you mean with "initial"? In wires, net electron motions are always along the direction of the wire. You can rotate the wire to change the flow direction.
"stays the same path" as what?
3. Jun 6, 2014
### Staff: Mentor
Apply a second force to the wire (e.g. from your hands!) so the net force on the wire is zero and it doesn't move.
4. Jun 6, 2014
### Dash-IQ
edited.
Wouldn't the Lorentz force change the direction of the electron's velocity?
What would happen to the wire if the velocity direction is changed?
5. Jun 6, 2014
### Staff: Mentor
If you have a beam of electrons going through a vacuum without being inside a wire, a magnetic field causes the beam to curve (change direction):
If the electrons are moving along a straight wire, then they can't follow that path, but are instead constrained to a straight path by the electrical attraction of the positive charges in the wire. And of course the electrons pull back on the positive charges (Newton's Third Law!). The net force on the wire (the magnetic force on the electrons) stays in one direction, perpendicular to the wire.
Last edited by a moderator: May 3, 2017
6. Jun 6, 2014
### Dash-IQ
Thank you for the help, that cleared A LOT of things...
Last edited by a moderator: May 3, 2017
7. Jun 7, 2014
### Dash-IQ
jtbell, what about comparing the Hall effect to the beam system presented above?
Don't the electrons move similar to that beam due to that conductor being fixed and not able to move?
8. Jun 11, 2014
### Dash-IQ
Is it even possible that when a large conducting plate is placed in a magnetic field, and current flow where the plate starts to accelerate due to the Lorentz force... will the charges within the plate change direction and alter the Lorentz force path in any way?
What other forms of deceleration might that plate experience aside from the eddy currents and the mechanical resistances and the reduced current do to Faraday's law?
9. Jun 11, 2014
### Staff: Mentor
You get a small, short-term net motion of charges, but this is completely negligible.
Can you clarify which setup you have in mind here and where you expect a deceleration?
10. Jun 11, 2014
### Dash-IQ
Like a rail gun concept:
Or any other similar idea, where a conductor passes such a magnetic field.
11. Jun 11, 2014
### Dash-IQ
I don't think its true that the Lorentz force direction will change as the conductor starts to accelerate?
12. Jun 11, 2014
### Staff: Mentor
The force does not change its direction. The motion of the bar does not do anything: electrons and protons move with the same velocity. The only force you get is from the current flow, and that acts along the rails.
13. Jun 11, 2014
### Dash-IQ
What about the motion of the bar induced motional emf = -vBL?
14. Jun 14, 2014
### Staff: Mentor
I don't understand that question.
15. Jun 14, 2014
### Jano L.
When the conductor changes velocity, the current density vector in the wire changes direction too and this makes the force
$$\int \mathbf j \times\mathbf B dV$$
change too. Besides the x component of that force which translates into accelerating force on the rod, there will be also y component of this force that counteracts the flow of current in the circuit.
16. Jun 15, 2014
### Dash-IQ
I don't understand what you mean, the conductor is a restrictor to the electrons they flow in a straight path as stated above, how can it's velocity change direction?
17. Jun 15, 2014
### Jano L.
When the rod on the picture above stands still, the electrons move along the axis y. When the rod moves in the direction of axis x. The electrons move diagonally, in between the direction of y and x.
18. Jun 15, 2014
### Dash-IQ
That will result in the change of the force's direction from the x axis?
What about the "constraints" that does not allow the electrons to do so? The constraint being the wire of course.
19. Jun 16, 2014
### Jano L.
It is just vector addition of velocities. Wire moves along x. In the frame of the wire, the electrons move along y. Hence, in the lab frame the electrons move diagonally, partially along x, partially along y.
20. Jun 16, 2014
### Dash-IQ
Not sure, what you point is honestly. As long as the wire is moving in the x direction due to the Lorentz force, that is the most important thing to me.
I doubt it changes direction, since the current will flow in the y axis, and can't "change" due to the conductor being it's constraint. | 1,229 | 4,909 | {"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} | 2.84375 | 3 | CC-MAIN-2018-13 | longest | en | 0.926584 |
http://reference.wolfram.com/legacy/language/v10.3/ref/FindShortestTour.html | 1,508,746,306,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00348.warc.gz | 275,362,504 | 10,921 | Wolfram Language & System 10.3 (2015)|Legacy Documentation
This is documentation for an earlier version of the Wolfram Language.
BUILT-IN WOLFRAM LANGUAGE SYMBOL
FindShortestTour
FindShortestTour[{v1,v2,}]
attempts to find an ordering of the that minimizes the total distance on a tour that visits all the once.
FindShortestTour[graph]
attempts to find an ordering of the vertices in graph that minimizes the total length when visiting each vertex once.
FindShortestTour[{v1,v2,},j,k]
finds an ordering of the that minimizes the total distance on a path from to .
FindShortestTour[graph,s,t]
finds an ordering of the vertices that minimizes the total length on a path from s to t.
FindShortestTour[{vw,},]
uses rules to specify the graph g.
Details and OptionsDetails and Options
• FindShortestTour is also known as the traveling salesman problem (TSP).
• FindShortestTour returns a list of the form , where is the length of the tour found, and is the ordering.
• The following options can be given:
• DistanceFunction Automatic function to apply to pairs of objects Method Automatic method to use
• Automatic settings for DistanceFunction depending on the include:
• EuclideanDistance numbers of lists of numbers EditDistance strings GeoDistance geo positions
• For graph, the distance is taken to be GraphDistance, which is the shortest path length for an unweighted graph and the sum of weights for a weighted graph.
ExamplesExamplesopen allclose all
Basic Examples (3)Basic Examples (3)
Find the length and ordering of the shortest tour through points in the plane:
Out[2]=
Order the points according to the tour found:
Out[3]=
Plot the tour:
Out[4]=
Find the shortest tour in a graph:
Out[2]=
Highlight the tour:
Out[3]=
Find a shortest tour of Europe from Albania to Spain:
Out[4]=
Show the tour:
Out[5]= | 438 | 1,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | latest | en | 0.788445 |
http://oeis.org/A331605/internal | 1,669,742,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00345.warc.gz | 39,681,319 | 3,812 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A331605 Positive integers k such that k = (a^2 + b^2 + c^2)/(a*b + b*c + c*a) for some integers a, b and c. 4
%I
%S 1,2,5,10,14,17,26,29,37,50,62,65,74,77,82,98,101,109,110,122,125,145,
%T 149,170,173,190,194,197,209,226,242,245,257,269,290,302,305,314,325,
%U 334,362,365,398,401,410,434,437,442,469,482,485,497,509,514,530,554,557
%N Positive integers k such that k = (a^2 + b^2 + c^2)/(a*b + b*c + c*a) for some integers a, b and c.
%C This sequence is infinite because k = x^2 + 1 is a term, where a = x + 1, b = x^2 + 1 and c = x^4 + x^3 + 3*x^2 + 2*x + 1. There are other forms of k:
%C k = (a^2-a+2)^2 - 2 when a + b = 1 and c = a^2 - a + 1.
%C k = x^4 + 2*x^3 + 5*x^2 + 4*x + 2 when a = k*(b+c) + x, b = x^2 + x + 1 and c = x + 1.
%C k = ((a^2+a*b+b^2)^2 - 2*a*b + 1)/(a + b)^2 when a = Fibonacci(2*m-1), b = Fibonacci(2*m) and c = ((a^2+a*b+b^2)^2 - a*b)/(a + b).
%C a(n) == 1 or 2 (mod 4). Proof: a^2 - k*(b+c)*a + (b^2+c^2-k*b*c) = 0, hence discriminant D = (k^2-4)*(b^2+c^2) + (2*k^2+4*k)*b*c is a square. Because (a/g, b/g, c/g) is also a set of solution if k = (a^2 + b^2 + c^2)/(a*b + b*c + c*a) and gcd(a, b, c) = g, we only need to consider the case of gcd(a,b,c) = 1.
%C Case (i). k = 4*r, then D/4 = (4*r^2-1)*(b^2+c^2) + (8*r^2+4*r)*b*c == 2 or 3 (mod 4), hence D is not a square, a contradiction.
%C Case (ii). k = 4*r - 1, then (a+b)^2 + (b+c)^2 + (c+a)^2 = 8*r*(a*b + b*c + c*a) is divisible by 4, hence a, b and c are odd numbers. Therefore, ((a+b)^2 + (b+c)^2 + (c+a)^2)/4 == 1 (mod 2), a contradiction.
%H Jinyuan Wang, <a href="/A331605/a331605.txt">PARI program and details with k less than 100</a>
%e a(4) = 10 because 10 = ((-1)^2 + 2^2 + 5^2)/((-1)*2 + 2*5 + 5*(-1)).
%K nonn
%O 1,2
%A _Jinyuan Wang_, Jan 22 2020
%E a(22)-a(57) from _Giovanni Resta_, Jan 29 2020
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Last modified November 29 11:57 EST 2022. Contains 358424 sequences. (Running on oeis4.) | 1,053 | 2,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-49 | latest | en | 0.729041 |
https://www.aqua-calc.com/one-to-one/luminous-intensity/attocandela/kilocandela/1-point-0eplus18 | 1,632,171,779,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00382.warc.gz | 676,149,233 | 5,963 | # 1 × 10+18 attocandelas [acd] in kilocandelas
## attocandelas to kilocandela unit converter of luminous intensity
1.0 × 10+18 attocandelas [acd] = 0.001 kilocandela [kcd]
#### Foods, Nutrients and Calories
SEA SALT ROASTED MACADAMIA NUTS, SEA SALT, UPC: 085239087794 contain(s) 714 calories per 100 grams (≈3.53 ounces) [ price ]
4979 foods that contain Alanine. List of these foods starting with the highest contents of Alanine and the lowest contents of Alanine
#### Gravels, Substances and Oils
Substrate, Clay/Laterite weighs 1 019 kg/m³ (63.61409 lb/ft³) with specific gravity of 1.019 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Yttrium oxide [Y2O3 or O3Y2] weighs 5 010 kg/m³ (312.76408 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-408A, liquid (R408A) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
A dekaampere is a SI-multiple (see prefix deka) of the electric current unit ampere and equal to ten amperes (10 A)
The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?"
mg/dm³ to long tn/in³ conversion table, mg/dm³ to long tn/in³ unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a hexagonal prism and its surface area | 473 | 1,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-39 | latest | en | 0.783032 |
https://ccrma.stanford.edu/~jos/pasp/Linear_Interpolation_Frequency_Response.html | 1,719,292,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00812.warc.gz | 137,324,185 | 5,266 | Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
#### Linear Interpolation Frequency Response
Since linear interpolation is a convolution of the samples with a triangular pulse (from Eq.(4.5)), the frequency response of the interpolation is given by the Fourier transform , which yields a sinc function. This frequency response applies to linear interpolation from discrete time to continuous time. If the output of the interpolator is also sampled, this can be modeled by sampling the continuous-time interpolation result in Eq.(4.5), thereby aliasing the sinc frequency response, as shown in Fig.4.9.
In slightly more detail, from , and sinc , we have
sinc
where we used the convolution theorem for Fourier transforms, and the fact that sinc .
The Fourier transform of is the same function aliased on a block of size Hz. Both and its alias are plotted in Fig.4.9. The example in this figure pertains to an output sampling rate which is times that of the input signal. In other words, the input signal is upsampled by a factor of using linear interpolation. The main lobe'' of the interpolation frequency response contains the original signal bandwidth; note how it is attenuated near half the original sampling rate ( in Fig.4.9). The sidelobes'' of the frequency response contain attenuated copies of the original signal bandwidth (see the DFT stretch theorem), and thus constitute spectral imaging distortion in the final output (sometimes also referred to as a kind of aliasing,'' but, for clarity, that term will not be used for imaging distortion in this book). We see that the frequency response of linear interpolation is less than ideal in two ways:
• The spectrum is rolled'' off near half the sampling rate. In fact, it is nowhere flat within the passband'' (-1 to 1 in Fig.4.9).
• Spectral imaging distortion is suppressed by only 26 dB (the level of the first sidelobe in Fig.4.9.
These qualitative remarks apply to all upsampling factors using linear interpolation. The case is considered in the next section.
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https://www.physicsforums.com/threads/verifying-gravity-from-our-lab.288653/ | 1,726,482,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00799.warc.gz | 874,289,331 | 19,353 | # Verifying gravity from our lab.
• agnosticjoe
In summary, the student calculated the gravity by using the general solution to the position equation of motion and using the slope of a linear fit. However, the slope was incorrect and he needed to use a more sophisticated analysis to find the gravity.
agnosticjoe
## Homework Statement
We had our lab yesterday in engineering physics where we had a car with nearly no friction descending down an incline plane. The incline was at 16 degrees, and we used data studio to record a crazy amount of trials at various distances. The point of this is to calculate gravity.
I should also add, that there was a time taken from when the car passed two gates.. so we have an average of distances vs times (10 sets of data).. that's what the slope is.
2. Homework Equations and what I've tried
Originally for the prelab i solved what i considered to be the correct general solution to find acceleration from the position equation of motion.
I came up with -2( r sin 16) / t^2. This is obviously wrong because I'm not getting anywhere near gravity with this.
I'm pretty sure I'm supposed to use the slope of a linear fit of the data points somehow into this, but for some reason i don't really understand how it relates to this equation. I thought the slope for the distance vs time^2 was supposed to be the acceleration.. guess I'm wrong.. anything to help me understand this and get me to the desired solution would be amazing.
Last edited:
It isn't clear what has been measured.
Did you measure the time for the car to roll down a certain distance from an initial position where the speed was zero? (Was the first gate practically at the starting position?)
If so, you could apply the formula d = 0.5at^2 and solve for the acceleration down the ramp. This would be less than 9.81 m/s^2 because of the angle of the ramp, so it would have to be divided by sin(16). Pretty much what your formula says.
You could calculate the acceleration from each of your data sets and average them.
A more sophisticated analysis would show the data on a graph so that the trend of the data can be seen and found with a best fit line. To do this, you must see
d = 0.5at^2 as a linear equation corresponding to y = mx + b.
What will you choose as the "y" to put on the vertical axis of the graph?
What will you choose as the "x" to put on the horizontal axis?
You want the slope or "m" to be the thing you want to find - acceleration.
I'm sorry i was too vague. I measured distances between two gates with a photo lens. The time it took for the car to roll down from point a to point b was also measured. Then those are the graphs i did the fit on.
I've been trying things for the last hour non stop.. so far I'm here.
2(dx/dt^2) = g sin 16 dx/dt^2 = 1.373x+.112 x = .8900s x=t where t was the average time value of the ten trials. dx/dt^2 was the slope of the fit on d vs. t^2
I get 9.8m/s^2 if i do this.. but i don't know if how i set it up is valid.
y axis is distance x-axis is time squared
I measured distances between two gates
Is the first gate at the top, or starting point, where the speed is zero?
Do points a and b refer to the locations of these two gates?
I have no way of understanding how you found 2(dx/dt^2) so I can't say if it is correct. Your derivation should begin with some commonly known formula. Is dx/dt a derivative?
delta x= v naught *t +.5at^2
initial velocity was indeed zero
a=2(change of x/ chang of t^2)
then change of x/ change of y=1.373(.8900)+.112 i let x = the average time it took on all the runs. that's the equation for the slope so i figured since the fit line was on the distance vs. time squared
that is equal to acceleration.
then i found g sin 16 = a from a free body diagram
g sin 16 = 2(1.373(.8900)+.112)
then i did 2(1.373(.8900)+.112)/ sin 16
Sorry for being so crude I've never tried talking geek on a forum :). I'll learn how to do the real symbols soon. And learn to be more clear on what I'm trying to get across.
## 1. How is gravity verified in a lab?
Gravity can be verified in a lab using various experiments and equipment such as pendulums, mass and weight measurements, and free fall experiments.
## 2. Can gravity be manipulated in a lab setting?
No, gravity cannot be manipulated in a lab. It is a natural force that cannot be controlled or altered.
## 3. How does verifying gravity in a lab contribute to scientific understanding?
By verifying gravity in a lab, scientists can confirm the mathematical equations and theories that describe the force of gravity. This contributes to our overall understanding of the laws of physics and the behavior of objects in the universe.
## 4. What are some potential sources of error in verifying gravity in a lab?
Potential sources of error in verifying gravity in a lab could include inaccuracies in equipment, external factors such as air resistance, and human error in conducting experiments or recording data.
## 5. Are lab experiments the only way to verify gravity?
No, lab experiments are not the only way to verify gravity. Other methods such as astronomical observations and theoretical calculations can also be used to verify the existence and behavior of gravity.
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6K | 1,335 | 5,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-38 | latest | en | 0.973659 |
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A278018 Number of 5Xn 0..2 arrays with every element plus 1 mod 3 equal to some element at offset (-1,-1) (-1,0) (-1,1) (0,-1) (0,1) or (1,0), with upper left element zero. 1
0, 1230, 121826, 14786448, 1772657688, 212235186204, 25411730935962, 3042585288805250, 364290594311156506, 43616685036230806738, 5222245347063515784452, 625261742374783468880692 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Row 5 of A278014. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 EXAMPLE Some solutions for n=3 ..0..1..0. .0..1..2. .0..1..2. .0..1..2. .0..2..1. .0..1..2. .0..0..2 ..0..2..1. .0..0..0. .0..2..0. .1..2..0. .1..0..2. .0..0..0. .1..1..0 ..2..1..0. .1..2..1. .2..1..2. .0..0..1. .2..2..2. .0..1..0. .2..0..0 ..1..2..0. .0..1..0. .2..0..0. .1..2..1. .2..1..0. .2..2..1. .0..1..0 ..1..2..2. .0..1..2. .2..1..0. .0..1..1. .0..2..2. .0..1..1. .2..0..0 CROSSREFS Cf. A278014. Sequence in context: A368637 A250894 A251405 * A182721 A183881 A270540 Adjacent sequences: A278015 A278016 A278017 * A278019 A278020 A278021 KEYWORD nonn AUTHOR R. H. Hardin, Nov 08 2016 STATUS approved
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Last modified June 19 03:10 EDT 2024. Contains 373492 sequences. (Running on oeis4.) | 680 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.681461 |
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