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https://tura-msu.ru/literature/fisica-universitaria-sears-zemansky-solucionario-espaol.php | 1,627,390,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00645.warc.gz | 580,543,474 | 8,719 | # FISICA UNIVERSITARIA SEARS ZEMANSKY SOLUCIONARIO ESPAOL PDF
Units, Physical Quantities and Vectors. We know the vector sum and want to find the magnitude of the vectors. Use the method of. The two vectors A.
Author: Kiran Karr Country: Dominica Language: English (Spanish) Genre: Photos Published (Last): 19 March 2017 Pages: 407 PDF File Size: 3.94 Mb ePub File Size: 18.13 Mb ISBN: 871-4-88985-969-1 Downloads: 9204 Price: Free* [*Free Regsitration Required] Uploader: Zulkizshura
The problem says instead to assume a day year. The radius is then found from the volume equation for a sphere and the result for the volume. Compare that number to the number of seconds in a year. We know the total distance to only three significant figures. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures.
When we add or subtract numbers it is the location of the decimal that matters. The sum of these fractional uncertainties in the length and width are 5.
SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target variables and from these we get the uncertainty. We can find the uncertainty in the volume as follows. By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8.
The smallest possible value of the ratio is 8. Thus the uncertainty is 20 and we write the ratio as Express each of m, cm and mm in inches. Express months in years. This is much larger than the typical weight of a man. This is much greater than the height of a person. Some people are this tall, but not an ordinary man. This is much too short. This is the age of a teenager; a middle-aged man is much older than this. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed.
There are ft in a mile but only ft in a km. Convert 1. SET UP: 1. Embed Size px x x x x Fisica universitaria-sears-zemanskyava-edicion-vol1 - copia Documents. Solucionario Fisica universitaria Sears Zemansky 12va edicion Education. Fisica Universitaria Sears Zemansky 11 Ed. Vol 2. Fisica universitaria sears zemansky 11 edicion volumen 2 cap 30 superior Education. Fisica universitaria vol. Capitulo 2 Sears Zemansky Documents. Fisica universitaria sears zemansky vol 1 Education.
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HILLFOLK RPG PDF
## solucionario fisica universitaria sears zemansky 12 edicion
.
EINFHRUNG IN AUTOMATENTHEORIE FORMALE SPRACHEN UND BERECHENBARKEIT PDF
. | 705 | 2,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-31 | latest | en | 0.893724 |
http://rjlipton.wordpress.com/2011/12/06/theorems-are-forever-a-great-one-from-1492/ | 1,419,160,903,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802771091.111/warc/CC-MAIN-20141217075251-00132-ip-10-231-17-201.ec2.internal.warc.gz | 239,266,490 | 22,446 | Make that 1942
Raphaël Salem was a great mathematician who worked mainly on Fourier analysis. He will always be remembered for his many beautiful contributions and for the Salem Prize that is named after him. Also he has his own class of real numbers: a real algebraic number ${\alpha>1}$ is a Salem number if all its conjugates have absolute value at most ${1}$ and one has absolute value exactly one. See this for a list of known Salem numbers.
Today I want to discuss a results that is simple to state, that is widely used, and that has a clever proof.
One of the great things about proving mathematical theorems is that they are forever. We still talk about results from Euclid, 2300 years later; perhaps we will talk about your result years from now. Franchino Gaffurio noted the importance of arithmetic progressions to harmony in his treatise Theorica musicae in 1492. The result in question is a bit more recent—it was proved in 1942 by Salem and Donald Spencer.
What made me think about their theorem is that it is used in very diverse areas of mathematics. For example, it plays a role in the study of ${\omega}$, the matrix product exponent, and also was used recently by Dieter van Melkebeek in a proof of a beautiful result on the complexity of ${\mathsf{SAT}}$.
These applications of the theorem are quite different: one is looking for patterns in trilinear forms, and one is looking for the existence of a protocol that Alice and Bob can use to solve instances of boolean satisfaction. Very different problems, I think anyone would agree. Yet the same theorem is used.
Let’s turn to the statement of this wonderful theorem. Perhaps you will be able to use it in some proof in the future.
The Theorem
The paper of Salem and Spencer is titled: On Sets Of Integers Which Contain No Three Terms In Arithmetical Progression. The title kind-of gives away the whole result. This was probably more important in 1942, since a paper with a title like: A Result On Patterns In Numbers would be hard to find. Recall there was no search inside papers in those days. Nowadays we can use any title at all and people can still search and find your result. They can even search to see if you’ve worked on a particular equation. But then a good title was worth many citations.
Let ${S}$ be a set of integers. Say that the set is progression-free if for any three distinct integers ${x,y,z}$ in the set,
$\displaystyle x + y \neq 2z.$
Note that this is the same as saying that there is no progression of length three in the set: if
$\displaystyle x, x + \Delta, x +2\Delta$
are all in the set with ${\Delta > 0}$, then
$\displaystyle x + (x + 2\Delta) = 2(x + \Delta).$
Salem and Spencer defined ${\nu(N)}$ to be the maximum number of elements that can be in a subset of ${\{1,\dots,N\}}$ that is progression-free. They proved:
Theorem: For any ${\epsilon>0}$,
$\displaystyle \nu(N) > N^{1-\frac{\log 2 + \epsilon}{\log \log N}}.$
Note this means that ${\nu(N)}$ is extremely close to ${N}$, and that there always exists a set ${S}$ that is very dense yet avoids every three-term progression.
Unconventional Wisdom, Again
In their introduction they pointed out that at the time the result was surprising, since then the conventional wisdom was that ${\nu(N)}$ had to behave like ${N^{\alpha}}$ where ${\alpha < 1}$. Clearly their result is much stronger than this—again people guessed wrong. Here the people are Paul Erdös and Paul Turán who made the guess. In that same paper Erdös and Turán made a small calculation error that was pointed out in one of the shortest papers I have every seen. It was in the Journal of the London Mathematical Society:
As I have quoted before, from Roger Brockett:
“It is hard to argue with a counter-example.”
The Proof
I will not give the full proof as it is quite nicely given in the original paper, or take a look at a more modern treatment here.
The main idea of the proof is very simple. Let ${x}$ be a number in the range ${1,\dots,N}$ and use ${v_{x}}$ to denote the vector ${v_{0},\dots,v_{m}}$ where
$\displaystyle x = \sum_{k=0}^{m} v_{k}10^{k}.$
The vector ${v_{x}}$ is just the vector of the digits of ${x}$ in decimal notation. Look at the set ${S}$ of such ${x}$ so that all their digits restricted to be in the set ${\{0,1,2,3,4\}}$. The the critical observation is that for ${x}$ and ${y}$ in the set ${S}$,
$\displaystyle x + y = z \text{ if and only if } v_{x} + v_{y} = v_{z},$
where the latter sum is the vector sum of the digits. This follows since there is no possible carry in adding ${x}$ and ${y}$. For example,
$\displaystyle \begin{array}{rcl} &1234231301 \\ +&4023141312 \\ &----------------\\ &5257372613. \end{array}$
The reason this is an important observation is: this replaces looking for a set without a three progression to a set of vectors that contains no line. This is much easier, as you might expect. The remainder of the proof is to replace decimal by a higher radix and to replace the restriction on the digits by a more complex one. The idea is to restrict the digits so that the sum of integers with the restriction are mostly like a vector sum.
See the paper for the details—it is less than three pages long and not hard to follow. Also see this recent paper by Bill Gasarch, James Glenn, and Clyde Kruskal, which covers constructions of 3-free sets for small ${n}$ with tables for ${n \leq 250}$, and their followup on theory and heuristics for higher ${n}$. See also this 2010 post by Bill referencing also this by Gil Kalai on a “breakthrough” and “barrier.”
Open Problems
Can you use this wonderful theorem in your work? Do you know of another theorem like this that is widely used in very different parts of mathematics?
1. December 6, 2011 9:47 am
I ought to know the answer to this, but still. Your description of the proof could be a description of the Behrend construction of 1946. Was Behrend’s contribution to this story just a better optimization of the parameters or was the idea of taking a sphere his too? If the latter, then what sort of line-avoiding set did Salem and Spencer go for?
• December 6, 2011 10:27 am
Behrend introduced the use of a sphere and got better asymptotics. In their original paper, Salem and Spencer just used two properties for numbers written in base d with d large: having small enough digits that no carrying can occur when doing the addition, and having each of those digits occur exactly the same number of times.
December 6, 2011 6:01 pm
The linked paper is an improvement by Behrend (the same one mentioned in gower’s comment?). I think this is the Salem and Spencer paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1078539/pdf/pnas01647-0055.pdf
December 8, 2011 10:38 am
It is unusual that the representation of numbers is useful. I know of
two brainteaser like situation where it is.
Firstly:
\sum_{n=0}^\infinity a^n
But if a is a positive integer there is an unusual way to compute the
series: it is equal to 1.11111.. in base a. And by doing long hand
division it is easy to check that this is equal to 10/(a-1) (where 10
is in a in base a).
Secondly:
Suppose you have a biased coin that gives head 20% of the times, and
tail 80% of the time. You can flip this coin to make a draw of a
binary random variable that has an equal probability to be 0 or 1. But
can you do so with a bounded number of flips? The answer is no.
Suppose you could do so in at most N flips. Then you could partition
the 2^N possible sequence of flips into two sets of probability 1/2
each. The probability of each partition is equal to the sum of the
probabilities of the sequences comprising it, each of which is equal
to 0.2^n 0.8^{N-n} for some n. It is clear this is impossible if you
write everything in base 5: the probability of each sequence has a
finite expansion in base five, so their sum can never be equal to
1/2 = 0.22222222… which has an infinite expansion.
I wonder if there is a way to generalize either of these to more
general cases?
4. December 8, 2011 12:50 pm
Amusingly, the 1942 work of Salem and Spencer is the topic of the very first English language mathematical article that was ever called a “breakthrough” by any MathSciNet reviewer. Was this intentional on the part of Gödel’s Lost Letter and P=NP … or was it simply good taste in choosing “breakthrough” topics?
Specifically, the first-ever-theorem-to-be-called by MathSciNet reviewers a “breakthrough” was Robert Rankin’s improvement and extension of the Salem-Spencer bound as stated in Rankin’s 1960 article “Sets of integers containing not more than a given number of terms in arithmetical progression”, which was reviewed by Tom Apostol (MR0142526).
Since Apostol’s “breakthrough” in first applying the term “breakthrough” to mathematics, more than eight hundred subsequent MathSciNet reviews have followed suit. ;)
In aggregate, what does this rising tide of mathematical breakthroughs mean, and what does it portend for the 21st century? This I take to be among the most natural and important questions that are associated to the GASARCH/Fortnow Computational Complexity weblog’s presently active topic “What is a Breakthrough?” | 2,272 | 9,175 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.973175 |
http://www.jiskha.com/display.cgi?id=1352865775 | 1,498,549,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321025.86/warc/CC-MAIN-20170627064714-20170627084714-00261.warc.gz | 549,887,739 | 3,637 | # Calculus
posted by .
Find the absolute maximum & minimum of the function f (x)=e^x for any closed interval [a,b] Justify your answer.
• Calculus -
since f' = e^x > 0, f(x) is always increasing.
So, on [a,b] min=e^a and max is e^b | 76 | 236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-26 | latest | en | 0.891684 |
https://everything.explained.today/Lift-to-drag_ratio/ | 1,695,885,694,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00042.warc.gz | 272,355,818 | 10,076 | # Lift-to-drag ratio explained
In aerodynamics, the lift-to-drag ratio (or L/D ratio) is the lift generated by an aerodynamic body such as an aerofoil or aircraft, divided by the aerodynamic drag caused by moving through air. It describes the aerodynamic efficiency under given flight conditions. The L/D ratio for any given body will vary according to these flight conditions.
For an aerofoil wing or powered aircraft, the L/D is specified when in straight and level flight. For a glider it determines the glide ratio, of distance travelled against loss of height.
The term is calculated for any particular airspeed by measuring the lift generated, then dividing by the drag at that speed. These vary with speed, so the results are typically plotted on a 2-dimensional graph. In almost all cases the graph forms a U-shape, due to the two main components of drag. The L/D may be calculated using computational fluid dynamics or computer simulation. It is measured empirically by testing in a wind tunnel or in free flight test.[1] [2] [3]
The L/D ratio is affected by both the form drag of the body and by the induced drag associated with creating a lifting force. It depends principally on the lift and drag coefficients, angle of attack to the airflow and the wing aspect ratio.
The L/D ratio is inversely proportional to the energy required for a given flightpath, so that doubling the L/D ratio will require only half of the energy for the same distance travelled. This results directly in better fuel economy.
The L/D ratio can also be used for water craft and land vehicles. The L/D ratios for hydrofoil boats and displacement craft are determined similarly to aircraft.
## Lift and drag
Lift can be created when an aerofoil-shaped body travels through a viscous fluid such as air. The aerofoil is often cambered and/or set at an angle of attack to the airflow. The lift then increases as the square of the airspeed.
Whenever an aerodynamic body generates lift, this also creates lift-induced drag or induced drag. At low speeds an aircraft has to generate lift with a higher angle of attack, which results in a greater induced drag. This term dominates the low-speed side of the graph of lift versus velocity.Form drag is caused by movement of the body through air. This type of drag, known also as air resistance or profile drag varies with the square of speed (see drag equation). For this reason profile drag is more pronounced at greater speeds, forming the right side of the lift/velocity graph's U shape. Profile drag is lowered primarily by streamlining and reducing cross section.
The total drag on any aerodynamic body thus has two components, induced drag and form drag.
### Lift and drag coefficients
The rates of change of lift and drag with angle of attack (AoA) are called respectively the lift and drag coefficients CL and CD. The varying ratio of lift to drag with AoA is often plotted in terms of these coefficients.
For any given value of lift, the AoA varies with speed. Graphs of CL and CD vs. speed are referred to as drag curves. Speed is shown increasing from left to right. The lift/drag ratio is given by the slope from the origin to some point on the curve and so the maximum L/D ratio does not occur at the point of least drag coefficient, the leftmost point. Instead, it occurs at a slightly greater speed. Designers will typically select a wing design which produces an L/D peak at the chosen cruising speed for a powered fixed-wing aircraft, thereby maximizing economy. Like all things in aeronautical engineering, the lift-to-drag ratio is not the only consideration for wing design. Performance at a high angle of attack and a gentle stall are also important.
## Glide ratio
As the aircraft fuselage and control surfaces will also add drag and possibly some lift, it is fair to consider the L/D of the aircraft as a whole. The glide ratio, which is the ratio of an (unpowered) aircraft's forward motion to its descent, is (when flown at constant speed) numerically equal to the aircraft's L/D. This is especially of interest in the design and operation of high performance sailplanes, which can have glide ratios almost 60 to 1 (60 units of distance forward for each unit of descent) in the best cases, but with 30:1 being considered good performance for general recreational use. Achieving a glider's best L/D in practice requires precise control of airspeed and smooth and restrained operation of the controls to reduce drag from deflected control surfaces. In zero wind conditions, L/D will equal distance traveled divided by altitude lost. Achieving the maximum distance for altitude lost in wind conditions requires further modification of the best airspeed, as does alternating cruising and thermaling. To achieve high speed across country, glider pilots anticipating strong thermals often load their gliders (sailplanes) with water ballast: the increased wing loading means optimum glide ratio at greater airspeed, but at the cost of climbing more slowly in thermals. As noted below, the maximum L/D is not dependent on weight or wing loading, but with greater wing loading the maximum L/D occurs at a faster airspeed. Also, the faster airspeed means the aircraft will fly at greater Reynolds number and this will usually bring about a lower zero-lift drag coefficient.
## Theory
### Subsonic
Mathematically, the maximum lift-to-drag ratio can be estimated as:
(L/D)max=
1 \sqrt{ 2
\pi\varepsilonAR CD,0
},[4]
where AR is the aspect ratio,
\varepsilon
the span efficiency factor, a number less than but close to unity for long, straight edged wings, and
CD,0
the zero-lift drag coefficient.
Most importantly, the maximum lift-to-drag ratio is independent of the weight of the aircraft, the area of the wing, or the wing loading.
It can be shown that two main drivers of maximum lift-to-drag ratio for a fixed wing aircraft are wingspan and total wetted area. One method for estimating the zero-lift drag coefficient of an aircraft is the equivalent skin-friction method. For a well designed aircraft, zero-lift drag (or parasite drag) is mostly made up of skin friction drag plus a small percentage of pressure drag caused by flow separation. The method uses the equation:
CD,0
=C
fe Swet Sref
,
[5]
where
Cfe
is the equivalent skin friction coefficient,
Swet
is the wetted area and
Sref
is the wing reference area. The equivalent skin friction coefficient accounts for both separation drag and skin friction drag and is a fairly consistent value for aircraft types of the same class. Substituting this into the equation for maximum lift-to-drag ratio, along with the equation for aspect ratio (
2/S b ref
), yields the equation:
(L/D)max=
1 \sqrt{ 2
\pi\varepsilon Cfe
b2 Swet
}
where b is wingspan. The term
2/S b wet
is known as the wetted aspect ratio. The equation demonstrates the importance of wetted aspect ratio in achieving an aerodynamically efficient design.
### Supersonic
At very great speeds, lift to drag ratios tend to be lower. Concorde had a lift/drag ratio of about 7 at Mach 2, whereas a 747 is about 17 at about mach 0.85.
Dietrich Küchemann developed an empirical relationship for predicting L/D ratio for high Mach:[6]
L/Dmax=
4(M+3) M
where M is the Mach number. Windtunnel tests have shown this to be approximately accurate.
## Examples of L/D ratios
L1011-100 14.5 Nov 16, 1970 DC-10-40 13.8 Aug 29, 1970 A300-600 15.2 Oct 28, 1972 16.1 Jan 10, 1990 B767-200ER 16.1 Sep 26, 1981 A310-300 15.3 Apr 3, 1982 B747-200 15.3 Feb 9, 1969 15.5 Apr 29, 1988 B757-200 15 Feb 19, 1982 A320-200 16.3 Feb 22, 1987 A310-300 18.1 Nov 2, 1992 A340-200 19.2 Apr 1, 1992 A340-300 19.1 Oct 25, 1991 B777-200 19.3 Jun 12, 1994
## References
[13]
1. https://journals.sagepub.com/doi/pdf/10.1177/1729881418766190 Accurate calculation of aerodynamic coefficients of parafoil airdrop system based on computational fluid dynamic
2. https://www.diva-portal.org/smash/get/diva2:342732/fulltext01.pdf Validation of software for the calculation of aerodynamic coefficients
3. https://www.mdpi.com/2226-4310/8/2/43/pdf In-flight Lift and Drag Estimation of an Unmanned Propeller-Driven Aircraft
4. Web site: Loftin, LK Jr.. Quest for performance: The evolution of modern aircraft. NASA SP-468. 2006-04-22.
5. Book: Raymer. Daniel. Aircraft Design: A Conceptual Approach. 2012. AIAA. New York. 5th.
6. http://www.aerospaceweb.org/design/waverider/design.shtml Aerospaceweb.org Hypersonic Vehicle Design
7. Web site: Lift-to-Drag Ratios . Antonio Filippone . Advanced topics in aerodynamics . dead . https://web.archive.org/web/20080328133906/http://aerodyn.org/HighLift/ld-tables.html . March 28, 2008 .
8. Book: Cumpsty, Nicholas. Jet Propulsion. Cambridge University Press. 2003. 4.
9. Book: The Concorde Story . Christopher Orlebar . 116 . Osprey Publishing . 1997 . 9781855326675 .
10. Book: Leishman, J. Gordon . Principles of helicopter aerodynamics . 230 . Cambridge University Press . 24 April 2006 . 0521858607 . The maximum lift-to-drag ratio of the complete helicopter is about 4.5.
11. U2 Developments transcript . . 1960 . YouTube . 2016-03-05 . 2022-06-19 . https://web.archive.org/web/20220619041725/https://www.youtube.com/watch?v=nQnBJrj_-l8 . dead .
• Web site: June 4, 2013 . U2 Developments . https://web.archive.org/web/20130816051125/https://www.cia.gov/library/video-center/video-transcripts/u2-developments.html . 2013-08-16 . Central Intelligence Agency .
12. David Noland . The Ultimate Solo . Popular Mechanics . February 2005 .
13. Cessna Skyhawk II Performance Assessment http://temporal.com.au/c172.pdf | 2,363 | 9,637 | {"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-2023-40 | latest | en | 0.947135 |
https://solvedlib.com/n/suemmissions-useduse-green-theorcmlvuluatesv-dx-1rlv-x27,14895095 | 1,722,998,145,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00585.warc.gz | 441,146,018 | 20,125 | #### Similar Solved Questions
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##### Problem 3. Show the formula P((An B)U(A n B))- P(A) +P(B)-2P(AnB), which givgs the probability that...
Problem 3. Show the formula P((An B)U(A n B))- P(A) +P(B)-2P(AnB), which givgs the probability that exactly one of the events A and B will occur. [Compare with the formula P(AU B) P(A) P(B) - P(AnB), which gives the probability that at least one of the events A and B will occur.]... | 483 | 1,698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.835631 |
https://keep.lib.asu.edu/search?f%5B0%5D=all_subjects%3A44094&f%5B1%5D=all_subjects%3A115047&f%5B2%5D=genre%3ADoctoral%20Dissertation&f%5B3%5D=linked_agents%3ACarlson%2C%20Marilyn&f%5B4%5D=linked_agents%3AJoshua%2C%20Surani%20Ashanthi&f%5B5%5D=member_of%3AASU%20Electronic%20Theses%20and%20Dissertations&f%5B6%5D=member_of%3ATheses%20and%20Dissertations | 1,701,850,538,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00758.warc.gz | 368,397,439 | 17,374 | Matching Items (3)
#### Students' ways of thinking about two-variable functions and rate of change in space
Description
This dissertation describes an investigation of four students' ways of thinking about functions of two variables and rate of change of those two-variable functions. Most secondary, introductory algebra, pre-calculus, and first and second semester calculus courses do not require students to think about functions of more than one variable. Yet
This dissertation describes an investigation of four students' ways of thinking about functions of two variables and rate of change of those two-variable functions. Most secondary, introductory algebra, pre-calculus, and first and second semester calculus courses do not require students to think about functions of more than one variable. Yet vector calculus, calculus on manifolds, linear algebra, and differential equations all rest upon the idea of functions of two (or more) variables. This dissertation contributes to understanding productive ways of thinking that can support students in thinking about functions of two or more variables as they describe complex systems with multiple variables interacting. This dissertation focuses on modeling the way of thinking of four students who participated in a specific instructional sequence designed to explore the limits of their ways of thinking and in turn, develop a robust model that could explain, describe, and predict students' actions relative to specific tasks. The data was collected using a teaching experiment methodology, and the tasks within the teaching experiment leveraged quantitative reasoning and covariation as foundations of students developing a coherent understanding of two-variable functions and their rates of change. The findings of this study indicated that I could characterize students' ways of thinking about two-variable functions by focusing on their use of novice and/or expert shape thinking, and the students' ways of thinking about rate of change by focusing on their quantitative reasoning. The findings suggested that quantitative and covariational reasoning were foundational to a student's ability to generalize their understanding of a single-variable function to two or more variables, and their conception of rate of change to rate of change at a point in space. These results created a need to better understand how experts in the field, such as mathematicians and mathematics educators, thinking about multivariable functions and their rates of change.
ContributorsWeber, Eric David (Author) / Thompson, Patrick (Thesis advisor) / Middleton, James (Committee member) / Carlson, Marilyn (Committee member) / Saldanha, Luis (Committee member) / Milner, Fabio (Committee member) / Van de Sande, Carla (Committee member) / Arizona State University (Publisher)
Created2012
#### Investigating the Role of Relative Size Reasoning in Students’ Understanding of Precalculus Ideas
Description
The ideas of measurement and measurement comparisons (e.g., fractions, ratios, quotients) are introduced to students in elementary school. However, studies report that students of all ages have difficulty comparing two quantities in terms of their relative size. Students often understand fractions such as 3/7 as part-whole relationships or “three out
The ideas of measurement and measurement comparisons (e.g., fractions, ratios, quotients) are introduced to students in elementary school. However, studies report that students of all ages have difficulty comparing two quantities in terms of their relative size. Students often understand fractions such as 3/7 as part-whole relationships or “three out of seven.” These limited conceptions have been documented to have implications for understanding the quotient as a measure of relative size and when learning other foundational ideas in mathematics (e.g., rate of change). Many scholars have identified students’ ability to conceptualize the relative size of two quantities values as important for learning specific ideas such as constant rate of change, exponential growth, and derivative. However, few researchers have focused on students’ ways of thinking about multiplicatively comparing two quantities’ values as they vary together across select topics in precalculus. Relative size reasoning is a way of thinking one has developed when conceptualizing the comparison of two quantities’ values multiplicatively, as their values vary in tandem. This document reviews literature related to relative size reasoning and presents a conceptual analysis that leverages this research in describing what I mean by a relative size comparison and what it means to engage in relative size reasoning. I further illustrate the role of relative size reasoning in understanding rate of change, multiplicative growth, rational functions, and what a graph’s concavity conveys about how two quantities’ values vary together. This study reports on three beginning calculus students’ ways of thinking as they completed tasks designed to elicit students’ relative size reasoning. The data revealed 4 ways of conceptualizing the idea of quotient and highlights the affordances of conceptualizing a quotient as a measure of the relative size of two quantities’ values. The study also reports data from investigating the validity of a collection of multiple-choice items designed to assess students’ relative size reasoning (RSR) abilities. Analysis of this data provided insights for refining the questions and answer choices for these assessment items.
ContributorsLock, Kayla Ashley (Author) / Carlson, Marilyn (Thesis advisor) / Apkarian, Naneh (Thesis advisor) / Strom, April (Committee member) / Byerley, Cameron (Committee member) / Roh, Kyeong-Hah (Committee member) / Arizona State University (Publisher)
Created2023
#### Conceptualizing and Reasoning with Frames of Reference in Three Studies
Description
This dissertation reports three studies about what it means for teachers and students to reason with frames of reference: to conceptualize a reference frame, to coordinate multiple frames of reference, and to combine multiple frames of reference. Each paper expands on the previous one to illustrate and utilize the construct
This dissertation reports three studies about what it means for teachers and students to reason with frames of reference: to conceptualize a reference frame, to coordinate multiple frames of reference, and to combine multiple frames of reference. Each paper expands on the previous one to illustrate and utilize the construct of frame of reference. The first paper is a theory paper that introduces the mental actions involved in reasoning with frames of reference. The concept of frames of reference, though commonly used in mathematics and physics, is not described cognitively in any literature. The paper offers a theoretical model of mental actions involved in conceptualizing a frame of reference. Additionally, it posits mental actions that are necessary for a student to reason with multiple frames of reference. It also extends the theory of quantitative reasoning with the construct of a ‘framed quantity’. The second paper investigates how two introductory calculus students who participated in teaching experiments reasoned about changes (variations). The data was analyzed to see to what extent each student conceptualized the variations within a conceptualized frame of reference as described in the first paper. The study found that the extent to which each student conceptualized, coordinated, and combined reference frames significantly affected his ability to reason productively about variations and to make sense of his own answers. The paper ends by analyzing 123 calculus students’ written responses to one of the tasks to build hypotheses about how calculus students reason about variations within frames of reference. The third paper reports how U.S. and Korean secondary mathematics teachers reason with frame of reference on open-response items. An assessment with five frame of reference tasks was given to 539 teachers in the US and Korea, and the responses were coded with rubrics intended to categorize responses by the extent to which they demonstrated conceptualized and coordinated frames of reference. The results show that the theory in the first study is useful in analyzing teachers’ reasoning with frames of reference, and that the items and rubrics function as useful tools in investigating teachers’ meanings for quantities within a frame of reference.
ContributorsJoshua, Surani Ashanthi (Author) / Thompson, Patrick W (Thesis advisor) / Carlson, Marilyn (Committee member) / Roh, Kyeong Hah (Committee member) / Middleton, James (Committee member) / Culbertson, Robert (Committee member) / Arizona State University (Publisher)
Created2019 | 1,624 | 8,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.934122 |
http://www.sediemoderne.net/srw8hc/distributive-property-of-whole-numbers-16da6a | 1,632,504,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00665.warc.gz | 100,769,855 | 11,233 | # distributive property of whole numbers
In these worksheets, students use the distributive property to multiply 1x2 digit numbers. 2 x (3 + 4) = 2x (7) = 14. Class 6 math (India) Unit: Whole numbers. Subtraction: a-b ≠ b-a. Keep whichever one is in the parentheses. Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. This is the currently selected item. Khan Academy is a 501(c)(3) nonprofit organization. Note: It doesn’t matter if the operation is plus or minus. This is called the distributive property of multiplication over subtraction. If a, b and c are any three whole numbers, then a x (b + c) = ab + ac. Distributive Properties of Multiplication: The Distributive Property of Multiplication over Addition of Whole Numbers (the Distributive Property of Multiplication over Subtraction of Whole Numbers) shows us how multiplying a value times a sum (difference) may be broken into the sum (difference) of … The distributive property of multiplication tells us that 5 x (2 + 3) is the same as 5 x 2 + 5 x 3. 0. Skill Summary Legend (Opens a modal) Whole numbers on the number line. Distributive property of multiplication over subtraction is a very useful property that lets us simplify expressions in which we are multiplying a number by the difference of two other numbers. The different properties are associative property, commutative property, distributive property, inverse property, identity property and so on. This indicates that real numbers include natural numbers, whole numbers, integers, rational numbers, and irrational numbers. But take heart and know that fractions are just as easy to work with as any other number type. Apply properties of operations as strategies to multiply and divide. Do not do any calculations now, just make a choice. CCSS.Math.Content.6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Here is another way of showing the same thing, using bundles of ten. Intro to whole numbers (Opens a modal) Intro to the number line (Opens a modal) Practice. They actually use the distributive property, but we do not need to explain that to 4th grade students. So, distributive property over subtraction is proved. Otherwise subtraction is not possible in whole numbers. Our mission is to provide a free, world-class education to anyone, anywhere. Negative numbers. Distributive Property (i) Distributive property of multiplication over addition : Multiplication of whole numbers is distributive over addition. Hence, closure property holds good for multiplication of whole numbers. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. Distributive property. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers … For example, express 36 + 8 as 4 (9 + 2). Property 2. Commutative property: Commutative property states that there is no change in result though the numbers in an expression are interchanged. Donate or volunteer today! About. Some of the worksheets for this concept are Natural and whole numbers, Natural and whole numbers, Grade 4 supplement, Whole numbers using an area model to explain multiplication, Grade 5 supplement, Multiplying mixed numbers, Sample work from, Exercise work. Free interactive exercises to practice online or download as pdf to print. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. The distributive property is given by: a(b+c) = ab + ac. Property 1. For example, express 36 + 8 as 4 (9 + 2). (Distributive property of multiplication) (CCSS 3.OA.B.5) Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Property 4. The distributive property helps in making difficult problems simpler. $$W$$ is closed, associative and commutative under both addition and multiplication (but not under subtraction and division). Example: 1+2 = 2+1. In general, it refers to the distributive property of multiplication over addition or subtraction. Practice: Distributive property of whole numbers. The distributive property is the one which allows us to multiply the number by a group of numbers, which are added together. 6th standard ncert maths / cbse syllabus, chapter: 2 / whole numbers Successor and predecessor on the number line Get 3 of 4 questions to level up! The distributive property is one of the most frequently used properties in basic Mathematics. 4) Distributive property of multiplication over addition. Property 3. Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Break 46 into two parts: 40 and 6. The numbers that are neither rational nor irrational, say $$\sqrt{-1}$$, are NOT real numbers. If A and B are two whole numbers such that A > B or A = B, then A − B is a whole number. Properties and patterns for multiplication. Named the ‘Distributive Property (sometimes referred to as the distributive law) because in essence, you are distributing something as you separate or break it into parts. 6.NS.B.4: Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Then multiply those two parts separately by 3: 3 × 40 is 120, and 3 × 6 is 18. Then add these two partial results: 120 + 18 = 138. If A and B are two whole numbers, then in general A − B is not equal to B − A. If A is any whole number, then A − 0 = A but 0 − A is not defined. The distributive property makes numbers easier to work with. I can find the greatest common factor and least common multiple. Whole Numbers Distributive Property. We can use this to transform a difficult multiplication (3 x 27) into the sum of two easy multiplications (3x20 + 3x7). (a) Closure Property: If a and b are two whole numbers, then a × b = c will always be a whole number. The Distributive Property is easy to remember, if you recall that "multiplication distributes over addition". Which of the following calculations would you choose to calculate the number of yellow beads in this pattern? Example. Legend (Opens a modal) Possible mastery points. Site Navigation. The distributive property also can be used to simplify algebraic equations by eliminating the parenthetical portion of the equation. (CCSS 4.NBT.B.5) Closure property : 5 + 6 = 11 9 + 8 = 17 36 + 0 = 36 9 x 8 = 72 6 x 11 = 66 0 x 84 = 0 From the example we can conclude that when we add or multiply any two whole numbers we get a whole number. Improve your math knowledge with free questions in "Multiply a decimal by a 1-digit whole number using the distributive property" and thousands of other math skills. We will learn about the distributive property and its examples. 5) Identity for addition and multiplication. Distributive property in integer powers relative to multiplication. The property states that the product of a number and the difference of two other numbers is equal to the difference of the products. Properties of whole numbers The commutative property of addition and multiplication. Distributive Property worksheets and online activities. Fractions can strike fear in the hearts of many students. Hence, 2 x (3 + 4) = 2x3 + 2x4 Addition: a+b = b+a. Learn . $$0$$ is a whole number but it is NOT a natural number. Next lesson. Displaying top 8 worksheets found for - Whole Numbers Distributive Property. 3 × 40 = 120. Some of the worksheets for this concept are Natural and whole numbers, Sample work from, Distributive property whole number coefficients work 5, Grade 4 supplement, Whole numbers, Using order of operations, Exercise work, Natural and whole numbers. This leads us to the next question; which numbers are NOT real numbers? Distributive Property & Fractions. You can use the distributive property of multiplication to rewrite expression by distributing or breaking down a factor as a sum or difference of two numbers. If a, b and c are any two whole numbers, then a(b–c) = a×b – a×c. 46: 46: 46. Distributive property worksheets. Apply and extend previous understandings of numbers to the system of rational numbers. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. Commutative property holds for addition and multiplication but not for subtraction and division. 3=3, which is true. Distributive property of multiplication over subtraction Similar to the operation above, performing the distributive property with subtraction follows the same rules -- except you’re finding the difference instead of the sum. The product of a whole number with the difference of the two other whole numbers is equal to the difference of the products of the whole number with other two whole numbers. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. Multiply 3 × 46. More clearly, Practice Questions. Distributive Property Whole Numbers - Displaying top 8 worksheets found for this concept.. In algebra when we use the distributive property, we’re expanding (distributing). Common Core: 6.NS.4. Example : 2 x (3 + 4) = 2x3 + 2x4 = 6 + 8 = 14. Distributive Property: If x,y and z are three whole numbers, the distributive property of multiplication over addition is x*(y+z)=(x*y)+(x*z), similarly the distributive property of multiplication over subtraction is x*(y-z)=(x*y)-(x*z) Multiplication by zero: When a whole number is multiplied to 0, … Here, for instance, calculating 8 … Take for instance the equation a(b + c), which also can be written as (ab) + (ac) because the distributive property dictates that a, which is outside the parenthetical, must be multiplied by both b and c. For example, express 36 + 8 as 4 (9 + 2). Having reviewed these definitions, it may then be a little easier to cover the property pointed to by Mathematics in reference to the Distributive Property that can take place in any operation of multiplication of powers of integers. Suggested Learning Targets. These numbers include the set of complex numbers, $$C$$. It is also known as the distributive law of multiplication. Unit: Whole numbers. For example, express 36 + 8 as 4 (9 + 2). Negative numbers, fractions and decimals are neither natural numbers nor whole numbers unless they can be simplified as a natural number or whole number. Both addition and multiplication ( 0\ ) is a whole number, then in general a b! Is distributive over addition re expanding ( distributing ) are two whole numbers - top., distributive property is given by: a ( b+c ) = ab ac. − 0 = a but 0 − a is not defined is closed, associative and commutative both. Under both addition and multiplication but not for subtraction and division ) 120! This indicates that real numbers of multiplication over addition '' algebra when we use the distributive property numbers. Two whole numbers, then a − b is not equal to b − a is any number! 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( 0\ ) is closed, associative and commutative under both addition and multiplication over addition to whole the! Factor and least common multiple or minus = a but 0 − a is not a natural.. Previous understandings of numbers to the system of rational numbers, then a x 3! Nor irrational, say \ ( W\ ) is a 501 ( c ) = 14 property also can used. / whole numbers distributive property also can be used to simplify algebraic equations by eliminating the portion! Fear in the hearts of many students subtraction and division ) a,! Whole number but it is also known as the distributive property is given by: a ( b+c =! Any whole number, then a ( b–c ) = 2x3 + =!: 2 / whole numbers, \ ( \sqrt { -1 } \ ), are not real numbers result. Inverse property, but we do not do any calculations now, just make a choice numbers. 18 = 138 given by: a ( b+c ) = ab +.. Easy to work with as any distributive property of whole numbers number type in general a 0! 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A x ( b + c ) ( 3 ) nonprofit organization property worksheets, express 36 8! Heart and know that fractions are just as easy to work with / syllabus! Holds good for multiplication of whole numbers a whole number, then a − b is not.! 36 + 8 = 14 of ten: 120 + 18 = 138 = 2x3 + =! Irrational numbers, but we do not need to explain that to grade! Note: it doesn ’ t matter if the operation is plus or minus, distributive property of multiplication addition... Used to simplify algebraic equations by eliminating the parenthetical portion of the.! A natural number to whole numbers, integers, rational numbers, then −! Irrational, say \ ( C\ ) to work with as any other number.. Over addition or subtraction of complex numbers, which are added together which are added together learn about distributive... / whole numbers is distributive over addition ) intro to whole numbers on the of. To calculate the number of yellow beads in this pattern using bundles of ten the same thing, bundles! That the product of a number and the difference of two other numbers is equal to the number (! An expression are interchanged found for this concept beads in this pattern India ) Unit: numbers! Closure property holds for addition and multiplication ( but not for subtraction and division ) equal to b a! 8 worksheets found for this concept frequently used properties in basic Mathematics number! Then a − b is not equal to the distributive property worksheets =.. Numbers in an expression are interchanged = a×b – a×c = 6 + 8 as 4 9!, calculating 8 … the distributive property distributive property of whole numbers multiplication over addition: of... Distributive property, we ’ re expanding ( distributing ) just make a choice of complex,! ( W\ ) is a 501 ( c ) = 2x3 + 2x4 = +... ’ t matter if the operation is plus or minus now, make.: multiplication of whole numbers, which are added together those two parts separately by 3: ×. 2X ( 7 ) = 14 whole numbers on the number line Get 3 4. 3 of 4 questions to level up you recall that multiplication distributes over addition { -1 \... 40 and 6 next question ; which numbers are not real numbers × 40 is 120, 3! The most frequently used properties in basic Mathematics mastery points 3 of 4 questions to level up to. Mission is to provide a free, world-class education to anyone, anywhere ( distributing.... Math ( India ) Unit: whole numbers, and 3 × is... I can find the greatest common factor and least common multiple to calculate the line. Multiply those two parts separately by 3: 3 × 40 is 120, and irrational numbers by a! A choice for this concept a group of numbers to the next question ; which numbers not! | 4,352 | 19,139 | {"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} | 4.34375 | 4 | CC-MAIN-2021-39 | latest | en | 0.878779 |
http://whatis.techtarget.com/definition/degree-per-second | 1,521,919,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650993.91/warc/CC-MAIN-20180324190917-20180324210917-00431.warc.gz | 327,877,674 | 21,712 | Browse Definitions:
Definition
# degree per second
The degree per second (symbolized deg/s or deg · s-1) is a unit of angular (rotational) speed. This quantity can be defined in either of two senses: average or instantaneous.
Average angular speed is obtained by measuring the angle in degrees through which an object rotates in a certain number of seconds, and then dividing the total angle by the time. If uavg represents the average angular speed of an object (in degrees per second) during a time interval t (in seconds), and the angle through which the object rotates in that time is equal to q (in degrees), then:
uavg = q / t
Instantaneous angular speed is more difficult to intuit, because it involves an expression of motion over an "infinitely short" interval of time. Let p represent a specific point in time. Suppose an object is in rotational motion at about that time. The average angular speed can be measured over increasingly short time intervals centered at p, for example:
[p-4, p+4]
[p-3, p+3]
[p-2 , p+2]
[p-1, p+1]
[p-0.5, p+0.5]
[p-0.25, p+0.25]
.
.
.
[p-x, p+x]
.
.
.
where the added and subtracted numbers represent seconds. The instantaneous angular speed, uinst, is the limit of the measured average speed as x approaches zero. This is a theoretical value, because it cannot be obtained except by inference from measurements made over progressively shorter time spans.
Also see angular degree, angular speed, angular velocity, radian per second, International System of Units (SI), and Table of Physical Units.
This was last updated in May 2008
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A crisis management plan (CMP) is a document that outlines the processes an organization will use to respond to a critical ...
• ### business continuity and disaster recovery (BCDR)
Business continuity and disaster recovery (BCDR) are closely related practices that describe an organization's preparation for ...
• ### business continuity plan (BCP)
A business continuity plan (BCP) is a document that consists of the critical information an organization needs to continue ...
## SearchStorage
A bad block is an area of storage media that is no longer reliable for storing and retrieving data because it has been physically...
• ### all-flash array (AFA)
An all-flash array (AFA), also known as a solid-state storage disk system, is an external storage array that uses only flash ...
• ### volume manager
A volume manager is software within an operating system (OS) that controls capacity allocation for storage arrays.
## SearchSolidStateStorage
• ### hybrid hard disk drive (HDD)
A hybrid hard disk drive is an electromechanical spinning hard disk that contains some amount of NAND Flash memory.
Close | 1,035 | 4,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-13 | latest | en | 0.912787 |
https://amon.in/quiz/2479-finding-patterns-in-text-algebric-laws-and-derivatives | 1,638,255,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00252.warc.gz | 164,310,538 | 4,057 | # Finding Patterns in Text,Algebric Laws and Derivatives Online Exam Quiz
Finding Patterns in Text,Algebric Laws and Derivatives GK Quiz. Question and Answers related to Finding Patterns in Text,Algebric Laws and Derivatives. MCQ (Multiple Choice Questions with answers about Finding Patterns in Text,Algebric Laws and Derivatives
### A CFG is ambiguous if
Options
A : It has more than one rightmost derivations
B : It has more than one leftmost derivations
C : No parse tree can be generated for the CFG
D : None of the mentioned
### A CFG is not closed under
Options
A : Dot operation
B : Union Operation
C : Concatenation
D : Iteration
### A->aAa|bAb|a|b|e^ Which among the following is the correct option for the given production?
Options
A : Left most derivation
B : Right most derivation
C : Recursive Inference
D : None of the mentioned
### Choose the incorrect process to check whether the string belongs to the language of certain variable or not?
Options
A : recursive inference
B : derivations
C : head to body method
D : All of the mentioned
Options
A : TRUE
B : FALSE
C : -
D : -
### Statement: Left most derivations are lengthy as compared to Right most derivations.^ Choose the correct option:
Options
A : correct statement
B : incorrect statement
C : may or may not be correct
D : depends on the language of the grammar
Options
A : 0
B : 1
C : 100
D : 22/7
### Which of the following are always unambiguous?
Options
A : Deterministic Context free grammars
B : Non-Deterministic Regular grammars
C : Context sensitive grammar
D : None of the mentioned
### Which of the following is a parser for an ambiguous grammar?
Options
A : GLR parser
B : Chart parser
C : All of the mentioned
D : None of the mentioned
### Which of the following is an real-world programming language ambiguity?
Options
A : dangling else problem
B : halting problem
C : maze problem
D : none of the mentioned
### Ambiguous Grammar more Online Exam Quiz
Applications of DFA
Applications of NFA
Applications of Pumping Lemma/Pigeonhole principle
CFG-Eliminating Useless Symbols
Deterministic Finite Automata-Introduction and Definition
From Grammars to Push Down Automata
Node-Cover Problem, Hamilton Circuit Problem
Problem Solvable in Polynomial Time
Properties-Non Regular Languages
Regular Language & Expression - 1 | 567 | 2,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.780695 |
https://oeis.org/A004969 | 1,670,086,554,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00240.warc.gz | 462,336,967 | 4,189 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A004969 a(n) = ceiling(n*phi^14), where phi is the golden ratio, A001622. 0
0, 843, 1686, 2529, 3372, 4215, 5058, 5901, 6744, 7587, 8430, 9273, 10116, 10959, 11802, 12645, 13488, 14331, 15174, 16017, 16860, 17703, 18546, 19389, 20232, 21075, 21918, 22761, 23604, 24447 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Tanya Khovanova, Non Recursions FORMULA a(n) >= A004949(n). - R. J. Mathar, Oct 28 2008 MATHEMATICA With[{c=GoldenRatio^14}, Ceiling[c*Range[0, 30]]] (* Harvey P. Dale, Jul 08 2019 *) CROSSREFS Sequence in context: A252568 A045243 A096025 * A004949 A252061 A031707 Adjacent sequences: A004966 A004967 A004968 * A004970 A004971 A004972 KEYWORD nonn AUTHOR STATUS approved
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Last modified December 3 11:54 EST 2022. Contains 358521 sequences. (Running on oeis4.) | 487 | 1,489 | {"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} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.656348 |
https://www.fredjones.com/anysubjeopardy | 1,718,329,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00277.warc.gz | 716,752,563 | 138,409 | top of page
## PAT Bank
Author: Patti Anderson, Westerly Hills Elementary, North Carolina
Subject: Any
Objective: Recalling Facts about any subject
Materials and Preparation: I use a (sentence strip) pocket chart and 3x5 note cards. For the older students, use the answer format (they have to phrase the "answer" as a question), but this is too hard for younger students.
I make however many question cards as I need with anwers on the back. Along with the answer on the back, I determine how hard the question is. An easy answer is a 100 card. The harder they are, the higher the number. You need 3 - 100 cards, 3 - 200 cards, etc. A good number of cards to make is 12 - 15. There are three colums of cards. Place all the cards on the pocket chart, in three columns. Place the 100 cards in the first row, the 200 cards in the second row, etc. You can place 4 - 5 cards in each column. I place 3 cards at the top of each column (I label them A B C but you can get creative). I make another set of cards labeled 100, 200, 300 etc. These are placed on top of the question cards. The first row has 100 placed over them, the second row has 200, etc.
Student Grouping: Two Teams
The Play:
1. The class is divided into 2 teams. I put team 1 and team 2 on the board, so we can keep score. The students rotate turns so that everyone gets to play.
2. The "play" goes from one team then over to the other team - alternating questions. Team 1 player chooses a column (ie:column B) and a number (200). Remind the students that the higher the number, the harder the question will be.
3. The teacher asks the question.
• If the students gives the right answer in a set amount of time, they get points for their team.
• If the students gets the wrong answer, the same question goes over to the player on the other team.
• If both players get it wrong, no points are awarded.
4. You can choose to place it back on the board, or not. The question only gets asked to both teams once. The next student chooses a card, etc.
Scoring: If a student gets a 100 card correct, I give 1 point to that team. If they get a 400 card correct, they get 4 points for their team. I play until all the cards are used up and the team with the highest score wins.
Examples:
I'm a music teacher, so I've used this for composer questions, music history, instruments, etc. As a music teacher, you can even do it with sight reading - the student has to sight read the card to get points. For other subjects, make up questions for anything you're studying - science units, history units, math problems, grammar, reading comprehension. | 639 | 2,610 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.941168 |
http://squaring.net/history_theory/brooks_smith_stone_tutte_II.html | 1,723,784,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00575.warc.gz | 27,159,112 | 12,802 | Brooks, Smith, Stone, Tutte (Part II)
A number of accounts have related how Arthur Stone was the first of the four undergraduates to raise the problem of the squared square. CAB Smith in personal communication to Skinner states;
"W.R. Dean is largely responsible for the development of the idea of squaring the square - he visited Arthur Stone's school before Arthur came to Cambridge and said that an unsolved problem was to show that a square can not be dissected into a finite number of unequal squares. He later became Professor of Applied Mathematics at University College, London."
We know that this time was at or prior to 1935. The extant publications were few (the Dehn paper (1903), Loyd's(1914) and Dudeney's(1902, 1907-19) Puzzles, the Moron' paper(1925), the Abe papers (1930,31) and the Kraitchik paper(1930). Kraitchik published a personal communication from Russian mathematician N.N. Lusin which stated that it was impossible to dissect a square into a finite number of different elements. This became known as Lusin's conjecture.
The paper they co-authored at the conclusion of their investigation, "The Dissection of Rectangles into Squares;" Ranged across a variety of mathematical disciplines; Electrical network theory, planar graphs, 3-connection, duality, the matrix-tree theorem, number theory, and determinants.
The main results were;
- Every squared rectangle has commensurable sides and elements.
- Every rectangle with commensurable sides is perfectible in an infinity of different ways.
- Discovery of spss order 39 (not shown), and cpss order 26 (empirical).
- No perfect rectangles order less than 9. There are 2 perfect rectangles of order 9.
- Generalisations of the problem; rectangled rectangles, squared cylinders and tori, triangled equilateral triangles and a proof of no cubed cubes.
The story of how the four Cambridge undergraduates solved the problem has been told by various authors. In my view the best account is given by W.T. Tutte himself in the article he wrote for Martin Gardners book 'More Mathematical Puzzles and Diversions'[1].
Squaring the Square, W. T. Tutte
William Thomas Tutte
This is the story of a mathematical research conducted by four students of Trinity College, Cambridge, in the years 1936-8. One was the author of this article. Another was Cedric Smith, now a statistical geneticist at the University of London. He is also well known as a writer on the theory of games and the counterfeit-coin problem. Another was A. H. Stone, now researching at Manchester into recondite regions of point set topology. He is one of the inventors of the flexagons described in Mathematical Puzzles and Diversions. The fourth was R.L. Brooks. He has now left the academic world for the Civil Service. But he retains an enthusiasm for mathematical recreations, and an important theorem in the theory of graph colourings bears his name. These four students referred to themselves, with characteristic modesty, as the 'Important Members' of the Trinity Mathematical Society.
Figure 70; 33x32 Squared Rectangle
In 1936 there were a few references in the literature to the problem of cutting up a rectangle into unequal squares. Thus it was known that a rectangle of sides 32 and 33 can be dissected into nine squares with sides of 1,4,7,8,9,10,14,15, and 18 units (Figure 70). Stone was intrigued by a statement in Dudeney's Canterbury Puzzles which seemed to imply that is is impossible to cut up a square into unequal smaller squares. He tried to prove the impossibility for himself, but without success. He did, however, discover a dissection of the rectangle of sides 176 and 177 into 11 unequal squares (Figure 72).
This partial success fired the imaginations of Stone and his three friends and soon they were spending much time constructing, and arguing about, dissections of rectangles into squares. Any rectangle cut up into unequal squares was called by them a 'perfect' rectangle. Years later the term 'squared rectangle' was introduced to describe any rectangle cut up into two or more squares, not necessarily unequal.
Figure 71; Algebraic Method
The construction of perfect rectangles proved to be quite easy. The method used was as follows. First we sketch a rectangle cut up into rectangles, as in Figure 71. We then think of the diagram a bad drawing of a squared rectangle, the small rectangles being really squares, and we work out by elementary algebra what the relative sizes of the squares must be on this assumption. Thus in Figure 71 we have denoted the sides of two adjacent small squares by x and y and then that the side of the square next on the left is x+2y, and so on. Proceeding in this way we get the formulae shown in Figure 71 for the sides of the 11 small squares. These formulae make the squares fit together exactly except along one segment AB. But we can make them fit on AB too by choosing x and y to satisfy the equation (3x + y) + (3x - 3y) = (14y - 3x), that is 16y = 9x . Accordingly we put x = 16 and y = 9 . This gives the perfect rectangle of Figure 72, which is the one first found by Stone.
Figure 72; 177 x 176 Squared Rectangle
Sometimes this method gave negative values for the sides of some small squares. It was found, however, that such negative squares could always be converted into positive ones by minor modifications of the original diagram. They therefore gave no trouble. In some of the more complicated diagrams it proved necessary to start with three unknown squares, with sides x, y and z, and solve two linear equations instead of one at the end of the algebraic computations. Sometimes the squared rectangle finally obtained proved not to be perfect, and the attempt was considered a failure. Fortunately this did not happen very often. We recorded only 'simple' perfect rectangles, that is perfect rectangles containing no smaller ones. For example, the perfect rectangle obtained from Figure 70 by erecting a new component square of side 32 on the upper horizontal side is not simple, and we did not include it in our catalogue.
In this first stage of the research, large numbers of perfect rectangles were constructed in which the number of component squares ranged from 9 to 26. In the final form of each rectangle the sides the sides of the component squares were represented as integers without a common factor. Of course we all hoped that if we constructed enough perfect rectangles by this method we would eventually obtain one which was a 'perfect square'. But as the list of perfect rectangles lengthened this hope faded. Production slowed down accordingly.
Inspection of the catalogue we had constructed revealed some very odd phenomena. We had classified our rectangles according to their 'order', that is, the number of component squares. We noticed a tendency for numbers representing sides to be repeated in any one order. Moreover the semi-perimeter of a rectangle in one order often reappeared several times as a side in the next order. For example, using the information now available, one finds that four of the six simple perfect rectangles of order 10 have semi-perimeter 209, and that five of the 22 simple perfect rectangles of order 11 have 209 as a side. There was much discussion of this "Law of Unaccountable Recurrence", but it led to no satisfactory explanation.
In the next stage of the research we abandoned experiment in favour of theory. We tried to represent squared rectangles by diagrams of different kinds. The last of these diagrams, introduced by Smith, was a really big step forward. The other three researchers called it the Smith diagram. But Smith objected to this name arguing that his diagram was only a minor modification of one of the earlier ones. However that may be, Smith's diagram suddenly made our problem part of the theory of electrical networks.
Figure 73; 69 x 61 Squared Rectangle and Smith Diagram
Figure 73 shows a perfect rectangle together with its Smith diagram. Each horizontal line-segment in the drawing is represented in the Smith diagram by a dot, or 'terminal'. In the Smith diagram the terminal is made to lie on a continuation to the right of its corresponding horizontal segment in the rectangle. Any component square of the rectangle is bounded above and below by two of the horizontal segments. Accordingly it is represented by a line or 'wire' in the diagram joining the two corresponding terminals. We imagine an electric current flowing in each wire. The magnitude of the current is numerically equal to the side of the corresponding square, and its direction is from the terminal representing the lower one.
The terminals corresponding to the upper and lower horizontal sides of the rectangle may conveniently be called the positive and negative poles, respectively, of the electrical network.
Surprisingly enough the electric currents assigned by the above rule really do obey Kirchhoff's Laws for the flow of current in a network, provided that we take each wire to be of unit resistance. Kirchhoff's First Law states that, except at a pole, the algebraic sum of the currents flowing to any terminal is zero. This corresponds to the fact that the sum of the sides of the squares bounded below by a given horizontal segment is equal to the sum of the squares bounded above by the same segment, provided of course that the segment is not one of the horizontal sides of the rectangle. The second law says that the algebraic sum of the currents in any circuit is zero. This is equivalent to saying that when we describe the circuit the net corresponding change in level must be zero.
The total current entering the network at the positive pole, or leaving it at the negative pole is evidently equal to the horizontal side of the rectangle, and the potential difference between the two poles is equal to the vertical side.
The discovery of this electrical analogy was important to us because it linked our problem with an established theory. We could now borrow from the theory of electrical networks and obtain formulae for the currents in a general Smith diagram and the sizes of the corresponding component squares. The main results of this borrowing can be summarized as follows. With each electrical network there is a number calculated from the structure of the network, without any reference to which particular pair of terminals is chosen as poles. We called this number the complexity of the network. If the units of measurement for the corresponding rectangle are chosen so that the horizontal side is equal to the complexity, then the sides of the component squares are all integers. Moreover, the vertical side is equal to the complexity of another network obtained from the first by identifying the two poles.
The numbers giving the side of the rectangle and its component squares in this system of measurement were called the full sides and full elements of the rectangle respectively. For some rectangles the full elements have common factor greater than unity. In any case division by their common factor gives the 'reduced' sides and elements. It was the reduced sides and elements that had been recorded in our catalogue.
These results imply that if two squared rectangles correspond to networks of the same structure, differing only in the choice of poles, then the full horizontal sides are equal. Further if two rectangles have networks which acquire the same structure when the two poles of each are identified, then the full horizontal sides are equal. Further, if two rectangles have networks which acquire the same structure when the poles of each are identified, then the two vertical sides are equal. These two facts explained all the cases of 'unaccountable recurrence' which we had encountered.
The discovery of the Smith diagram simplified the procedure for producing and classifying simple squared rectangles. It was an easy matter to list all the permissible electrical networks of up to 11 wires and to calculate all the squared rectangles. We then found that there were no perfect rectangles below the 9th order, and only 2 of the ninth(Figures 70 and 73). There were six of the 10th order and 22 of the 11th. The catalogue then advanced, though more slowly, through the 12th order (67 simple perfect rectangles) and into the 13th.
Figure 74; 112 x 75A Smith Diagram
It was a pleasing recreation to work out perfect rectangles corresponding to networks with a high degree of symmetry. We considered, for example, the network defined by a cube, with corners for terminals and edges for wires. This failed to give any perfect rectangles. However, when complicated by a diagonal wire across one face, and flattened into a plane, it gave the Smith diagram of Figure 74 and the corresponding squared rectangle of Figure 75. This rectangle is especially interesting because its reduced elements are unusually small for the thirteenth order. The common factor of the elements is six. Brooks was so pleased with this rectangle that he made a jigsaw puzzle of it, each of the pieces being one of the component squares.
Figure 75; 112 x 75A Squared Rectangle
It was at this stage that Brooks's mother made the key discovery of the whole research. She tackled Brooks's puzzle and eventually succeeded in putting the pieces together to form a rectangle. But it was not the squared rectangle which Brooks had cut up!
Brooks returned to Cambridge to report the existence of two different perfect rectangles with the same reduced sides and the same reduced elements. Here was unaccountable recurrence with a vengeance! the Important Members met in emergency session.
Figure 76; Perfect Square, Two Rectangles
We had sometimes wondered whether it was possible for different perfect rectangles to have the same shape. We would have liked to obtain two such rectangles with no common reduced element, and thus get a perfect square by the construction shown in Figure 76. The shaded regions in this diagram represent the two perfect rectangles. Two unequal squares were then added to make the large perfect square. But no rectangles of the same shape had hitherto appeared in our catalogue, and we had reluctantly come to believe that the phenomenon was impossible. Mrs Brooks's discovery renewed our hopes, even though her rectangles failed in the worst possible way to have no common reduced element.
Figure 77; 112 x 75B Squared Rectangle
There was much excited discussion at the emergency session. Eventually the Important Members calmed down sufficiently to draw the Smith diagrams of the two rectangles. Inspection of these soon made clear the relationship between them.
The second rectangle is shown in Figure 77 and its Smith diagram in Figure 78. It is evident that the network of Figure 78 can be obtained from that of Figure 74 by identifying the terminals p and p'. As p and p' happen to have the same electrical potential in figure 74 this operation causes no change in the currents in the individual wires, no change in the total current, and no change in the potential difference between the poles. We thus have a simple electrical explanation of the fact that the two rectangles have the same reduced sides and the same reduced elements.
Figure 78; 112 x 75B Smith Diagram
But why do p and p' have the same potential in figure 74? Before the emergency session broke up it had obtained an answer to this question also. The explanation depends on the fact that the network can be decomposed into three parts meeting only at the poles A1 and A2 and the terminal A3. One of these parts consists solely of the wire joining A2 and A3. A second part is made up of the three wires meeting at p', and a third is constituted by the remaining nine wires. Now the third part has threefold rotational symmetry with p as the centre of rotation. Moreover, current enters or leaves this part of the network only at A1, A2 and A3, which are equivalent under the symmetry. This is enough to ensure that if any potentials whatever are applied to A1, A2, and A3 the potential of p will be their average. The same argument applied to the second part of the network shows that the potential of p' must also be the average of the potentials A1, A2 and A3. Hence p and p' have the same potential, whatever potentials are applied to A1, A2 and A3, and in particular they have the same potential when A1 and A2 are taken as poles in the complete network, and the potential of A3 is fixed by Kirchhoffs Laws.
The next advance was made accidently by the present writer. We had just seen Mrs Brook's discovery completely explained in terms of a simple property of symmetrical networks. It seemed to me that it should be possible to use this property to construct other examples of pairs of perfect rectangles with the same reduced elements. I could not have explained how this would help us in our main object of of constructing, or proving the impossibility of, a perfect square. But I thought we should explore the possibilites of the new idea before abandoning them.
The obvious thing to do was to replace the third part of the network of Figure 74 by another network having threefold rotational symmetry about a central terminal. But this can be done only under severe limitation, which should now be explained.
It can be shown that the Smith diagram of a squared rectangle is always planar, that is, it can be drawn in the plane with no crossing wires. And the drawing can always be made so that no circuit separates the two poles. There is also a converse theorem which states that if an electrical network of unit resistances can be drawn in the plane in this way, then it is the Smith diagram of some squared rectangle. It would not be proper to take up space in this book with rigorous proofs of these theorems. It would not even be historically accurate; the four researchers did without rigorous proofs right up to the time when they began to prepare their technical paper for publication.
It is not always advisable to disregard rigour in this way in the course of a mathematical research. In research aiming at a proof of the Four Colour Theorem, for example, such an attitude would be, and often is, disastrous. But our research was largely experimental, and its experimental results were perfect rectangles. Our methods were justified, for the time being, by the rectangles they produced, even when their theory had not been precisely worked out.
But let us return to Figure 74 and the replacement of its third part by a new symmetrical network with centre p. The complete network obtained in this way must not only be planar but it must remain planar when p and p' are identified.
After a few trials I found two closely related networks satisfying these conditions. The corresponding Smith diagrams are shown in figures 79 and 80. As expected, each diagram allowed of the identification of p and p', and so gave rise to two squared rectangles with the same reduced elements. But all four rectangles had the same reduced sides, and this result was unexpected.
Figure 79; Smith Diagram
Essentially the new discovery was that the rectangles corresponding to Figures 79 and 80 have the same shape, though they do not have their reduced elements all the same. A simple theoretical explanation of this was soon found. The two networks have the same structure, apart from the choice of poles, and therefore the rectangles have the same full horizontal side. Moreover the networks remain identical when poles are coalesced, and therefore the two rectangles have the same vertical side. We felt, however, that this explanation did not probe sufficiently deep, since it made no reference to rotational symmetry.
Figure 80; Smith Diagram
We eventually agreed to refer to the new phenomenon as 'rotor-stator' equivalence. It was always associated with a network which could be decomposed into two parts, the 'rotor' and the 'stator', with the following properties. The rotor had rotational symmetry, the terminals common to the rotor and stator were all equivalent under the symmetry of the rotor, and the poles were terminals of the stator. In Figure 79, for example, the stator is made up of the three wires joining p' to A1, A2, and A3, and the wire linking A2 with A3. A second network could then be obtained by an operation called 'reversing' the rotor. With a properly drawn figure this could be explained as a reflection of the rotor in the line pA3 and so obtain the network of Figure 80.
After studying a few examples of rotor-stator equivalence the researchers convinced themselves that reversing the rotor made no difference to the currents in the wire of the stator. But the currents in the rotor might change. Satisfactory proofs of these results were obtained only at a much later stage.
Rotor-stator equivalence proved to have no very close relationship with the phenomenon discovered by Mrs Brooks. It was merely another one associated with networks having a part with rotational symmetry. The importance to us of Mrs Brooks's discovery was that it led us to study such networks.
Figure 81; Perfect Square, Shared Corner
A very tantalizing question now arose. What was the least possible number of common elements in a rotor-stator pair of perfect rectangles? Those of Figures 79 and 80 had seven common elements, three of which corresponded to currents in the rotor. The same rotor with a stator consisting of a single wire A2A3 gave two perfect rectangles of the 16th order with four common elements. Using a one-wire stator there seemed no theoretical reason why we should not obtain a pair of perfect rectangles having only one element, corresponding to the stator, in common. But we saw that if we could do this we could also obtain a perfect square. For with the rotors of threefold symmetry which we were studying, a one wire stator always represented a corner element of each corresponding rectangle. From the two perfect rectangles with only a corner element in common we can expect to obtain a perfect square by the construction illustrated in Figure 81. Here the shaded regions represent the two rectangles. The square in which they overlap is the common corner element.
Naturally we got to work calculating rotor-stator pairs. We made the rotors as simple as we could, partly to save labour and partly in the hope of getting a perfect square with small reduced elements. But one construction after another failed, because of common elements in the rotor, and we became discouraged. Was there some theoretical barrier yet still to be explored?
It occurred to some of us that perhaps our rotors were too simple. Something more complicated might be better. the numbers involved would be bigger and the likelihood of a chance coincidence would be reduced. So it came to pass that Smith and Stone sat down to compute a complicated rotor-stator pair while Brooks, unknown to them worked on another in a different part of the College. After some hours Smith and Stone burst into Brooks's room crying 'We have a perfect square!' To which Brooks replied 'So have I!'
Figure 82; Brooks Perfect Square Rotor
Both these squares were of the 69th order. But Brooks went on to experiment with simpler rotors and obtained a perfect square of the 39th order. This corresponds to the rotor shown in Figure 82. A brief description of it is provided by the following formula:
(2378,1163,1098) (65,1033) (737,491) (249,242) (7,235) (478,259) (256) (324,944) (219,296) (1030,829,519,697) (620) (341,178) (163,712,1564) (201,440,157,31) (126,409) (283) (1231) (992,140) (852)
In this formula each pair of brackets represents one of the horizontal segments in the subdivision pattern of the perfect square. These segments are taken in vertical order, beginning with the upper horizontal side of the square, and the lower horizontal side is omitted. the numbers enclosed by a pair of brackets are the sides of those component squares which have their upper horizontal sides in the corresponding segment. They are taken in order from left to right. The reduced side of the perfect square is the sum of the numbers in the first pair of brackets, that is 4,639.
The notation we have just used is that of C.J. Bouwkamp. He has employed it in his published list of the simple squared rectangles up to the 13th order.
This really completes the story of how this particular team solved the problem of the perfect square. We did more work on the problem, it is true. All the perfect squares obtained by the rotor-stator method had certain properties which we regarded as blemishes. Each contained a smaller perfect rectangle; that is, was not simple. Each had a point at which four of the component squares met; that is, was 'crossed'. Finally, each had a component square, not one of the four corner squares, which was bisected by a diagonal of the complete figure. Using a more advanced theory of rotors we were able to get perfect squares without the first two blemishes. Years later, by a method based on a completely different kind of symmetry, I obtained a perfect square of the 69th order free of all three kinds of blemish. But for an account of this work I must refer the interested reader to our technical papers. [references 2.]
There are three more episodes in the history of the perfect square which ought to be mentioned, though each one may seem like an anti-climax.
To begin with, we kept adding to the list of simple perfect rectangles of the 13th order. Then one day we found that two of these rectangles had the same shape and no common element. They gave rise to a perfect square of the 28th order by the construction of Figure 76. Later we found a 13th order perfect rectangle which could be combined with one of the 12th order and one extra component square to give a perfect square of the 26th order. If the merit of a perfect square is measured by the smallness of its order, then the empirical method of cataloguing the perfect rectangles had proved superior to our beautiful theoretical method.
Figure 83; Willcocks 175 Compound Perfect Square
Other researchers have used the empirical method with spectacular results. R. Sprague of Berlin fitted a number of perfect rectangles together in a most ingenious way to produce a perfect square of the 55th order. This was the first perfect square to be published (1939). More recently T.H. Willcocks of Bristol, who did not confine his catalogue to simple and perfect squared rectangles, obtained a perfect square of the 24th order (Figure 83). Its formula is as follows: (55,39,81) (16,9,14) (4,5) (3,1) (20) (56,1) (38) (30,51) (64,31,29) (8,43) (2,35) (33). This perfect square still holds the low-order record. [Perfect Compound Square record, see footnote 1.]
Unlike the theoretical method, the empirical one has not yet given rise to any simple perfect square.
In case any reader should like to do some work on perfect rectangles himself, here are two unsolved problems. The first is to determine the smallest possible order for a perfect square [solved, see footnote 2.]. The second is to find a simple perfect rectangle whose horizontal side is twice the vertical side. [solved, see footnote 3.]
W.T.TUTTE
REFERENCES
1. "More Mathematical Puzzles and Diversions", 1961, collection by Martin Gardner. Ch 17. 'Squaring the Square' by William T. Tutte , from 'Mathematical Games' column Nov 1958, Scientific American.
2. 'Beispiel Einer Zerlegung des Quadrats in Lauter Verschiendene Quadrate' by R. Sprague in Mathematische Zeitschrift, Vol. 45, pages 607-8, 1939.
3. 'A Class of Self-Dual maps' by C.A.B. Smith and W.T.Tutte in Canadian Journal of Mathematics, Vol. 2, pages 179-96, 1950.
4. 'On the Construction of Simple Perfect Squared Squares' by C.J. Bouwkamp in Koninklinjke Nederlandsche Akademie van Wetenschappen, Proceedings, Vol. 50, pages 1296-99, 1947.
5. 'The Dissection of Rectangles into Squares' by R.L. Brooks, C.A.B. Smith, A.H. Stone, and W.T. Tutte in Duke Mathematical Journal, vol. 7, pages 312-40, 1940.
6. 'On the Dissection of Rectangles into Squares (I-III)' by C.J. Bouwkamp in Koninklinjke Nederlandsche Akademie van Wetenschappen, Proceedings, Vol. 49, pages 1176-88, 1946, and Vol. 50, pages 58-78, 1947.
7. 'A Note on Some Perfect Squared Squares' by T.H. Willcocks in Canadian Journal of Mathematics, Vol. 3, pages 304-8, 1951.
8. 'Question E401 and Solution' by A.H. Stone in American Mathematical Monthly, Vol. 47, pages 570-2, 1940.
9. 'A Simple Perfect Square' by R.L. Brooks, C.A.B. Smith, A.H. Stone, and W.T. Tutte in Koninklinjke Nederlandsche Akademie van Wetenschappen, Proceedings, vol 50, pages 1300-1, 1947.
10. 'Squaring the Square' by W.T. Tutte in Canadian Journal of Mathematics, Vol 2, pages 197-209, 1950.
11. Catalog of Simple Squared Rectangles of Orders Nine Through Fourteen and their Elements by C.J. Bouwkamp, A.J.W. Duijvestijn, and P. Medema. Department of Mathematics, Technische Hogeschool, Eindhoven, Netherlands, 1960.
12. 'The Dissection of Equilateral Triangles into Equilateral Triangles' by W.T.Tutte in Proceedings of the Cambridge Philosophical Society, Vol. 44, pages 464-82, 1948.
WEBMASTER FOOTNOTES
1. The low order record for a perfect square is AJW Duijvesijn's order 21 square, but this order 24 square of Willcocks (Figure 83) is the lowest possible order for a compound perfect square. See 'Compound Perfect Squares' by A. J. W. Duijvestijn, P. J. Federico, P. Leeuw in The American Mathematical Monthly, Vol. 89, No. 1 (Jan., 1982), pp. 15-32
2. Order 21 simple perfect squared square, found in March 1978 by AJW Duijvestijn, published 'Simple perfect squared square of lowest order', J. Combin. Theory Ser. B 25 (1978), no. 2, 240-243.
3. Brooks, R. L. "A procedure for dissecting a rectangle into squares, and an example for the rectangle whose sides are in the ratio 2:1", J. Combinatorial Theory, 10 (1971) 206-211
Order 23 2x1 simple perfect squared rectangle, discovered by P. J. Federico. Exercise 35, Ch 5, 'Tiling', 'Mathematics, The Man-Made Universe', Sherman Stein, 3rd ed., p 96.
Order 22 2x1 simple perfect squared rectangle of lowest order, found in August 1978 by AJW Duijvestijn, published 'A lowest order simple perfect 2x1 squared rectangle', J. Combin. Theory Ser. B 26 (1979), no. 3, 372-374. | 6,687 | 30,234 | {"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-2024-33 | latest | en | 0.950875 |
https://dsp.stackexchange.com/questions/32448/psd-using-matlabs-tt-pwelch-algorithm-window-length-and-measure-units | 1,712,954,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00587.warc.gz | 198,374,464 | 41,183 | # PSD using MATLAB's $\tt pwelch$ algorithm window length and measure units
When using pwelch algorithm in MATLAB for a signal with $N$ samples,
• Is there a good rule of thumb for determining the length $L$ of the sections (second variable for the function)?
• And assuming my input is a voltage signal , what are the measure units I get for $10\log_{10}(P_{xx})$ where $P_{xx}$ is the output signal?
Is there a good rule of thumb for determining the length $$L$$ of the sections (second variable for the function)?
$$L$$ (or window argument) specifies a window function to apply, prior to FFT computation of the signal. If $$L$$ is a vector, that vector is considered as window coefficients and is applied on (multiplied to) signal, for example:
Pxx=pwelch(x,[0 0 1 2 3 3 2 1 0 ]); % Means compute spectrum of x.*[0 0 1 2 3 3 2 1 0]
if it is an integer, then a Ham. or Han. window with length of $$L$$ is generated and then this Hamming window is applied to the signal:
Pxx=pwelch(x,10); % Means compute spectrum of x.*(10 point hamming window)
pwelch is based on FFT, and since the spectrum (DFT) of segmented signals shows frequency leakage, I think it would depend on how much resolution you you want and how much frequency leakage, you can tolerate, but generally The more longer the window for frequency static signals is the more leakage is reduced. See this great tutorial for more detail:
http://www.fhnw.ch/technik/ime/publikationen/2012/how-to-use-the-fft-and-matlab2019s-pwelch-function-for-signal-and-noise-simulations-and-measurements
A side note for example codes above is that if the length of your signal x is greater than specified length for the window, then your signal is segmented to vectors with length of $$L$$
assuming my input is a voltage signal , what are the measure units I get in the output?
Since your signal is pure voltage (unknown impedance and hence no current (A) is associated), it is not actual "power" or "energy", however you might use $$\mathrm{V}^2/\mathrm{H}z$$ for PSD. This link might shed more light on it: https://en.wikipedia.org/wiki/Spectral_density#Electrical_engineering In case of logarithmic PSD, you've mentioned it is generally regarded as $$\mathrm{dB}/\mathrm{Hz}$$.
• First of all thanks a lot @Mimsaad. I did look at the [wiki link] (en.wikipedia.org/wiki/Spectral_density#Electrical_engineering) you posted prior to my question post and I guess I have to clarify the question(edited). What are the measure units if I take $10*log_{10}(P_{xx})$? is it $\frac{dBv}{Hz}$? or do I have to use sqrt() before the log extraction? Aug 4, 2016 at 6:43
• The answer to this question is already given here: stackoverflow.com/questions/15255632/… Aug 4, 2016 at 14:29
• Aug 4, 2016 at 14:44 | 725 | 2,757 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.864217 |
https://avatto.com/data-scientist/interview-questions/tricky/page/18/ | 1,624,473,078,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00402.warc.gz | 123,455,177 | 30,894 | >>>>>>Data Scientist Tricky Interview Questions
What is R2? What are some other metrics that could be better than R2 and why?
The goodness of fit measure. Variance explained by the regression / total variance Remember, the more predictors you add the higher R^2 becomes. Hence use adjusted R^2 which adjusts for the degrees of freedom or train error metrics
Is more data always better?
Statistically, It depends on the quality of your data, for example, if your data is biased, just getting more data won’t help. It depends on your model. If your model suffers from high bias, getting more data won’t improve your test results beyond a point. You’d need to add more features, etc.
Practically, Also there’s a tradeoff between having more data and the additional storage, computational power, memory it requires. Hence, always think about the cost of having more data.
You have several variables that are positively correlated with your response, and you think combining all of the variables could give you a good prediction of your response. However, you see that in the multiple linear regression, one of the weights on the predictors is negative. What could be the issue?
Multicollinearity refers to a situation in which two or more explanatory variables in a multiple regression model are highly linearly related.
Leave the model as is, despite multicollinearity. The presence of multicollinearity doesn’t affect the efficiency of extrapolating the fitted model to new data provided that the predictor variables follow the same pattern of multicollinearity in the new data as in the data on which the regression model is based.
principal component regression
What are the tests which are performed on data sets?
There are many tests, few are:-
-A/B Test
-Student’s T Test
-Chi Square Test
-Fisher’s Exact Test
-Mann-Whitney Test | 386 | 1,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.924204 |
https://community.quickbase.com/discussion/formula-to-calculate-weekdays?ReturnUrl=%2Fcommunities%2Fcommunity-home%2Fdigestviewer%3FCommunityKey%3Dd860b0f8-6a48-487b-b346-44c47a19a804 | 1,708,961,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00138.warc.gz | 181,866,589 | 151,793 | # Discussions
View Only
## Formula to calculate weekdays
• #### 1. Formula to calculate weekdays
Posted 01-25-2023 12:02
I am trying to figure out total "Days on Orientation and "Days Overdue". My current formulas are able to give me the total days, and not business days. What do I need to add to the below formula to count only business days?
Total days on orientation:
If([Process Status]="Completed",([Orientation Completed]-[Effective Date]),Today()-[Effective Date])
Total Days Overdue:
Today()-[OVT Due Date]
------------------------------
Lynda Schutter
------------------------------
• #### 2. RE: Formula to calculate weekdays
Posted 01-25-2023 12:08
Edited by Mark Shnier (Your Quickbase Coach) 01-25-2023 12:10
Try this
Total days on orientation:
If(
[Process Status]="Completed",WeekDaySub([
Orientation Completed], [Effective Date]),
WeekDaySub(Today(),[Effective Date]))
Total Days Overdue:
WeekDaySub(Today(),[OVT Due Date])
------------------------------
mark.shnier@gmail.com
------------------------------
• #### 3. RE: Formula to calculate weekdays
Posted 01-25-2023 12:27
Thank you so much for the quick response!
in the first formula I get an error message "expecting duration but found number"
My Orientation Complete and Effective Date are both date field types and my field to display the numbers are Duration-Formula.
Should those be something else?
------------------------------
Lynda Schutter
------------------------------
• #### 4. RE: Formula to calculate weekdays
Posted 01-25-2023 12:31
I suggest that you change the field type to formula numeric. The is no need for the extra complexity of a duration field type when you are really just counting the integer number of days.
------------------------------ | 406 | 1,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.701951 |
https://solvedlib.com/n/q4-a-solve-the-followingl-et-j-v-sin-v-dv-b-solve-the-followingl,15454193 | 1,653,790,398,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00610.warc.gz | 608,155,792 | 19,311 | Similar Solved Questions
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https://plainmath.net/pre-algebra/80128-how-do-you-write-an-inequality-and-solve | 1,675,199,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00199.warc.gz | 491,840,942 | 21,580 | Villaretq0
2022-06-28
How do you write an inequality and solve given "three fourths of a number decreased by nine is at least forty two"?
tennispopj8
Expert
Step 1
First, let's call "a number" $n$
"three fourths of a number" then can be written as:
$\frac{3}{4}n$
This, "decreased by nine" then can be written as:
$\frac{3}{4}n-9$
"is at least" is the same as $\ge$ so we can now write:
$\frac{3}{4}n-9\ge$
and what it is "at least" is "forty two" so:
$\frac{3}{4}n-9\ge 42$
To solve this we first add $9$ to each side of the inequality to isolate the n term while keeping the inequality balanced:
$\frac{3}{4}n-9+9\ge 42+9$
$\frac{3}{4}n-0\ge 51$
$\frac{3}{4}n\ge 51$
Now, we multiply each side of the inequality by $\frac{4}{3}$ tosolve for n while keeping the inequality balanced:
$\frac{4}{3}×\frac{3}{4}n\ge \frac{4}{3}×51$
$\frac{\overline{)4}}{\overline{)3}}×\frac{\overline{)3}}{\overline{)4}}n\ge 4×17$
$n\ge 68$
Expert
Step 1
Let's use some punctuation so show exactly what is meant: note how the placement of a comma gives a different meaning.
$\text{three fourths of a number},\phantom{\rule{1ex}{0ex}}\text{decreased by 9}\phantom{\rule{1ex}{0ex}}\text{is at least 42}$
Let the number be x
$\frac{3}{4}x-9\ge 42$
Solving gives:
$\frac{3}{4}x\ge 51$
$\frac{4}{3}×\frac{3}{4}x\ge 51×\frac{4}{3}$
$x\ge 56$
However, if we interpret it differently we could have:
$\text{three fourths of,}\phantom{\rule{1ex}{0ex}}\text{a number decreased by 9,}\phantom{\rule{1ex}{0ex}}\text{is at least 42}$
$\frac{3}{4}\left(x-9\right)\ge 42$
Solving gives:
$\frac{4}{3}×\frac{3}{4}\left(x-9\right)\ge 42×\frac{4}{3}$
$x-9\ge 56$
$x\ge 65$
Do you have a similar question? | 644 | 1,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2023-06 | latest | en | 0.746656 |
https://www.begalileo.com/math-games/Grade-3/Multiplication/Multiply-by-multiples-of-10,-100,-and-1000/1151 | 1,686,201,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654097.42/warc/CC-MAIN-20230608035801-20230608065801-00728.warc.gz | 720,004,011 | 47,171 | # Multiply by multiples of 10, 100, and 1000
Which of the following statements is true?
Statement 1: 800 × 20 - 80 × 20 = 14400
Statement 2: 900 × 10 - 90 × 10 = 8100
Next | 66 | 172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-23 | longest | en | 0.839969 |
https://betapython.com/queue/ | 1,642,357,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00110.warc.gz | 205,842,804 | 20,383 | # QUEUE
In this post we will study about “QUEUE” and its application.
# “QUEUE”
• Like stack queue is a linear structure.
• which follows a particular order in which the operations are performed.
• The order is First In First Out (FIFO).
• Unlike stacks, a queue is open at both its ends.
• One end is always used to insert data (enqueue) and the other is used to remove data (dequeue).
#### Examples of QUEUE:
• examplee of queue can be a single-lane one-way road where the vehicle enters first exits first.
• queue at the ticket windows and bus-stop.
### Operations on Queue:
There are 4 type of operation performed in queue:
1. Enqueue
2. Dequeue
3. Front
4. Rear.
## ENQUEUE:
1. The word means add an item to the queue.
2. If the queue is full, then it is said to be an Overflow condition.
### Algorithm for enqueue operation:
1. Create a new node
ptr=(qType*)malloc(sizeof(qType))
2. Assign the item to info
set ptr->info = item
set ptr->next= NULL
4. Check for existing queue
if ( rear==NULL)
{
set rear=ptr
set front=ptr
}
else goto 5.
5. Adjust rear pointer
Set rear -> next =ptr
Set rear= ptr
6.exit
before enqueue
after enqueue
CODE:
## DEQUEUE:
• Removes an item from the queue.
• The items are popped in the same order in which they are pushed.
• If the queue is empty, then it is said to be an Underflow condition.
### Algorithm for dequeue operation:
1. Check if queue is Empty
if (front==NULL)
display “Stack Empty ” and exit else goto 2
2. Set temp=front
front=front->next
4. Display temp->info
5.Free(temp)
6.Exit
before dequeue
after dequeue
code
## FRONT:
• Get the front item from queue.
## REAR:
• Get the last item from queue | 442 | 1,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | longest | en | 0.784197 |
http://www.google.com/patents/US7587934?dq=7,546,338 | 1,405,219,324,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776435811.28/warc/CC-MAIN-20140707234035-00017-ip-10-180-212-248.ec2.internal.warc.gz | 326,550,254 | 18,669 | ## Patents
Publication number US7587934 B2 Publication type Grant Application number US 12/012,404 Publication date Sep 15, 2009 Filing date Feb 1, 2008 Priority date Feb 2, 2007 Fee status Paid Also published as Publication number 012404, 12012404, US 7587934 B2, US 7587934B2, US-B2-7587934, US7587934 B2, US7587934B2 Inventors David Bertrand Original Assignee Michelin Recherche Et Technique S.A. Export Citation Patent Citations (7), Classifications (6), Legal Events (2) External Links:
Method of quantifying the utilization of a maximum grip potential of a tire
US 7587934 B2
Abstract
A method of quantifying the utilization of a maximum grip potential of a tire running on the ground. At two distinct azimuth angles, at least two respective values are determined for a differential extension of the tread of the tire or for a shear of the tread that is homogeneous with said differential extension. The utilization is quantified with the help of a function of the two determined values.
Images(7)
Claims(10)
1. A method of quantifying the utilization of a maximum grip potential of a tire running on the ground, comprising the steps of:
determining at least two values for a differential extension of the tread of the tire or for a shear of the tread that is homogeneous with said differential extension, respectively at two distinct azimuth angles; and
wherein the value of the differential extension or of the shear, as determined at a given azimuth angle θ is written Δ(θ), and said utilization is quantified with the help of a value S that is defined as follows:
$S = ( ( Δ ( θ 1 ) - Δ ( θ 2 ) ) - O - P slip · ( Δ ( θ 1 ) + Δ ( θ 2 ) ) ( P grip - P slip ) · ( Δ ( θ 1 ) + Δ ( θ 2 ) )$
where Pgrip, Pslip, and O are predetermined constants, and θ1 and θ2 are the two distinct azimuth angle values.
2. The method according to claim 1, wherein the two azimuth angles define an acute angular sector containing the contact area of the tread with the ground.
3. The method according to claim 2, wherein the angular sector is about 50°.
4. The method according to claim 1, wherein Pslip=0.
5. The method according to claim 1, wherein the quantification of said utilization is also a function of a length of the contact area of the tread with the ground.
6. The method according to claim 1, wherein, when the determined values are shear values, each shear value of the tread is determined substantially in an equatorial plane of the tread.
7. The method according to claim 1, wherein the differential extension corresponds to the difference between two extension values of the tread measured substantially symmetrically about an equatorial plane of the tire.
8. The method according to claim 7, wherein the measured extension values are extension values in a direction that is substantially circumferential relative to the tire.
9. The method according to claim 1, wherein the extension or the shear of the tread is determined respectively by means of at least one extension or shear sensor, preferably implanted between a carcass ply and inside rubber of the tire or on a face of the inside rubber that is in contact with the air inside the tire.
10. The method according to claim 1, wherein said utilization is quantified with the help of a function of the difference between the two determined values, and of the sum of said two determined values.
Description
RELATED APPLICATION
This application claims the priority of French patent application no. 07/53025 filed Feb. 2, 2007, the content of which is hereby incorporated by reference.
FIELD OF THE INVENTION
The present invention relates to a method of quantifying the utilization of a maximum grip potential of a tire running on ground, and also to a tire.
BACKGROUND OF THE INVENTION
The grip potential μ of a tire is defined, at a given instant, as being the ratio of the resultant of the longitudinal and lateral forces over the resultant of the vertical forces:
$μ = F x 2 + F y 2 F z$
At a given instant, the maximum grip potential μmax of the tire on the ground is also defined as being the maximum value that the grip potential μ can take on. This maximum grip potential μmax depends on several factors including the nature of the ground (or the road) or its state of wear, the temperature of the ground and of the tire, or indeed weather factors involving, for example, the presence of water or snow on the ground.
The utilization percentage Pu of the maximum grip potential μmax of the tire is then defined by the following formula:
$P u = μ μ max · 100$
This utilization percentage Pu corresponds to the percentage of the grip potential that is actually being used relative to the maximum grip potential. This value varies over the range from 0 to 100%. Naturally, the closer this value is to 100%, the greater the risk of the tire losing grip. Thus, the utilization percentage Pu serves to quantify utilization of maximum grip potential.
It is advantageous to quantify in real time the extent to which the maximum grip potential of each tire of a motor vehicle is being utilized in order to determine whether or not one of the tires is close to losing its grip with the ground. This information concerning tire grip can be transmitted to the driver of the vehicle so as to adapt driving accordingly, or to an electronic device for controlling the road holding of the vehicle.
Document WO 03/066400 discloses that the maximum grip potential μmax of a tire depends in particular on the following parameters:
• the driving or braking force applied to the tire;
• the lateral thrust force applied to the tire;
• the load carried by the tire; and
• the self-alignment torque, i.e. the moment about a vertical axis that is exerted by the tire.
These various force parameters can be measured by means of sensors carried by the tire, in particular by means of sensors that measure the forces to which the walls or the rubbing strips of the tire are subjected.
The maximum grip potential can be deduced from those force measurements by training an approximation function, e.g. by training the weights of a neural network.
That known method of estimation turns out to be particularly complex to implement and requires complicated calibration of the tire and also close monitoring of variation in its parameters over time. As a result that method is expensive to implement in practice.
OBJECTS AND SUMMARY OF THE INVENTION
One object of the invention is to provide a method of quantifying the utilization of the maximum grip potential of a tire running on ground, which method is particularly simple to implement.
To this end, one aspect of the invention is directed to a method of quantifying the utilization of a maximum grip potential of a tire running on the ground, the method comprising the following steps:
• determining at least two values for a differential extension of the tread of the tire or for a shear of the tread that is homogeneous with said differential extension, respectively at two distinct azimuth angles; and
• quantifying said utilization with the help of a function of these two determined values.
By means of the invention, the utilization of the maximum grip potential of the tire is quantified in a manner that is particularly simple and fast. The values used for making this estimate are very easy to measure by means of a conventional sensor incorporated in the tread of the tire, and the utilization function makes use of simple algebraic operators such as addition or division. Furthermore, the method makes it possible to obtain directly a measure that quantifies said utilization, without it being necessary to calculate beforehand the maximum grip potential of the tire, thereby achieving a considerable reduction in the resources needed to perform said quantification.
The result of the quantification may, for example, be given as a number lying in the range 0 to 1, or else as a percentage.
A quantification method according to an embodiment of the invention may also include one or more of the following characteristics.
• The two azimuth angles define an acute-angle sector containing the contact area of the tread with the ground. The deformation of the tire is due essentially to contact with the ground. Consequently, it is preferable for the extension or shear values used for quantifying said utilization to be determined for azimuth angles that are close to the contact area so that these values are influenced by the tire making contact with the ground and thus by the grip conditions of the tire.
• The angular sector is about 50°. Quantification is of better quality if the values are measured away from the contact area (so as not to be subjected to the influence of the surface state of the road), but as close as possible to the contact area (in order to maximize the influence of the grip conditions of the tire on the ground). It has been found that an angular sector of about 50° centered on the contact area is optimum for satisfying these two criteria.
• The value of the differential extension or of the shear, as determined at a given azimuth angle θ is written Δ(θ), and said utilization is quantified with the help of a value S that is defined as follows:
$S = ( ( Δ ( θ 1 ) - Δ ( θ 2 ) ) - O - P slip · ( Δ ( θ 1 ) + Δ ( θ 2 ) ) ( P grip - P slip ) · ( Δ ( θ 1 ) + Δ ( θ 2 ) )$
where Pgrip, Pslip, and O are predetermined constants, and θ1 and θ2 are the two distinct azimuth angle values.
The value of this quotient is directly connected to the utilization being made of the maximum grip potential. This calculation, which can be performed particularly fast, thus makes it possible to obtain a value that can be interpreted very easily since it varies over the range 0 to 1.
• In the above formula, Pslip=0. Generally Pslip can be ignored, thereby simplifying the formula for calculating S.
• The quantification of said utilization is also a function of a length of the tread contact area with the ground.
• When the determined values are shear values, each shear value of the tread is determined substantially in an equatorial plane of the tread.
• The differential extension corresponds to the difference between two extension values of the tread measured substantially symmetrically about an equatorial plane of the tire.
• The measured extension values are extension values in a direction that is substantially circumferential relative to the tire.
• The extension or the shear of the tread is determined respectively by means of at least one extension or shear sensor, preferably implanted between a carcass ply and inside rubber of the tire or on a face of the inside rubber that is in contact with the air inside the tire.
• Said utilization is quantified with the help of a function of the difference between the two determined values, and of the sum of said two determined values.
Another aspect of the invention is directed to a tire including at least two extension sensors and wherein the extension sensors are carried by the tread of the tire and are arranged to measure the extension of two portions of the tread for a given azimuth angle.
In the state of the art, tires are fitted with force sensors disposed in each of the walls of a tire. For the sensors to be powered electrically, it is necessary to have wires inside the tire interconnecting its two walls. By means of an embodiment of the invention, the two sensors are placed on the tread. This makes it possible to integrate both sensors within a single module that is disposed on the tread and that includes an electrical power supply that is fully integrated.
BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is a perspective diagram of a tire according to an embodiment of the invention, fitted with two extension sensors;
FIGS. 2 and 3 are graphs showing the variation respectively in Δe+Δs and Δs−Δe as a function of a lateral force;
FIGS. 4 and 5 are graphs showing how Δs−Δe varies as a function of Δs+Δe;
FIGS. 6 and 7 show respectively how Δs+Δe and Δs−Δe vary as a function of lateral force, for different vertical loads;
FIG. 8 shows how the length of the contact area varies as a function of the deflection of the tire;
FIGS. 9 and 10 are graphs showing how the parameters P and O vary as a function of the length of the contact area;
FIG. 11 shows how the values of Δs+Δe and Δs−Δe vary when a driving torque is applied to the tire;
FIG. 12 is a graph of the signals delivered by the two extension sensors under normal conditions of use;
FIGS. 13 and 14 are respectively a diagram of a tire subjected to a vertical force and a graph of the signals supplied by the extension sensors of the tire;
FIGS. 15 and 16 are respectively a diagram of a tire subjected to a driving torque and the graph of the signals delivered by the extension sensors of the tire; and
FIGS. 17 and 18 are respectively a diagram of a tire subjected to a side force and a graph showing the signals delivered by the extension sensors of the tire.
DETAILED DESCRIPTION OF THE DRAWINGS
FIG. 1 shows a tire given overall reference 10, having two walls 12 and a tread 14. The portion of the tread 14 in contact with the ground is referred to as the contact area 16. The contact area 16 is visible in FIG. 13, which is described below.
The tire 10 is provided with two extension sensors 18 and 20 arranged on the tread 14 in such a manner as to measure the extension of the tread at two points in a substantially circumferential direction of the tire.
The two extension sensors 18 and 20 are positioned relative to the tire 10 at the same azimuth angle. As can be seen in FIG. 1, the two sensors 18 and 20 are disposed substantially symmetrically about an equatorial plane of the tire 10, e.g. at 30 millimeters (mm) from said plane. This characteristic is nevertheless optional when implementing the invention.
A tire is considered while it is rotating. At a given instant, the azimuth angle at which the two sensors 18 and 20 are to be found relative to a frame of reference external to the tire is written to θ. The azimuth angle θ=180° corresponds to the sensors 18 and 20 passing vertically under the axis of rotation of the tire.
For a given azimuth angle θ, a differential extension Δ(θ) is defined as follows:
Δ(θ)=ε1(θ)−ε2(θ)
The value Δ(θ) corresponds to the difference between the extension ε measured by the two extension sensors 18 and 20. It is thus representative of the state of bending in the plane of the tread 16 at azimuth angle θ.
Two azimuth angles defining an acute angular sector containing the contact area 16 of the tread 14 on the ground are written θe and θs. Azimuth angle θe is selected to be at the point of entry into the contact area and azimuth angle θs is selected at the exit point from the contact area. By way of example, the selected values are θe=156° and θs=204°, such that the angular sector is of the order of 50°.
In the description below, a tire running on the ground is considered as being subjected to two different types of stress:
• a first type of stress corresponding to normal running on ground providing perfect grip. This stress is referred to as “grip stress”; and
• a second type of stress corresponds to running on ground having characteristics that are selected such that the forces generated correspond to the grip limit. This stress is referred to as “slip stress”.
Two these two kinds of stress correspond to extreme circumstances and stress as really encountered is usually intermediate therebetween.
Use is made below more particularly of the following values Δe=Δ(θe), Δs=Δ(θs), Δes and Δs−Δe, and to the way in which they vary as a function of the forces applied to the tire 10 under three circumstances.
First Circumstance: Tire Subjected Solely to a Lateral Force Fy and to a Constant Vertical Load
FIG. 2 is a graph showing the values taken by Δes for a given load as a function of the lateral thrust Fy. In these figures, solid lozenges correspond to grip stress and open squares correspond to slip stress. It can be seen from the graph that independently of the type of stress, there is a simple affine type relationship between Δes and the value of the lateral thrust Fy.
FIG. 3 shows values of Δs−Δe for a given load as a function of the lateral thrust Fy. It can be seen that an affine relationship exists for the two magnitudes, both for grip stress and for slip stress, but that the gradient of this relationship depends on the type of stress.
By combining the results given in FIGS. 2 and 3, it is possible to form FIG. 4 that shows the relationship between Δs−Δe and Δes depending on the type of stress. It can be seen that for a given type of stress, these two values are associated by a proportionality coefficient. This is represented by two straight lines plotted in FIG. 4.
During real stress, the points representing the pairs (Δes, Δs−Δe) lie in the space between the two straight lines shown in FIG. 4 that correspond to two extreme kinds of stress. A value S is then defined such that:
• S=0 when the point corresponding to the current pair is situated on the straight line representing slip stress; and
• S=1 when the point corresponding to the current pair is situated on the straight line representing grip stress.
S can then be written as follows:
$S = ( Δ s - Δ e ) - ( Δ s - Δ e ) slip ( Δ s - Δ e ) grip - ( Δ s - Δ e ) slip$
Using the notation O for the ordinate value at the origin of the two straight lines, Pgrip for the slope of the straight line corresponding to grip stress, and Pslip for the slope corresponding to slip stress, the expression for S becomes:
$S = ( Δ s - Δ e ) - O - P slip · ( Δ s + Δ e ) ( P grip - P slip ) · ( Δ s + Δ e )$
In a simplified version in which the slope Pslip is ignored, the following expression is obtained:
$S = ( Δ s - Δ e ) - O P grip · ( Δ e + Δ s )$
Thus, calculating S and comparing it with the values 0 and 1 makes it possible to determine whether the stress on the tire is closer to perfect grip stress or to slip stress, and to quantify how close. By calculating the value S, it is thus possible to quantify utilization of the maximum grip potential of the tire while it is running on the ground.
To summarize, calculating the value S consists in determining at least two values Δe and Δs for differential extension of the tread of the tire at two distinct azimuth angle values θe and θs, and in calculating the value of S so as to quantify the extent to which the maximum potential is being utilized.
Second Circumstance: Tire Also Subjected to Variations in Load
The above reasoning assumes that the tire is subjected solely to a given vertical load and to a given lateral thrust.
In reality, the load borne by the tire varies continuously. It is therefore necessary to take account of variations in the load borne by the tire in the method used for calculating S in order to improve the model.
FIGS. 6 and 7 show the effects of varying the load borne by the tire on the values of Δes and Δs−Δe. Load variation and thus deflection have no significant effect on the relationship between Δes and lateral thrust (FIG. 6). In contrast, FIG. 7 shows that load does have an influence on the relationship between Δs−Δe and lateral thrust.
In order to introduce a correction into the affine relationship connecting the Δs−Δe to Δes as a function of the deflection of the tire, a criterion is defined that makes it possible to estimate the length of the contact area. This value can be used for parameterizing the slope Pgrip and the ordinate at the origin O.
A value Σ(θ) is then defined as follows:
Σ(θ)=ε1(θ)+ε2(θ)
The value Σ(θ) is characteristic of the radius of curvature of the tread. When the radius of curvature increases, the signals from the two extension sensors both increase by the same amount. Consequently, during a revolution of the wheel, the value Σ(θ) presents two characteristic points θe and θs corresponding respectively to the sensors entering and exiting the contact area. A criterion that is characteristic of the length of the contact area is then defined as follows: Lcas−θe. Under such circumstances, the criterion is expressed in degrees and not in meters.
FIG. 8 shows how the length of the contact area varies as a function of the deflection of a tire as obtained for different loads. The length of the contact area is manifestly proportional to the deflection.
FIGS. 9 and 10 show how Pgrip and O vary as a function of Lca, as defined above. These two graphs show that it is possible to associate Pgrip and O simply with Lca, e.g. in affined manner. Pgrip and O are thus defined as follows:
P grip =a p L ca +b p
O=a 0 L ca +b 0
Naturally, the length of the contact area can be measured using data other than the data provided by the extension sensors.
Third Circumstance: Tire Also Subjected to a Driving or Braking Torque
While in use on a vehicle, a tire is also used to transmit a driving or braking torque. In this respect, it is subjected to deformations that can interact with the above-described operation. As shown below when describing FIGS. 15 and 16, the effect of driving or braking torque gives rise to difference between entry to and exit from the contact area.
FIG. 11 shows how the values Δes and Δs−Δe vary, firstly without driving toque (open marks), and secondly with driving torque having a magnitude of about 45 meter-decanewtons (m·daN) (solid marks). It can be seen that the model presents very little sensitivity to the effects generated by a driving or braking torque. Thus, the model as described above does not need to be modified in order to take account of a driving torque effect.
The above description of the method of the invention thus shows that determining the value S serves in particularly satisfactory manner to quantify the extent to which the maximum grip potential of the tire is being utilized. This determination takes account of the various forces applied to the tire, in particular the lateral thrust force, the vertical load, and also drive and braking torque. In addition, the value of S may also vary a little as a function of the pressure to which the tire is inflated and of its camber angle.
It should be observed that the description above relies on determining a differential extension by means of two extension sensors carried by the tread of the tire. Nevertheless, it would not go beyond the ambit of the invention to use a single shear sensor disposed in the equatorial plane of the tire in order to determine a shear in the tread. The shear in the equatorial plane of the tire is uniform at differential extension Δ(θ). Consequently, the formulae for calculating the value of S remain valid in the event of it being a shear that is determined, since it suffices to replace the differential extension value Δ(θ) by a shear value as measured by means of a shear sensor.
Examples of Signals Delivered by the Extension Sensors
By way of illustration, the description below shows how a tire deforms when subjected to various forces. It also shows the influence of these various deformations on the signals delivered by the two extension sensors carried by the tread of the tire.
Consideration is given to the tire shown in FIG. 1 and provided with two extension sensors 18 and 20 positioned respectively at +30 mm and at −30 mm from the equatorial plane of the tire. When the sensors measure extension, the value of the signal increases, and when they measure compression, the value of the signal decreases.
FIG. 12 shows the signals generated by the two sensors over one revolution of the wheel (i.e. during rotation through 360°) when the tire is subjected to a vertical load of 542 decanewtons (daN). The signal delivered by the first sensor is represented by dots and by a continuous line, while the signal delivered by the second sensor is represented by lozenges and by a discontinuous line.
It can be seen that the signals delivered by the two sensors are substantially identical. These two signals differ only in the non-isotropic effects of the materials constituting the tire.
It can be seen that in the vicinity of azimuth angle 180°, the signals delivered by the sensors increase. This is due to the sensors passing into the contact area of the tire with the ground. While the tread is in contact with the ground, it deforms and its radius of curvature increases. In fact, its radius of curvature tends towards infinity since the tread flattens. This increase in radius of curvature then causes the tread to stretch and therefore stretches the extension sensor.
This flattening of the contact area is shown in FIG. 13 which shows two tires, one (in gray) that is not subjected to any vertical load, and the other (in black) that is subjected to a vertical load. The deformation of the tread creates a contact area (16) of length Lca.
The greater the vertical force applied to the tire, the more the tire flattens and thus the greater the length of the contact area. FIG. 14 shows the effect of variation in deflection on the appearance of the measurement signals. Variation in deflection is obtained by varying the vertical load. FIG. 14 shows how a signal as delivered by a sensor during a revolution of the tire varies for three given load values: the continuous curve corresponds to a load of 326 daN, the dotted curve to a load with a value of 542 daN, and the chain-dotted curve to a load with a value of 758 daN.
From these signals, it can be seen that the length of the zone corresponding to the contact area increases with increasing deflection. It can also be seen that the values of the signal on entry and exit to or from the contact area decrease with increasing deflection, which means that the radius of curvature is smaller.
FIGS. 15 and 16 show the effect of driving torque on the tire and on the signals delivered by the extension sensors.
In FIG. 15, it can be seen that when the tire is subjected to a driving torque, the contact area of the tread with the ground is shifted forwards.
This shift can also be deduced from the graph of FIG. 16 that shows how the signals delivered by the sensors vary when the tire is additionally subjected to a driving torque that generates a longitudinal force of 150 daN. From this graph, it can be seen that the presence of a driving torque gives rise to the following effects:
• a reduction in the extension measured on entry into the contact area and an increase in the extension measured on exit from the contact area. This is the result of a decrease in the radius of curvature on entry into the contact area and an increase in the radius of curvature on exit from the contact area, as can be seen in FIG. 15; and
• a small shift of the zone corresponding to the contact area in the forward direction (i.e. towards smaller azimuth angles), which corresponds to a longitudinal offset.
Consideration is now given to circumstances in which the tire is subjected to a lateral force. For this purpose, FIG. 17 shows the trace of the tread of the tire, when not subjected to any lateral force (left-hand figure) and when it is subjected to a lateral force (right-hand figure). The contact area of the tread with the ground is shown on the trace.
When the tire is subjected to a lateral force, the contact area is offset sideways, thereby causing bending in the plane of the tread that is in contact with the ground. In addition, because the shear stresses are not distributed uniformly throughout the contact area, the portion of the tire tread pattern situated at the exit from the contact area is subjected to greater shear than is the portion situated at the entry. This generates a torque about the vertical axis that corresponds to the self-alignment torque. This torque tends to cause the contact area to turn about the vertical axis, with this turning then giving rise to a difference between the bending in the plane of the tread on entry to, and on exit from, the contact area, as can be seen in FIG. 7.
The effects of these deformations on the signals delivered by the sensors can be seen clearly in FIG. 11 in which it is assumed that the tire is subjected to a lateral thrust of 300 daN. The signals can be analyzed as follows:
• the lateral thrust applies at a point disposed towards the rear of the contact area relative to the center of the contact area such that the signals delivered by the two sensors differ, particularly within the contact area;
• the signal delivered by one of the two sensors presents an amplitude greater than the signal delivered by the other sensor, which is representative of a difference in extension and thus of bending in the plane of the tread of the tire; and finally
• the difference between the values of the two signals on entry to (156°) and on exit from (204°) the contact area is not the same, which means that there is a difference in bending between entry to and exit from the contact area. As shown above with reference to FIG. 17, this difference in bending is due to the turning of the contact area imposed by the self-alignment torque.
FIGS. 12 to 18 as described above show in particular that because of the way the two extension sensors are positioned in the tread of the tire, it is possible to know accurately the deformation to which the tire is subjected. This deformation is due to the forces applied to the tire, which forces are themselves associated with the grip performance of the tire. This therefore makes it possible to show why the method of the invention is effective in quantifying the extent to which the maximum grip potential of a tire running on the ground is being utilized, by measuring deformation of the tread.
Patent Citations
Cited PatentFiling datePublication dateApplicantTitle
US20030056579 *Jul 8, 2002Mar 27, 2003Valery PoulbotTire comprising a measurement device
US20040158414Feb 5, 2004Aug 12, 2004David BertrandMethod of determining components of forces exerted on a tyre and the self-alignment torque
US20050065699Aug 4, 2004Mar 24, 2005David BertrandEstimating maximum friction coefficient based on knowledge of the loads and self-alignment torque generated in a tire contact zone
US20050188756 *Feb 25, 2005Sep 1, 2005Kenji MorikawaDevice and method for detecting force acting on tire
US20080202657 *May 29, 2006Aug 28, 2008Electrovac AgArrangement for Pressure Measurement
US20080245456 *Apr 3, 2008Oct 9, 2008Michelin Recherche Et Technique S.A.Method of detecting hydroplaning and estimating an intensity of hydroplaning of a tire on a wet road
WO2006010680A1Jun 27, 2005Feb 2, 2006Michelin Soc TechSystem for estimating the maximum adherence coefficient by measuring stresses in a tyre tread
Non-Patent Citations
Reference
1Search Report dated Oct. 10, 2007 issued for the corresponding French Patent Application No. 0753025.
Classifications
U.S. Classification73/146, 152/152.1
International ClassificationG01M17/02
Cooperative ClassificationB60T8/1725, G01N19/02
European ClassificationB60T8/172C
Legal Events
DateCodeEventDescription
Mar 7, 2013FPAYFee payment
Year of fee payment: 4
Apr 16, 2008ASAssignment
Owner name: MICHELIN RECHERCHE ET TECHNIQUE S.A., SWITZERLAND
Free format text: ASSIGNMENT OF ASSIGNORS INTEREST;ASSIGNOR:BERTRAND, DAVID;REEL/FRAME:020812/0088
Effective date: 20080404 | 6,933 | 31,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-23 | latest | en | 0.862887 |
https://romantonumber.com/dcccxxxi-in-arabic-numerals | 1,708,631,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00416.warc.gz | 508,643,036 | 22,362 | # DCCCXXXI in Hindu Arabic Numerals
DCCCXXXI = 831
8 3 1 M C X I MM CC XX II MMM CCC XXX III CD XL IV D L V DC LX VI DCC LXX VII DCCC LXXX VIII CM XC IX
DCCCXXXI is valid Roman numeral. Here we will explain how to read, write and convert the Roman numeral DCCCXXXI into the correct Arabic numeral format. Please have a look over the Roman numeral table given below for better understanding of Roman numeral system. As you can see, each letter is associated with specific value.
Symbol Value
I1
V5
X10
L50
C100
D500
M1000
## How to write Roman Numeral DCCCXXXI in Arabic Numeral?
The Arabic numeral representation of Roman numeral DCCCXXXI is 831.
## How to convert Roman numeral DCCCXXXI to Arabic numeral?
If you are aware of Roman numeral system, then converting DCCCXXXI Roman numeral to Arabic numeral is very easy. Converting DCCCXXXI to Arabic numeral representation involves splitting up the numeral into place values as shown below.
DCCCXXXI D + C + C + C + X + X + X + I 500 + 100 + 100 + 100 + 10 + 10 + 10 + 1 831
As per the rule highest numeral should always precede the lowest numeral to get correct representation. We need to add all converted roman numerals values to get our correct Arabic numeral. The Roman numeral DCCCXXXI should be used when you are representing an ordinal value. In any other case, you can use 831 instead of DCCCXXXI. For any numeral conversion, you can also use our roman to number converter tool given above.
## Current Date and Time in Roman Numerals
The current date and time written in roman numerals is given below. Romans used the word nulla to denote zero because the roman number system did not have a zero, so there is a possibility that you might see nulla or nothing when the value is zero.
2024-02-14
04:16:57
MMXXIV-II-XIV
IV:XVI:LVII
Disclaimer:We make a reasonable effort in making sure that conversion results are as accurate as possible, but we cannot guarantee that. Before using any details provided here, you must validate its correctness from other reliable sources on internet. | 508 | 2,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.782078 |
https://www.coursehero.com/file/5856425/Chap13/ | 1,498,178,699,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319933.33/warc/CC-MAIN-20170622234435-20170623014435-00010.warc.gz | 854,075,710 | 96,735 | # Chap13 - Chapter 13 Equity Valuation CHAPTER 13 EQUITY...
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Chapter 13 - Equity Valuation CHAPTER 13 EQUITY VALUATION 1. Theoretically, dividend discount models can be used to value the stock of rapidly growing companies that do not currently pay dividends; in this scenario, we would be valuing expected dividends in the relatively more distant future. However, as a practical matter, such estimates of payments to be made in the more distant future are notoriously inaccurate, rendering dividend discount models problematic for valuation of such companies; free cash flow models are more likely to be appropriate. At the other extreme, one would be more likely to choose a dividend discount model to value a mature firm paying a relatively stable dividend. 2. It is most important to use multi-stage dividend discount models when valuing companies with temporarily high growth rates. These companies tend to be companies in the early phases of their life cycles, when they have numerous opportunities for reinvestment, resulting in relatively rapid growth and relatively low dividends (or, in many cases, no dividends at all). As these firms mature, attractive investment opportunities are less numerous so that growth rates slow. 3. The intrinsic value of a share of stock is the individual investor’s assessment of the true worth of the stock. The market capitalization rate is the market consensus for the required rate of return for the stock. If the intrinsic value of the stock is equal to its price, then the market capitalization rate is equal to the expected rate of return. On the other hand, if the individual investor believes the stock is underpriced (i.e., intrinsic value < price), then that investor’s expected rate of return is greater than the market capitalization rate. 4. MV = 10 + 90 = 100 mil BV = 10 + 60 – 40 = 30 mil MV / BV = 100 / 30 = 3.33 5. g = .6 x .10 = .06 price = 2 / (.08 - .06) = 100 P/E = 100 / 5 = 50 6. k = .04 + .75 (.12-.04) = .10 Price = 4 / (.10 - .04) = 66.66 7. Price with no growth = 6 / .10 = 60 g = .60 x .15 = .09 Price with growth = 2.40 / (.10 - .09) = 240 PVGO = 240 – 60 = 180 13-1
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Chapter 13 - Equity Valuation 8. 300 EBIT 105 -Taxes 195 Net Income 20 + depreciation 60 - CapEx 30 - increase in WC 125 FCF 9. FCFE = 205 – 22 x (1 - .35) + 3 = 193.70 Value = 193.70 / (.12 - .03) = \$2,152.22 10. P = \$2.10/0.11 = \$19.09 11. High-Flyer stock k = r f + β (k M – r f ) = 5% + 1.5(10% – 5%) = 12.5% Therefore: 41 . 29 \$ 04 . 0 125 . 0 50 . 2 \$ 1 0 = - = - = g k D P 12. a. False. Higher beta means that the risk of the firm is higher and the discount rate applied to value cash flows is higher. For any expected path of earnings and cash flows, the present value of the cash flows, and therefore, the price of the firm will be lower when risk is higher. Thus the ratio of price to earnings will be lower. b.
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## This note was uploaded on 04/11/2010 for the course BUSINESS FIN taught by Professor Sata during the Spring '10 term at A.T. Still University.
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Chap13 - Chapter 13 Equity Valuation CHAPTER 13 EQUITY...
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Ask a homework question - tutors are online | 935 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-26 | longest | en | 0.914048 |
https://www.restauranterobinesbinissalem.es/DNuzM9xf | 1,603,407,906,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00280.warc.gz | 869,750,617 | 9,255 | # 1000 kg per hour boiler
### 750, 1000, 1500, 2000, 2500, 3000, 5000, and 6000 MBH Boilers
BMK 6000 40 gallons (151 L ) per hour Condensate drain systems must be sized for full condensing mode. In multiple boiler applications, it is common to manifold these drains together in a plastic pipe
### THERMAX STEAM BOILERS - Thermax Thermion Semi IBR Small
Wholesale Trader of THERMAX STEAM BOILERS - Thermax Thermion Semi IBR Small Boiler - 300/500/750 Kg/hr., Thermax Thermion IBR Packaged Boiler - 1& 1.5 Tph Coal Wood, Thermax Combipac Fbc Boiler - Ibr 3/4/5/6/8 Tph Husk Coal Wood and Thermax Combloc Multi Fuel IBR Steam Boiler 2/3/4/5/6 TPH Coal Wood offered by Nikhil Technochem Private Limited, Kanpur, Uttar Pradesh.
### price boiler 1000 kg hr – best boiler for sale
1000 kg hr steam boiler – industrial boiler for sale. 1000 Kg/ hr Boiler, 1000 Kg/ hr Boiler Suppliers 1000 kg/hr steam boiler / boiler supplier / industrial steam boiler price hot sale on the international .
### How much fuel required for 1TPH boiler? - Quora
Mar 10, 2018 · Fuel Required = 798857.14/2272 = 351.6 kg/hr. Fuel Required for 1 TPH Boiler is 351.6 kg/hr. That means by burning 351.6 kg/hr of of Bagasse, you can generate 1000 kg/hr saturated steam at 10.5 kg/cm2(g) pressure.
### MMBTU to PPH - OnlineConversion Forums
Since the amount of steam delivered varies with temperature and pressure, a common expression of the boiler capacity is the heat transferred over time expressed as British Thermal Units per hour. A boilers capacity is usually expressed as kBtu/hour (1000 Btu/hour) and can be calculated as W = (hg - hf) m (1)
### How to Calculate the Coal Quantity Used in a Power Plant
A 100 MW unit produces 100,000 units of electricity. So the cost of coal per unit of electricity is (3497/100,000) 3.5 cents per unit.
### kg/hour - Kilogram Per Hour. Conversion Chart / Flow Rate by
Flow Rate By Mass Converter / Metric / Kilogram Per Hour [kg/hour] Online converter page for a specific unit. Here you can make instant conversion from this unit to all other compatible units.
### Steam Capacity Conversion Table | Industrial Controls
How many lbs per hour of steam is need to heat 40 gallons per minute flow rate of water from 60 ° F to 180 ° F? Sizing a Boiler Feed Pump Receiver. The receiver tank should be able to hold five minutes worth of condensate for boilers up to 200 BHP and 10 minutes of condensate for boilers over 200 BHP.
### Horsepower - Wikipedia
The average steam consumption of those engines (per output horsepower) was determined to be the evaporation of 30 pounds of water per hour, based on feed water at 100 °F, and saturated steam generated at 70 psi. This original definition is equivalent to a boiler heat output of 33,485 Btu/h.
### CONVERSION OF BOILER IN KG/H AND KW - cni.co.th
CONVERSION OF BOILER IN KG/H AND KW (Approximate) G.B. and U.S.A. Steam from and at 100°C Boiler HP Kg/h 2.26MJ/kg Metric System Steam from 0°C to 100°C Kg/h 2.26MJ/kg Power In kW 1 5 10 15 20 30 40 50 60 70 80 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 15.65 78 156 235 313 470 626 783 940 1096 1252 1566 2350 3130 3915 4700 5480
### average cost of steam per 1000 lbs – Chain-Grate-Boiler
Dec 28, 2017 · average cost of steam per 1000 lbs is very important for food processing industry, it can be used for drying, disinfect, curing, distillation, and sterilization. It is suggested to use oil, gas, solid biomass fuel, and electric as the boiler fuel when generate steam, as the advantages of low pollution, low cost, and clean.
### CONVERSION US · UK · METRIC · SI UNITS FOR THERMAL ENGINEERS
CONVERSION US · UK · METRIC · SI UNITS FOR THERMAL ENGINEERS FLUIDS & FLOW Normal cubicmeter per hour (m n 3/h) / 0°C / 1 1 Boiler HP = 15,65 kg steam /h
### Converting Boiler steam rating from Kg/hr to Liters of water
As cold water, 15000 kg is about 15000 L. However, I assume this is a closed system, the steam flows to a radiator or other load, is condensed and returned to the boiler. So the same water may make several trips per hour through the steam system and return being counted (as kg of steam) several times.
### Calculate How To - US Department of Energy
CG, is the sum of all these individual contributions, expressed as dollars per thousand pounds (\$/Klb) of steam generated: CG = CF + CW + CBFW + CP + CA + CB + CD + CE + CM Fuel cost is usually the dominant component, accounting for as much as 90% of the total. It is given by: CF = aF x (HS – hW)/1000/η B where aF = fuel cost, (\$/MMBtu)
### Knowing the Cost of Steam - Swagelok
or condensate return. Also, the calculation uses only one boiler efficiency. Steam is being generated at 100 psig and is being returned to a deaerator tank operating at 10 psig. Fuel cost is \$9.50/MMBtu, and boiler efficiency is 80%. \$11.67 per 1,000 lbs. IMPROVING THE CALCULATION OF THE UNLOADED COST OF STEAM Condensate Return to the Boiler Operation
### How much coal is required to generate 1 TPH of steam in
It is little bit tough to answer any how i will The factors affecting to calculate the quantity of coal required to produce the 1TPH steam 1. Boiler efficiency 2. GCV ( Gross calorific value ) of the coal 3.
### Steam Calculators: Boiler Calculator
Steam, Boiler, and Blowdown Pressure are the same. Combustion Efficiency is the % of fuel energy that is directly added to the feedwater and not otherwise lost or used. Blowdown Rate is the % of incoming feedwater mass flow rate that leaves the boiler as a saturated liquid at boiler pressure.
### What does THIS STANDS FOR &1000 POUNDS PER HOUR& WHEN USED AS
find out what is the full meaning of this stands for "1000 pounds per hour" when used as measuring steam from a boiler. k is a thousand, as in \$100k = one hundred thousand dollars. and pph is pounds per hour. mass per unit time as a measure of rate of operation of a steam boiler on abbreviations.com!
### Convert bhp to kgf m/h | Boiler horsepower to kilograms force
Diferent power units conversion from Boiler horsepower to kilograms force meter/hour. Between bhp and kgf m/h measurements conversion chart page. Convert 1 bhp into kilogram metre per hour and Boiler horsepower to kgf m/h.
### Boiler Formulas
Boiler HP x.069 Boiler HP x 33.4 Boiler HP x 139 BTU/Hour Output ч 240 EDR/1000 x 0.5 Lbs of Steam /Hr ч 500 = Lbs. of Steam per Hour = Evaporation Rate Gallons per Minute (GPM) = MBTU per Hour Output (MBH) = Sq. Feet of Equivalent Direct Radiation (EDR) = Sq. Feet of Equivalent Direct Radiation (EDR) = Evaporation Rate Gallons per Minute (GPM)
### Conversion Factors - Heating Help
Jun 17, 2014 · Boiler Horsepower to Pounds of Condensate per hour (Multiply BHP by 34.5) • Boiler steaming rate (regardless of type or manufacturer) = 1/2 GPM per 1,000 sq. ft. EDR F uel Values: Maximum copper tubbing flow rates and BTUH loads (at 20° Δ T)
### Electric Steam Boilers - ThomasNet
Manufacturer of electric or electrically heated steam boilers available in various types. Boiler output capacities range from 23 kg/hour to 480 kg/hour. Various features include pressure vessel, electric Incoloy® elements, pre-piped feed water pump, front mounted control panel casing, splash proof electrical enclosure & switches.
### Used Boiler Package Water Tube, Used Boilers
1600 Kg/Hour 0.9 MPA Sigma Slatina Boiler. Stock #HG59629. Used boiler made by sigma Slatina, model VSP 1600. Capable of making up to 1600 kg of steam per hour, heating medium is gas. Boiler is new 1982. The boiler is offered including control panel, burner PRVNÍ Brněnská Strojírna Třebíč, Location: Central Europe. Add to Cart View Details
### Boiler Horsepower to Btus Per Hour | Kyle's Converter
800 Boiler Horsepower to Btus Per Hour = 26780287.3706: 20 Boiler Horsepower to Btus Per Hour = 669507.1843: 900 Boiler Horsepower to Btus Per Hour = 30127823.292: 30 Boiler Horsepower to Btus Per Hour = 1004260.7764: 1,000 Boiler Horsepower to Btus Per Hour = 33475359.2133: 40 Boiler Horsepower to Btus Per Hour = 1339014.3685
### Steam Boilers at Best Price in India
Capacity (kg/hr): 500-1000 , 1000-2000 , 2000-3000 , As per requirement. Fuel: Gas Fired, As per requirement. Working Pressure (kg/cm2g): 15-20 , 20-25 , As per requirement. We have gained vast industry experience in this realm for offering a wide array of Steam Boiler for our
### How to Calculate Boiler Steam Flows | Hunker
A boiler burning 20 gallons per hour would produce [(150,000 btu/gal x 20 gal/hr) / 970.28 btu/lb] x 0.85 (efficiency) for a total of 2,628.1 lb/hr of steam. Step 5 Calculate the combustion-air mass required to burn the 20 gallons of No. 6 fuel oil above.
### Does anyone know how to calculate steam Lbs/Hr? — Heating
1000 PPH steam/500 = 2 gpm into the boiler. If the feed water temperature is less than 212F and the pressure is above 0 psig then the number of PPH of steam you can produce per boiler HP is less than 34.5 PPH. If your feedwater temperature is above 212F you can produce more than 34.5 PPH/BHP still at 0 psig.
### Biocom 30-100kW Wood Pellet Biomass Boiler - Wye Valley Energy
If you get bought-in wood chips, you can expect to pay £120 per tonne. Alternatively, if you have your own wood supply, you can arrange for contract chipping for around £50 a time. Contract chippers charge about £150 per hour and they can chip around 15 tonnes in an hour.
### Unit Conversion--How to Calculate Boiler Horsepower
The Boiler Horsepower (BHP) is the amount of energy required to produce 34.5 pounds of steam per hour at a pressure and temperature of 0 Psig and 212 ℉, with feedwater at 0 Psig and 212 ℉. “Boiler horsepower, a unit of measurement of power of steam boilers” from Wiki.
### Heating Oil
This calculator is offered as a means of converting heating oil price per litre to price per kWh so that prices can be compared with that for other fuels such as gas, electricity or wood (using our Wood Fuel Cost Calcutator).
### How much coal is required to generate 1 TPH of steam in
Every boiler has different design and rating, however I have seen 200 Mw, 500 Mw and 600 Mw boilers of double pass, tilt tangential design type. 600 Mw boiler can generate 1850 TPH of steam per hour at 423 TPH coal.
### Forbes Marshall Marshall B Series Industrial Boilers (1000
Marshall B Series Industrial Boilers (1000-5000kg/hr) This website uses cookies. By continuing to browse, you are agreeing to our use of cookies as explained in our, Cookies Policy .
### in steam boiler convert lb hr to kg hr – CFBC Boiler Manufacturer
Instant units and measurements conversion, metric conversion and other systems. Many units kilogram per minute (kg/min). 0.00756. horsepower (boiler) to MMBtu (IT)/hour (hp (boiler)—MMBtu/h measurement units to evaporate 34.5 lb (15.65 kg) of water at 212 °F (100 °C) in one hour.
### Instant Steam Boiler | Capacities 500 kg/h to 2,000 kg/h
For quick steam generation, CERTUSS offer you their steam boiler model, Universal TC, with 5 fuel firing options e.g. natural gas or light oil fired. Instant Steam Boiler | Capacities 500 kg/h to 2,000 kg/h
### 25000 KG Per Hour Steam Boiler and 500 KG Per Hour Steam
Shree Mahalaxmi Boilers - Manufacturer of 25000 kg per hour steam boiler, 500 kg per hour steam boiler & 1000 kg per hour steam boiler in Kolhapur, Maharashtra.
### Steam Generation Suggested Actions - Department of Energy
Steam Cost = (\$8.00/MMBtu/106 Btu/MMBtu) x 1,000 lb x 1,006 (Btu/lb)/0.857 = \$9.39/1,000 lb Effective Cost of Steam The effective cost of steam depends on the path it follows from the boiler to the point of use. Take a systems approach and consider the entire boiler island,
### Pounds Per Hour to Kilograms Per Hour | Kyle's Converter
1 Kilogram per Hour: Mass flow of kilograms across a threshold per unit time of an hour. A hour containing 3600 seconds. 1 Kilogram per hour = 1 / 3600 kilograms per second (SI base unit). 1 kg/h ? 0.000 277 778 kg/s. Link to Your Exact Conversion
### Calculating the Cost of Steam | Thermaxx Jackets
Jan 31, 2013 · ηB = true boiler efficiency (per ASME PTC 4.1 method) 1,000 = cost measured in units of 1,000 lbs. per hour In this simplified calculation method, the formula assumes that one boiler is being used, one fuel and a single steam pressure.
### 1 kg of coal produces how much steam – oil fired boiler for sale
Boiler efficiency: 87.44 % Equivalent evaporation per kg of dry coal: 12.02 kg .. sg = 8 #Steam generated (kg per kg of coal): mh = 0.04*9 #H2O produced during . collects and controls the steam produced in the boiler. Steam is . Most coal-fired power station boilers use pulverized coal, and many of the larger industrial .
### Solved: 11. A Certain Boiler Generates 1000 Kg Of Steam Pe
A Certain Boiler Generates 1000 Kg Of Steam Per Hour At 10 Bar And 0.97 Dry. This Steam Is Then Taken To A Superheater And Is Heated To A Temperature Of 280°C Keeping The Pressure Constant. The Feed Water Is Available At 30°C. | 3,529 | 12,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | latest | en | 0.769642 |
https://www.teacherspayteachers.com/Product/Valentines-Math-Skills-Learning-Center-Multiply-by-1-2-Digit-Numbers-1688151 | 1,524,521,347,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946199.72/warc/CC-MAIN-20180423203935-20180423223935-00349.warc.gz | 872,930,369 | 18,466 | Total:
\$0.00
# Valentine's Math Skills & Learning Center (Multiply by 1- & 2-Digit Numbers)
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Do you need a fun way for your students to practice the skill (4.NBT.B.5) of multiplying by one-digit and two-digit whole numbers? How about a Valentine's math activity or learning center perfect for the month of February? Try Cupid's Candy Heart Dart, the simple game on this file that students will talking about for months... just print and play!
Three different sets of recording sheets and game cards are included on this file. I prefer to use the first set as an introductory, whole-class activity, and then use the second and third sets as a learning center for the remainder of the season or month.
There are two methods of play; solely dependent upon the amount of time you have available in preparation of this fun activity. One method involves game cards, where one side of a paper card has a multiplication problem, and the other side will have the correct product. Students play by solving the problems that are face-up on a card, record their answers on a recording sheet, self-check the answer on the back, and if he or she is correct, they get to spin a paper-clip spinner and move down through the candy hearts.
The second method of play eliminates the use of the game cards, thus reducing the overall preparation time. Students use a single die to determine problems to solve on their recording sheets, independently self-check the answer with the corresponding answer key, and if he or she is correct, they get to spin a paper-clip spinner and move down through the candy hearts.
This file contains:
●suggested procedures and management for both methods of play (whole-class and learning center)
●2 Cupid’s Candy Heart Dart game boards… one with festive messages, one without.
●2 paper-clip spinners with 6 and 8 outcomes
●3 multiplying by one-digit numbers recording sheets with corresponding answer keys for the students to record their answers to the various questions they will solve during the game (and for teachers to have answer documentation)
●3 multiplying by two-digit numbers recording sheets with corresponding answer keys for the students to record their answers to the various questions they will solve during the game (and for teachers to have answer documentation)
●3 sets of 16 multiplying by one-digit numbers question cards and 16 corresponding answer cards
●3 sets of 16 multiplying by two-digit numbers question cards and 16 corresponding answer cards
●1 set of blank recording sheets and answer keys
●1 set of 32 blank question cards and 32 blank answer cards
Your students are certain to have a blast playing this lovely game, all the while sharpening their ability to solve products of whole numbers multiplied by one-digit and two-digit numbers.
Enjoy!
Total Pages
60 pages
Included
Teaching Duration
45 minutes
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\$3.00 | 654 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-17 | latest | en | 0.941817 |
https://socratic.org/questions/what-is-the-slope-of-the-line-passing-through-the-points-4-3-and-5-2#280432 | 1,723,404,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00771.warc.gz | 420,581,802 | 5,736 | # What is the slope of the line passing through the points (4,3) and (-5,-2)?
Jun 23, 2016
$\frac{5}{9}$
#### Explanation:
Slope = rise (change in height) over run (lateral distance). In this case the change in height is:
$3 - \left(- 2\right) = 3 + 2 = 5$
The run is:
$4 - \left(- 5\right) = 4 + 5 = 9$
Therefore, the slope is $\frac{5}{9}$. | 127 | 350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.790623 |
https://www.analyticsvidhya.com/blog/2021/05/machine-learning-model-evaluation/ | 1,702,254,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00045.warc.gz | 701,229,412 | 58,140 | Shiv Maharaj — Published On May 6, 2021
## What is Machine Learning?
Machine Learning is a popular topic in Information Technology in the present day. Machine Learning allows our computer to gain insight from data and experience just as a human being would. In Machine Learning, programmers teach the computer how to use its past experiences with different entities to perform better in future scenarios.
Machine Learning involves constructing mathematical models to help us understand the data at hand. Once these models have been fitted to previously seen data, they can be used to predict newly observed data.
In Machine Learning, models are only as useful as their quality of predictions; hence, fundamentally our goal is not to create models but to create high-quality models with promising predictive power. We shall now examine strategies for evaluating the quality of models that are generated by our Machine Learning Algorithms.
Source: Nokia Corporation.
## Evaluating Binary Classifier Predictions.
When it comes to evaluating a Binary Classifier, Accuracy is a well-known performance metric that is used to tell a strong classification model from one that is weak. Accuracy is, simply put, the total proportion of observations that have been correctly predicted. There are four (4) main components that comprise the mathematical formula for calculating Accuracy, viz. TP, TN, FP, FN, and these components grant us the ability to explore other ML Model Evaluation Metrics. The formula for calculating accuracy is as follows:
Source: My PC
Where:
• TP represents the number of True Positives. This refers to the total number of observations that belong to the positive class and have been predicted correctly.
• TN represents the number of True Negatives. This is the total number of observations that belong to the negative class and have been predicted correctly.
• FP is the number of False Positives. It is also known as a Type 1 Error. This is the total number of observations that have been predicted to belong to the positive class, but instead, actually, belong to the negative class.
• FN is the number of False Negatives. It may be
referred to as a Type 2 Error. This is the total number of observations that
have been predicted to be a part of the negative class but instead belong to
the positive class.
The main reason for individuals to utilize the Accuracy Evaluation Metric is for ease of use. This Evaluation Metric has a simple approach and explanation. It is, as discussed before, simply the total proportion (total number) of observations that have been predicted correctly. Accuracy, however, is an Evaluation Metric that does not perform well when the presence of imbalanced classes-when in the presence of imbalanced classes, Accuracy suffers from a paradox; i.e., where the Accuracy value is high but the model lacks predictive power and most, if not all, predictions are going to be incorrect.
For the above reason, when we are unable to use the Accuracy Evaluation Metric, we are compelled to turn to other evaluation metrics in the scikit-learn arsenal. These include, but are not limited to, the following Evaluation Metrics:
#### Precision
This refers to the proportion (total number) of all observations that have been predicted to belong to the positive class and are actually positive. The formula for Precision Evaluation Metric is as follows:
Source: My PC
#### Recall
This is the proportion of observation predicted to belong to the positive class, that truly belongs to the positive class. It indirectly tells us the model’s ability to randomly identify an observation that belongs to the positive class. The formula for Recall is as follows:
Source: My PC
#### F1 Score.
This is an averaging Evaluation Metric that is used to generate a ratio. The F1 Score is also known as the Harmonic Mean of the precision and recall Evaluation Metrics. This Evaluation Metric is a measure of overall correctness that our model has achieved in a positive prediction environment-
i.e., Of all observations that our model has labeled as positive, how many of these observations are actually positive. The formula for the F1 Score
Evaluation Metric is as follows:
Source: My PC
## Evaluating Multiclass Classifier Predictions.
As we have learned from earlier information in the article, in Machine Learning, all input data is not balanced, hence the issue of Imbalanced Classes. With the Accuracy Evaluation Metric removed from our options, we specifically turn to Precision, Recall, and F1 Scores. We use parameter
options in Python, which are used for aggregating the evaluation values by averaging them. The three main options that we have available to us are:
1. _macro – Here we specify to the compiler to calculate the mean of metric scores for each class in the dataset, weighting each class equally.
2. _weighted – We calculate the mean of metric scores for each class, and we weigh each class directly proportional to its size in the dataset.
3. _micro – Here we calculate the mean of metric
scores for each OBSERVATION in the dataset.
Source: Medium.
## Visualizing a Classifier’s Performance.
Currently, the most popular way to visualize a classifier’s performance is through a Confusion Matrix. A Confusion Matrix may be referred to as an Error Matrix. A Confusion Matrix has a high level of interpretability. It comprises a simple tabular format, which is often generated and visualized as a Heatmap. Each Column of the Confusion Matrix represents the predicted classes, while every row shows the true (or actual) classes.
There are three important facts to be aware of about a Confusion Matrix:
1. A Perfect Confusion Matrix will have values along the main diagonal (from left to right), and there will zeroes (0) everywhere else in the confusion matrix.
2. A Confusion Matrix does not only show us where the Machine Learning Model faltered but also how it reached those conclusions.
3. A Confusion Matrix will function with any number
of classes, i.e., Having a dataset containing 50 classes, will not affect model
performance nor the Confusion Matrix- it just means your Visualized Matrix will
be very large in size.
Source: ResearchGate.
## Evaluating a Regression Model’s Performance.
For a Regressor, you will find that one of the most used and well-known Evaluation Metrics is MSE. MSE stands for Mean Squared Error. Put into a mathematical representation, MSE is calculated as follows:
Source: My PC.
Where:
• n represents the number of observations in the dataset.
• yi is the true value of the target value we are trying to predict for observation I.
• ŷi is the model’s predicted value for yi.
MSE is a calculation that involves finding the squared sum of all the distances between predicted and true values. The higher the output value for MSE, the greater the sum of squared error present in the model, and hence, the worse the quality of model predictions. There are advantages of squaring the error margins, as seen in the model:
• Firstly, squaring the error constrains all error values to be positive.
• Secondly, this means the model will penalize few
large error values, more than it will penalize many small error values.
Source: Statistics How To.
This concludes my article on Machine Learning Model Evaluation. | 1,459 | 7,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | latest | en | 0.946201 |
https://numbersystem.net/rational-number/ | 1,628,007,379,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00063.warc.gz | 417,775,374 | 13,480 | # Complete Detail About Rational Number
Table of Contents
## Rational Numbers
The numbers of the form \frac { p }{ q } , where p and q are integers and q \neq 0, are known as Rational Numbers.
Remarks-: (i) 0 is a rational number, since we can write, 0=\frac { 0 }{ 1 }.
(ii) Every natural number is a rational number, since we can write, 1=\frac { 1 }{ 1 }, 2=\frac { 2 }{ 1 }, 3=\frac { 3 }{ 1 }, etc.
(iii) Every integer is a rational number, since an integer a can be written as \frac { a }{ 1 }.
### Equivalent Rational Numbers
We know that \frac { 1 }{ 2 } =\frac { 2 }{ 4 } =\frac { 3 }{ 6 } =....\frac { 15 }{ 30 } =\frac { 16 }{ 32 } =....\frac { 144 }{ 288 } =…
These are known as equivalent rational numbers.
### Simplest form of a Rational Number
A rational number \frac { p }{ q } is said to be in simplest form, if p and q are integers having no common factor other than 1 and q \neq 0.
Thus, the simplest form of each of \frac { 2 }{ 4 } ,\frac { 3 }{ 6 } ,\frac { 4 }{ 8 } ,\frac { 5 }{ 10 } ,etc.,is\frac { 1 }{ 2 }.
Similarly, the simplest form of \frac { 6 }{ 9 } is\frac { 2 }{ 3 } and\quad that\quad of\frac { 95 }{ 133 } is\frac { 5 }{ 7 }.
### Representation of Rational Number on real line
Draw a line XY which extends endlessly in both the directions. Take a point O\quad on it and let it represent 0 (zero).
Taking a fixed length, called unit length, marks off OA = 1 unit.
The midpoint B of OA denotes the rational number \frac { 1 }{ 2 } . Starting from O, set off equal distance each equal to OB = \frac { 1 }{ 2 } unit.
From the pont O, on its right, the points at distance equal to OB, OB 2OB, 3OB, 4OB, etc., denote respectively the rational numbers \frac { 1 }{ 2 } ,\frac { 2 }{ 2 } ,\frac { 3 }{ 2 } ,\frac { 4 }{ 2 }, etc.
Similarly, from the pont O, on its left, the points at distances equal to OB, 2OB, 3OB, 4OB, etc., enote respectively the rational numbers \frac { -1 }{ 2 } ,\frac { -2 }{ 2 } ,\frac { -3 }{ 2 } ,\frac { -4 }{ 2 }, etc.
Thus, each rational number with 2 as its denominator can be represented by some point on the number line.
Next, draw the line XY. Take a point O on is representing 0. Let OA= 1 unit. Divide OA into three equal parts with OC as the first part.
Then C represents the rational number \frac { 1 }{ 3 } .
From the point O, set off equal distance, each equal to OC = \frac { 1 }{ 3 } unit on both sides of O.
The points at distances equal to OC, 2OC, 30C, 4OC,etc., from the point O on its right denote respectively the rational numbers \frac { 1 }{ 3 } ,\frac { 2 }{ 3 } ,\frac { 3 }{ 3 } ,\frac { 4 }{ 3 } ,etc.
Similarly, the points at distances equal to OC, 2OC, 30C, 4OC,etc., from the point O on its right denote respectively the rational numbers \frac { -1 }{ 3 } ,\frac { -2 }{ 3 } ,\frac { -3 }{ 3 } ,\frac { -4 }{ 3 } ,etc.
Thus, each rational number with 3 as its denominator can be represented by some point on the number line.
Proceeding in this manner, we can represent each and every rational number by some point on the line.
Example- Represent
(i) 2\frac { 3 }{ 8 } \quad and\quad (ii)\quad -1\frac { 5 }{ 7 }
Solution:- Draw a line XY and taking a fixed length as unit length, represent integers on this line.
(i) On the right of O, take OA=1 unit. Then, OB=2 units.
Divide the 3rd unit BC into 8 equal parts.
BP represents \frac { 3 }{ 8 } of a unit. Therefore, P represents 2\frac { 3 }{ 8 } .
(ii) On the left of O, take OD=1 unit.
Divide the 2nd unit DE into 7 equal parts.
DQ represents \frac { 5 }{ 7 } of a unit. Therefore, Q represents -1\frac { 5 }{ 7 } .
### Two Important Results
(i) Let x and y be two rational numbers such that x < y.
Then, \frac { 1 }{ 2 }\left( x \right +\left y \right) is a rational number lying between x and y.
(ii) let x and y be two rational numbers such that x< y.
Suppose we want to find n rational numbers between x and y.
Let d=\frac { \left( { y }-{ x } \right) }{ \left( { n }+{ 1 } \right) }
Then, n rational numbers lying between x and y are:
\left( { x }+{ d } \right) ,\left( { x }+2{ d } \right) ,\left( { x }+{ 3d } \right) ,...,\left( { x }+{ nd } \right) .
## TERMINATING AND RECURRING DECIMALS
### TERMINATING DECIMAL-:
Every fraction \frac { p }{ q } can be expressed as a decimal. If the decimal expression of \frac { p }{ q } terminates, i.e., comes to an end, then the decimal so obtained is called a terminating decimal.
Examples-: We have: (i)\frac { 1 }{ 4 } =0.25, (ii) \frac { 5 }{ 8 } =0.625, (iii) 2\frac { 3 }{ 5 } =\frac { 13 }{ 5 } =2.6 .
Thus, each of the number \frac { 1 }{ 4 }, \frac { 5 }{ 8 } and 2\frac { 3 }{ 5 } can be expressed in the form of a terminating decimal.
An Important Observation-: A fraction \frac { p }{ q } is a terminating decimal only, when prime factors of q are 2 and 5 only.
Example-: Each one of the fractions \frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,\frac { 7 }{ 20 } ,\frac { 13 }{ 25 } is a terminating decimal, since the denominator of each has no prime factor other than 2 and 5.
### Repeating (or recurring) Decimals
A decimal in which a digit or a set of digits repeats periodically, is called a repeating or a recurring decimal.
In a recurring decimal, we place a bar over the first block of the repeating part and omit the other repeating block.
Examples-: We have:
(i) \frac { 2 }{ 3 } =0.666...=\bar { 0.6. }
(ii) \frac { 3 }{ 11 } =0.272727...=\bar { 0.27. }
(iii) \frac { 15 }{ 7 } =2.142857142857...=\bar { 2.142857. }
(iv) \frac { 11 }{ 6 } =1.18333...=\bar { 1.183. }
### Length of a period
Repeated number of decimal places in a rational number is called the length of its period.
Example-\frac { 15 }{ 7 } =\bar { 2.14257. }.
So, the length of its period is 6.
### Special Characteristics of Rational Numbers
(i) Every rational number is expressible either as a terminating decimal or as a repeating decimal.
(ii) Every terminating decimal is a rational number.
(iii) Every repeating decimal is a rational number.
## How to find rational number between two integral numbers?
Suppose we have to find 4 rational numbers between 2 and 3.
Method:
1. Write 2 and 3 with denominator (4+1)
2. 2= \frac { 2\times (4+1) }{ (4+1) } =\frac { 10 }{ 5 } \quad and\quad \frac { 3\times (4+1) }{ (4+1) } =\frac { 15 }{ 5 }
3. So, the four required numbers are: \frac { 11 }{ 5 } ,\frac { 12 }{ 5 } ,\frac { 13 }{ 5 } ,\frac { 14 }{ 5 }
\Rightarrow \frac { 11 }{ 5 } <\frac { 12 }{ 5 } <\frac { 13 }{ 5 } <\frac { 14 }{ 5 }
## Insertion of Two or More Rational Numbers Between any Two Rational Numbers
THEOREM-: Prove that between any two distinct rational numbers a and b, there exists another rational number.
Proof-: Let a and b be the two rational numbers, such that a < b.
Now, a < b
\Rightarrow a+a,b+a
\Rightarrow 2a<a+b
\Rightarrow a<\frac { a+b }{ 2 }
Again, a<b
\Rightarrow a+b<b+b
\Rightarrow a+b<2b
\Rightarrow \frac { a+b }{ 2 } <b
Combining (1) and (2), a<\frac { a+b }{ 2 } <b
Since a, b and 2 are non-zero rational numbers, therefore \frac { a+b }{ 2 } is a rational number.
Hence, between two distinct rational numbers a and b, there exists another rational number \frac { a+b }{ 2 }.
## Final Wordings:
So Guys, You have made it to the end of our complete detail about Rational Numbers. And hope you like it. If yes, then share it with your friends using the sharing tools at your end. If you have questions and comments ask us by using the form at the bottom, or send us a mail.
Source-: Ts aggarwal & R.L Arora References
Top | 2,481 | 7,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-31 | latest | en | 0.754517 |
http://mathhelpforum.com/calculus/70315-calculus-print.html | 1,519,469,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00080.warc.gz | 213,139,894 | 2,745 | # Calculus
• Jan 27th 2009, 10:41 PM
juicysharpie
Calculus
I need help on this problem. The answer is 36, but I'm no sure how the process works in order to get that answer. Will someone please help? Thank you!!!!!
Evaluate:
13
4
k=5
• Jan 27th 2009, 11:12 PM
earboth
Quote:
Originally Posted by juicysharpie
I need help on this problem. The answer is 36, but I'm no sure how the process works in order to get that answer. Will someone please help? Thank you!!!!!
Evaluate:
13
4
k=5
You have a sum of constant summands:
$4+4+4+4+\underbrace{4+4+4+4+4+4+4+4+4}_{\text{summ ands with indices 5 to 13}}+4+4+4+4+...$
There are 9 summands of 4. | 227 | 645 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-09 | longest | en | 0.884957 |
http://pballew.blogspot.com/2012/05/on-this-day-in-math-may-15.html | 1,531,952,021,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00298.warc.gz | 264,951,568 | 26,086 | ## Tuesday, 15 May 2012
### On This Day in Math - May 15
The Moving Finger writes; and, having writ,
Moves on : nor all thy Piety nor Wit
Shall lure it back to cancel half a Line,
Nor all thy Tears wash out a Word of it.
The Rubaiyat by Omar Khayyam
The 136th day of the year; 136 is the sum of the cubes of the digits of the sum of the cubes of its digits. ( 13 + 33 + 63 = 244 and 23 + 43 + 43 = 136) *Tanya Khovanova, Number Gossip
EVENTS
1618 Kepler Writes his third law of planetary motion. He was teaching at the Landschaftsschule in Linz (1612 - 1630) and also continuing to be Court Mathematician. During this period, he married Susanna Reuttinger while he was here (1613) and produced Harmonices Mundi, (1618) giving his third law.[Kepler's third law says that the square of the time it takes a planet to travel its path around the sun is proportional to the cube of the average distance from the sun] He also became acquainted with the techniques of measuring wine casks here, a practical art for the 17th century as barrels were not uniform in design. He wrote "Stereometria Doliorum Vinariorum” in 1615 and gave the dimensions of the “ideal” cask.
He attempted to explain proportions and geometry in planetary motions by relating them to musical scales and intervals (an extension of what Pythagoras had described as the "harmony of the spheres".) Kepler said each planet produces musical tones during its revolution about the sun, and the pitch of the tones varies with the angular velocities of those planets as measured from the sun. The Earth sings Mi, Fa, Mi. At very rare intervals all planets would sing in perfect concord. Kepler proposed that this may have happened only once in history, perhaps at the time of creation. (assorted sources). For more detail, including Kepler's own announcement of his third law, see "The Renaissance Mathematicus"
1834 After first rejecting Whewell's suggestions on May 3, Faraday writes, "I have taken your advice, and the names used are anode cathode anions cations and ions; the last I shall have but little occasion for. I had some hot objections made to them here and found myself very much in the condition of the man with his son and ass who tried to please every body; but when I held up the shield of your authority, it was wonderful to observe how the tone of objection melted away." *Frank James (ed.), The Correspondence of Michael Faraday (1993), Vol. 2, 186.
1836 Francis Baily observed "Baily's Beads" during an annular solar eclipse. His vivid description aroused new interest in the study of eclipses. Baily's Beads are the bright points of light, that appear around the edge of the moon during a solar eclipse. The beads are created by sunlight passing through the moon's valleys. The last bead is the brightest, resembling a diamond on a brilliant ring. After retiring from a successful business career (1825), Baily turned to science. He revised several star catalogs, repeated Henry Cavendish's experiments to determine the density of the Earth, and measured its elliptical shape. His protests regarding the British Nautical Almanac, then notorious for its errors, were instrumental in bringing about its reform.
1910 Halley's comet was big news during its visible period in New York City. Beginning with the Saturday edition of May 14 and continuing on through the Sunday edition of May 22, the comet was given top billing in the New York Times. This was the period when the comet was at the height of its brilliance and activity and the coverage clearly reflected this.
May 15: (Sunday edition) – Speculation on probability of Earth passing through comet’s tail. Article on those still living who remember Halley’s Comet visit of 75 years ago. *Joseph M. Laufer, Halley's Comet Society - USA
1921 First record of Aurora Borealis observation during day time? Aurora seen in New Zealand and surrounding islands. NASA Eclipse Calendar
1935, at the Franklin Institute, Philadelphia, Albert Einstein was awarded the Benjamin Franklin Medal for his outstanding fundamental contributions to theoretical physics, especially his relativity theory. According to Time magazine, "A throng of scientists and dignitaries was assembled to hear what the medalist had to say. Einstein genially informed the chairman that he had nothing to say, that inspiration which he had awaited until the last moment had failed him. The chairman, much more embarrassed than the medalist, conveyed this information to the audience." In atonement, Einstein wrote a 44-page essay entitled "Physics and Reality," published in the Mar 1936 issue of their Journal of the Franklin Institute. *TIS
1948 The independent State of Israel established. In 1952 the Israeli government asked Einstein, who had labored for the creation of the State, to accept the presidency of the country. He sadly declined the honor, insisting that he was not fitted for such a position. *VFR
1971 Nicaragua issued a series of stamps showing “mathematical equations which changed the world.” They range from 1 + 1 = 2 (Egyptians counting on their fingers) to Napier’s law of logarithms and the Pythagorean Theorem. On the back of each stamp is a descriptive paragraph. [Scott #877–881, C761–5] all ten are here
1985 At a Columbia University graduation Benoit B. Mandelbrot received the Barnard Medal for Meritorious Service to Science, an award made every five years. He is noted for his work on fractals.*VFR Here is a nice link to a web page showing the Mandelbrot set as a “catalogue\ of Julia Sets”. Mandelbrot has been at the IBM research institute in Yorktown Heights since 1958. He is now emeritus and also holds a professorship at Yale since 2000.
2004 Josh Findley discovered the 41st Mersenne prime, 224,036,583 - 1. He found it using a 2.4-GHz Pentium 4 computer. A Mersenne prime number is one less than a power of two expressed as Mn = 2n - 1. For this to be true, the exponent n must also be prime. Mersenne primes have a close connection to perfect numbers, which are equal to the sum of their proper divisors. The study of Mersenne primes was motivated by this connection. In the 4th century B.C. Euclid demonstrated that if M is a Mersenne prime, then M(M+1)/2 is a perfect number. In the 18th century, Leonhard Euler proved that all even perfect numbers have this form. No odd perfect numbers are known and it is suspected that none exist. It is currently unknown whether an infinite number of Mersenne primes exist. *CHM
BIRTHS
1048 Omar Khayyam (15 May, 1048 - 1131)
mathematician and poet, He was the first to claim cubic equations—and hence angle trisection—could not be solved with straightedge and compass. P. Wantzel gave a proof in 1837. *VFR Omar Khayyám (1048–1131; Persian: عمر خیام) was a Persian polymath: philosopher, mathematician, astronomer and poet. He also wrote treatises on mechanics, geography, mineralogy, music, climatology and Islamic theology.
Born in Nishapur, at a young age he moved to Samarkand and obtained his education there, afterwards he moved to Bukhara and became established as one of the major mathematicians and astronomers of the medieval period. He is the author of one of the most important treatises on algebra written before modern times, the Treatise on Demonstration of Problems of Algebra, which includes a geometric method for solving cubic equations by intersecting a hyperbola with a circle. He also contributed to a calendar reform.*Wik
Khayyam , produced a work on algebra that was used as a textbook in Persia until this century. In geometry, he studied generalities of Euclid and contributed to the theory of parallel lines. Around 1074, he set up an observatory and led work on compiling astronomical tables, and also contributed to the reform of the Persian calendar. His contributions to other fields of science included developing methods for the accurate determination of specific gravity. He is known to English-speaking readers for his "quatrains" as The Rubáiyát of Omar Khayyám, published in 1859 by Edward Fitzgerald, though it is now regarded as an anthology of which little or nothing may be by Omar. *TIS
Cubic equation and intersection of conic sections" the first page of two-chaptered manuscript kept in Tehran University *Wik
1615 Frans van Schooten (1615 in Leiden – 29 May 1660 in Leiden) was a Dutch mathematician who was one of the main people to promote the spread of Cartesian geometry. *Wikipedia His group of students extended Descartes work and created a calculus without limits. Hudde in particular was highly rated by Leibniz; "Leibniz in particular was impressed with Hudde’s work, and when Johann Bernoulli proposed the brachistochrone problem, Leibniz lamented: If Huygens lived and was healthy, the man would rest, except to solve your problem. Now there is no one to expect a quick solution from, except for the Marquis de l’Hopital, your brother [Jacob Bernoulli], and Newton, and to this list we might add Hudde, the Mayor of Amsterdam, except that some time ago he put aside these pursuits ."
1637 Valentin Heins (May 15th 1637 in Hamburg - November 17 1704 ) was a German arithmetician (Reckoner)
The son of a linen weaver, the source of his education is unknown. From 1651 Heins was licensed to provide instruction in commercial computing (accounting, bookkeeping, arithmetic, etc). In the years 1658 and 1659 Heins studied theology for several semesters at the universities of Jena and Leipzig , but then returned to Hamburg. There he married and had a vicariate (financial endoument) in 1661 at the Cathedral. Whether Heins performed for a service is not known.
In 1670 he became writing and arithmetic master of the German Church School St. Michaelis . He was also from 1663-1672 accountant of the Guinean-African Company.
He wrote several textbooks, which made him known beyond national boundaries. They were reprinted up to the beginning of the 19th Century. Particularly popular was his tyrocinium mercatorio arithmeticum, a commercial arithmetic and accounting book.
Heins founded in 1690, with the calculation of the parish school master of St. Jacobi Henry Meissner , the art-loving Societät billing. This later became the Mathematical Society of Hamburg, the worlds oldest existing mathematical society. *Wik
1835 Émile Léonard Mathieu (May 15, 1835, Metz – October 19, 1890, Nancy) is remembered especially for his discovery (in 1860 and 1873) of five sporadic simple groups named after him*SAU The Mathieu group is related to the solutions of the fifteen puzzle, and the more recent Rubix Cube.
1857 Hermann Ludwig Gustav Wiener (15 May 1857 in Karlsruhe, Germany - 13 June 1939 in Darmstadt, Germany) was a German mathematician who worked on the foundations of geometry. Although Wiener is not explicitly credited with influencing Hilbert in his championing of the axiomatic method, it is still worth noting that he gave the talk Über Grundlagen und Aufbau der Geometrie to the German Mathematical Society which was published in the first volume of the Jahresberichte der Deutschen Mathematiker vereinigung (1892). Wiener proposed that geometry be studied without using visual images, but rather by abstract axiomatic methods. He also joined his father in the creation of mathematical models of geometric surfaces, constructed from plaster and wire. *SAU
1857 Williamina Paton Stevens Fleming (May 15, 1857 – May 21, 1911) Scottish-born American astronomer who pioneered in the classification of stellar spectra and the first to discover stars called "white dwarfs." She emigrated to Boston at age 21. Prof. Edward Pickering, director of the Harvard Observatory first employed Fleming as a maid, but in 1881 hired her to do clerical work and some mathematical calculations at the Observatory. She further proved capable of doing science. After devising her system of classifying stars by their spectra, she cataloged over 10,000 stars within the next nine years. Her duties were expanded and she was put in charge of dozens of young women hired to do mathematical computations (as now done by computers).*TIS
1863 Frank Hornby (15 May 1863; 21 Sep 1936 at age 73) English inventor and manufacturer who patented the Meccano construction set in 1901. This toy used perforated metal strips, wheels, roods, brackets, clips and assembly nuts and bolts to build unlimited numbers of models. His original sets, marketed as "Mechanics Made Easy" produced in a rented room, were initially sold at only one Liverpool toy shop. By 1908, he had formed his company, Meccano Ltd., and within five more years had established manufacturing in France, Germany, Spain and the U.S. He introduced Hornby model trains in 1920, originally clockwork and eventually electrically powered with tracks and scale replicas of associated buildings. The "Dinky" range of miniature cars and other motor vehicles was added in 1933. *TIS
1899 Joseph Berkson, (15 May, 1899 - 12 Sep, 1982) Dr. Berkson became Head of Biometry and Medical Statistics at the renowned Mayo Clinic in 1933, which he held until his retirement in 1964.
His research interests covered all aspects of medical statistics, resulting in 118 scientific papers from 1928 to 1980. He was involved in a number of controversies, particularly that involving the rate of cigarette smoking in lung cancer.
Two well-known coinages of Berkson that became common in Statistics are from two of his articles: “Rao-Blackwellization” (1955 article in JASA) and “logit” (1944 article in JASA).
Also, a 1946 paper by Berkson introduced what later became known as “Berkson’s Fallacy”, which is now part of Biostat 101 courses.
[Note: Berkson's Fallacy would make for a good post-APStatExam lesson as it involves a common chi-square test in a Simpson's Paradox-like setting...] *David Bee
1903 Maria Reiche ( 15 May 1903 - 8 June, 1998) German-born Peruvian mathematician and archaeologist who was the self-appointed keeper of the Nazca Lines, a series of desert ground drawings over 1,000 years old, near Nazcain in southern Peru. For 50 years the "Lady of the Lines" studied and protected these etchings of animals and geometric patterns in 60 km (35 mi) of desert. Protected by a lack of wind and rain, the figures are hundreds of feet long best seen from the air. She investigated the Nazca lines from a mathematical point of view. Death at age 95 interrupted her new mathematical calculations: the possibility that the lines predicted cyclical natural phenomena like El Nino, a weather system that for centuries has periodically caused disastrous flooding along the Peruvian coast. *TIS
1939 Brian Hartley (15 May 1939-8 October 1994) was a British Mathematician specialising in group theory.
Hartley's Ph.D. thesis was completed in 1964 at the University of Cambridge under Philip Hall's supervision. He spent a year at the University of Chicago, and another at MIT before being appointed as a lecturer at the newly established University of Warwick in 1966, and was promoted to reader in 1973. He moved to a chair at Manchester in 1977 where he served as head of the Mathematics department between 1982 and 1984.
He published more than 100 papers, mostly on group theory, and collaborated widely with other mathematicians. His main interest was locally finite groups where he used his wide knowledge of finite groups to prove properties of infinite groups which shared some of the features of finite groups. One recurrent theme appearing in his work was the relationship between the structure of groups and their subgroups consisting of elements fixed by particular automorphisms.
Hartley is perhaps best known by undergraduates for his book Rings Modules and Linear Algebra, with Trevor Hawkes (ISBN 9780412098109).
Hartley was a keen hill walker, and it was while descending Helvellyn in the English Lake District that he collapsed with a heart attack and died.
The 'Brian Hartley Room' at the School of Mathematics at Manchester is named in his honour.
*Wik
1964 Sijue Wu (May 15, 1964 - ) She received her B.S. (1983) and M.S. (1986) from Beijing University, Beijing, China, and her Ph.D. (1990) [Abstract] from Yale University. Since then she has held the following position: Courant Instructor at Courant Institute, New York University (2 years); assistant professor at Northwestern University (4 years); and assistant, then associate professor at the University of Iowa (2 years). She was also a member at the Institute for Advanced Study in the fall of 1992 and during the year 1996-97. She has been as associate professor at the University of Maryland, College Park, since 1998.
Sijue Wu was awarded the 2001 Ruth Lyttle Satter Prize by the American Mathematics Society. This prize is awarded every two years to recognize an outstanding contribution to mathematics research by a woman in the previous five years. Following is the selection committee's citation:
The Ruth Lyttle Satter Prize in Mathematics is awarded to Sijue Wu for her work on a long-standing problem in the water wave equation, in particular for the results in her papers (1) "Well-posedness in Sovolev spaces of the full water wave problem in 2-D", Invent. Math. 130 (1997), 39-72; and (2) "Well-posedness in Sobolev spaces of the full water wave problem in 3-D", J. Amer. Math. Soc. 12, no. 2 (1999), 445-495. By applying tools from harmonic analysis (singular integrals and Clifford algebra), she proves that the Taylor sign condition always holds and that there exists a unique solution to the water wave equations for a finite time interval when the initial wave profile is a Jordon surface. *Women Mathematicians, Agnes Scott College
DEATHS
1923 Arthur Gordon Webster (November 28, 1863 - May 15, 1923) was the founder of the American Physical Society. A group of twenty physicists, invited by Webster, founded the American Physical Society at a meeting at Fayerweather Hall in Columbia University on 20 May 1899. In 1903, Webster became president of the American Physical Society and was elected to the National Academy of Sciences.
Webster committed suicide in 1923, following the closure of the mathematics department at Clark, after it was rumored that the physics department would be the next to be closed by the new president. With a revolver he had bought a few hours before, Webster shot himself twice in the head in his private office while a class waited for him next door. He left a note to his son which read;
Dear Gordon: This is the only way. For years I have been a failure - my research is worth nothing. Everyone else knows it, and S.N. physics has got away from me and I cannot come back. Everything I have started has stalled. Students will not come and they will put me out. Your mother will not see. She will get over this. Take care of her. I am sorry for the trouble I have caused you. Am sorry to make so much trouble. Do your best and tell the truth. With my best love, "Papa"*Wik (with thanks to Arjen Dijksman).
1975 Taira Honda (本田 平 Honda Taira?, 2 June 1932 Fukui, Japan – 15 May 1975 Osaka, Japan) was a Japanese mathematician working on number theory who proved the Honda–Tate theorem classifying abelian varieties over finite fields. *Wik His mathematical research was mainly devoted to the investigation of the arithmetic properties of commutative formal groups. A brilliant career was cut short when he took his own life.*SAU
1991 Andreas Floer (23 Aug 1956 in Duisburg, Germany - 15 May 1991 in Bochum, Germany) was a German mathematician who made seminal contributions to the areas of geometry, topology, and mathematical physics, in particular the invention of Floer homology.*Wik
Credits
*WM = Women of Mathematics, Grinstein & Campbell
*VFR = V Frederick Rickey, USMA
*TIS= Today in Science History
*Wik = Wikipedia
*SAU=St Andrews Univ. Math History
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts | 4,626 | 19,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-30 | latest | en | 0.964549 |
https://chemistry.stackexchange.com/questions/90964/is-it-possible-to-solve-the-equation-for-the-diffusion-of-gas-through-a-sphere-a | 1,708,950,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00477.warc.gz | 163,183,975 | 40,330 | # Is it possible to solve the equation for the diffusion of gas through a sphere analytically by applying e.g. "combination of variables"?
In the case of diffusion through a thin film it is possible to combine the variables in the following way: $\sigma = \frac{x}{\sqrt{4Dt}}$
where
• $x$=length
• $D$=diffusion coefficient
• $t$=time
• $\sigma$=a function of x and t
Thereby, x and t is substituted by the single variable sigma
Is there a similar expression for a sphere?
E. L. Cussler discribes in detail how to solve the equation for a thin film by applying "combination of variables".
• Good, but still unclear. You can combine (say, multiply) any variable with any other variable. What is the significance of this particular combination? Feb 19, 2018 at 20:24
• @Sigils while not off topic here, you might get a better answer on Physics SE.
– Tyberius
Feb 19, 2018 at 20:25
• Are you looking for a solution which gives you for example in how much time a sphere of sugar will dissolve in water by only diffusion? Feb 19, 2018 at 22:19
Given the diffusion equation
$D(\frac{d^2y}{dr^2}+(\frac{2}{r})\frac{dy}{dr})=\frac{dy}{dt}$
You can put in
$y=z/r$
And then
$\frac{dy}{dr}=(\frac{1}{r})(\frac{dz}{dr}-\frac{z}{r})$
$\frac{d^2y}{dr^2}=(\frac{1}{r})(\frac{d^2z}{dr^2}-(\frac{2}{r})\frac{dz}{dr}+\frac{2z}{r^2})$
$\frac{dy}{dt}=(\frac{1}{r})\frac{dz}{dt}$
Plug these into your diffusion equation and a lot of terms cancel leaving the same equation you would have in rectangular coordinates:
$D\frac{d^2z}{dr^2}=\frac{dz}{dt}$
So you then use the same combination of variables as in the rectangular case except you put $z=ry$ as your dependent variable.
• Then I get a result of the type: z(sigma) = C1+erf(sigma) * C2 Which leads to an equation of the following type with respect to c: c(r,t) = (C1 + erf(x/(sqrt(4Dt))) * C2)/(r) Is this correct? Feb 20, 2018 at 18:18
• Your boundary conditions on $z$ will not be analogous to those on $y$ in the rectangular case. You can reduce your problem to one independent variable but you might get different functional forms depending on the form of the boundary conditions for $z$. So treat that carefully. Also remember to use $r$ for your coordinate throughout. Feb 20, 2018 at 19:04
Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are $$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)
And $$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$
With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).
Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.
Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$. | 990 | 3,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-10 | latest | en | 0.837455 |
http://mathhelpforum.com/calculus/200788-limit-two-sequences.html | 1,513,264,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544677.45/warc/CC-MAIN-20171214144324-20171214164324-00594.warc.gz | 169,815,219 | 11,682 | # Thread: Limit of two sequences
1. ## Limit of two sequences
Let an and bn be two convergent sequences. Prove: if for every even n: an<=bn, and for every odd n: an>=bn, then lim an = lim bn.
Any ideas? I'm looking for a formal proof.
Thanks!
2. ## Re: Limit of two sequences
This can be proved using the following facts.
(1) The limit of a subsequence of a converging sequence equals the limit of the sequence.
(2) If an and bn are converging sequences and an ≤ bn for all n, then lim an ≤ lim bn.
3. ## Re: Limit of two sequences
The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
"If a sequence converges, then its limit is unique."
"Every convergent sequence is bounded."
4. ## Re: Limit of two sequences
Let A = lim aₙ and B = lim bₙ. For a given ε > 0, choose N such that for all n > N we have |aₙ - A| < ε and |bₙ - B| < ε. Pick any even n > N; then inequalities in the previous sentence imply that A < aₙ + ε and bₙ < B + ε. Therefore, A < aₙ + ε ≤ bₙ + ε < B + 2ε. Thus, for the given ε we showed that A < B + 2ε. Since this holds for any positive ε, it must be that A ≤ B. Similarly, you can show that A ≥ B.
5. ## Re: Limit of two sequences
Originally Posted by loui1410
The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
"If a sequence converges, then its limit is unique."
"Every convergent sequence is bounded."
Suppose that ${\lim _{n \to \infty }}{a_n} = A\;\& \;{\lim _{n \to \infty }}{b_n} = B$. Going for a contradiction, suppose that $A. Now define $\varepsilon = \frac{{B - A}}{2} > 0$.
So we have $\left( {A - \varepsilon ,A + \varepsilon } \right) \cap \left( {B - \varepsilon ,B + \varepsilon } \right) = \emptyset$.
Because of convergence, almost all the $a_n's$ are in $\left( {A - \varepsilon ,A + \varepsilon } \right)$ and almost all the $b_n's$ are in $\left( {B - \varepsilon ,B + \varepsilon } \right)$.
But that is impossible because of the alteration of odd and even terms.
6. ## Re: Limit of two sequences
emakarov, thank you - that's a great solution, I wouldn't have thought about it.
Plato's solution, however, seems more natural to me as it's similar to some other examples I've already seen.
Thank you both. | 759 | 2,627 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-51 | longest | en | 0.939254 |
https://www.edalive.com/curriculum/structure/nsw_maths_s-3_nums-algebra_patterns-algebra_ma3-8na_patterns-algebra-1_equivalent~-d~s-acmna121_complete-n~c~ec-7-x-7-7 | 1,718,389,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00189.warc.gz | 679,345,733 | 70,005 | Complete number sentences involving multiplication and division, including those involving simple fractions or decimals, eg 7 x ( ) = 7.7
1 Number production: Put the correct ending on each multiplication sentence 2 Number production: Complete each multiplication sentence. 3 Maths production: Complete each multiplication sentence. 4 Number detectives: ? / 7 = 60 | 74 | 366 | {"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-2024-26 | latest | en | 0.819069 |
https://www.asvabtestbank.com/math-knowledge/practice-test/104563/5 | 1,560,793,757,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998513.14/warc/CC-MAIN-20190617163111-20190617185111-00372.warc.gz | 692,268,758 | 4,264 | ASVAB Math Knowledge Practice Test 104563
Questions 5 Topics Cylinders, Operations Involving Monomials, Quadratic Equations, Right Angle
Study Guide
Cylinders
A cylinder is a solid figure with straight parallel sides and a circular or oval cross section with a radius (r) and a height (h). The volume of a cylinder is π r2h and the surface area is 2(π r2) + 2π rh.
Operations Involving Monomials
You can only add or subtract monomials that have the same variable and the same exponent. However, you can multiply and divide monomials with unlike terms. | 141 | 558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-26 | latest | en | 0.83798 |
http://mathhelpforum.com/algebra/1548-animals-heads-legs-chickens.html | 1,521,581,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00140.warc.gz | 195,571,116 | 10,797 | # Thread: animals heads legs chickens
1. ## ratios
my 11yr old has been given some homework on ratios that i'm finding hard,
A farm has some chickens & some rabbits.
Altogether there are:
35 heads
94 legs
How many chickens are there?
and how many rabbits are there?
plus show your workings
Any help would be great
2. Originally Posted by lacourse
my 11yr old has been given some homework on ratios that i'm finding hard,
A farm has some chickens & some rabbits.
Altogether there are:
35 heads
94 legs
How many chickens are there?
and how many rabbits are there?
plus show your workings
Any help would be great
So it is about heads and legs.
Chicken has one head; rabbit has one head.
Chicken has two legs; rabbit has 4 legs.
Let c = number of chickens in there; and r = number of rabbits.
Re heads:
c*1 +r*1 = 35
c +r = 35 -------------(1)
Re legs:
c*2 +r*4 = 94
2c +4r = 94
Reduce that to its "lowest term", divide both sides by 2,
c +2r = 47 ------------(2)
One way to continue is by substitution.
From (1), c = 35-r
Substitute that into (2),
35-r +2r = 47
2r -r = 47 -35
r = 12
So, c = 35 -12 = 23
Therefore, there are 23 chickens and 12 rabbits in there. ------answer.
---------
Check.
Re heads,
c +r = 35
23 +12 =? 35
35 =? 35
Yes, so, OK.
Re legs,
2c +4r = 94
2*23 +4*12 =? 94
46 +48 =? 94
94 =? 94
Yes, so, OK.
3. Originally Posted by lacourse
my 11yr old has been given some homework on ratios that i'm finding hard,
A farm has some chickens & some rabbits.
Altogether there are:
35 heads
94 legs
How many chickens are there?
and how many rabbits are there?
plus show your workings
Any help would be great
As there are 35 heads we have 35 creatures altogether.
Counting two legs of each would account for a total of
70 legs, leaving 24 legs unaccounted for, which must
belong to 12 rabbits (to give each rabbit four legs)
So we have 12 rabbits and 23 chickens.
RonL
4. Thanks,
My daughter will be pleased
,
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# 35 heads and 94 legs program in c
Click on a term to search for related topics. | 607 | 2,053 | {"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} | 4.28125 | 4 | CC-MAIN-2018-13 | latest | en | 0.959241 |
https://www.transtutors.com/questions/you-are-the-contestant-on-the-monty-hall-show-monty-is-trying-out-a-new-version-of-h-2084147.htm | 1,642,860,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00633.warc.gz | 1,087,584,954 | 13,144 | # You are the contestant on the Monty Hall show. Monty is trying out a new version of his game,...
You are the contestant on the Monty Hall show. Monty is trying out a new version of his game, with rules as follows. You get to choose one of three doors. One door has a car behind it, another has a computer, and the other door has a goat (with all permutations equally likely). Monty, who knows which prize is behind each door, will open a door (but not the one you chose) and then let you choose whether to switch from your current choice to the other unopened door. Assume that you prefer the car to the computer, the computer to the goat, and (by transitivity) the car to the goat.
(a) Suppose for this part only that Monty always opens the door that reveals your less preferred prize out of the two alternatives, e.g., if he is faced with the choice between revealing the goat or the computer, he will reveal the goat. Monty opens a door, revealing a goat (this is again for this part only). Given this information, should you switch? If you do switch, what is your probability of success in getting the car?
(b) Now suppose that Monty reveals your less preferred prize with probability p, and your more preferred prize with probability q = 1 p. Monty opens a door, revealing a computer. Given this information, should you switch (your answer can depend on p)? If you do switch, what is your probability of success in getting the car (in terms of p)?
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker | 352 | 1,559 | {"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-2022-05 | latest | en | 0.957209 |
https://algorithms.tutorialhorizon.com/swap-kth-node-from-the-front-with-the-kth-node-from-the-end/ | 1,579,270,016,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589560.16/warc/CC-MAIN-20200117123339-20200117151339-00098.warc.gz | 329,214,474 | 20,193 | # Swap Kth Node from the front with the Kth Node from the End
Objective: Given a Linked List and a number k, Swap Kth Node from the front with the Kth Node from the End
Example:
```->10->20->30->40->50->60->70
Swapping 1 Node from the Front and from the End
->70->20->30->40->50->60->10
Swapping 2 Node from the Front and from the End
->70->60->30->40->50->20->10
Swapping 3 Node from the Front and from the End
->70->60->50->40->30->20->10
k = 4, Nodes are same from front and at the end, no swapping
->70->60->50->40->30->20->10
Swapping 5 Node from the Front and from the End
->70->60->30->40->50->20->10
Swapping 6 Node from the Front and from the End
->70->20->30->40->50->60->10
Swapping 7 Node from the Front and from the End
->10->20->30->40->50->60->70
INVALID NUMBER, No Swapping, k>length of list
->10->20->30->40->50->60->70
```
Approach:
• Find the length of the list, say it is ‘Len’.
• If k>Len, No Swapping.
• If kth node from the front and the end are same (2*k-1=Len), No Swapping.
• If above two steps are not true then we need swapping of the elements.
• Take a pointer left, move it by k nodes. Keep track of node prior to left( call it as left_prev, we need it for the swapping).
• Set left_prev = null if k=1.
• Take a pointer right, move it by len-k+1 nodes(it will be the kth node from the end). Keep track of node prior to left( call it as right_prev, we need it for the swapping).
• Set right_prev = null if k=Len.
• If left_prev!=NULL means left node is not the first node, so make left_prev will point to right
• If right_prev!=NULL means right node is not the first node, so right_prev will point to left node.
• Now just swap the next and right.next to complete the swapping.
• NOTE:We need to change the head of list if k =1 (head = right) or k = len (head = left).
Complete Code:
Output:
```->10->20->30->40->50->60->70
Swapping 1 Node from the Front and from the End
->70->20->30->40->50->60->10
Swapping 2 Node from the Front and from the End
->70->60->30->40->50->20->10
Swapping 3 Node from the Front and from the End
->70->60->50->40->30->20->10
k = 4, Nodes are same from front and at the end, no swapping
->70->60->50->40->30->20->10
Swapping 5 Node from the Front and from the End
->70->60->30->40->50->20->10
Swapping 6 Node from the Front and from the End
->70->20->30->40->50->60->10
Swapping 7 Node from the Front and from the End
->10->20->30->40->50->60->70
INVALID NUMBER, No Swapping, k>length of list
->10->20->30->40->50->60->70
```
__________________________________________________
Top Companies Interview Questions..-
If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
__________________________________________________ | 843 | 2,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-05 | longest | en | 0.648367 |
https://quantum-computing.ibm.com/lab/docs/iqx/operations_glossary | 1,685,852,512,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00228.warc.gz | 513,822,174 | 98,664 | # Operations glossary¶
## Overview¶
This page is a reference that defines the various classical and quantum operations you can use to manipulate qubits in a quantum circuit. Quantum operations include quantum gates, such as the Hadamard gate, as well as operations that are not quantum gates, such as the measurement operation.
Each entry below provides details and the OpenQASM reference for each operation. See the topic on Build your circuit with OpenQASM code in the IBM Quantum Composer docs for more information.
The q-sphere image in each gate entry below shows the state after the gate operates on the initial equal superposition state , where is the number of qubits needed to support the gate. See the q-sphere topic in the Quantum Composer docs for more information on this visualization.
You can define a custom operation to use in IBM Quantum Composer. For instructions, see the Create a custom operation in OpenQASM topic in the Quantum Composer docs.
To learn more about using operations to create quantum algorithms, see the single- and multi-qubit gates chapter of the Qiskit textbook, Learn Quantum Computation using Qiskit.
Note
The gate colors are slightly different in the light and dark themes. The colors from the light theme are shown here.
Click a quantum operation below to view its definition. Operations no longer used in Circuit Composer are listed in the Obsolete operations section as a historical reference.
Classical gates
Phase gates
Non-unitary operators and modifiers
Quantum gates
## Classical gates¶
### NOT gate¶
The NOT gate, also known as the Pauli X gate, flips the state to , and vice versa. The NOT gate is equivalent to RX for the angle or to ‘HZH’.
For more information about the NOT gate, see XGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
x q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### CNOT gate¶
The controlled-NOT gate, also known as the controlled-x (CX) gate, acts on a pair of qubits, with one acting as ‘control’ and the other as ‘target’. It performs a NOT on the target whenever the control is in state . If the control qubit is in a superposition, this gate creates entanglement.
All unitary circuits can be decomposed into single qubit gates and CNOT gates. Because the two-qubit CNOT gate costs much more time to execute on real hardware than single qubit gates, circuit cost is sometimes measured in the number of CNOT gates.
For more information about the CNOT gate, see CXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
cx q[0], q[1];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### Toffoli gate¶
The Toffoli gate, also known as the double controlled-NOT gate (CCX), has two control qubits and one target. It applies a NOT to the target only when both controls are in state .
The Toffoli gate with the Hadamard gate is a universal gate set for quantum computing.
For more information about the Toffoli gate, see CCXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
ccx q[0], q[1], q[2];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### SWAP gate¶
The SWAP gate swaps the states of two qubits.
For more information about the SWAP gate, see SwapGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
swap q[0], q[1];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### Identity gate¶
The identity gate (sometimes called the Id or the I gate) is actually the absence of a gate. It ensures that nothing is applied to a qubit for one unit of gate time.
Composer reference
Qasm reference
id q[0];
## Phase gates¶
### T gate¶
The T gate is equivalent to RZ for the angle . Fault-tolerant quantum computers will compile all quantum programs down to just the T gate and its inverse, as well as the Clifford gates.
For more information about the T gate, see TGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
t q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### S gate¶
The S gate applies a phase of to the state. It is equivalent to RZ for the angle . Note that S=P().
For more information about the S gate, see SGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
s q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### Z gate¶
The Pauli Z gate acts as identity on the state and multiplies the sign of the state by -1. It therefore flips the and states. In the +/- basis, it plays the same role as the NOT gate in the / basis.
For more information about the Z gate, see ZGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
z q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### T ^{\dagger} gate¶
Also known as the Tdg or T-dagger gate.
The inverse of the T gate.
For more information about the T gate, see TdgGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
tdg q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### S ^{\dagger} gate¶
Also known as the Sdg or S-dagger gate.
The inverse of the S gate.
For more information about the S gate, see SdgGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
sdg q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### Phase gate¶
The Phase gate (previously called the U1 gate) applies a phase of to the state. For certain values of , it is equivalent to other gates. For example, P()=Z, P(/)=S, and P()=T. Up to a global phase of , it is equivalent to RZ().
For more information about the Phase gate, see PhaseGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
p(theta) q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for theta is .
### RZ gate¶
The RZ gate implements . On the Bloch sphere, this gate corresponds to rotating the qubit state around the z axis by the given angle.
For more information about the RZ gate, see RZGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rz(angle) q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is . Therefore, this is the angle used in the q-sphere visualization.
## Non-unitary operators and modifiers¶
### Reset operation¶
The reset operation returns a qubit to state , irrespective of its state before the operation was applied. It is not a reversible operation.
Composer reference
OpenQASM reference
reset q[0];
### Measurement¶
Measurement in the standard basis, also known as the z basis or computational basis. Can be used to implement any kind of measurement when combined with gates. It is not a reversible operation.
Composer reference
OpenQASM reference
measure q[0];
### Control modifier¶
A control modifier yields a gate whose original operation is now contingent on the state of the control qubit. When the control is in the state, the target qubit(s) undergo the specified unitary evolution. In contrast, no operation is performed if the control is in the state. If the control is in a superposition state, then the resulting operation follows from linearity.
Drag the control modifier to a gate in order to add a control to it. Dots will appear above and below the gate, on the qubit wires that can be targets that control; click one or more dots to assign the target to one or more qubits. You can also assign a control by right-clicking a gate.
To remove a control, right-click the gate and select the option to remove control.
Composer reference
OpenQASM reference
c
### IF operation¶
The IF operation allows quantum gates to be conditionally applied, depending on the state of a classical register.
Composer reference
OpenQASM reference
if (c==0) x q[0];
### Barrier operation¶
To make your quantum program more efficient, the compiler will try to combine gates. The barrier is an instruction to the compiler to prevent these combinations being made. Additionally, it is useful for visualizations.
For more information about the Barrier instruction, see Barrier in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
barrier q;
### H gate¶
The H, or Hadamard, gate rotates the states and to and , respectively. It is useful for making superpositions. If you have a universal gate set on a classical computer and add the Hadamard gate, it becomes a universal gate set on a quantum computer.
For more information about the H gate, see HGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
h q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
## Quantum gates¶
### \sqrt{X} gate¶
Also known as the square-root NOT gate.
This gate implements the square-root of X, . Applying this gate twice in a row produces the standard Pauli-X gate (NOT gate). Like the Hadamard gate, creates an equal superposition state if the qubit is in the state , but with a different relative phase. On some hardwares, it is a native gate that can be implemented with a or X90 pulse.
For more information about the gate, see SXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
sx q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### \sqrt{X} ^{\dagger} gate¶
Also known as the SXdg or square-root NOT-dagger gate.
This is the inverse of the gate. Applying it twice in a row produces the Pauli-X gate (NOT gate), since the NOT gate is its own inverse. Like the gate, this gate can be used to create an equal superposition state, and it too is natively implemented on some hardwares using an X90 pulse.
For more information about the gate, see SXdgGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
sxdg q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### Y gate¶
The Pauli Y gate is equivalent to Ry for the angle . It is equivalent to applying X and Z, up to a global phase factor.
For more information about the Y gate, see YGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
y q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### RX gate¶
The RX gate implements . On the Bloch sphere, this gate corresponds to rotating the qubit state around the x axis by the given angle.
For more information about the RX gate, see RXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rx(angle) q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is . Therefore, this is the angle used in the q-sphere visualization.
### RY gate¶
The RY gate implements . On the Bloch sphere, this gate corresponds to rotating the qubit state around the y axis by the given angle and does not introduce complex amplitudes.
For more information about the RY gate, see RYGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
ry(angle) q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is . Therefore, this is the angle used in the q-sphere visualization below.
### RXX gate¶
The RXX gate implements . The Mølmer–Sørensen gate, the native gate on ion-trap systems, can be expressed as a sum of RXX gates.
For more information about the RXX gate, see RXXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rxx(angle) q[0], q[1];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is .
### RZZ gate¶
The RZZ gate requires a single parameter: an angle expressed in radians. This gate is symmetric; swapping the two qubits it acts on doesn’t change anything.
For more information about the RZZ gate, see RZZGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rzz(angle) q[0], q[1];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is .
### U gate¶
(Previously called the U3 gate) The three parameters allow the construction of any single-qubit gate. Has a duration of one unit of gate time.
For more information about the U gate, see UGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
u(theta, phi, lam) q[0];
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
In IBM Quantum Composer, the default value for angle is .
### RCCX gate¶
The simplified Toffoli gate, also referred to as Margolus gate.
The simplified Toffoli gate implements the Toffoli gate up to relative phases. This implementation requires three CX gates, which is the minimal amount possible, as shown in https://arxiv.org/abs/quant-ph/0312225. Note that the simplified Toffoli is not equivalent to the Toffoli, but can be used in places where the Toffoli gate is uncomputed again.
For more information about the RCCX gate, see RCCXGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rccx a, b, c;
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
### RC3X gate¶
The simplified 3-controlled Toffoli gate.
The simplified Toffoli gate implements the Toffoli gate up to relative phases. Note that the simplified Toffoli is not equivalent to the Toffoli, but can be used in places where the Toffoli gate is uncomputed again.
For more information about the RC3X gate, see RC3XGate in the Qiskit Circuit Library.
Composer reference
OpenQASM reference
Q-sphere
Note about q-sphere representations
rc3x a, b, c, d;
The q-sphere representation shows the state after the gate operates on the initial equal superposition state where is the number of qubits needed to support the gate.
## Obsolete operations¶
These operations are no longer used in IBM Quantum Composer; we list them here for historical purposes.
### CSWAP gate¶
Composer reference
OpenQASM reference
cswap q[0], q[1], q[2];
### U1 gate¶
The U1 gate has been renamed the phase gate.
Composer reference
OpenQASM reference
u1(theta) q[0];
### U3 gate¶
The U3 gate has been renamed the U gate.
Composer reference
OpenQASM reference
u3(theta, phi, lam) q[0];
### U2 gate¶
Composer reference
OpenQASM reference
u2(theta, phi) q[0];
### CU1 gate¶
To recreate this gate, add the control modifier to the phase gate (formerly the U1 gate).
Composer reference
OpenQASM reference
cu1(angle) q[0], q[1];
### CU3 gate¶
To recreate this gate, add the control modifier to the U gate (formerly the U3 gate).
Composer reference
OpenQASM reference
cu3(angle) q[0], q[1];
### CH gate¶
Composer reference
OpenQASM reference
ch q[0], q[1];
### CY gate¶
Composer reference
OpenQASM reference
cy q[0], q[1];
### CZ gate¶
Composer reference
OpenQASM reference
cz q[0], q[1];
### CRX gate¶
Composer reference
OpenQASM reference
crx(angle) q[0], q[1];
### CRY gate¶
Composer reference
OpenQASM reference
cry(angle) q[0], q[1];
### CRZ gate¶
Composer reference
OpenQASM reference
crz(angle) q[0], q[1]; | 4,289 | 18,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.887945 |
https://cr.openjdk.java.net/~ceisserer/7179526/webrev.00/src/solaris/classes/sun/java2d/xr/MaskTile.java.cdiff.html | 1,653,691,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00292.warc.gz | 236,465,682 | 2,923 | ```*** 39,140 ****
{
rects = new GrowableRectArray(128);
dirtyArea = new DirtyRegion();
}
- public void addRect(int x, int y, int width, int height) {
- int index = rects.getNextIndex();
- rects.setX(index, x);
- rects.setY(index, y);
- rects.setWidth(index, width);
- rects.setHeight(index, height);
- }
-
- public void addLine(int x1, int y1, int x2, int y2) {
- /*
- * EXA is not able to accalerate diagonal lines, we try to "guide" it a
- * bit to avoid excessive migration See project documentation for an
- * detailed explanation
- */
- DirtyRegion region = new DirtyRegion();
- region.setDirtyLineRegion(x1, y1, x2, y2);
- int xDiff = region.x2 - region.x;
- int yDiff = region.y2 - region.y;
-
- if (xDiff == 0 || yDiff == 0) {
- region.x2 - region.x + 1, region.y2 - region.y + 1);
- } else if (xDiff == 1 && yDiff == 1) {
- } else {
- lineToRects(x1, y1, x2, y2);
- }
- }
-
- private void lineToRects(int xstart, int ystart, int xend, int yend) {
- int x, y, t, dx, dy, incx, incy, pdx, pdy, ddx, ddy, es, el, err;
-
- /* Entfernung in beiden Dimensionen berechnen */
- dx = xend - xstart;
- dy = yend - ystart;
-
- /* Vorzeichen des Inkrements bestimmen */
- incx = dx > 0 ? 1 : (dx < 0) ? -1 : 0;
- incy = dy > 0 ? 1 : (dy < 0) ? -1 : 0;
- if (dx < 0)
- dx = -dx;
- if (dy < 0)
- dy = -dy;
-
- /* feststellen, welche Entfernung groesser ist */
- if (dx > dy) {
- /* x ist schnelle Richtung */
- pdx = incx;
- pdy = 0; /* pd. ist Parallelschritt */
- ddx = incx;
- ddy = incy; /* dd. ist Diagonalschritt */
- es = dy;
- el = dx; /* Fehlerschritte schnell, langsam */
- } else {
- /* y ist schnelle Richtung */
- pdx = 0;
- pdy = incy; /* pd. ist Parallelschritt */
- ddx = incx;
- ddy = incy; /* dd. ist Diagonalschritt */
- es = dx;
- el = dy; /* Fehlerschritte schnell, langsam */
- }
-
- /* Initialisierungen vor Schleifenbeginn */
- x = xstart;
- y = ystart;
- err = el / 2;
-
- /* Pixel berechnen */
- for (t = 0; t < el; ++t) /* t zaehlt die Pixel, el ist auch Anzahl */
- {
- /* Aktualisierung Fehlerterm */
- err -= es;
- if (err < 0) {
- /* Fehlerterm wieder positiv (>=0) machen */
- err += el;
- /* Schritt in langsame Richtung, Diagonalschritt */
- x += ddx;
- y += ddy;
- } else {
- /* Schritt in schnelle Richtung, Parallelschritt */
- x += pdx;
- y += pdy;
- }
- // SetPixel(x,y);
- // System.out.println(x+":"+y);
- }
- }
-
public void calculateDirtyAreas()
{
for (int i=0; i < rects.getSize(); i++) {
int x = rects.getX(i);
int y = rects.getY(i);
--- 39,48 ----
``` | 974 | 3,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | latest | en | 0.148055 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=56151 | 1,582,136,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00498.warc.gz | 443,278,805 | 11,638 | ## Concentration affecting equilibrium
Clarice Chui 2C
Posts: 81
Joined: Fri Aug 30, 2019 12:16 am
### Concentration affecting equilibrium
How does changing concentration of the products affect an equilibrium reaction?
Aliya Jain 2B
Posts: 80
Joined: Wed Sep 11, 2019 12:16 am
### Re: Concentration affecting equilibrium
I think that decreasing the concentration of products will decrease the reaction quotient while increasing the concentration pf products will increase the reaction quotient
kevinolvera1j
Posts: 79
Joined: Fri Aug 02, 2019 12:15 am
### Re: Concentration affecting equilibrium
Changing the concentrations does not change the K value. However, in the short term it will change the value of Q. Adding more products will make the value of Q larger and if Q>K then the system will favor the formation of reactants and vice versa.
ShastaB4C
Posts: 79
Joined: Thu Sep 26, 2019 12:18 am
### Re: Concentration affecting equilibrium
Can someone clarify how taking out some of the product can increase the amount of product made without considering temperature or additional input?
Justin Quan 4I
Posts: 84
Joined: Sat Sep 14, 2019 12:17 am
### Re: Concentration affecting equilibrium
ShastaB4C wrote:Can someone clarify how taking out some of the product can increase the amount of product made without considering temperature or additional input?
To clarify, when you remove some of the product, the concentration of the product will decrease, while the amount of reactants will stay the same. Thus, as you remove more product, Q will be less than K. When Q<K, the reaction proceeds forward and more product will be made.
Keerthana Sivathasan 2E
Posts: 80
Joined: Wed Sep 18, 2019 12:22 am
### Re: Concentration affecting equilibrium
Keeping in mind that the equilibrium constant is the ratio between concentration of products and concentration of reactants, if the concentration of products is increased, then the reaction's Q>K, and if the concentration of products is decreased then Q<K and Chatelier's principle will continue to try to bring it back into equilibrium by minimizing what has happened.
Return to “Equilibrium Constants & Calculating Concentrations”
### Who is online
Users browsing this forum: No registered users and 2 guests | 527 | 2,276 | {"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} | 2.78125 | 3 | CC-MAIN-2020-10 | longest | en | 0.879333 |
http://mathoverflow.net/questions/79273?sort=newest | 1,371,597,138,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707436332/warc/CC-MAIN-20130516123036-00083-ip-10-60-113-184.ec2.internal.warc.gz | 160,727,060 | 10,532 | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Intersection positivity for curves and surfaces
Let $X$ be a smooth complete variety over an algebraically closed field of dimension $\geq3$. Given a divisor $D_1$ on $X$ with $D_1 \cdot C>0$ for every curve $C \subset X$, and a divisor $D_2$ on $X$ satisfying $D_2^2 \cdot S>0$ for every surface $S \subset X$, does there exist a divisor $D$ on $X$ satisfying $D \cdot C>0$ and $D^2 \cdot S>0$ for every curve and surface respectively? I am willing to make any assumption on $X$, except that $X$ be projective.
As I understand it, since $D_1$ is nef, we have that $D_1^2 \cdot S\geq 0$, so even if for $m>>0$ we manage to have $(mD_1+D_2) \cdot C>0$ for every curve, the other requirement becomes $(mD_1+D_2)^2 \cdot S = m^2(D_1^2 \cdot S) + 2m(D_1\cdot D_2 \cdot S) + D_2^2 \cdot S >0$. The first term is non-negative, the last term is positive, but what needs to happen to ensure the middle term is non-negative also?
-
I am not sure whether your original question is true, but it seems to me that your proposed solution does not work. Here is why:
The main problem is that you know very little about $D_2$. Given what we know, it does not even have to be effective!
Example Let $X$ be a smooth complete variety with divisors $A,B$ such
• $A\cdot B=0$
• $(A^2+B^2)\cdot S>0$ for any surface $S$, and
• $B$ and $B^2$ are effective classes.
Then $D_2=A-B$ has the property that $D_2^2\cdot S>0$ for any $S$, but $D_1\cdot D_2\cdot B\cdot H <0$ for any $D_1$ divisor satisfying the criterion given in the question and an appropriate $H$ such that $B\cdot S$ is an effective surface and $B^2\cdot H$ is an effective curve.
To make this complete I should give an example that such an $X$ with $A,B$ exists. I think I can construct an example (a product of two surfaces blown up along a surface) satisfying these properties, but I don't know a simple one. In any case I think it is reasonable to expect that such an example exists which shows that $D_1\cdot D_2\cdot S$ is not necessarily non-negative even on a projective variety.
-
Yes, you're right Sandor. It seems one cannot get around this without making some critical assumptions on $D_1 \cdot D_2$. Thank you. – Parsa Oct 28 2011 at 11:27 | 710 | 2,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2013-20 | latest | en | 0.904968 |
https://www.teachoo.com/8259/2722/Ex-5.5--2/category/Perpendicular-Lines/ | 1,685,839,022,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00446.warc.gz | 1,114,975,815 | 32,567 | Perpendicular Lines
Chapter 5 Class 6 Understanding Elementary Shapes
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 5.5, 2 Let (ππ) Μ
be the perpendicular to the line segment (ππ) Μ
. Let (ππ) Μ
and (ππ) Μ
intersect in the point A. What is the measure of β PAY ? Since, (ππ) Μ
β₯ PQ Angle between them is a right angle. β΄ β PAY = 90Β° | 162 | 439 | {"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-2023-23 | longest | en | 0.724061 |
http://www.slate.com/blogs/browbeat/2012/08/28/ryan_hall_at_t_commercial_how_far_does_the_marathoner_run_while_listening_to_the_odyssey_.html | 1,411,231,469,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657133455.98/warc/CC-MAIN-20140914011213-00014-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 828,760,917 | 80,957 | # An AudiobookMarathon: How Far Does Ryan Hall Travel While Listening to the Unabridged Odyssey?
Slate's Culture Blog
Aug. 28 2012 12:46 PM
# An AudiobookMarathon: How Far Does Ryan Hall Travel While Listening to the Unabridged Odyssey?
In a recent commercial you've seen 115 times, American marathoner Ryan Hall listens to The Odyssey as he runs, all thanks to AT&T's impressively expansive 4G network.*
Hall’s jog begins on an unidentified campus, then continues under some train tracks, past a giant blue gorilla, and toward a setting sun. At the ad’s 16-second mark, Hall briefly pauses as The Odyssey reaches its conclusion, then starts to run again after queuing up Moby Dick. The audiobook version of Robert Fagles' translation of The Odyssey—I'm currently listening to it and Book Eleven is great—clocks in at 13 hours, 18 minutes, and 46 seconds (or 798.77 minutes). The inevitable question: How many miles did Ryan Hall cover while listening to Homer’s epic poem?
First, we must determine Hall’s stride rate. During the commercial’s fifth through tenth seconds, the U.S. record holder in the half-marathon takes 16 strides—that’s 3.2 strides per second, or 192 strides per minute. He doesn't seem to slow up at any point during the advertisement either, so let’s stipulate that he runs at that rate the entire, hypothetical day.
Second, we need Hall's stride length. In his record-setting half-marathon in Houston, Hall traveled 69,218 feet (13.1 miles) in 59 minutes and 43 seconds, taking approximately 10,868 steps. That gives him an average of around 6 feet 4 1/2 inches per stride. (While some sources peg Hall's stride length at 6 feet 10 inches, we'll use the measurement from his best race.)
At 6.37 feet per stride, and at 192 strides per minute, that means he's running 1,223 feet per minute in the commercial. If Hall runs at that pace for all of those mythical minutes—perhaps quickening his step when Odysseus de-ocularizes the Cyclops, but slowing when the men discuss the prevailing winds for the 48th time—how much ground does he cover in almost 800 minutes?
If only the answer were that simple. A close reading reveals that Hall, not known in running circles for taking shortcuts, seems to have skipped some of the performance. When the commercial starts, the time on his phone is 8:07 a.m. When he begins listening to Moby Dick (a 21-hour audiobook in its own right), the phone reads 7:12pm. That means that Hall, if he never hit pause along the way, listened to 11:05:00 of The Odyssey. What happened to those missing 2 hours, 13 minutes, and 46 seconds? Given that the gap in the Watergate tapes ran just 18.5 minutes, this AT&T commercial leaves us to ponder whether Ryan Hall is seven times trickier than Nixon.
But let’s give Hall the benefit of the doubt. If he changed time zones, running, say, from Mountain to Pacific, we would add an hour to his exercise time and assume that he’d heard 12:05:00 of The Odyssey. A time zone switch is not impossible. This is a long journey we're talking about. An expedition, a voyage, a peregrination.
So, on to our calculations. If he stays in one time zone, Hall moves at 1,223 feet per minute for 665 minutes. That's 813,190 feet, or 154 miles in the course of his run—in the neighborhood of six jaunts from Marathon to Athens. If we roll with the Mountain-to-Pacific theory and add another hour to Hall’s run, that gives us 886,561 feet, or 167.9 miles.
And if Hall had listened to The Odyssey the whole way through? He’d have covered 976,767 feet, or 185 miles. By chopping out 2 hours, 13 minutes, and 46 seconds of Greek poetry, then, Hall saved himself more than a marathon of work. But considering that the record for distance covered on a treadmill in 24 hours is a mere 153.76 miles, maybe we can cut Hall some slack for taking it easy.
Meanwhile, one final computation: A line of dactylic hexameter, the meter of The Odyssey, contains six poetic feet. There are 12,110 lines in Homer’s epic, which gives us about 72,660 units of rhythm. By this measure, Hall runs 10 feet per foot, give or take a couple of feet in either direction.
*Correction, Aug. 28: This post originally said that Ryan Hall was listening to Ian McKellen read Robert Fagles' translation of The Odyssey. Hall actually seems to be listening to a non-McKellen voice actor read Samuel Butler's translation.
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Health & Science | 1,390 | 5,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2014-41 | longest | en | 0.918812 |
https://www.gamedev.net/forums/topic/682719-finding-the-instantaneous-position-on-a-helix/ | 1,544,571,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823705.4/warc/CC-MAIN-20181211215732-20181212001232-00163.warc.gz | 900,183,233 | 29,997 | # Finding the instantaneous position on a helix
This topic is 803 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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x = startBase.x + path->R * sin(t);
y = startBase.y + path->R * cos(t);
z = startBase.z + path->R * t;
Let's say I'm counting off from 0, 0, 0
and the turning radius is 1000 meters.
at time 0.0, cos(t) is at R, which wasn't quite right.
Because at time 0.0, the agent should be close to 0.
What should be corrected?
Thanks
Jack
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Subtract R from y maybe? What exactly is supposed to happen?
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Thanks kolrabi,
I've just sort of figured out what should be the correct way.
I should be looking into the center of the rotation instead?
But really Thanks, appreciated!
Jack
Edited by lucky6969b
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Back to it,
So I am needing to use the Polar coordinate system to find the instantaneous position.
I already had the R and the theta, is it just simply do
x = R * sin(t)
y = R * cos(t)
without respect to z (up)
And the final product should be
x = center_of_rotation.x + R * cos(t)
y = center_of_rotation.y + R * sin(t)
depending on whether I am passing the quadrant marks?
Thanks
Jack
Edited by lucky6969b
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You aim to move around on a circle in the xy plane, while going up in z? I think all your fomulas do that.
The only difference is the initial starting point in the xy plane.
You can somewhat tweak that by swapping cos and sin around (cos and sin are basically the same function, they are just shifted by 90 degrees (or pi/2 if you use radians).
A more flexible approach is to add a little extra rotation, like at time 0, you already move to the right start angle in the xy plane.
x = center_of_rotation.x + R * cos(t + start_angle)
y = center_of_rotation.y + R * sin(t + start_angle)
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Hello,
I finally got some credible results.
But Only one thing
When the agent travels from say radian of 0
and when it elevates to a directional vector of which it has a -ve x component (I meant just going left)
The right solution comes up to my mind was the agent turns right and starts to elevate and goes around
eventually hitting a sort of upwards vector which pointing to the left and going into the screen somewhat.
But With Slerp or Lerp which comes from D3DX, the interpolation always wants to choose the shortest path,
I want the interpolation function takes the longer path, which indeed has to go around a large helix
before hitting the straight line segment, In my case, it chooses a shorter path and then connects with
the straight line with an abrupt turn which I believe is wrong.
When the dubins got some ideas knowing it is a right turn, can I force the Slerp or Lerp to go around
and choose the longer path?
Or In the picture, the agent actually has to go the other way round, but it *must* choose a longer path before
it stops the interpolation.
Hang on, I double checked, looks like the directional vector is wrong, the right turn ends at 0.6,xx,xx What the heck
So the interpolation will stop very quickly
Thanks
Jack[attachment=33558:2016-09-30 16_31_13-TestSB.jpg]
But wait a minute
Why with a right turn, the center of rotation should be in the positive x range
What did I do wrong?
[attachment=33559:2016-09-30 16_59_18-TestOpenTissue2 (Debugging) - Microsoft Visual Studio (Administrator).jpg]
Oh hell yeah.... rotZ(PI/2) instead of rotZ(-PI/2)
D3DXVECTOR3 dot(D3DXVECTOR3 t1, D3DXMATRIX t2)
{
// x = ax + by + cz
// y = px + qy + rz
// z = ux + vy + wz
//double x = t1.x * t2._11 + t1.y * t2._21 + t1.z * t2._31;
//double y = t1.x * t2._12 + t1.y * t2._22 + t1.z * t2._32;
//double z = t1.x * t2._13 + t1.y * t2._23 + t1.z * t2._33;
double x = t1.x * t2._11 + t1.y * t2._12 + t1.z * t2._13;
double y = t1.x * t2._21 + t1.y * t2._22 + t1.z * t2._23;
double z = t1.x * t2._31 + t1.y * t2._32 + t1.z * t2._33;
return D3DXVECTOR3(x, y, z);
}
D3DXMATRIX rotZ(float angle)
{
D3DXMATRIX rZ;
D3DXMatrixIdentity(&rZ);
D3DXMatrixRotationZ(&rZ, angle);
return rZ;
}
Okay, I know now, I messed up the code before, I just did too much to the code[attachment=33560:2016-09-30 17_31_22-TestSB.jpg]
Edited by lucky6969b
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https://cosmoesferas.it/feet-to-cubic-feet-calculator.html | 1,618,321,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00162.warc.gz | 296,892,907 | 9,801 | • Consider the following to help you choose a refrigerator with enough cubic feet (cu. ft.) for your family's needs: Each adult will need a minimum of 4–6 cu. ft. of refrigerator space, with some extra space for holidays and other gatherings.
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Feet to cubic feet calculator
Cubic feet calculator formula The first step in working out the amount of space your shape occupies is to measure its length, width and depth in feet. When you enter these measurements, our cubic feet calculator first figures out the amount of space it occupies using the equation: $$Volume = Length \times Width \times Height$$ Calculate how many cubic feet/yards needed by using the following formulas: Bulk loads of mulch measure 80–120 cubic yards per semi-load, depending on weight and trailer size. In addition to that, it helps build a barrier against weeds that start to grow and potentially destroy or harm your plants. How much is 89 l to cubic foot (UK)? +> with much ♥ by CalculatePlus. Free online Volume conversion. Convert 89 l to cubic foot (UK) (liters to ft3.uk).To calculate the volume of a given item or space in cubic feet, measure the length, width and height in feet and multiply the results together. For example, a storage unit 10 ft long, 6 ft wide and 8 ft high could be described as having a capacity of 480 cubic feet (10 x 6 x 8 = 480). Online visual tool to calculate the cubic feet volume of an object, from length, width and hieght, accept measure unit of inches, feet, yards, mm, cm or meter, easy to understand. How to use this CFT calculator.Flow calculator is a small, compact and useful tool which helps the user make very fast flow related calculations . It is a simple calculation without too many options for users who only want to see what they might expect . Online CFM to FPS unit converter calculator helps to calculate feet per second (FPS) from cubic feet per minute (CFM) area values as input. To calculate cubic feet using this cubic foot calculator, enter length, width, and height in the given input boxes. After entering the measurements of each side, select the unit of the measurement. It offers multiple units to facilitate your calculations.Whether you're using a cubic feet calculator or doing the calculations yourself, you need three pieces of information to find the volume of a cuboid: That shape's length, width and height. There's just one catch: If you want the result to be in cubic feet, every dimension must be measured in feet too.Example: 456 cubic feet equal how many tons (freight)? To solve this, multiply 456 cubic feet with the conversion factor from cubic feet to tons (freight).Capacity: 12.5 cubic feet EZ Dump™ Hopper for Collection System The EZ Dump™ increases productivity by giving the operator the ability to release grass clippings from the riding mower seat with the use of a toggle switch. Weight or cubic feet based? Customers are usually caught in the middle of two moving company salespeople. Cubic footage estimates are perfectly legal but are generally frowned upon because they can't be verified by the DOT. We also see scammy movers often quoting in terms of cubic feet...Do a quick conversion: 1 cubic meters = 35.314666572222 cubic feet using the online calculator for metric conversions. Check the chart for more details. Cubic Yards of Mulch: Number of 2 Cubic Foot BagsLength 48", Width 40", Height 28" (20" plus 8" of pallet) equals 53,760 cubic inches or 31.1 cubic feet. The density equals the weight, 110 lbs (80 lbs for the carton and approximately 30 lbs for the pallet), divided by the cubic dimension, 3.54 lbs per cubic ft Mar 29, 2019 · Convert cubic feet back into square footage by dividing the height from the cubic footage. Since it is such a simple conversion to get to square footage, it is simple to undo. For example, you might have a heater capable of filling a 6800 f t 3 {\displaystyle 6800ft^{3}} , but you want to know how much floor-space that covers in your room. Use the footing calculator to calculate the sides of the porch and the steps; Here is an example: This porch has 9 sq. feet of porch surface, so enter in the slab calculator 4" thickness by 3' width by 3' length. This totals .11 cubic yards. The porch also has 9 linear feet of 6" step. Trinity Trailer Mfg., Inc.® began designing and building self unloading belt trailers in 1975. With agriculture as the primary focus of our business, we excel at designing and manufacturing conveyor trailers.
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Here is a concrete cubic yard calculator for cylinders, sono-tubes, walls, sidewalks and floors. Just enter in your dimensions, make sure you choose feet An estimated cubic yardage will show in the box. Use these calculators for estimating purposes, there's a formula below to calculate concrete for...CBM Calculator allows you to calculate volume in cubic feet. On this Cubic Feet Calculator page you can check cbm calculation for multiple products In this Cubic Feet Calculator we have six unit of measurement for dimension Centimeter, Meter, Millimeter, Feet, Inch and Yard. For weight you can use Kilogram and Pound. 22000 Cubic Meters (m³) = 776,923 Cubic Feet (ft³) 1 m³ = 35.315 ft³. 1 ft³ = 0.028317 m³ Definition of cubic foots of water provided by WikiPedia. The cubic foot is an imperial and US customary (non-metric) unit of volume, used in the United States and the United Kingdom. It is defined as the volume of a cube with sides of one foot (0.3048 m) in length. Its volume is 28.3168 litres or about 1/35 of a cubic metre. Use this calculator to convert from centimeters or meters to feet and inches (feet plus inches). An answer like "5.74 feet" might not mean much to you because you may want to express the decimal part, which is in feet, in inches once its is a smaller unit.cubic yards to cubic feet Conversion Table:: cu yd to cu ft 1.0 = 27 2.0 = 54 3.0 = 81 4.0 = 108 5.0 = 135: cubic yards to cubic feet 6.0 = 162 7.0 = 189 8.0 = 216 | 4,227 | 17,550 | {"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} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.811518 |
bernico.net | 1,511,199,119,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806086.13/warc/CC-MAIN-20171120164823-20171120184823-00135.warc.gz | 31,727,784 | 6,305 | # Data Science Class: Week 1
http://www.dataists.com/2010/09/a-taxonomy-of-data-science/
Statistical Modeling: The Two Cultures
A Discussion on Terminology:
When I was making the videos this week, I quickly found myself using some terminology that may be unfamiliar to you. Here are some of the terms I want you to be familiar with:
Observations/Variables
First, lets talk about observations and variables. Observations are just individual data elements (datum) in data. They could be individual experiments (imagine coin flips), which is how they got their name. They are an instance of something happening. Variables are bits of information we have for each observation. Variables can also be called features.
Variable 1 Variable 2 Observation 1 Observation 2 Observation 3
Continuous/Categorical
Some variables are continuous, and some are categorical. Imagine a variable that contains the result of a coin flip. Lets call that variable flip. Flip could equal 1 for a head, or 0 for a tail. It’s important to note though, that for categorical variables, heads isn’t ‘greater than’ tails, just because 1 > 0. It’s also important to note that the distance between a head and a tail isn’t |1-0| = 1. Categorical variables don’t have any ordinal properties. They’re just labels.
Independent/Dependent variables
When we’re building models, we typically care to try to predict some dependent variable, based on the independent variables in the dataset.
Independence is pretty rigorously defined. Simply though, if A and B are independent then P(A,B) = P(A) * P(B) and P(B|A) = P(B). In real life, it’s really hard to know if A and B are truly independent, and we have to make assumptions about those variables. Careful though, A and B are dependent, and both find our way into our model, that’s called collinearity. If you have too much collinearity in your dataset, you’re going to have a bad time. We will talk about that more later.
Lets put this all together in an example:
obs. num bedrooms basement price house1 3 1 200k house2 2 1 150k house3 3 0 180k
In this dataset, there are two independent variables (bedrooms and basement) and three observations. Basement is categorical. The dependent variable would be price. | 543 | 2,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-47 | longest | en | 0.907107 |
https://www.dataindependent.com/python/python-glossary/exploratory-data-analysis/ | 1,632,203,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00364.warc.gz | 760,376,647 | 55,593 | Exploratory Data Analysis (EDA) is the act of getting intimate with your data.
This means you get a feeling for your data. You don’t simply know it’s characteristics (# rows, columns, distributions, etc.)…you actually feel it.
It may sound a bit corny, but after doing data for long enough, you gain the ability to understand a dataset on an intuition level.
EDA is the process of initial exploration. Imagine you are in a deep dark cave and all you have is a flash light. You illuminate sections of the walls, the ground, and head down passages. EDA is the same process for exploring data.
Whenever we do Exploratory Data Analysis, you can bet we are analyzing:
• # rows, #columns
• Column cardinality (how many unique elements are there in each group?)
• Correlations, which columns relate to each other?
• What are the min/max of each column?
• What do outliers (if any) say about the data?
There isn’t a right answer when doing EDA. The goal is for you to have a launching point that will lead to more analysis. You’ll know when you are done when you are sufficiently inspired to take the next step in your analysis.
Let’s take a look at a python EDA sample
## Python Exploratory Data Analysis¶
Let's get to know our dataset a bit more. We will perform basic analysis that describe our data. Let's start with:
• # Rows, # Columns
• Column cardinality (how many unique elements are there in each group?)
• What are the min/max of each column?
• What do outliers (if any) say about the data?
In [14]:
```import pandas as pd
```
### First, let's look at a few sample rows¶
In [2]:
```df.head()
```
Out[2]:
TreeIDqLegalStatusqSpeciesqAddressSiteOrderqSiteInfoPlantTypeqCaretakerqCareAssistantPlantDate...XCoordYCoordLatitudeLongitudeLocationFire Prevention DistrictsPolice DistrictsSupervisor DistrictsZip CodesNeighborhoods (old)
046534Permitted SiteTree(s) ::73 Summer St7.0Sidewalk: Curb side : CutoutTreePrivateNaN04/01/2002 12:00:00 AM...NaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
1121399DPW MaintainedCorymbia ficifolia :: Red Flowering Gum349X Cargo Way1.0Sidewalk: Curb side : CutoutTreeDPWNaNNaN...NaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
285269Permitted SiteArbutus 'Marina' :: Hybrid Strawberry Tree1000 Edinburgh St3.0Sidewalk: Curb side : CutoutTreePrivateNaN07/24/2007 12:00:00 AM...NaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
3121227DPW MaintainedSequoia sempervirens :: Coast Redwood4299x 17th St3.0Front Yard : YardTreeDPWNaNNaN...NaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
445986Permitted SiteTree(s) ::NaN226.0Sidewalk: Curb side : CutoutTreePrivateNaN12/06/2001 12:00:00 AM...NaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
5 rows × 23 columns
### Then lets get the count of rows and columns¶
In [3]:
```df.info()
```
```<class 'pandas.core.frame.DataFrame'>
RangeIndex: 193940 entries, 0 to 193939
Data columns (total 23 columns):
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 TreeID 193940 non-null int64
1 qLegalStatus 193883 non-null object
2 qSpecies 193940 non-null object
4 SiteOrder 192230 non-null float64
5 qSiteInfo 193940 non-null object
6 PlantType 193940 non-null object
7 qCaretaker 193940 non-null object
8 qCareAssistant 24478 non-null object
9 PlantDate 68911 non-null object
10 DBH 151614 non-null float64
11 PlotSize 143755 non-null object
12 PermitNotes 52455 non-null object
13 XCoord 191066 non-null float64
14 YCoord 191066 non-null float64
15 Latitude 191066 non-null float64
16 Longitude 191066 non-null float64
17 Location 191066 non-null object
18 Fire Prevention Districts 190815 non-null float64
19 Police Districts 190865 non-null float64
20 Supervisor Districts 190929 non-null float64
21 Zip Codes 190923 non-null float64
22 Neighborhoods (old) 190925 non-null float64
dtypes: float64(11), int64(1), object(11)
memory usage: 34.0+ MB
```
### Then let's find out how many values sit within each column¶
In [6]:
```df.apply(lambda x: [x.nunique()])
```
Out[6]:
TreeIDqLegalStatusqSpeciesqAddressSiteOrderqSiteInfoPlantTypeqCaretakerqCareAssistantPlantDate...XCoordYCoordLatitudeLongitudeLocationFire Prevention DistrictsPolice DistrictsSupervisor DistrictsZip CodesNeighborhoods (old)
0193940105718624231131322158945...1610691615101629481628811629591510112941
1 rows × 23 columns
### Then let's look at min and max of a few columns¶
In [11]:
```print ("Min Tree ID: {}".format(df['TreeID'].min()))
print ("Max Tree ID: {}".format(df['TreeID'].max()))
```
```Min Tree ID: 1
Max Tree ID: 262465
```
In [15]:
```print ("Min Tree Date: {}".format(df['PlantDate'].min()))
print ("Max Tree Date: {}".format(df['PlantDate'].max()))
```
```Min Tree Date: 1955-09-19 00:00:00
Max Tree Date: 2020-07-30 00:00:00
```
### Finally, let's get a brief feel on the outliers of location.¶
I like to start this off with a simple box plot. It'll show me the percentiles + outliers.
Without going further, I can already tell I'm going to need to take care of these distracting data points...
In [17]:
```df['Latitude'].plot.box();
```
In [18]:
```df['Longitude'].plot.box();
``` | 1,530 | 5,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.825612 |
https://metric-calculator.com/convert-inches-per-second-to-miles-per-hour.htm | 1,701,470,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00425.warc.gz | 452,431,210 | 7,887 | # Inches Per Second to Miles Per Hour Converter
Select conversion type:
Rounding options:
Convert Miles Per Hour to Inches Per Second (mph to in/s) ▶
## Conversion Table
inches per second to miles per hour in/s mph 10 in/s 0.5682 mph 20 in/s 1.1364 mph 30 in/s 1.7045 mph 40 in/s 2.2727 mph 50 in/s 2.8409 mph 60 in/s 3.4091 mph 70 in/s 3.9773 mph 80 in/s 4.5455 mph 90 in/s 5.1136 mph 100 in/s 5.6818 mph 110 in/s 6.25 mph 120 in/s 6.8182 mph 130 in/s 7.3864 mph 140 in/s 7.9545 mph 150 in/s 8.5227 mph 160 in/s 9.0909 mph 170 in/s 9.6591 mph 180 in/s 10.2273 mph 190 in/s 10.7955 mph 200 in/s 11.3636 mph
## How to convert
1 inch per second (in/s) = 0.056818182 mile per hour (mph). Inch Per Second (in/s) is a unit of Speed used in Standard system. Mile Per Hour (mph) is a unit of Speed used in Standard system.
## Inches per second: A unit of speed
Inches per second ( in/s) is a unit of speed or velocity in the US customary and imperial systems. It measures how fast an object is moving by calculating the distance traveled in inches divided by the time taken in seconds. For example, if a worm travels 2 inches in 1 second, its speed is 2 in/s.
## How to convert inches per second
Inches per second can be converted to other units of speed or velocity by using simple conversion factors. Here are some common units and their conversion factors:
• Meters per second (m/s): To convert from in/s to m/s, multiply by 0.0254. To convert from m/s to in/s, divide by 0.0254. For example, 1 in/s is equal to 0.0254 m/s, and 10 m/s is equal to 393.701 in/s.
• Kilometers per hour (km/h): To convert from in/s to km/h, multiply by 0.09144. To convert from km/h to in/s, divide by 0.09144. For example, 1 in/s is equal to 0.09144 km/h, and 50 km/h is equal to 547.867 in/s.
• Miles per hour (mph): To convert from in/s to mph, multiply by 0.0568182. To convert from mph to in/s, divide by 0.0568182. For example, 1 in/s is equal to 0.0568182 mph, and 30 mph is equal to 528 in/s.
• Knots (kn): To convert from in/s to kn, multiply by 0.0493737. To convert from kn to in/s, divide by 0.0493737. For example, 1 in/s is equal to 0.0493737 kn, and 15 kn is equal to 303.795 in/s.
• Feet per second (ft/s): To convert from in/s to ft/s, divide by 12. To convert from ft/s to in/s, multiply by 12. For example, 1 in/s is equal to 0.0833333 ft/s, and 10 ft/s is equal to 120 in/s .
• Miles per second (mi/s): To convert from in/s to mi/s , multiply by 1.5783x10 -5 . To convert from mi/s to in/s, divide by 1.5783x10 -5 . For example,1 in/s is equal to 1.5783x10 -5 mi/s, and 1 mi/s is equal to 63360 in/s .
## Where inches per second are used
Inches per second are mainly used in engineering and manufacturing to measure the speed or velocity of machines and processes.
For example, the cutting speed of a lathe or a milling machine is often given in inches per minute, which is equivalent to inches per second divided by 60.
The rotational speed of a hard disk drive is often given in revolutions per minute, which can be converted to inches per second by multiplying by the circumference of the disk.
The flow rate of a fluid through a pipe or a nozzle is often given in gallons per minute, which can be converted to inches per second by multiplying by the cross-sectional area of the pipe or the nozzle.
## Definition of inches per second
According to the US customary and imperial systems definition, one inch per second is the speed of a body that covers a distance of one inch in a time of one second.
Mathematically, it can be expressed as:
where v is the speed or velocity in inches per second, s is the distance traveled in inches, and t is the time taken in seconds.
## History of inches per second
The concept of speed or velocity has been studied since ancient times by philosophers and scientists such as Aristotle, Galileo, Newton, etc.
The inch was originally derived from the Roman uncia which was one twelfth of a Roman foot or about 0.97 inches .
The second was originally defined in terms of the Earth’s rotation as one eighty-six thousand four hundredth of a mean solar day.
The combination of these two units resulted in the inch per second as a unit of speed or velocity.
The inch per second was officially adopted as part of the US customary and imperial systems in the late 18th century.
## Example conversions of inches per second to other units
Here are some examples of converting inches per second to other units of speed or velocity:
1 in/s = 0.0254 m/s = 0.09144 km/h = 0.0568182 mph = 0.0493737 kn = 0.0833333 ft/s = 1.5783x10 -5 mi/s
2 in/s = 0.0508 m/s = 0.18288 km/h = 0.113636 mph = 0.0987474 kn = 0.166667 ft/s = 3.1566x10 -5 mi/s
5 in/s = 0.127 m/s = 0.4572 km/h = 0.284091 mph = 0.246869 kn = 0.416667 ft/s = 7.8915x10 -5 mi/s
10 in/s = 0.254 m/s = 0.9144 km/h = 0.568182 mph = 0.493737 kn = 0.833333 ft/s = 1.5783x10 -4 mi/s
20 in/s = 0.508 m/s = 1.8288 km/h = 1.13636 mph = 0.987475 kn = 1.66667 ft/s = 3.1566x10 -4 mi/s
50 in/s = 1.27 m/s = 4.572 km/h = 2.84091 mph = 2.46869 kn = 4.16667 ft/s = 7.8915x10 -4 mi/s
100 in/s = 2.54 m/s = 9.144 km/h = 5.68182 mph = 4.93737 kn = 8.33333 ft/s = 1.5783x10 -3 mi/s
Inch per second also can be marked as ips.
## Definition of Miles Per Hour
Miles per hour (mph, m.p.h., MPH, or mi/h) is a unit of speed that measures how fast something is moving in terms of miles per unit of time. It is commonly used in the United States, the United Kingdom, and some other countries that use the US Standard system of measurement or have close historical ties with them. The speed limit signs on roads and highways are usually expressed in mph.
One mile per hour is equal to 1.609344 kilometers per hour, or 0.44704 meters per second. It can also be converted to other units of speed, such as feet per second, knots, or meters per second, by using appropriate conversion factors.
## How to Convert Miles Per Hour
To convert miles per hour to other units of speed, we need to multiply or divide by the corresponding conversion factors. For example, to convert miles per hour to kilometers per hour, we need to multiply by 1.609344, since there are 1.609344 kilometers in one mile. To convert miles per hour to meters per second, we need to multiply by 0.44704, since there are 0.44704 meters in one mile.
Here are some examples of how to convert miles per hour to other units of length in the US Standard system and the SI system:
• To convert 10 mph to kilometers per hour (km/h), we multiply by 1.609344: 10 mph x 1.609344 = 16.09344 km/h
• To convert 20 mph to feet per second (fps), we multiply by 1.466666667, since there are 1.466666667 feet in one mile: 20 mph x 1.466666667 = 29.33333334 fps
• To convert 30 mph to knots (kn), we multiply by 0.8689762421, since there are 0.8689762421 nautical miles in one mile: 30 mph x 0.8689762421 = 26.069287263 kn
• To convert 40 mph to meters per second (m/s), we multiply by 0.44704: 40 mph x 0.44704 = 17.8816 m/s
• To convert 50 mph to meters per minute (m/min), we multiply by 26.8224, since there are 26.8224 minutes in one hour: 50 mph x 26.8224 = 1341.12 m/min
• To convert 60 mph to centimeters per second (cm/s), we multiply by 100, since there are 100 centimeters in one meter, and multiply by 0.44704: 60 mph x 100 x 0.44704 = 2682 cm/s
## Where Miles Per Hour Are Used
Miles per hour are mainly used in the United States, the United Kingdom, and some other countries that use the US Standard system of measurement or have close historical ties with them. They are often used for car speeds and road signs, as well as for trains, planes, boats, and bicycles.
For example, a car may have a speed of 60 mph, which means that it travels a distance of 60 miles in one hour. A road sign may indicate a speed limit of 55 mph, which means that drivers should not exceed this speed on that road section. A train may have a speed of up to 150 mph, which means that it covers a distance of up to 150 miles in one hour.
Miles per hour are also used in some other fields, such as sports, weather, and astronomy, where the speed or velocity of an object or phenomenon is measured in miles per unit of time.
## History of Miles Per Hour
The origin of miles per hour as a unit of speed can be traced back to the development of the US Standard system and the mile as a unit of length in the late eighteenth century. The mile was defined as one thousand six hundred and nine meters (or eight furlongs) based on the English statute mile.
The term mile first appeared in English in the late fourteenth century, but it was not until later in the early nineteenth century that the use of miles per hour became more common; before that, other units such as feet per second or leagues per hour were preferred for expressing speed.
Miles per hour have been used ever since as a standard unit of speed for the United States, the United Kingdom, and some other countries that use the US Standard system or have close historical ties with them. They have also been adopted by some international organizations and conventions, such as the International Civil Aviation Organization and the International System of Units.
## Example Conversions of Miles Per Hour to Other Units
Here are some examples of how to convert miles per hour to other units of speed, using the conversion factors given above:
• To convert 100 mph to kilometers per hour, we multiply by 1.609344: 100 mph x 1.609344 = 160.9344 km/h
• To convert 200 mph to feet per second, we multiply by 1.466666667: 200 mph x 1.466666667 = 293.3333334 fps
• To convert 300 mph to knots, we multiply by 0.8689762421: 300 mph x 0.8689762421 = 260.69287263 kn
• To convert 400 mph to meters per second, we multiply by 0.44704: 400 mph x 0.44704 = 178.816 m/s
• To convert 500 mph to meters per minute, we multiply by 26.8224: 500 mph x 26.8224 = 13411.2 m/min
• To convert 600 mph to centimeters per second, we multiply by 100 and multiply by 0.44704: 600 mph x 100 x 0.44704 = 268200 cm/s
• To convert 700 mph to millimeters per minute, we multiply by 1000, since there are 1000 millimeters in one meter, and multiply by 26.8224: 700 mph x 1000 x 26.8224 = 18774880 mm/min
Miles per hour also can be marked as mile/hour and mi/h.
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### Eight Hidden Squares
##### Age 7 to 14 Challenge Level:
On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares?
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Troublesome Dice
##### Age 11 to 14 Challenge Level:
When dice land edge-up, we usually roll again. But what if we didn't...?
### Auditorium Steps
##### Age 7 to 14 Challenge Level:
What is the shape of wrapping paper that you would need to completely wrap this model?
### 3D Stacks
##### Age 7 to 14 Challenge Level:
Can you find a way of representing these arrangements of balls?
### Cubist Cuts
##### Age 11 to 14 Challenge Level:
A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer?
### Königsberg
##### Age 11 to 14 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Soma - So Good
##### Age 11 to 14 Challenge Level:
Can you mentally fit the 7 SOMA pieces together to make a cube? Can you do it in more than one way?
### Tic Tac Toe
##### Age 11 to 14 Challenge Level:
In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells?
### Drilling Many Cubes
##### Age 7 to 14 Challenge Level:
A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet.
### Screwed-up
##### Age 11 to 14 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
### Getting an Angle
##### Age 11 to 14 Challenge Level:
How can you make an angle of 60 degrees by folding a sheet of paper twice?
### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### On Time
##### Age 11 to 14 Challenge Level:
On a clock the three hands - the second, minute and hour hands - are on the same axis. How often in a 24 hour day will the second hand be parallel to either of the two other hands?
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Counting Triangles
##### Age 11 to 14 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Dissect
##### Age 11 to 14 Challenge Level:
What is the minimum number of squares a 13 by 13 square can be dissected into?
### Hello Again
##### Age 11 to 14 Challenge Level:
Anne completes a circuit around a circular track in 40 seconds. Brenda runs in the opposite direction and meets Anne every 15 seconds. How long does it take Brenda to run around the track?
### Is There a Theorem?
##### Age 11 to 14 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Cubes Within Cubes
##### Age 7 to 14 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Seven Squares
##### Age 11 to 14 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Khun Phaen Escapes to Freedom
##### Age 11 to 14 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Cogs
##### Age 11 to 14 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Flight of the Flibbins
##### Age 11 to 14 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### How Many Dice?
##### Age 11 to 14 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . . | 2,456 | 10,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-35 | latest | en | 0.907664 |
https://kadbano.se/dzev3b/2a3660-all-real-numbers-are-complex-numbers | 1,618,329,993,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00579.warc.gz | 442,434,877 | 15,510 | Let M_m,n (R) be the set of all mxn matrices over R. We denote by M_m,n (R) by M_n (R). All the points in the plane are called complex numbers, because they are more complicated -- they have both a real part and an imaginary part. Complex Numbers are considered to be an extension of the real number system. True or False: All real numbers are complex numbers. can be used in place of a to indicate multiplication): Imagine that you have a group of x bananas and a group of y bananas; it doesn't matter how you put them together, you will always end up with the same total number of bananas, which is either x + y or y + x. r+i0.... are all complex numbers. real, imaginary, imaginary unit. For example, let's say that I had the number. We can understand this property by again looking at groups of bananas. In situations where one is dealing only with real numbers, as in everyday life, there is of course no need to insist on each real number to be put in the form a+bi, eg. For the second equality, we can also write it as follows: Thus, this example illustrates the use of associativity. Complex numbers are an important part of algebra, and they do have relevance to such things as solutions to polynomial equations. The real function acts on Z element-wise. The "a" is said to be the real part of the complex number and b the imaginary part. Follow answered 34 mins ago. A real number is any number which can be represented by a point on the number line. 5+ 9ὶ: Complex Number. They are widely used in electronics and also in telecommunications. For example, the rational numbers and integers are all in the real numbers. This is because they have the ability to represent electric current and different electromagnetic waves. Find the real part of each element in vector Z. Complex numbers are points in the plane endowed with additional structure. It just so happens that many complex numbers have 0 as their imaginary part. We can write any real number in this form simply by taking b to equal 0. But there is … (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0i, which is a complex representation.) Note that complex numbers consist of both real numbers ($$a+0i$$, such as 3) and non-real numbers ($$a+bi,\,\,\,b\ne 0$$, such as $$3+i$$); thus, all real numbers are also complex. Therefore a complex number contains two 'parts': one that is real What if I combined imaginary and real numbers? The system of complex numbers consists of all numbers of the form a + bi where a and b are real numbers. The number 0 is both real and imaginary. Associativity states that the order in which three numbers are added or the order in which they are multiplied does not affect the result. The set of real numbers is a proper subset of the set of complex numbers. I have not thought about that, I think you right. COMPOSITE NUMBERS Multiplying complex numbers is much like multiplying binomials. Ask specific questions about the challenge or the steps in somebody's explanation. Now that you know a bit more about the real numbers and some of its subsets, we can move on to a discussion of some of the properties of real numbers (and operations on real numbers). To me, all real numbers $$r$$ are complex numbers of the form $$r + 0i$$. To avoid such e-mails from students, it is a good idea to define what you want to mean by a complex number under the details and assumption section. So you can do something like that. Real Part of Complex Number. Every real number is a complex number, but not every complex number is a real number. Examples include 4 + 6i, 2 + (-5)i, (often written as 2 - 5i), 3.2 + 0i, and 0 + 2i. They are made up of all of the rational and irrational numbers put together. The complex numbers consist of all numbers of the form + where a and b are real numbers. I'm wondering about the extent to which I would expand this list, and if I would need to add a line stating. Children first learn the "counting" numbers: 1, 2, 3, etc. False. In the complex number 5+2i, the number 5 is called the _____ part, the number 2 is called the _____ part and the number i is called the _____. In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also. However, you can use imaginary numbers. The Real Number Line. $$i^{2}=-1$$ or $$i=\sqrt{−1}$$. A rational number is a number that can be equivalently expressed as a fraction , where a and b are both integers and b does not equal 0. have no real part) and so is referred to as the imaginary axis.-4 -2 2 4-3-2-1 1 2 3 +2i 2−3i −3+i An Argand diagram 4 standard form A complex number is in standard form when written as $$a+bi$$, where $$a, b$$ are real numbers. For example, the set of all numbers $x$ satisfying $0 \leq x \leq 1$ is an interval that contains 0 and 1, as well as all the numbers between them. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. The set of real numbers is divided into two fundamentally different types of numbers: rational numbers and irrational numbers. If we consider real numbers x, y, and z, then. A set of complex numbers is a set of all ordered pairs of real numbers, ie. This gives the idea ‘Complex’ stands out and holds a huge set of numbers than ‘Real’. explain the steps and thinking strategies that you used to obtain the solution. If we add to this set the number 0, we get the whole numbers. The Real Numbers had no name before Imaginary Numbers were thought of. The real number rrr is also a complex number of the form r+0i r + 0i r+0i. The complex number $a+bi$ can be identified with the point $(a,b)$. The set of real numbers is a proper subset of the set of complex numbers. But I think there are Brilliant users (including myself) who would be happy to help and contribute. The symbol is often used for the set of complex numbers. We can write any real number in this form simply by taking b to equal 0. Commutativity states that the order of two numbers being multiplied or added does not affect the result. I'll add a comment. The real numbers are complex numbers with an imaginary part of zero. Explanations are more than just a solution — they should For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. Recall that operations in parentheses are performed before those that are outside parentheses. Imaginary numbers: Numbers that equal the product of a real number and the square root of −1. We consider the set R 2 = {(x, y): x, y R}, i.e., the set of ordered pairs of real numbers. In fact, all real numbers and all imaginary numbers are complex. They have been designed in order to solve the problems, that cannot be solved using real numbers. Is 1 a rational number?". For example, 2 + 3i is a complex number. An imaginary number is the “$$i$$” part of a real number, and exists when we have to take the square root of a negative number. doesn't help anyone. For early access to new videos and other perks: https://www.patreon.com/welchlabsWant to learn more or teach this series? I have a suggestion for that. If I also always have to add lines like. I can't speak for other countries or school systems but we are taught that all real numbers are complex numbers. basically the combination of a real number and an imaginary number 7 years, 6 months ago. Every real number is a complex number, but not every complex number is a real number. There are rational and irrational numbers, positive and negative numbers, integers, natural numbers and real or imaginary numbers. imaginary unit The imaginary unit $$i$$ is the number whose square is $$–1$$. real numbers, and so is termed the real axis, and the y-axis contains all those complex numbers which are purely imaginary (i.e. Therefore, the combination of both the real number and imaginary number is a complex number.. How about writing a mathematics definition list for Brilliant? (Note that there is no real number whose square is 1.) The points on the horizontal axis are (by contrast) called real numbers. Complex Number can be considered as the super-set of all the other different types of number. No BUT --- ALL REAL numbers ARE COMPLEX numbers. For example:(3 + 2i) + (4 - 4i)(3 + 4) = 7(2i - 4i) = -2iThe result is 7-2i.For multiplication, you employ the FOIL method for polynomial multiplication: multiply the First, multiply the Outer, multiply the Inner, multiply the Last, and then add. In addition to the integers, the set of real numbers also includes fractional (or decimal) numbers. Indeed. Comments Learn what complex numbers are, and about their real and imaginary parts. Mathematicians also play with some special numbers that aren't Real Numbers. The numbers 3.5, 0.003, 2/3, π, and are all real numbers. In the special case that b = 0 you get pure real numbers which are a subset of complex numbers. The set of integers is often referred to using the symbol . Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. You can add them, subtract them, multiply them, and divide them (except division by 0 is not defined), and the result is another complex number. should further the discussion of math and science. Many of the real-world applications involve very advanced mathematics, but without complex numbers the computations would be nearly impossible. Even in this discussion I've had to skip all the math that explains why the complex numbers to the quadratic equation We can write this symbolically below, where x and y are two real numbers (note that a . All real numbers can be written as complex numbers by setting b = 0. I've been receiving several emails in which students seem to think that complex numbers expressively exclude the real numbers, instead of including them. This might mean I'd have to use "strictly positive numbers", which would begin to get cumbersome. Complex numbers are numbers in the form a+bia+bia+bi where a,b∈Ra,b\in \mathbb{R}a,b∈R. Let's say, for instance, that we have 3 groups of 6 bananas and 3 groups of 5 bananas. By … By now you should be relatively familiar with the set of real numbers denoted $\mathbb{R}$ which includes numbers such as $2$, $-4$, $\displaystyle{\frac{6}{13}}$, $\pi$, $\sqrt{3}$, …. Why not take an. Example: 1. The reverse is true however - The set of real numbers is contained in the set of complex numbers. The number i is imaginary, so it doesn't belong to the real numbers. So the imaginaries are a subset of complex numbers. 2. I agree with you Mursalin, a list of mathematics definitions and assumptions will be very apreciated on Brilliant, mainly by begginers at Math at olympic level. marcelo marcelo. For example, etc. Solution: If a number can be written as where a and b are integers, then that number is rational (i.e., it is in the set ). (A small aside: The textbook defines a complex number to be imaginary if its imaginary part is non-zero. Whenever we get a problem about three digit numbers, we always get the example that 012012012 is not a three digit number. I also get questions like "Is 0 an integer? Irrational numbers: Real numbers that are not rational. are all complex numbers. If we combine these groups one for one (one group of 6 with one group of 5), we end up with 3 groups of 11 bananas. The system of complex numbers consists of all numbers of the … Multiplying a Complex Number by a Real Number. This number line is illustrated below with the number 4.5 marked with a closed dot as an example. Real and Imaginary parts of Complex Number. Note the following: Thus, each of these numbers is rational. As you know, all complex numbers can be written in the form a + bi where a and b are real numbers. The last example is justified by the property of inverses. I've never heard about people considering 000 a positive number but not a strictly positive number, but on the Dutch IMO 2013 paper (problem 6) they say "[…], and let NNN be the number of ordered pairs (x,y)(x,y)(x,y) of (strictly) positive integers such that […]". Although when taken completely out of context they may seem to be less than useful, it does turn out that you will use them regularly, even if you don't explicitly acknowledge this in each case. A “real interval” is a set of real numbers such that any number that lies between two numbers in the set is also included in the set. The real numbers include the rational numbers, which are those which can be expressed as the ratio of two integers, and the irrational numbers… They got called "Real" because they were not Imaginary. I know you are busy. Distributivity is another property of real numbers that, in this case, relates to combination of multiplication and addition. It's like saying that screwdrivers are a subset of toolboxes. The set of real numbers is composed entirely of rational and irrational numbers. So, for example, These are formally called natural numbers, and the set of natural numbers is often denoted by the symbol . complex number system The complex number system is made up of both the real numbers and the imaginary numbers. Cite. The reverse is true however - The set of real numbers is contained in the set of complex numbers. True or False: The conjugate of 2+5i is -2-5i. I think yes....as a real no. Complex numbers are ordered pairs therefore real numbers cannot be a subset of complex numbers. The identity property simply states that the addition of any number x with 0 is simply x, and the multiplication of any number x with 1 is likewise x. We will now introduce the set of complex numbers. However, it has recently come to my attention, that the Belgians consider 0 a positive number, but not a strictly positive number. We distribute the real number just as we would with a binomial. These properties, by themselves, may seem a bit esoteric. 0 is a rational number. The complex numbers include all real numbers and all real numbers multiplied by the imaginary number i=sqrt(-1) and all the sums of these. But then again, some people like to keep number systems separate to make things clearer (especially for younger students, where the concept of a complex number is rather counterintuitive), so those school systems may do this. Complex numbers, such as 2+3i, have the form z = x + iy, where x and y are real numbers. R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. There are also more complicated number systems than the real numbers, such as the complex numbers. A useful identity satisfied by complex numbers is r2 +s2 = (r +is)(r −is). An irrational number, on the other hand, is a non-repeating decimal with no termination. © Copyright 1999-2021 Universal Class™ All rights reserved. Remember: variables are simply unknown values, so they act in the same manner as numbers when you add, subtract, multiply, divide, and so on. 2. The word 'strictly' is not mentioned on the English paper. The set of all the complex numbers are generally represented by ‘C’. Another property, which is similar to commutativity, is associativity. Some simpler number systems are inside the real numbers. This particularity allows complex numbers to be used in different fields of mathematics, engineering and mathematical physics. Thus, a complex number is defined as an ordered pair of real numbers and written as where and . A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i = −1. Real and Imaginary parts of Complex Number. Real-life quantities which, though they're described by real numbers, are nevertheless best understood through the mathematics of complex numbers. Rational numbers thus include the integers as well as finite decimals and repeating decimals (such as 0.126126126.). All rational numbers are real, but the converse is not true. The real part is a, and b is called the imaginary part. Are there any countries / school systems in which the term "complex numbers" refer to numbers of the form a+bia+bia+bi where aaa and bbb are real numbers and b≠0b \neq 0 b=0? The major difference is that we work with the real and imaginary parts separately. One property is that multiplication and addition of real numbers is commutative. A complex number is any number that includes i. We denote R and C the field of real numbers and the field of complex numbers respectively. If your students keep misunderstanding this concept, you can create a kind of nomenclature for complex numbers of the form a + bi ; where b is different from zero. There is disagreement about whether 0 is considered natural. What if I had numbers that were essentially sums or differences of real or imaginary numbers? Real numbers are incapable of encompassing all the roots of the set of negative numbers, a characteristic that can be performed by complex numbers. A point is chosen on the line to be the "origin". The numbers we deal with in the real world (ignoring any units that go along with them, such as dollars, inches, degrees, etc.) 0 is an integer. Complex numbers introduction. Imaginary numbers have the form bi and can also be written as complex numbers by setting a = 0. Complex numbers are formed by the addition of a real number and an imaginary number, the general form of which is a + bi where i = = the imaginary number and a and b are real numbers. The Set of Complex Numbers. Yes, all real numbers are also complex numbers. Real numbers are simply the combination of rational and irrational numbers, in the number system. they are of a different nature. Likewise, imaginary numbers are a subset of the complex numbers. But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. Share. 7: Real Number, … Complex Number can be considered as the super-set of all the other different types of number. They are not called "Real" because they show the value of something real. Every real number is a complex number. Complex numbers are an important part of algebra, and they do have relevance to such things as solutions to polynomial equations. Hint: If the field of complex numbers were isomorphic to the field of real numbers, there would be no reason to define the notion of complex numbers when we already have the real numbers. Note that a, b, c, and d are assumed to be real. Open Live Script. The construction of the system of complex numbers begins by appending to the system of real numbers a number which we call i with the property that i2 = 1. In addition to positive numbers, there are also negative numbers: if we include the negative values of each whole number in the set, we get the so-called integers. Complex numbers are ubiquitous in modern science, yet it took mathematicians a long time to accept their existence. For that reason, I (almost entirely) avoid the phrase "natural numbers" and use the term "positive numbers" instead. 1 is a rational number. I read that both real and imaginary numbers are complex numbers so I … Main Article: Complex Plane Complex numbers are often represented on the complex plane, sometimes known as the Argand plane or Argand diagram.In the complex plane, there are a real axis and a perpendicular, imaginary axis.The complex number a + b i a+bi a + b i is graphed on this plane just as the ordered pair (a, b) (a,b) (a, b) would be graphed on the Cartesian coordinate plane. Calvin Lin So, too, is $3+4i\sqrt{3}$. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part.For example, $5+2i$ is a complex number. I can't speak for other countries or school systems but we are taught that all real numbers are complex numbers. Show transcribed image text. Classifying complex numbers. Let’s begin by multiplying a complex number by a real number. The first part is a real number, and the second part is an imaginary number. Although some of the properties are obvious, they are nonetheless helpful in justifying the various steps required to solve problems or to prove theorems. Note that Belgians living in the northern part of Belgium speak Dutch. Can be written as There isn't a standardized set of terms which mathematicians around the world uses. Eventually all the ‘Real Numbers’ can be derived from ‘Complex Numbers’ by having ‘Imaginary Numbers’ Null. If $b^{2}-4ac<0$, then the number underneath the radical will be a negative value. A complex number is made up using two numbers combined together. At the same time, the imaginary numbers are the un-real numbers, which cannot be expressed in the number line and is commonly used to represent a complex number. As a brief aside, let's define the imaginary number (so called because there is no equivalent "real number") using the letter i; we can then create a new set of numbers called the complex numbers. The set of real numbers is often referred to using the symbol . Complex numbers are the numbers which are expressed in the form of a+ib where ‘i’ is an imaginary number called iota and has the value of (√-1).For example, 2+3i is a complex number, where 2 is a real number and 3i is an imaginary number. Real Numbers. related to those challenges. Let's look at some of the subsets of the real numbers, starting with the most basic. Note by Often, it is heavily influenced by historical / cultural developments. 1. One can represent complex numbers as an ordered pair of real numbers (a,b), so that real numbers are complex numbers whose second members b are zero. For example, etc. And real numbers are numbers where the imaginary part, b=0b=0b=0. Understanding Real and Complex Numbers in Algebra, Interested in learning more? You can still include the definitions for the less known terms under the details section. Thus, 3i, 2 + 5.4i, and –πi are all complex numbers. The last two properties that we will discuss are identity and inverse. x is called the real part and y is called the imaginary part. This leads to a method of expressing the ratio of two complex numbers in the form x+iy, where x and y are real complex numbers. Similarly, if you have a rectangle with length x and width y, it doesn't matter if you multiply x by y or y by x; the area of the rectangle is always the same, as shown below. Email. Sign up, Existing user? o Learn what is the set of real numbers, o Recognize some of the main subsets of the real numbers, o Know the properties of real numbers and why they are applicable. Improve this answer. Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. I've always been taught that the complex numbers include the reals as well. True. A complex number can be written in the form a + bi where a and b are real numbers (including 0) and i is an imaginary number. numbers that can written in the form a+bi, where a and b are real numbers and i=square root of -1 is the imaginary unit the real number a is called the real part of the complex number Practice Problem: Identify the property of real numbers that justifies each equality: a + i = i + a; ; 5r + 3s - (5r + 3s) = 0. A) I understand that complex numbers come in the form z= a+ib where a and b are real numbers. It can be difficult to keep them all straight. Intro to complex numbers. Complex numbers must be treated in many ways like binomials; below are the rules for basic math (addition and multiplication) using complex numbers. Intro to complex numbers. For example, the rational numbers and integers are all in the real numbers. A real number is any number that can be placed on a number line that extends to infinity in both the positive and negative directions. in our school we used to define a complex number sa the superset of real no.s .. that is R is a subset of C. Use the emojis to react to an explanation, whether you're congratulating a job well done. Real numbers include a range of apparently different numbers: for example, numbers that have no decimals, numbers with a finite number of decimal places, and numbers with an infinite number of decimal places. The most important imaginary number is called {\displaystyle i}, defined as a number that will be -1 when squared ("squared" means "multiplied by itself"): Both numbers are complex. of complex numbers is performed just as for real numbers, replacing i2 by −1, whenever it occurs. The problem is that most people are looking for examples of the first kind, which are fairly rare, whereas examples of the second kind occur all the time. Because i is not a real number, complex numbers cannot generally be placed on the real line (except when b is equal to zero). Some simpler number systems are inside the real numbers. Open Live Script. Expert Answer . Google Classroom Facebook Twitter. Then you can write something like this under the details and assumptions section: "If you have any problem with a mathematical term, click here (a link to the definition list).". Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. That is an interesting fact. A complex number is any number that includes i. In a complex number when the real part is zero or when , then the number is said to be purely imaginary. While this looks good as a start, it might lead to a lot of extraneous definitions of basic terms. Real does not mean they are in the real world . Applying Algebra to Statistics and Probability, Algebra Terminology: Operations, Variables, Functions, and Graphs, Understanding Particle Movement and Behavior, Deductive Reasoning and Measurements in Geometry, How to Use Inverse Trigonometric Functions to Solve Problems, How to Add, Subtract, Multiply, and Divide Positive and Negative Numbers, How to Calculate the Chi-Square Statistic for a Cross Tabulation, Geometry 101 Beginner to Intermediate Level, Math All-In-One (Arithmetic, Algebra, and Geometry Review), Physics 101 Beginner to Intermediate Concepts. Image Text from this question are Brilliant users ( including myself ) would. By historical / cultural developments ensure you get pure real numbers have not thought about that, think! Have 3 groups of bananas, by themselves, may seem a esoteric! ( including myself ) who would be happy to help and contribute ask specific questions about the extent to i... 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all real numbers are complex numbers 2021 | 7,793 | 35,266 | {"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": 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} | 4.0625 | 4 | CC-MAIN-2021-17 | latest | en | 0.953166 |
http://statwonk.com/chances-supreme-court-turnover.html | 1,548,072,025,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583792338.50/warc/CC-MAIN-20190121111139-20190121133139-00066.warc.gz | 215,581,135 | 6,289 | Caveat emptor: I make a lot of assumptions to answer this question (during an afternoon recovering from a cold and hanging out with a toddler). Any analysis can be improved. If you like this one, copy the code and share improvements freely!
supreme_court_wiki_page <- read_html(
"https://en.wikipedia.org/wiki/List_of_United_States_Supreme_Court_Justices_by_time_in_office")
First I’ll download the data and prepare if for analysis. I’d specifically like to answer the question: “what are the chances the Supreme Court sees turnover in the remaining portion of Trump’s first term?” … if the future plays out similar to the past. Note, for example, that this analysis currently assumes life expectencies are all the same among justices and other unreasonable things.
supreme_court_wiki_page %>%
html_nodes("#justices") %>%
html_table() %>%
{ .[[1]] } %>%
tibble::as_tibble() %>%
select(rank = Rank, justice = Justice, start_on = 4, end_on = 5) %>%
mutate_at(vars(start_on, end_on),
funs(as.POSIXct(., origin = "1970-01-01", format = "%B %d, %Y"))) %>%
mutate(censor = ifelse(is.na(end_on), 0, 1),
time_to = ifelse(censor == 1,
difftime(end_on, start_on, units = "days"),
difftime(Sys.time(), start_on, units = "days")),
time_to = time_to/365.25) -> d
d %>%
survfit(Surv(time_to, censor) ~ 1, data = .) %>%
plot(main = "Supreme Court tenure durations",
xlab = "Years on Court",
ylab = "Chance still on Court")
Let’s assume an independent weakest-link process using the Weibull distribution. The idea is that when each justice joins the Court, they start a set of timers. The timers have imaginary labels like, “for family,” “for a medical condition,” “retirement”, etc. If for each justice we observe the first of these timers ringing, then we have a Weibull process.
A visual inspection of the fit to data suggests this is a reasonable story.
d %>%
select(time_to, censor) %>%
mutate(left = time_to,
right = ifelse(censor == 0, NA, left)) %>%
as.data.frame() %>%
fitdistrplus::fitdistcens(distr = "weibull") %>%
plot()
d %>%
select(time_to, censor) %>%
mutate(left = time_to,
right = ifelse(censor == 0, NA, left)) %>%
as.data.frame() %>%
fitdistrplus::fitdistcens(distr = "weibull") -> wfit
wfit
## Fitting of the distribution ' weibull ' on censored data by maximum likelihood
## Parameters:
## estimate
## shape 1.680964
## scale 19.203452
## Fixed parameters:
## data frame with 0 columns and 0 rows
d %>% filter(censor == 0) %>% rename(years = time_to)
Time until next presidential election,
difftime(as.POSIXct("2020-11-03", origin = "1970-01-01"),
Sys.time(),
units = "days") %>%
as.numeric() %>%
{ . / 365.25 }
## [1] 1.837955
So we’re about 1.83 years until Trump’s ability to nominate a new justice closes without a new election (barring impeachment and removal from office!).
Let’s check the longest-serving member, Clarence Thomas’s chances of remaining through November 3rd, 2020:
chance_of_tenure_longer_than_x <- function(x) {
pweibull(x, shape = 1.68, scale = 19.2, lower.tail = FALSE)
}
# This reads: P( chance Thomas tenure goes beyond next election )
(p <- chance_of_tenure_longer_than_x(27.2 + 1.83) / chance_of_tenure_longer_than_x(27.2))
## [1] 0.8125909
(1/(1 - p))
## [1] 5.335922
Yikes, a one in five chance that Clarance Thomas’ tenure ends during Trump’s current term.
Okay, but there’s more than Thomas at-risk,
d %>%
filter(censor == 0) %>%
rename(Years = time_to) %>%
mutate(Chance tenure ends before election = purrr::map_dbl(Years, function(x) {
1 - (chance_of_tenure_longer_than_x(x + 1.83) / chance_of_tenure_longer_than_x(x))
})) %>%
select(-start_on, -end_on, -censor)
How about the chance that one of the justice’s tenures ends before the next election? To answer this we need to combine the chances of each individual justice. I’ll also assume the tenures are indepenedent (they’re not! This assumption isn’t very accurate).
d %>%
filter(censor == 0) %>%
mutate(chance_tenure_ends_before_election = purrr::map_dbl(time_to, function(x) {
1 - (chance_of_tenure_longer_than_x(x + 1.83) / chance_of_tenure_longer_than_x(x))
})) %>%
mutate(tenure_ends_before_election = purrr::map(
chance_tenure_ends_before_election, ~ rbinom(1e5, 1, .x))) %>%
select(justice, tenure_ends_before_election) %>%
tidyr::unnest() %>%
group_by(justice) %>%
mutate(sim = 1:n()) %>% ungroup() %>%
group_by(sim) %>%
summarise(justices_w_tenure_ending = sum(tenure_ends_before_election)) %>%
count(chance_one_or_more_tenure_ends_before_election = justices_w_tenure_ending > 0) %>%
mutate(p = n/sum(n)) %>%
filter(chance_one_or_more_tenure_ends_before_election) %>%
pull(p) %>%
scales::percent()
## [1] "67.6%"
Double yikes, there’s a better chance than not that another Supreme Court justice’s tenure ends before next election!
@statwonk | 1,376 | 4,772 | {"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} | 2.9375 | 3 | CC-MAIN-2019-04 | latest | en | 0.782009 |
http://mathoverflow.net/questions/48248/what-do-named-tricks-share/48250 | 1,469,501,324,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824570.25/warc/CC-MAIN-20160723071024-00016-ip-10-185-27-174.ec2.internal.warc.gz | 164,461,451 | 26,637 | What do named “tricks” share?
There are a number of theorems or lemmas or mathematical ideas that come to be known as eponymous tricks, a term which in this context is in no sense derogatory. Here is a list of 10 such tricks (the last of which I learned at MO):
Further Edit. And although my original interest was in eponymous (=named-after-someone) tricks, several non-eponymous tricks have been mentioned, so I'll gather those here as well:
Some of those listed above do not yet have Wikipedia pages (hint, hint—Thierry).
I (JOR) am not seeking to extend this list (although I would be incidentally interested to learn of prominent omissions), but rather I am wondering:
Is there some aspect or trait shared by the mathematical ideas or techniques that, over time, come to be named "tricks"?
I am aware this is a borderline question; feel free to close if it unduly distracts.
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"An idea which can be used only once is a trick. If one can use it more than once it becomes a method." Quoted from books.google.co.uk/… – Andrey Rekalo Dec 4 '10 at 4:45
That construction of Whitney ought to have a more dignified name. – Tom Goodwillie Dec 4 '10 at 4:58
The Eilenberg Swindle is another good one. – Sean Tilson Dec 4 '10 at 5:36
You could add the Eilenberg Swindle to your list. (A Swindle sounds even more disreputable than a Trick.) – Charles Rezk Dec 4 '10 at 5:36
Why is the page linking to the Whitney trick linking to a "Global Oneness" site? <br> <br> Why do they even have a page on Whitney embedding on a site about spirituality? I'm very confused. – Simon Rose Dec 4 '10 at 5:42
How about the following (which I think applies to some of these tricks but not others): a trick is something whose usefulness is not fully captured by any particular set of hypotheses, so it would limit its usefulness to write it down as a lemma.
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This is my favorite answer. It captures the notion that a trick is one of the essentially human contributions to math, a creative touch that isn't part of some axiomatic or algorithmic approach that something more mechanical might invent. It's generalizable but not categorizable. – Ryan Reich Dec 4 '10 at 22:04
I like this answer, but I'm not sure it really distinguishes between tricks and methods. For instance, I don't know of a lemma one could write down that would do justice to the probabilistic method in combinatorics. But actually I'm tempted to say that "Take a random X" feels like a trick that can be applied over and over again, perhaps for this very reason. – gowers Dec 5 '10 at 22:24
@Ryan: I must also leap to the defence of the poor old computer. I don't see any reason in principle that a clever program could not invent mathematical tricks. (I think a proof that a program couldn't invent tricks would prove that humans couldn't either.) – gowers Dec 5 '10 at 22:26
I'll take a stab at this.
I think that the term "trick" is used to connote a technique that achieves something as if by magic. If I make a cake by combining flour, sugar, and eggs and baking, that is simply a standard technique, but if I make the cake by putting the ingredients into a top hat and waving a wand over it, that is a magic trick. The way that the Weyl unitary trick makes complex groups behave like compact ones seems like a magic trick. (For those of you trying to follow this half baked analogy, the cake is complete reducibility of representations, the oven is integration, and the hat is ... uhhh.... )
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Ha ha, half baked :) – Zev Chonoles Dec 4 '10 at 7:56
I think this is the answer. – timur Dec 4 '10 at 15:29
The hat is orthogonality? – Qiaochu Yuan Dec 4 '10 at 16:12
The hat is Fourier transform, of course! – Paul Siegel Dec 6 '10 at 12:20
One well-known trick is a way to evaluate the Gaussian integral $G = \int_\mathbb{R} e^{-x^2}dx = \sqrt{\pi}$ by writing $$G^2 = \left(\int_\mathbb{R} e^{-x^2}dx\right)\left(\int_\mathbb{R} e^{-y^2}dy\right) = \int_{\mathbb{R}^2} e^{-(x^2+y^2)}dxdy$$ which when transformed to polar coordinates becomes $$G^2 = 2\pi \int_0^\infty e^{-r^2} r dr = \pi \int_0^\infty e^{-u} du = \pi$$ via the substitution $u=r^2$. It appears this idea is due to Poisson.
In a 2005 note in the American Mathematical MONTHLY, R. Dawson has observed that this is a trick that only works once; there are no other integrals that can be evaluated by this method. Specifically:
Theorem. Any Riemann-integrable function $f$ on $\mathbb{R}$, such that $f(x)f(y) = g(\sqrt{x^2+y^2})$ for some $g$, is of the form $f(x)=ke^{ax^2}$.
See: Dawson, Robert J. MacG. On a "singular" integration technique of Poisson. American Mathematical Monthly 112 (2005), 270-272.
So if a technique is a trick that works twice, this one is definitely still a trick.
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Note that this trick has something in common with the Rabinowitsch, Cayley and Eilenberg tricks and probably some others on the list: in order to solve a $k$-dimensional problem, you go into more than $k$ dimensions. This also seems to be a distinguishing feature of things called tricks. – darij grinberg Dec 5 '10 at 22:04
I think Dawson was anticipated by James Clerk Maxwell. A result called Maxwell's theorem says that if $X_1, \dots, X_n$ are independent real-valued random variables and their joint density is spherically symmetric, then all of them are normally distributed, i.e. the probability density of each of them is a Gaussian function. – Michael Hardy Dec 5 '10 at 22:17
Constantine Georgakis, "A Note on the Gaussian Integral", Mathematics Magazine, February, 1994, page 47 This paper gives what its author considers "a better alternative to the usual method of reduction to polar coordinates" for evaluating this integral. See en.wikipedia.org/wiki/Gaussian_integral . – Michael Hardy Dec 5 '10 at 22:19
While the specific form $f(x) f(y) = g(\sqrt{x^2+y^2})$ applies only to Gaussians, there are further uses of this kind of transformation: in one direction, to the volumes of Euclidean spheres in higher dimension (imagine you already know $\Gamma(1/2)$ but not the area of a circle); in another direction, to the usual proof of $B(x,y) = \Gamma(x) \Gamma(y) / \Gamma(x+y)$; combining these, to Dirichlet integrals; and by analogue, to the relation between Gauss sums and Jacobi sums — and probably others that don't come to mind right now. – Noam D. Elkies Jul 9 '11 at 6:11
@Michael Hardy: that "better alternative" in the paper by Georgakis in fact is due to Laplace. See york.ac.uk/depts/maths/histstat/normal_history.pdf. – KConrad Mar 26 '12 at 3:15
http://en.wikipedia.org/wiki/Rosser%27s_trick
"A technique is a trick that works twice"
Note that Grothendieck never published his proof of the Grothendieck-Riemann-Roch theorem because he felt that the proof depended on an "astuce" (trick) rather than flowing naturally.
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It is interesting that the accepted English translation of "astuce" is "trick", whereas that of the adjectival form "astucieux/euse" is "clever" or "astute". This seems to give the concept of a trick a better connotation in French than in English. (I am tempted to load up Lord of the Rings with the French subtitles on to see whether Gollum accuses Frodo and Sam of being "astucieux".) – Pete L. Clark Dec 4 '10 at 16:32
@Pete: you're absolutely right. For instance, to translate the expression "dirty trick" into French, you would not use the word "astuce" because "astuce" has a uniformly positive connotation of praise (yes, even in that Gollum context). – Thierry Zell Dec 4 '10 at 16:53
In French, one says "un artifice de calcul", "l'artifice de Legendre". By the way, there is also the van der Waerden trick (associativity of the composition of reduced words). – Panurge Feb 11 at 14:26
To my way of thinking, the other answers are missing an important element, a necessary feature for a mathematical tool or method to be called "trick."
Namely, in order to be called a "trick," a method or technique must involve artifice or misdirection of some kind. When we treat a mathematical object as something that it isn't really or when we pretend that something is other than it is in order to advance an argument (which is not to suggest that the mathematics is not correct), then we are using trickery. When we solve a problem by placing our focus on something else, in which we aren't actually interested as such and which may even be silly in some way—a kind of misdirection—but by doing so we become successful in the original problem, then we are using trickery. When we replace a robust concept, in which we are really interested, with a modified version of it, perhaps even an absurd version of it, but which makes the argument work, then we are using trickery.
For example, with Craig's trick, we replace a formula $\varphi$ with the conjunction with itself $\varphi\wedge\varphi\wedge\dots\wedge\varphi$ repeated many times over. The new assertion is just silly and we don't actually care about it as such, although of course it is logically equivalent to $\varphi$. How could it possibly help? The point is that we can use the new assertion to code some extra information into an axiomatization or presentation: the number of times it was repeated. By this artifice, we can deduce that every computably enumerable theory has a computable set of axioms. The same idea works in many other contexts. For example, every c.e. presentable group has a computable presentation, by sufficiently repeating relations suitably in the presentation.
With Scott's trick, the issue to be solved is that the equivalence class of an object forms a proper class, which can cause certain problems, and so we replace that equivalence class with the set of rank-minimal members of the class. If we think of this fake equivalence class as the real thing, then everything works great! This trick is surprisingly robust, and can be used to find small canonical sets of representing structures in almost any situation. For example, in ZFC there is a definable manner of choosing a set of groups from each group isomorphism class: the rank-minimal groups from that class. This is a trick, because we don't really care much about that particular collection as such.
With Rosser's trick, we replace the concept of a theory $T$ proving a sentence $\sigma$, with: $T$ proves $\sigma$ by a proof for which there is no shorter proof of $\neg\sigma$. When you think of "proof" using this concept, then Gödel's incompleteness theorem is improved to the Gödel-Rosser theorem, where one can drop Gödel's extra hypotheses about $\omega$-consistency. This is a trick, because we don't actually want to think about "proof" using Rosser's concept, except that it makes the argument work.
In many of the other tricks, we do something that seems a little absurd at first, misdirecting our attention from the original problem to this other thing, which may seem irrelevant at first, but when we follow it more fully it provides the answer we seek.
In each case, we replace the concepts or objects in which we are truly interested by concepts or objects that we don't actually care about as such and which in several cases are comical versions of the original, except that they make the argument work.
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In particular, I don't agree with the "used only once" joke idea for distinguishing tricks and methods. We have already many counterexamples to that view mentioned on this thread of posts. Some of these tricks are robust mathematical methods; but they are still tricks, because they involve artifice. – Joel David Hamkins Feb 11 at 14:39
While tricks have names because they wind up being associated with with some particular mathematician, tricks are tricks because something important goes on "behind the curtain."
For instance, to prove $$(a_1 b_1 + \cdots + a_n b_n)^2 \leq ({a_1}^2 + \cdots + {a_n}^2) ({b_1}^2 + \cdots + {b_n}^2),$$ write \begin{align*} A &= ({a_1}^2 + \cdots + {a_n}^2)\\\ B &= (a_1 b_1 + \cdots + a_n b_n)\\\ C &= ({b_1}^2 + \cdots + {b_n}^2), \end{align*} then we must show $$B^2 \leq AC.$$ Equality clearly holds when $A = 0$. Otherwise, since $\mathbb{R}$ has no negative squares, for all $x \in \mathbb{R}$, $$0 \leq (a_1 x - b_1)^2 + \cdots + (a_n x - b_n)^2.$$ Expanding the squares, $$0 \leq Ax^2 - 2Bx + C.$$
The quadratic expression vanishes whenever $$x = \frac{B}{A} \pm > \sqrt{\left(\frac{B}{A}\right)^2 - > \frac{C}{A}}.$$
If $x = \dfrac{B}{A}$, then $$0 \leq A\left(\frac{B}{A}\right)^2 - 2 B\left(\frac{B}{A}\right) + C = \frac{B^2}{A} - 2 \frac{B^2}{A} + C = - \frac{B^2}{A} + C,$$ thus $$B^2 \leq AC.$$
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Scott's Trick is called a "trick" because it is not actually necessary for the completion of the proof in which it is involved; however, without the trick the proof is massively more tedious. Although the other tricks may not have a widely-agreed-upon-reason for being a trick, I suspect that they may be called such for similar reasons.
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Same for the Rabinowitsch trick. Also as pointed in another answer, the Rabinowitsch trick works only once (although similar ideas must be used, albeit in less famous circumstances). – Thierry Zell Dec 4 '10 at 16:10
Well, it’s only unnecessary if you’re assuming AC (which was not, if I understand right, quite as entrenchedly automatic among set theorists back in the ’60s as it is now). It’s necessary if you want to talk about cardinalities — and more generally, construct quotients of classes by equivalence relations — in ZF and many other set theories. – Peter LeFanu Lumsdaine Dec 6 '10 at 5:31
I've long known the adage that a "trick" works only once whereas a "method" works in multiple instances, or maybe is expected to work in yet unanticipated future instances.
But there's another POV: a trick is something whose efficacy cannot be anticipated, but only by hindsight is seen to work. All methods I've seen of finding $\int \sec x \ dx$ are "tricks". I've always leaned toward viewing unanticipatability as the essence of trickhood.
But I also like Qiaochu Yuan's answer.
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There is an algorithm to find the integral of any rational function of trigonometric functions: use a rational parameterization (e.g. using tan x/2), then use partial fractions (or a residue computation, etc.). I don't see how this is a "trick" in any sense. It is a direct corollary of the fact that the circle is birational to the projective line. – Qiaochu Yuan Dec 4 '10 at 17:37
Yes, I agree that that argument is trick-like. The rational parameterization of the circle is not. It is a natural and beautiful geometric argument (just taking the slope of a line between two points) and I think it could easily be presented to a first-year calculus class. – Qiaochu Yuan Dec 4 '10 at 18:05
I had a student come up with a different way to do this, on the fly, on the final exam. Write $$\sec(x) = {1\over \sqrt{1 - \sin^2(x)}},$$ then substitute $u = \sin(x)$ and use partial fractions! – Jeff Strom Dec 5 '10 at 2:31
I usually integrate this function differently (using something trick-like, but natural one): $$\frac{dx}{\cos(x)} = \frac{\cos(x)dx}{{\cos^2(x)}} = \frac{d(\sin(x)}{1 - {\sin^2(x)}}$$ – Ostap Chervak Jun 14 '11 at 15:00
A fairly new Wikipedia article: en.wikipedia.org/wiki/Integral_of_the_secant_function Raise your hand if you knew that this was a famous conjecture in the 17th century. Or that it was originally done for the purposes of cartography (I think maybe lots of people know that). Or that Isaac Newton was aware of the conjecture and wrote about it, and Isaac Barrow was the first to prove it. – Michael Hardy Jun 14 '11 at 18:43
Would regarding a scalar as the trace of a $1\times1$ matrix be considered a "trick"?
Here's an occasion where that's useful:
http://en.wikipedia.org/wiki/Estimation_of_covariance_matrices#Maximum-likelihood_estimation_for_the_multivariate_normal_distribution
- | 4,138 | 15,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.949953 |
https://discourse.julialang.org/t/linearsolve-is-changing-their-solutions/95028 | 1,679,627,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00448.warc.gz | 261,309,740 | 5,796 | # LinearSolve is changing their solutions
I am solving some Linear Systems using LinearSolve.jl and its caching system. I was fine doing fine with this example
``````using LinearSolve
n = 4
A = rand(n, n)
b = [rand(n) for i ∈ 1:3]
## everything is 'true'
let
linsolve_1 = LinearProblem(A, b[1])
sol_1 = solve(linsolve_1)
x1 = A\b[1]
@info all(sol_1.u .≈ x1)
linsolve_2 = LinearSolve.set_b(sol_1.cache, b[2])
sol_2 = solve(linsolve_2)
x2 = A\b[2]
@info all(sol_2.u .≈ x2)
linsolve_3 = LinearSolve.set_b(sol_1.cache, b[3])
sol_3 = solve(linsolve_3)
x3 = A\b[3]
@info all(sol_3.u .≈ x3)
end
# returns
#[ Info: true
#[ Info: true
#[ Info: true
``````
Until, I just crop my verification of each solution, and paste them in the end of the code, Suddendly, I have some `false`.
``````let
linsolve_1 = LinearProblem(A, b[1])
sol_1 = solve(linsolve_1)
x1 = A\b[1]
println(sol_1.u)
linsolve_2 = LinearSolve.set_b(sol_1.cache, b[2])
sol_2 = solve(linsolve_2)
x2 = A\b[2]
println(sol_2.u)
linsolve_3 = LinearSolve.set_b(sol_1.cache, b[3])
sol_3 = solve(linsolve_3)
x3 = A\b[3]
println(sol_3.u)
println("---------")
println(sol_1.u)
println(sol_2.u)
println(sol_3.u)
@info all(sol_1.u .≈ x1)
@info all(sol_2.u .≈ x2)
@info all(sol_3.u .≈ x3)
end
# returns
#[-1.0672073084102143, -2.5268971546641725, 3.084186997230096, 1.1307573345753459]
#[-1.0438221505570733, -0.08229940246406439, 1.891254274527264, -0.13214959434703005]
#[-1.6954927011416538, -4.2716079880714295, 4.98481354047552, 0.9501886647386011]
#---------
#[-1.6954927011416538, -4.2716079880714295, 4.98481354047552, 0.9501886647386011]
#[-1.6954927011416538, -4.2716079880714295, 4.98481354047552, 0.9501886647386011]
#[-1.6954927011416538, -4.2716079880714295, 4.98481354047552, 0.9501886647386011]
#[ Info: false
#[ Info: false
#[ Info: true
``````
As we can see by the `println`, something had changed in `sol_1` and `sol_2` to have the value of `sol_3`.
Is this the expected behavior? Because I would consider a bug, or at least, some `!` symbol should appear somewhere.
Note: I tested with Julia v1.8.5 and 1.9.0-beta4
You’re reusing the same cache. That’s the purpose of the interface for reusing the cache! Don’t reuse the cache if you don’t want the cache reused .
For me it was not clear that I was using the cache. Let me point out what I though:
Here I setup the problem
``````linsolve_1 = LinearProblem(A, b[1])
sol_1 = solve(linsolve_1)
``````
Next, naively I just saw `sol_1.cache` being explicitly called, than I was expecting that I was copying it into a new object called `linsolve_2`.
``````linsolve_2 = LinearSolve.set_b(sol_1.cache, b[2])
sol_2 = solve(linsolve_2)
``````
Instead, if I would see a `!`, like `LinearSolve.set_b!`, I would see an indication that I am changing `sol_1.cache`, and in ultimately, that my results would be tied to `sol_1`
You can do `set_u` to swap out the `u` vector. | 1,063 | 2,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-14 | latest | en | 0.624565 |
https://blender.stackexchange.com/questions/100314/how-to-project-back-rendered-depth-maps-to-recover-the-ground-truth-3d-shape-usi | 1,653,676,980,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00540.warc.gz | 180,471,720 | 69,375 | # How to project back rendered depth maps to recover the ground-truth 3D shape using Blender?
You can discard the images I have uploaded and can simply use some camera angles of your choice to render depth maps of a 3D shape and then fuse the rendered depth maps to recover the underlying 3D shape
We recently published a paper on 3D shape generation in a computer vision conference (CVPR). My co-author wrote the code (in C++ and using OpenCV) for fusing the depth maps and getting the final 3D shapes from the produced multi-view outputs. The inputs to the code he wrote is 20 depth maps, ground-truth camera angles (posted below) and the distance from the centroid of the shapes to the camera (distance to the shape centroid=1.5 on a sphere). The centroid of shapes are calculated as follow:
First, the centers of faces (triangle) of a mesh centroid of a mesh is calculated. Then the faces areas are computed. The new centroid is the average of the mesh faces' centers, weighted by their area.
Here some people have written algorithms on how to compute the centroids.
Unfortunately my friend is not available to help me on this and I cannot use OpenCV and C++ for this new project that I'm beginning to work on. So any help would be appreciated. My goal is to write some code using Blender's Python API instead of using my co-author's C++ code to do the same thing. But before I move on, I wonder if Blender has some built-in functions that can generate the final 3D shape given rendered depth maps of that shape, camera angles and the distance to the camera? If not, can anyone give me some ideas on how I should do that and give me a code sample for it?
Here I have uploaded a set of rendered depth maps a headphone's 3D shape that you can use for backward projection (reconstructing the 3D shape). And if you prefer to start with a 3D shape directly, here you can download a 3D shape of a different headphone we used in our work before. You can render depth maps the 3D shape using the camera angles posted below.
FYI, here is my co-author's high-level description on how his approach works:
In the final step, all depth maps are projected back to the 3D space to create the final rendering. We reconstruct 3D shapes from multi-view silhouettes and depth maps by first generating a 3D point cloud from each depth image with its corresponding camera setting (x, y, z coordinates). The union of these point clouds from all views can be seen as an initial estimation of the shape.
And here are the camera angles we used for doing the rendering in the first place:
-0.57735 -0.57735 0.57735
0.934172 0.356822 0
0.934172 -0.356822 0
-0.934172 0.356822 0
-0.934172 -0.356822 0
0 0.934172 0.356822
0 0.934172 -0.356822
0.356822 0 -0.934172
-0.356822 0 -0.934172
0 -0.934172 -0.356822
0 -0.934172 0.356822
0.356822 0 0.934172
-0.356822 0 0.934172
0.57735 0.57735 -0.57735
0.57735 0.57735 0.57735
-0.57735 0.57735 -0.57735
-0.57735 0.57735 0.57735
0.57735 -0.57735 -0.57735
0.57735 -0.57735 0.57735
-0.57735 -0.57735 -0.57735
• That second link doesn't go anywhere. Preferably add the image(s) into the post directly. See: meta.stackexchange.com/questions/75491/… Feb 7, 2018 at 16:57
• @RayMairlot Sorry, I just fixed it. There are 20 depth maps so I did not want to attach them one by one here. Also because the compression methods are applied on uploaded images and you cannot use them in raw formats I have (probably)
– Amir
Feb 7, 2018 at 18:13
• @Amir are these the angles in correspondence to the sequence of the same images? Feb 7, 2018 at 21:51
• Yes Rick. I also uploaded an .obj file we used for rendering. The link is on the bottom of my post.
– Amir
Feb 7, 2018 at 21:55
• @Amir I think your second image must be rotated 180° on its own normal. Just positioning the first two images is showing this conflict. Is the original code doing this to the output? Also do you happen to know camera distance from the subject? Feb 7, 2018 at 22:34
After looking at the following image from your paper, it seems to me as if your image is basically a sprite map, so I propose the following workflow to get you there.
Proposed Workflow:
I'm certain that you could mimic the polyhedron shape in blender to set active planes (viewports) on programatically, and use an UV coordinate mapping per sprite location to get a depth map per the corresponding camera location/angle (position).
Set up a highly subdivided plane for each view, then attach the Displacement Modifier to it, along with a custom UV layout to the sprite sheet that corresponds with the appropriate viewpoint.
I would then add a global driver to attempt to control the displacement modifier strength per projection, and when they get close to each other...
Apply the modifier, Join the objects together, and remove duplicate verts. | 1,311 | 4,850 | {"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-2022-21 | longest | en | 0.937012 |
https://essaysmasters.net/free-sample/citibike-data-and-simulation-models-using-excel-or-risk-or-arena/ | 1,606,362,734,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186414.7/warc/CC-MAIN-20201126030729-20201126060729-00081.warc.gz | 287,839,876 | 8,622 | # CitiBike Data and Simulation Models using Excel or @RISK or Arena
First, conclude that bikers enter to Post 1 (post id =1) according to a nonstationary Poisson manner. Complete the beneath board (enclosed represent smooth) of manifestatlon rates at Post 1. Hint: Create a agetreasure column. Excel ‘Timetreasure (age quotation)’ returns the decimal number of the age represented by a quotation string. The decimal number is a treasure ranging from 0 (zero) to 0.99988426, representing the ages from 0:00:00 (12:00:00 AM) to 23:59:59 (11:59:59 P.M.). For stance, TIMEVALUE("1-June-2017 6:35 AM") = 0.2743. Use the histogram of the agetreasure to abuse the manifestatlon rates. Second, appoint a doom (end post id) to each manifestatlon at Post 1 by using a discrete verisimilitude arrangement of this form: DISC(p1,1, p2,2,…p12,12) Determine the treasure of p1, p2…p12 using the referring-to quantity bar chart of ‘end post id’. Copy and paste your Excel worksheet. Finally, plant a probabilistic example for the fall continuance from Post 1 to Post i, delay i=1,2,…12 i. First stride is to transport outliers. If a bike has been laceration out for over than 24 hours at a age, it is considered past or stolen. Transport any falls longer than 24 hours (86,400 seconds). You can besides transport over outliers if you imagine it is essential. ii. Considering the scatterplot beneath (Word instrument) and the subsidence of the 12 posts, mention the verisimilitude arrangement of the fall continuance from Post 1 to Post i, delay i =1, 2,…12. You can use Arena Input Analyzer, @RISK, or any other statistical software you enjoy. You may incorporate some (or flush all) end posts to mention input verisimilitude arrangements. Copy and paste your Arena Input analyzer results or Excel worksheet. Copy and paste your Excel worksheet. | 485 | 1,835 | {"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} | 3.03125 | 3 | CC-MAIN-2020-50 | latest | en | 0.790053 |
https://studyres.com/doc/8491453/nserc-applications | 1,628,100,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00604.warc.gz | 510,615,305 | 11,019 | Survey
Thank you for your participation!
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Document related concepts
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Transcript
```A Model for the Study of HIV/AIDS
Brandy L. Rapatski
Juan Tolosa
Richard Stockton
College of NJ
Classic Model – Compartment Model
S
Susceptible
I
Infected
R
Recovered
Removed/Recovered Class
•Immunity (can’t infect or be infected)
•Death
•Isolation
Assumptions
An average infective makes contact sufficient to transmit
infection with bN others per unit time. (N = total
population).
Probability a random contact by an infective is with a
susceptible is S/N
The number of new infections in unit time is
bN (S/N)I = bSI
There is no entry into or departure from the population
except possibly through death from the disease.
S
dS
bSI
dt
dI
bSI I
dt
dR
I
dt
I
R
Qualitative Approach
dS
bSI
dt
dI
( bS ) I
dt
dS
0 S(t) decreases for all time
dt
dI
0 if S , I(t) decreases to 0.
dt
b
dI
0 if S , I(t) increases to maximum then decreases.
dt
b
bS (0)
threshold value
Basic Reproduction Number
bS (0)
R0
The number of secondary infections caused by a
single infective introduced into a wholly susceptible
population over the course of the infection of this
single infective.
R0 1, epidemic dies out
R0 1, there is an epidemic
Progression of Infection by Stage
We investigate how infectivity varies with stage of infection.
1/6
Year
First
Stage
7 Years
Second
Stage
3 Years
Third Stage
including AIDS
Before modern medical intervention
AIDS is roughly the last year of the third stage
Death
Progression of Infection by Stage
an alternative set of durations
Infectivity varies by stage of the disease
1/6
Year
First
Stage
7 Years
Second
Stage
2 Years
AIDS & Death
Third
Stage*
Before modern medical intervention
*Here last year of AIDS is omitted due to sexual inactivity
[------------------2nd Stage-------------------]
[1st Stage]
[-----3rd Stage----]
Anderson, R. M. “The spread of HIV and sexual mixing patterns,” pp. 71-86 in Mann, J. and D. Tarantola (eds.), AIDS in
the World II: Global Dimensions, Social Roots, and Responses. The Global AIDS Policy Coalition. (New York: Oxford
University Press, 1996).
Modes of HIV Transmission
Sexual Contact
IV Drug Use
Vertical Transmission (Mother to Child)
Blood Transfusion
Modeling SF Gay Population: the
Data
HIV exploded in San Francisco between the years 1978
and 1984.
San Francisco obtained high quality data on 6875 gay
men: Infection rates and number of sexual contacts.
Blood samples were taken and frozen during the years
HIV was quietly breaking out – as part of a hepatitis
vaccine study.
San Francisco
San Francisco Transmission Dynamics
Analysis of 1978-1984
Six Activity Levels (from survey data)
Infectiousness depends on stage (3 stages)
Bathhouse Assumption
Men vary in how often they visit the
bathhouses but once inside choose partners at
random.
Model Compartments
Individuals can be in any of four stages (including susceptible)
and in any of six activity groups. So the model keeps track of
fraction of people in each of the 4 x 6 subpopulations.
Simple Transmission Dynamics
dS
bIS new case rate
dt
S fraction of susceptibl es, between 0 and 1
I fraction infected 1 S
b effective contact rate
Multiple Group Transmission Dynamics:
from Group j to group i
dSi
bijIjSi
dt
rate of new group i cases caused by group j
Si fraction of susceptibl es, between 0 and 1
Ij fraction infected 1 Sj
bij effective contact rate, which equals
numbers of contacts times infectious ness
Go to Six-Group Models
Six-Group Models
80
70
60
50
model
40
data
30
20
10
0
0
1
2
3
4
5
6
7
8
Current Work
Test and Treat US MSM HIV/AIDS population
Go to Test and Treat Model
Juan’s Stuff
Contact Information
[email protected]
[email protected]
Current MSM Epidemic in US
MSM accounted for 71% of all HIV infections among male adults and adolescents in
2005 even though only about 5% to 7% of male adults and adolescents in the United
States identify themselves as MSM
Through its National HIV Behavioral Surveillance system, CDC found that 25% of
the MSM surveyed in 5 large cities were infected with HIV and 48% of those infected
were unaware of their infections.
Young black MSM in this study were more likely to be unaware of their infection –
approximately 9 of 10 young black MSM compared to 6 of 10 young white men.
Of the men who tested positive, most (74%) had previously tested negative for HIV
infection and 59% believed that they were at low or very low risk.
```
Related documents | 1,421 | 4,994 | {"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-2021-31 | latest | en | 0.829412 |
https://metanumbers.com/25524 | 1,600,556,212,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00205.warc.gz | 521,957,627 | 7,510 | ## 25524
25,524 (twenty-five thousand five hundred twenty-four) is an even five-digits composite number following 25523 and preceding 25525. In scientific notation, it is written as 2.5524 × 104. The sum of its digits is 18. It has a total of 5 prime factors and 18 positive divisors. There are 8,496 positive integers (up to 25524) that are relatively prime to 25524.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 25 thousand 524 twenty-five thousand five hundred twenty-four
## Notation
Scientific notation 2.5524 × 104 25.524 × 103
## Prime Factorization of 25524
Prime Factorization 22 × 32 × 709
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 4254 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 25,524 is 22 × 32 × 709. Since it has a total of 5 prime factors, 25,524 is a composite number.
## Divisors of 25524
1, 2, 3, 4, 6, 9, 12, 18, 36, 709, 1418, 2127, 2836, 4254, 6381, 8508, 12762, 25524
18 divisors
Even divisors 12 6 4 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 18 Total number of the positive divisors of n σ(n) 64610 Sum of all the positive divisors of n s(n) 39086 Sum of the proper positive divisors of n A(n) 3589.44 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 159.762 Returns the nth root of the product of n divisors H(n) 7.11085 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 25,524 can be divided by 18 positive divisors (out of which 12 are even, and 6 are odd). The sum of these divisors (counting 25,524) is 64,610, the average is 35,89.,444.
## Other Arithmetic Functions (n = 25524)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 8496 Total number of positive integers not greater than n that are coprime to n λ(n) 708 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2814 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 8,496 positive integers (less than 25,524) that are coprime with 25,524. And there are approximately 2,814 prime numbers less than or equal to 25,524.
## Divisibility of 25524
m n mod m 2 3 4 5 6 7 8 9 0 0 0 4 0 2 4 0
The number 25,524 is divisible by 2, 3, 4, 6 and 9.
• Refactorable
• Abundant
• Polite
## Base conversion (25524)
Base System Value
2 Binary 110001110110100
3 Ternary 1022000100
4 Quaternary 12032310
5 Quinary 1304044
6 Senary 314100
8 Octal 61664
10 Decimal 25524
12 Duodecimal 12930
20 Vigesimal 33g4
36 Base36 jp0
## Basic calculations (n = 25524)
### Multiplication
n×i
n×2 51048 76572 102096 127620
### Division
ni
n⁄2 12762 8508 6381 5104.8
### Exponentiation
ni
n2 651474576 16628237077824 424419123174379776 10832873699902869402624
### Nth Root
i√n
2√n 159.762 29.4431 12.6397 7.61009
## 25524 as geometric shapes
### Circle
Diameter 51048 160372 2.04667e+09
### Sphere
Volume 6.96522e+13 8.18667e+09 160372
### Square
Length = n
Perimeter 102096 6.51475e+08 36096.4
### Cube
Length = n
Surface area 3.90885e+09 1.66282e+13 44208.9
### Equilateral Triangle
Length = n
Perimeter 76572 2.82097e+08 22104.4
### Triangular Pyramid
Length = n
Surface area 1.12839e+09 1.95966e+12 20840.3
## Cryptographic Hash Functions
md5 22269b8cb48a778bb09fdfe1d3a37911 229f11f3b416f5e6d18d7088c0918c2ab7f22d01 160406179c6da55afa728cb1a240de9a1c19300b7eeee016d5eeebd7140b0f37 06a3bdc49827da9a9442c0946ee67e498f325de574a7b458a7d1234329684aec348ba1f9d5abb3cd6b9477e3def4f05f5d3f3ad59b1a798fafcc14cc618197f4 93d1b73b4a4e7ae8bd2864c118c845e420112288 | 1,492 | 4,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-40 | latest | en | 0.813678 |
https://www.meritnation.com/ask-answer/question/what-is-square-law-device-what-is-the-function-of-this-pleas/communication-systems/512397 | 1,582,559,166,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00416.warc.gz | 780,565,193 | 10,051 | what is square law device? what is the function of this? Please suggest me any me hurry up!i am wating your reply?
A square law device is something where either current or voltage depends on the square of the other quantity.!
For e.g: for a saturated MOSFET current(I) is directly prooportional to square of voltage(V2).
A diode characteristics also has a square but its not as clean.It's exponential and through a taylor series we get a linear, square ir cubic terms...
but anyway its useful bcoz it can b used as a modulator or mixer.
adding voltage is easy but multiplying them is hard..with a square law device like MOSFET it could b done easily.!
I proportional to(P.T) V2
I P.T (sin(wct) + m(t))2
I P.T (sin(wct)) + m(t)2 + m(t)sin(wct).
Now if we follow it with just bandpass filter on wc we get simply m(t)sin(wct).
Amplitude modulation.
• 5
A
• -6
thanks to you for your valuable time
OK!
• -1
Hi Anup!
A square law device is one that produces an output voltage or current that is proportional to the square of its input voltage or current. Square law devices are mainly used in modulators and detectors. Example: The detector diode used in communication systems, FETs used in balanced modulator etc.
Cheers!!
• 14
THANKS TO YOU
• 0
thank u
• 0
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# Problem EHorror Film Night
Emma and Marcos are two friends who love horror films. This year, and possibly the years hereafter, they want to watch as many films together as possible. Unfortunately, they do not exactly have the same taste in films. So, inevitably, every now and then either Emma or Marcos has to watch a film she or he dislikes. When neither of them likes a film, they will not watch it. To make things fair they thought of the following rule: They can not watch two films in a row which are disliked by the same person. In other words, if one of them does not like the current film, then they are reassured they will like the next one. They open the TV guide and mark their preferred films. They only receive one channel which shows one film per day. Luckily, the TV guide has already been determined for the next $1$ million days.
Can you determine the maximal number of films they can watch in a fair way?
## Input
The input consists of two lines, one for each person. Each of these lines is of the following form:
• One integer $0 \leq k \leq 1000000$ for the number of films this person likes;
• followed by $k$ integers indicating all days (numbered by $0, \dots , 999999$) with a film this person likes.
## Output
Output a single line containing a single integer, the maximal number of films they can watch together in a fair way.
Sample Input 1 Sample Output 1
1 40
2 37 42
3
Sample Input 2 Sample Output 2
1 1
3 1 3 2
2
Sample Input 3 Sample Output 3
1 2
1 2
1
Hide
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# Friday 26th June
Morning Ahlberg class, today we have a dance but with days of the week and rhyming words, can you spot the rhyming words? https://www.youtube.com/watch?v=VD6SCq-OlhI
For today’s learning we have…
RWI
Remember to practise your sounds daily. RWI will be holding daily virtual lessons for children to practise their sounds:
Set 1 sounds at 9:30am
Set 2 sounds at 10:00am
Set 3 sounds at 10:30am
They have updated these lessons to include word time for set 1 speed sounds and spelling for set 2 and 3.
There is also “Storytime with Nick” three times a week (Mon, Wed & Fri at 2pm) and some poems that your child can learn and perform.
SPaG
Today is our spelling day, we’re going to concentrate on our common exception words with this fun activity mat. See if you can complete it independently this week.
Literacy
How did you get on yesterday?
Today you are going to put the final touches to your comic strip.
Under each picture is a caption box. A caption is the words that describe or explain the picture.
Today you are going to write sentences for each picture that tells the reader what happens in the story. Remember you need to use a capital letter to start a sentence and a full stop at the end of a sentence. You may need to use an exclamation mark (!) at the end of a sentence to show how someone is feeling. (angry, happy, sad, scared, excited)
When you have finished writing your caption boxes make sure you check you have finished the pictures, the sound / action words and captions. Don’t forget to add colour to your pictures too.
I would love to see your finished comic so please send them to me when you have finished.
If you enjoyed making your comic strip, you may want to write another comic adventure that Traction Man goes on
Maths
This morning we have a maths warm up to get our brains working, try to complete this independently.
G/Y/R/O groups:
This week we have learnt about fractions. We have found ½ of shapes, ½ of amounts, ¼ of shapes and ¼ of amounts. Click on the link to recap your understanding of fractions: https://www.youtube.com/watch?v=ZLxbPQRIyjw
Today we are going to use this knowledge of fractions to answer some questions. With the word problems remember to circle the important information like the number and fraction.
Remember to practise your times tables with Times Tables Rock Stars.
Blue group:
This week we have learnt about adding and taking away using objects to help us. Today I want you to read the maths questions carefully. Are you adding more or taking away? Say the maths sentence and say the actions out loud.
For take away we say “get ready to take away”
For equals we say “how many are there?”
Music
Last Music lesson we learnt about a string instrument called the violin, can you remember? Today we are learning about an interesting instrument called the accordion. Click on the link below to learn about it. After you have watched the link, click on the worksheet. There is an Accordion for you to colour in after you have written 3 facts that you have learnt about the accordion, I’ve done the first one for you.
PE
For PE this week we are going to be learning how to control our speed when we move. For this you will need a big open space, a garden if you have one or maybe walk to the local park. You could move all your furniture and use your living room if it is raining outside. Today we are going to jog, run and sprint and by the end of the lesson you will be able to describe how to do each of these. Click on the powerpoint below and follow the instructions, have fun staying active!
Here is the link for Joe Wick’s PE sessions if you would like to challenge yourselves.
https://www.thebodycoach.com/blog/pe-with-joe-1254.html
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ALL >> General >> View Article
Mastering Unit Conversions For Measurement: A Comprehensive Guide
By Author: shivansh nautiyal
Total Articles: 1
• Converting Square Feet to Square Meters: Converting between square feet (sq ft) and square meters (sq m) is a frequent requirement, especially in construction and real estate. The conversion factor is 1 square meter equals approximately 10.764 square feet. To convert square feet to square meters, divide the area in square feet by 10.764. Here’s the formula:
Area in square meters (sq m) = Area in square feet (sq ft) / 10.764
Example: Let’s convert an area of 500 square feet to square meters: 500 sq ft / 10.764 = 46.4516 sq m (rounded to four decimal places)
• Converting Square Meters to Square Feet: To convert square meters to square feet, multiply the area in square meters by 10.764. The formula is as follows:
Area in square feet (sq ft) = Area in square meters (sq m) * 10.764
Example: Let’s convert an area of 100 square meters to square feet: 100 sq m * 10.764 = 1076.4 sq ft (rounded to one decimal place)
• Converting Square Meters to Square Yards: Converting square meters to square yards is useful for understanding larger areas, such as parks or fields. One square yard is equivalent ...
... to 0.836 square meters. To convert square meters to square yards, divide the area in square meters by 0.836. The formula is as follows:
Area in square yards (sq yd) = Area in square meters (sq m) / 0.836
Example: Let’s convert an area of 200 square meters to square yards: 200 sq m / 0.836 = 239.23 sq yd (rounded to two decimal places)
• Converting Acres to Hectares: Converting between acres and hectares is often required in agriculture and land measurement. One acre is equivalent to 0.4047 hectares. To convert acres to hectares, multiply the number of acres by 0.4047. The formula is as follows:
Area in hectares (ha) = Area in acres * 0.4047
Example: Let’s convert an area of 5 acres to hectares: 5 acres * 0.4047 = 2.0235 ha (rounded to four decimal places)
• Converting Acres to Square Meters: Sometimes, you may need to convert acres to square meters for precise measurements. One acre is equivalent to 4,046.86 square meters. To convert acres to square meters, multiply the number of acres by 4,046.86. The formula is as follows:
Area in square meters (sq m) = Area in acres * 4046.86
Example: Let’s convert an area of 2 acres to square meters: 2 acres * 4046.86 = 8093.72 sq m (rounded to two decimal places)
Conclusion: Unit conversions for measurement are vital skills that simplify our interactions with diverse systems of measurement. In this article, we covered several common conversions, including square feet to square meters, square meters to square feet, square meters to square yards, acres to hectares, and acres to square meters. With a solid understanding of these conversions, you can confidently work with different units and ensure accuracy in your measurements. Happy converting!
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0
119
6
+79
Determine if the graph of the equation below is a parabola, circle, ellipse, hyperbola, point, line, two lines, or empty. Enter the most specific answer.
$$(x-3)^2 + y^2 = 10$$
Thank you for helping!
Feb 2, 2021
#1
+113118
+1
you should be able to work this out very easily for yourself.
Try putting it into here and see what happens
https://www.desmos.com/calculator
Feb 2, 2021
#2
+118505
+1
(x - 3)^2 + y^2 = 10
x^2 - 6x + 9 + y^2 = 10
1x^2 + 1 y^2 - 6x - 1 = 0
When we have the form
Ax^2 + By^2 + Cx + Dy + E = 0
If A = B (and E is negative) then we have a circle
Feb 2, 2021
#4
+79
0
Thank you!
StarsIsTheLimit Feb 3, 2021
#3
+113118
+1
It was already in the most obvious form Chris.
$$(x-3)^2+y^2 =\sqrt{10}^2\\ centre(3,0)\;\;radius \sqrt{10}$$
If the kid looked at his formula notes.
Or if he went to desmos and graphed it, he would have known that.
I give partial answers for good reasons.
Feb 2, 2021
#5
+79
+1
I will try to work it out by myself next time.
StarsIsTheLimit Feb 3, 2021
#6
+113118
0
Good, it pleases me to hear that.
Melody Feb 3, 2021 | 470 | 1,149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-21 | latest | en | 0.835441 |
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Problem 1 Assume the priority queue also must support the update DELETE(A,i), where i gives you the location in the data strucure where element i is located. In a binary heap, that would be A [ i ]. Write pseudocode to achieve DELETE(A,i) in O (log n ) time for binary heaps, where n is the size of the heap. Solution: 1, Delete a node from the array (this creates a vacant node and the tree is not complete) 2. Replace the deletion node with the fartest right node on the lowest level of the Binary Tree (To complete the tree) 3. Heapify (fix the heap): Algorithm for Max Heapify Procedure: MaxHeapify(A, n) 1. l left(n) 2. r right(n) 3. if l heap-size[A] and A[l] > A[n] 4. then largest l 5. else largest i 6. if r heap-size[A] and A[r] > A[largest] 7. then largest r 8. if largest n 9. then exchange A[n] A[largest] 10. MaxHeapify(A, largest) Running time( delete(n) ) = Running time heap delete in heap with n node = height of a heap with n nodes height of the heap containing n nodes =log(n + 1) Running time of delete into a heap of n nodes = O(log(n)) Complexity of deletion: O(logN) in the worst case.
Problem 2 Using only the definition of a binary search tree, show that if a node in a binary search tree has two children, then its successor has no left child and then its predecessor has no right child.
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# Transformable Boolean operation (Sweep with body as tool)
Member Posts: 23 ✭✭
Recently I had to design a guiding slot. The function was to transform a horizontal rotation of a ring into a vertical angle change of a pin. The basic concept looked like this:
In CAM or real world milling, such geometries are no problem since you can define all axis of a tool and move it around as you please. But in the CAD design itself, it's a real pain. I managed to create the geometry by using several rotation cuts, a lofted surface which got thickened afterwards and finally a Boolean subtraction.
I've added the above example so you can understand the origin of my idea and my motivation. I started to program a custom feature which does a Boolean subtraction by moving a body in 3D space. With this feature, very complex surfaces could be created in a short amount of time without having to play around with curves, extrudes and subtractions. Simply put, a sweep which is able to use a 3d body. This is where I stand right now:
The procedure is to move the tool by certain increments along the curve and do a subtraction after every transform. It works better than expected but the resulting surfaces aren't usable since they are not tangential to each other.
@Jake_Rosenfeld and @NeilCooke already helped me with the transform along the curve. Now comes the tricky part, has anyone any idea/concept of how to smoothen the cut surface? I really appreciate any help and I hope I can create a stable and reliable feature in the end that's also useful to others. I've just got into featurescript.
Feature plans are to add the possibility of using several curves to define the orientation fo the tool. For example one curve that guides the tool by its origin and aligns the tool axis normal to the curve, and a second curve that defines the orientation of the tool.
• Member Posts: 1,579 EDU
MB - I make FeatureScripts: view FS (My FS's have "Official" beside them)
• Moderator, Onshape Employees Posts: 2,270
edited December 2018
Neil Cooke, Director of Technical Marketing, Onshape Inc.
• Member Posts: 901 ✭✭✭✭✭
edited December 2018
Its all about envelopes, the idea is flying in the air, but seems like there are no general and robust solution with appropriate perfomance.
A few thoughts:
- envelope for the myltyface body is significanly harder to generate then the envelope for a particular face or a limited set of faces of that body.
- envelopes for cylinder faces may be made by sweeping cylinder axis along the path, and then creting offset of the resulting surface.
- envelope for rotational faces of the bodies may be created by sweeping their cross section along the path.
- envelopes for the outer sharp edges may be also created by sweeping those edges along the path.
- for the general shape faces you may need to generate a number of intermidiate positions of the face geometry, and try to pick the points, for cross sections for lofted surface with some method - for example using the secting lines as in my Envelope FS.
- for some general shape envelopes complex bodies should be decomposed into primitives so that you could use the rules from above for each of them separately.
• Member Posts: 1,579 EDU
For extruding parts you could use this feature:
MB - I make FeatureScripts: view FS (My FS's have "Official" beside them)
• Member Posts: 23 ✭✭
First of all, thank you for your participation!
@konstantin_shiriazdanov yes I agree! I've experimented with creating a loft using the outer edges of each section. This COULD result in a surface where all section edges are tangential.
Regarding the sweep method, I guess it is very perfomance eating if I create with every section a new sweep which has to find the outer edges of the cut, create a profile and do a sweep to the next section. The problem is also, a sweep alows no deformation of the profile like a loft. So it could happen, that the sweep end does not have the same profile than the starting profile of the next sweep. So you have edges again.
I've checked out your FS and it is very impressive! This already works very well! But in your case, you created different sititations of how to use certain tools. Like gears for example. I'd like to create a feature where any body can be moved along any path using any rotational movement. I hope I can manage that somehow one day. Let's see about that
@mbartlett21 Yes you're right. Here is the link. Please excuse my somewhat primitive programming skills
Regarding the other FS you've sent, it's not quite the feature I need. But thanks anyway!
• Member Posts: 1,579 EDU
You should be able to sweep each face individually @johannes_w
MB - I make FeatureScripts: view FS (My FS's have "Official" beside them)
• Member Posts: 901 ✭✭✭✭✭
when the envelope is made by sharp outer edges it can be easily made with sweep or loft like that:
the probmes begin when you had to deal with envelopes made from some variable internal regions of the faces
• Member Posts: 23 ✭✭
edited December 2018
I did the EXACT same thing before I've read this post. And I came to the same conclusion.
I have decided to limit the possibilities of this feature to cylindrical shapes. Like a milling cutter. This should cover nearly all CAM situations you may have when designing such geometries. Like in the first example I've posted.
I thought about do a subtraction on the begin of the path, transform the body to the end of the body with all angle parameters and do another subtraction. Then I'd use the path and the envelope of the body to make a loft along the path. Sweep isn't possible since sweep doesn't allow me to change the angle along the path with a second guiding curve.
There is one problem though. The end profile of the loft is not planar. If I may use the first example again:
As you can see, if I created an end profile that's normal to the end tangent of the guiding curve, the loft would cut too far on the bottom.
Also, the tangents from the cutting surface to the geometry of the body (yellow lines) isn't parallel to the axis of the tool which is normal to the viewing angle right now.
• Member Posts: 901 ✭✭✭✭✭
edited December 2018
so if you about rotational bodies, you can modify @mbartlett21 's Extrude body FS and try to change internal extrude of outline profile of the body to the sweep (or loft). the trick is to correctly make a plane for outline projection
• Member Posts: 23 ✭✭
Yes I agree this FS has the same concept. Pattern a part to the sought location and connect the outline projections of the two parts. But as soon as there is a change of orientation, you can't use a plane for the outline projection. If this were the case, everything would be much easier and such a FS wouldn't be needed in the first place. Let me show you an example:
In this case, a milling cutter rotates its axis vertically while performing the transform. On the starting point everything works out since the cutter starts horizontally. But when it comes to determining the end profile you can't use a plane to determine the outline projection. The inner side (right) would require a plane like the sketch I drew for the revolve. But the outer side (left) would require a plane that is normal to the tangent of the guiding curve.
• Member Posts: 901 ✭✭✭✭✭
edited December 2018
then what if you create a number of outlines for the parts in intermidiate positions and use them as profiles for loft?
• Member Posts: 23 ✭✭
That would be possible. However, I think that creating single sections of the outline projection would really complicate the process and make the loft very complex.
Right now I'm trying an approach of creating an extra surface that has that twist and use that surface for the end profile. It works but not 100% precise unfortunately.
• Member Posts: 199 PRO
What I'd be interested in, is to be able to use the rotated end of the swept body to represent what happens when your toolpath doesn't go all the way through a part.
A very simple example is correctly representing a blind chamfer:
As you know, a chamfer + a fillet won't work, since the fillet can't be applied in the correct "orientation" to accurately represent the spinning mill in this case.
I make a body (this is very simple because the edge I am using is straight) that represents the mill during motion AND at the end of the path, then do the Boolean to subtract:
Could this FeatureScript you guys are talking about include the "end condition" that I'm talking about?
• Member Posts: 199 PRO
I did try your Extrude Part FS @MBartlett21 but it didn't work in this case:
Could it be the vertex at the bottom of the part that makes it fail? (I did try it with a non-pointy bottom end and it still failed).
• Member Posts: 23 ✭✭
Exactly @romeo_graham392. The end condition you are talking about is exactly what I'm trying to achieve.
Unfortunately, my job didn't allow me to investigate and try this further since my last update but the goal is still the same. It would allow some sort of CAM modeling functionality. | 2,177 | 9,627 | {"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-2019-30 | latest | en | 0.945625 |
https://www.physicsforums.com/threads/problem-with-box-and-string.550060/ | 1,527,288,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00537.warc.gz | 797,058,916 | 16,629 | Homework Help: Problem with box and string
1. Nov 13, 2011
GingerKhan
1. The problem statement, all variables and given/known data
A spring with a force constant of 225 N/m is resting on a friction-less surface and mounted against a wall horizontally. A 1.5 kg box is pushed against the spring and compresses it 12 cm (0.12m) from equilibrium. When released the spring pushes the box across the surface.
2. Relevant equations
F = kx
W = FΔd
Ee = 1/2 k x^2
Ek = 1/2 m v^2
3. The attempt at a solution
a) How much force needs to be applied to the spring to compress it to 12 cm (0.12m)?
F = 225 x 0.12 = 27 N
b) How much work is done to compress the spring to 12 cm?
W = 27 x 0.12 = 3.24 J
c) How much elastic energy is stored in the spring when compressed?
Ee = 1/2 x 225 x 0.12^2 = 1.62 J
Question: Is it normal that Ee is less than the work energy applied to the system in b)?
d) What maximum speed will the box attain once released?
1.62 = 1/2 x 1.5 x v^2
1.62/0.75 = v^2
v = 1.47 m/s
** not sure about this because I might be using a wrong Ek obtained in c) **
2. Nov 13, 2011
grzz
3. Nov 13, 2011
Filip Larsen
This is not correct. You can only use "work equals force times distance" on the whole distance if both are constant. If they are not you must integrate the work up from the instantaneous force as a function of distance, e.g. W = ∫F(s)*ds, where s is distance.
In the context of this problem you may also solve b) by using conservation of energy. The original kinetic energy from the block must all be stored in the spring since there is no friction and nothing else moves, hence the block does the amount of work on the spring equal to its kinetic energy, which again is equal to the potential energy right after compression.
4. Nov 13, 2011
cepheid
Staff Emeritus
Welcome to PF GingerKhan!
Nope. These two numbers should be the same. The mistake is in assuming that W = FΔx. This is only true if the force is constant. The force is not constant here, but rather it varies with x. The more general formula is W = ∫F(x)dx where F(x) is the (position-varying) force function. But if you don't know integral calculus, don't worry. In this case, you've already (implicitly) been given the expression for the work done. Think about this: what is the work done by the spring force as the spring is compressed, (hint: how does this relate to the elastic potential energy stored)? Now, how does the work done by the applied force have to compare to the work done by the spring force?
5. Nov 13, 2011
grzz
Since the force F is directly proportional to the compression the average value of the force will be (final value of F)/2.
6. Nov 13, 2011
GingerKhan
Oh, I forgot about the spring becoming harder to compress as it is being compressed.
So with a graph that illustrates the scenario, I would need to find the area of the right triangle and divide the whole thing by two, giving me 1.62 J
I could also use Ee = 1/2 k x^2 to calculate the work done because W = ΔE.
If c) is correct then d) is correct as well, right?
Thanks guys.
Last edited: Nov 13, 2011 | 859 | 3,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-22 | longest | en | 0.923917 |
https://tw.tradingview.com/chart/EMC2BTC/Lg1QeD6e-Einsteinium-Bitcoin-Compelling-Buy/ | 1,558,613,098,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257243.19/warc/CC-MAIN-20190523103802-20190523125802-00275.warc.gz | 660,424,001 | 64,391 | BITTREX:EMC2BTC Einsteinium / Bitcoin
777瀏覽
777
Traders I recommend reading this entire post so that you fully understand what I have done, and how I have came to my analysis.
EMC2/BTC looks like a compelling long opportunity. I love this set up so far!
After analyzing this pair, I discovered that price appears to be moving in patterns. Price seems to move up 5539 points, have a corrective move to the downside, and then move up again, around 5539 each time.
With this logic, I was able to predict and plot the next move on the chart. I have calculated each cycles move by measuring the amount of points it moved up before correcting. After I found there was an approximate 5539 point move before each correction, I was then able to replicate a cycle at around 5539 points to see the future direction of the instrument.
I am expecting an up move of 5539 which will complete at around 0.00012800. Below I have illustrated the calculations as to how I came to this conclusion.
The principle of the cycles being used here is that every time there is an up move, there is a corrective move, and then an up-move which exceeds the previous move by approximately 5539. With this information we can calculate (using cycles 1+2 data) where cycle 3 is likely to end up. Using the calculations below we can make a mathematical assumption as to where price will lead.
Cycle 1 = Cycle + 5539 = 0.00008100
Cycle 2 - Cycle 1 + 5539 = 0.00009100
Cycle 3 - Cycle 2 + 5539 = 0.00012800
Following these calculations, we can expect a reliable up-move, using previous trend cycles to predict a reliable, yet conservative move.
Do remember that price may correct before cycle 3 initiates. So it is important that you watch this trade carefully before considering a position. I would also like to point out that there is a key side-wards resistance level at 0.00010000. Price needs to break this level and retest this level before Cycle #3 completes.
I will be posting updates on this trade below, so leave it a like or follow so that you can keep up to date with this analysis.
Price reached target level, and is now correcting to the downside. When price completes the corrective move to the downside, we must be open to the possibility of a cycle #4.
When correction completes I will post more updates below.
Price of the Crypto currency doubled in value in just 21 hours and we caught the entire move here!
I am glad you were all able to follow this trade with me. Congratulations to anyone who used this analysis to profit with me.
I am interestingly keeping the track your TA & FA and exciting by the movements of the graphs through you lines! Thank you Tom! You are teach me a lot since I follow you!
kwarty
@kwarty, Thanks for your compliments. Glad my content is useful to you. I would really appreciate it if you could like all of my recent ideas.
Except the hardfork EMC2 there also an information that EMC2 and APPLE will have and agreement, may be in paying function or donations on 19th of December. Any bets should be make in the sense of price increasing or the news are rumours ?
kwarty
@kwarty, I have already put my bets into this market. I believe the rumors with Apple have actually already been priced into this market and price shows as such.
Until 19th December, I would just assume that price will rally in an attempt to price in Apple. (Since if this is true AND happens, would be absolutely massive). Only when the raw fundamentals come out (as you mentioned above) come out, I will then reconsider my analysis with the new data.
wow this is really weird. Good catch
SDCVoltaic
@SDCVoltaic, Thank you. We have nailed virtually every single currency we have analyzed on TradingView lately. Let's hope for more wins!
@TomProTrader, good call, keep em coming!
magicpony
@magicpony, Many more to come :)
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HE עברית | 979 | 4,087 | {"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} | 3.40625 | 3 | CC-MAIN-2019-22 | latest | en | 0.950986 |
http://www.chegg.com/homework-help/college-algebra-8th-edition-solutions-9780073312620 | 1,469,713,797,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828282.32/warc/CC-MAIN-20160723071028-00241-ip-10-185-27-174.ec2.internal.warc.gz | 369,701,344 | 17,320 | View more editions
# TEXTBOOK SOLUTIONS FOR College Algebra 8th Edition
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Chapter: Problem:
The set of real numbers can be partitioned into two disjoint subsets, the set of rational numbers and the set of irrational numbers. In this activity, we explore the connections between two different methods for representing rational numbers, and we prove that is irrational.
Rational numbers can be represented in two ways: as fractions, where both the numerator and denominator are integers, and as decimal expansions that are either repeating or terminating. Consider the rational number r = a/b, where a and b are integers with no common factor and b ≠ 0.
If b = 10n for a positive integer n, what kind of decimal expansion will r have?
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9780073312620ISBN-13: 0073312622ISBN:
Alternate ISBN: 9780072917673, 9780073312620, 9780077232634 | 275 | 1,147 | {"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-2016-30 | longest | en | 0.855466 |
http://bricks.stackexchange.com/questions/2991/how-to-minimize-the-impact-of-erroneous-sensor-readings | 1,469,385,099,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00062-ip-10-185-27-174.ec2.internal.warc.gz | 33,589,226 | 18,248 | # How to minimize the impact of erroneous sensor readings?
A group including myself are using the Lego Mindstorm kit to design a robot that is going to compete in a school competition. However, the sensors aren't as reliable as we would like them to be, particularly the ultrasonic sensor, which is what we're using to judge how far away objects are from our robot. The problem that we face is that every so often the sensor returns a value that is just wrong: in a set of 3 datapoints for an object that is 45 units away, we'd possibly get values like 45, 45, and then 186. The erroneous values are not consistent, nor is the amount by which the error occurs. We also can't judge the probability of an erroneous value occurring. Our current algorithm takes the average of a small number (i.e. 3) of datapoints, but we've realized that if the number of datapoints is small, then the average could be distorted greatly by erroneous values. If the number of datapoints is large, then it takes a long time for the robot to get all of the sensor readings.
What do you recommend as a process to reduce the impact of erroneous sensor readings?
I'm sorry if I'm posting this question in the wrong location, but this problem seemed more closely linked to mathematics than to programming.
-
## migrated from math.stackexchange.comMar 9 '14 at 23:23
This question came from our site for people studying math at any level and professionals in related fields.
Might as well use Kalman filters. – TZakrevskiy Mar 6 '14 at 15:08
Are the items you're tracking moving at all (i.e. other robots) or are they static? If they are static you could just discard the results based on your current speed and previous changes in distance... – Zhaph - Ben Duguid Mar 10 '14 at 10:14
What is essentially happening with the robots is that we're using them to locate small, scattered, stationary canisters which we then need to capture and move to a corner of the 'field.' Consequently, the current algorithm is to basically move to somewhere in the center, rotate and take samples (non-continuous, btw) of the distances it sees, and then try to go for the best choice. Consequently, the objects that I'm trying to find are static (though there is another robot on the field at the time). However, the robot is not moving at the time either. We just want to reduce the error we get. – Amndeep7 Mar 10 '14 at 18:23 | 546 | 2,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-30 | latest | en | 0.962933 |
https://www.mathcounts.org/resources/salice-wonderland | 1,675,891,053,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00278.warc.gz | 878,488,075 | 15,063 | SALICE IN WONDERLAND
Original Handbook Problem: In Lewis Carroll’s Through the Looking-Glass, this conversation takes place between Tweedledee and Tweedledum. Tweedledum says, “The sum of your weight and twice mine is 361 pounds.” Tweedledee answers, “The sum of your weight and twice mine is 362 pounds.” What is the absolute difference in the weights of Tweedledee and Tweedledum? (2014-15 Problem #33)
Why Team Wonderland got high scores in creativity:
• Student-written humor like the silly antics of the characters and the parody song (1:26) to set up the problem make the story more interesting.
• Simple but creative costumes and props, plus intentionally over-the-top acting by the students (for example, 0:01-0:32 and 0:50-1:00) add to the scene of the story.
What else the judges liked:
• Correct use of math terminology during the solution.
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Math topic | 234 | 972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-06 | longest | en | 0.866864 |
https://www.analystforum.com/t/help-probability-of-going-3-3/22227 | 1,611,295,201,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00779.warc.gz | 663,360,348 | 5,753 | # Help! Probability of Going 3/3???
I know this has been discussed before, but does anyone have any source for the percentage of people that have passed the CFA while going 3/3? Other stats, such as average time to complete the CFA or percentage who enroll and eventually finish would be useful. Thanks for the help.
No. That information does not exist. If I had to guess though I’d say approximately half of everyone who passes Level 1 goes 3 for 3. That would put the % somewhere around 20% of level 1 candidates.
take the joint prob of the pass rates. around 7%
When I took the tests the pass rates were .40, .40, .53, respectively. But it’s hard for me to believe that only 8.4% of people that started with me went 3/3. I would think that the percentage would be higher…closer to Danteshek’s estimate of 20%. I’m smart enough to pass the CFA, but not smart enough to figure this stuff out…damn.
i had done this calc in another thread somewhere…search under my id for ‘probability’ or something.
Hmmm, I don’t think that you can take the joint prob because the events are not independent - ie repeaters on L2 & L3 are included in the 2nd & 3rd year %s. Need to dust off my stats books though…
if the events were random, then you coulod take joint probability. however there is conditional probabilty where if you pass level 1 first time, I would assume you are more likely to pass level 2 first time then someone who failed level 1 at least once. and even greater if you pass level 1&2 first time the probability is greater then passing level 3 first time then taking level 3 pass rate of 53%. I would guess, b/.c that’s what we are left with: 3for3 from all candidates that sign up for level 1 would be 8-12% and all charterholders would probably be 25-30% with a confidence interval of 100% side note: I work with 6 other charterholders of the 7 of us: 6 for 7 went 3/3
How many of the 6 went 3/3 in eighteen months? And do the six repeatedly remind the odd one out that he doesn’t have inferior ability?
I think DanteShek was about right. They say 20% of the people who start the program finish it, and for the most part, it seems like half of the people who get the charter get it without failing an exam. I am not sure what the big deal is, whether someone goes 3 for 3, or 3 for 5, or 3 for 9
Yes. I agree. The 3 events are not independent. So can not just multiply them like that. One key information missing is how likely some one will repeat at each level. They do annouce number of people taking level 1, but there is no way to know how many are new candidates and how many are repeating candidates. I guess we will never know for sure.
I think if you take 39% pass rate for Dec 06, 40% pass rate for June 07 , 53% for June 08 then the probability would be around 8% just my 2 cents on it any comments?
For some people, 0%. For others, 100%.
The 8% figure is a sensible start, but is almost certainly too low. If you pass L1 on the first try, the odds are probably higher that you pass L2 on the first try, and if you are 2/2 the odds are higher still that you pass L3 on the first try. So maybe 10-15 percent of those who sign up for L1 go 3/3. Remember, not all who pass L1 are doing so on the first try.
The 8% figure is a sensible start, but is almost certainly too low. If you pass L1 on the first try, the odds are probably higher that you pass L2 on the first try, and if you are 2/2 the odds are higher still that you pass L3 on the first try. So maybe 10-15 percent of those who sign up for L1 go 3/3. Remember, not all who pass L1 are doing so on the first try.
you wouldn’t be asking the question if you haven’t already gone 2/2. So if you’ve already passed 2/2, then the odds of going 3 for 3 are significantly better than the L3 pass rate (which includes L3 repeaters) - so probably at least 75%.
i would say that 10% of a random selection of people walking into the test center for Level 1 for the first time will go 3 for 3. of those people that actually passed all 3 exams, the probability is in the 25-35% range. | 1,055 | 4,058 | {"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-2021-04 | latest | en | 0.977797 |
https://www.gurufocus.com/term/Accts+Rec./WEN/Accounts+Receivable/Wendy%2527s+Co | 1,508,444,429,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00380.warc.gz | 927,443,778 | 40,401 | Switch to:
The Wendy's Co (NAS:WEN) Accounts Receivable: \$107 Mil (As of Jun. 2017)
Accounts Receivable are created when a customer has received a product but has not yet paid for that product. The Wendy's Co's accounts receivables for the quarter that ended in Jun. 2017 was \$107 Mil.
Accounts receivable can be measured by Days Sales Outstanding. The Wendy's Co's Days Sales Outstanding for the quarter that ended in Jun. 2017 was 30.38.
In Ben Graham's calculation of liquidation value, accounts receivable are only considered to be worth 75% of book value. The Wendy's Co's Net-Net Working Capital for the quarter that ended in Jun. 2017 was \$-3,336 Mil.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
The Wendy's Co Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Accounts Receivable 58.81 62.02 64.06 102.50 95.84
The Wendy's Co Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Accounts Receivable 115.30 109.84 95.84 90.32 106.65
Calculation
Accounts Receivable is money owed to a business by customers and shown on its Balance Sheet as an asset.
Explanation
1. Accounts Receivable are created when a customer has received a product but has not yet paid for that product. Days Sales Outstanding measures of the average number of days that a company takes to collect revenue after a sale has been made. It is a financial ratio that illustrates how well a company's accounts receivables are being managed.
The Wendy's Co's Days Sales Outstanding for the quarter that ended in Jun. 2017 is calculated as:
Days Sales Outstanding = Accounts Receivable / Revenue * Days in Period = 106.649 / 320.342 * 91 = 30.38
2. In Ben Graham's calculation of liquidation value, The Wendy's Co's accounts receivable are only considered to be worth 75% of book value:
The Wendy's Co's liquidation value for the quarter that ended in Jun. 2017 is calculated as:
Liquidation value = Cash And Cash Equivalents - Total Liabilities + (0.75 * Accounts Receivable) + (0.5 * Total Inventories) = 204.543 - 3621.614 + 0.75 * 106.649 + 0.5 * 2.922 = -3,336
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Be Aware
Net receivables tells us a great deal about the different competitors in the same industry. In competitive industries, some attempt to gain advantage by offering better credit terms, causing increase in sales and receivables.
If company consistently shows lower % Net receivables to gross sales than competitors, then it usually has some kind of competitive advantage which requires further digging.
Average Days Sales Outstanding is a good indicator for measuring a company's sales channel and customers. A company may book great revenue and earnings growth but never receive payment from their customers. This may force a write-off in the future and depress future earnings.
Related Terms | 744 | 3,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-43 | latest | en | 0.920454 |
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An I-section has the dimensions shown in Fig. Q4 and is subjected to an axial torque,...
An I-section has the dimensions shown in Fig. Q4 and is subjected to an axial torque, T = 5000 Nm A) Determine the maximum value of the shear stress (Tmax) in the material. (8 marks) B) Find the twist angle per meter length, (0/L). Assume the modulus of rigidity G for the material is 82 GN/m2. d1= 266 mm — b1 = 30 b2= 16.5 mm d2= 779 mm b3 = 30 Fig. Q4: An I-Section Structure.
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https://www.physicsforums.com/members/angelog.66714/recent-content | 1,623,655,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00497.warc.gz | 849,866,414 | 11,022 | # Recent content by AngeloG
1. ### Linear Dependence
Err, it was part of: 1, cos(pi x), sin(pi x). Those are the functions. 1 is linear independent, cos(pi x) and sin(pi x) I'm not sure about.
2. ### Linear Dependence
The question is: Check the linear dependency of the functions sin(pi x).
3. ### Linear Dependence
Check for Linear Dependence for: \sin \pi x [-1, 1] I'm thinking it's Linear Dependent. Since it says that any linear combination must be 0. a*x + b*y = 0, a = b = 0. So for any integer x, the value is 0. So [-1, 1] works.
4. ### At a crossroad; Civil or Mechanical Engineering
Well, I'm going to a big University this upcoming fall. I'm at a crossroads on what and where I want to go in terms of Engineering. I'm planning to shoot for my masters. Civil vs Mechanical. I'm a bit sad that Civil Engineers get paid crap and it takes so long to get paid well (not sure?)...
5. ### Slightly Confused About Tension
The best thing to do is draw a force body diagram (as said above by Archduke). It makes it a lot easier. The forces you see in the X or Y need to all equal out to M*A. Since Fx or Fy = M*A, also the sum of forces must equal m*a. So in the picture, you have the force of gravity pulling down and...
6. ### Work example wrong in my text? (it's driving me bonkers!)
Calling the X and Y on the diagonals; there is no acceleration in the Y. http://img87.imageshack.us/img87/6626/problemzl2.th.jpg [Broken]
7. ### Work example wrong in my text? (it's driving me bonkers!)
\ is the Y, / is the x. There is no acceleration in the Y, as this object is fixed on the ramp. If it had acceleration in the Y, it would be moving up or down. However, it is not. Through similar triangles, the {Fy = N = mgcos(theta).
8. ### Work example wrong in my text? (it's driving me bonkers!)
Through similar triangles. {Fx = mgsin(theta) = ma {Fy = N - Wcos(theta) = 0 (no accel in the y); N= Wcos(theta). The theta is 10 degrees.
9. ### Work example wrong in my text? (it's driving me bonkers!)
Horizontal distance from top to bottom; refers to the distance that the object travels on the incline. Rather than the bottom of the triangle. The actual distance it travels is 40m, however, we want the height. So we just do the cos(theta) of it, cos(theta) = adjacent/hypotenuse. adjacent =...
10. ### Work example wrong in my text? (it's driving me bonkers!)
No, you use the inclination of the hill =p. The work book is correct. d*cos(theta) = height. mgh -> (45)(-9.8)*[0-40cos(10)] = 1.7x10^4 J (two sig figs)
11. ### Two Resistance Problems (Circuits)
Doh! Haha, that could be the problem =p. It helps to read carefully :s.
12. ### Two Resistance Problems (Circuits)
Homework Statement 15. (II) A close inspection of an electrical circuit reveals that a 480-Ω resistor was inadvertently soldered in the place where a 320-Ω resistor is needed. How can this be fixed without removing anything from the existing circuit? 16. (II) Two resistors when connected in...
13. ### Forces and Speed
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14. ### Forces and Speed
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15. ### Forces and Speed
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College Algebra Exam Review 346
# College Algebra Exam Review 346 - M Then ± 1.A D A C N D f...
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356 8. MODULES Proof. Exercise 8.2.5 . n Proposition 8.2.9. (Factorization Theorem) Let ' W M ! M be a sur- jective homomorphism of R –modules with kernel K . Let N K be a submodule , and let W M ! M=N denote the quotient map. Then there is a surjective homomorphism Q ' W M=N ! M such that Q ' ı D ' . (See the following diagram.) The kernel of Q ' is K=N M=N . M ' q q M q q q q Q ' M=N Proof. Exercise 8.2.6 . n Proposition 8.2.10. (Diamond Isomorphism Theorem) Let ' W M ! M be a surjective homomorphism of R –modules with kernel N . Let A be a submodule of
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Unformatted text preview: M . Then ' ± 1 .'.A// D A C N D f a C n W a 2 A and n 2 N g : Moreover, A C N is a submodule of M containing N , and .A C N/=N Š '.A/ Š A=.A \ N/: Proof. Exercise 8.2.7 . n Exercises 8.2 R denotes a ring and M an R –module. 8.2.1. Prove Proposition 8.2.1 . 8.2.2. Prove Proposition 8.2.2 . 8.2.3. Let I be an ideal of R . Show that the quotient module M=IM has the structure of an R=I –module....
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https://topbitzlej.web.app/sprow21040riwy/how-many-gallons-of-oil-does-it-take-to-make-a-gallon-of-gas-so.html | 1,639,003,866,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00370.warc.gz | 651,263,955 | 6,836 | ## How many gallons of oil does it take to make a gallon of gas
So, roughly, one gallon of crude oil gives you half a gallon of gasoline. So it takes two gallons of crude oil to produce one gallon of gasoline. The same ratio applies to litres So, roughly, one At currently \$124 per barrel or \$2.95 per gallon of crude oil it would take \$5.95 (\$2.95/gallon x 1.68 gallons of crude/gallon of refined motor fuels) worth of crude to make 1 gallon of refined motor fuels. Gasoline is currently \$3.65/gallon and jet fuel is \$3.40/gallon (\$143.20/42 gallon barrel. All other motor fuels average about \$3.60/gallon. It takes about 2.14 gallons of crude oil to make one gallon of gasoline. Asked in Fuel and Engines, Chemical Engineering, Fossil Fuels How much oil does it take to make a gallon of gas?
## 22 Apr 2014 Sue T asks: Why do we measure oil in barrels instead of gallons or liters? How much oil is in a barrel? oil-pump. More than just gasoline for our cars, as alkylates” are added to the crude to create a “processing gain,” such that: of the container quickly became standardized around the 42-gallon barrel,
At currently \$124 per barrel or \$2.95 per gallon of crude oil it would take \$5.95 (\$2.95/gallon x 1.68 gallons of crude/gallon of refined motor fuels) worth of crude to make 1 gallon of refined motor fuels. Gasoline is currently \$3.65/gallon and jet fuel is \$3.40/gallon (\$143.20/42 gallon barrel. All other motor fuels average about \$3.60/gallon. It takes about 2.14 gallons of crude oil to make one gallon of gasoline. Asked in Fuel and Engines, Chemical Engineering, Fossil Fuels How much oil does it take to make a gallon of gas? The question being answered in the post is: if you decrease (or increase) gasoline consumption by 1 million gallons, how many barrels of oil does that translate to? Well, one barrel of crude produces 19.5 gal of gasoline plus 22.7 gallons of other petroleum products (or 2.2 gallons more than the original 42 gallons. It takes 8 gallons of oil to produce ONE TIRE. Can you imagine all the oil it would take to build one of those enormous Wind Turbines? How many gallons would it take to make a solar panel- when both combined only produce 2% of what we really need. Its BS! The real reason govt is banning Edison bulbs is they are too cheap and GE can't make profits so they will force us to buy those squiggly How many gallons of gasoline does one dinosaur make? —Doug Wescott, Adamstown, MarylandContrary to popular belief, none! Oil was created from organisms that lived long before dinosaurs roamed 1 gal gas to 2.6 oz oil 2 gal gas to 5.1 oz oil 3 gal gas to 7.7 oz oil 4 gal gas to 10.2 oz oil 5 gal gas to 12.8 oz oil 6 gal gas to 15.4 oz oil Most portable out board tanks are 6 gallon tanks. The most common practice is to mix 1 pint (16 oz) to the 6 gallons of fuel that they hold. 1 gal gas to 2.6 oz oil 2 gal gas to 5.1 oz oil 3 gal gas to 7.7 oz oil 4 gal gas to 10.2 oz oil 5 gal gas to 12.8 oz oil 6 gal gas to 15.4 oz oil Most portable out board tanks are 6 gallon tanks. The most common practice is to mix 1 pint (16 oz) to the 6 gallons of fuel that they hold.
### Knowing how to mix oil and gas at a 50 to 1 gas oil ratio is the first step in keeping your A gas to oil mix chart referencig the gallons of gasoline per fluid ounces of oil Storing of gas and oil should never be longer than three months.
15 Sep 2018 Know Gallons & History Averages; How Much Heating Oil Will I Use Anyone with a fuel oil system faces the challenge of making sure breakdown of heating oil prices, showing average dollars per gallon But assuming an average oil tank holds about 275 gallons, here are some steps you can take to You can make ads in the Engineering ToolBox more useful to you! Energy content Fuel Oil no.1, 1 Gallon, 137400 Residual Fuel Oil 1), 1 Barrel - 42 gallons, 6287000 An item using one kilowatt-hour of electricity will generate 3412 Btu. How many gallons does the average gas station's underground gas tank As far as markup goes, station owners only make about 5 cents per gallon or less. many stations are now owned by the oil companies themselves.
### 6 Feb 2020 Which country has the highest gas prices? At 5.95 U.S. dollars per gallon, gas prices in Germany were lower than capita basis, with approximately 1.22 gallons of gas per person daily. Make sure to contact us if you are interested in scientific citation. Statistics on "Oil and gasoline prices - UK Brent".
How many gallons of gasoline and diesel fuel are made from one barrel of oil? Petroleum refineries in the United States produce about 19 to 20 gallons of motor gasoline and 11 to 12 gallons of ultra-low sulfur distillate fuel oil (most of which is sold as diesel fuel and in several states as heating oil) from one 42-gallon barrel of crude oil. So, roughly, one gallon of crude oil gives you half a gallon of gasoline. So it takes two gallons of crude oil to produce one gallon of gasoline. The same ratio applies to litres So, roughly, one At currently \$124 per barrel or \$2.95 per gallon of crude oil it would take \$5.95 (\$2.95/gallon x 1.68 gallons of crude/gallon of refined motor fuels) worth of crude to make 1 gallon of refined motor fuels. Gasoline is currently \$3.65/gallon and jet fuel is \$3.40/gallon (\$143.20/42 gallon barrel. All other motor fuels average about \$3.60/gallon.
## Lucas Oil 10013-PK4 Fuel Treatment - 1 Gallon (Pack of 4) I was told I could get two for just over \$10.00; I said, "how much is one, in a shocked tone". and he
How many gallons of gasoline does one dinosaur make? —Doug Wescott, Adamstown, MarylandContrary to popular belief, none! Oil was created from organisms that lived long before dinosaurs roamed 1 gal gas to 2.6 oz oil 2 gal gas to 5.1 oz oil 3 gal gas to 7.7 oz oil 4 gal gas to 10.2 oz oil 5 gal gas to 12.8 oz oil 6 gal gas to 15.4 oz oil Most portable out board tanks are 6 gallon tanks. The most common practice is to mix 1 pint (16 oz) to the 6 gallons of fuel that they hold. 1 gal gas to 2.6 oz oil 2 gal gas to 5.1 oz oil 3 gal gas to 7.7 oz oil 4 gal gas to 10.2 oz oil 5 gal gas to 12.8 oz oil 6 gal gas to 15.4 oz oil Most portable out board tanks are 6 gallon tanks. The most common practice is to mix 1 pint (16 oz) to the 6 gallons of fuel that they hold.
6 Feb 2020 Which country has the highest gas prices? At 5.95 U.S. dollars per gallon, gas prices in Germany were lower than capita basis, with approximately 1.22 gallons of gas per person daily. Make sure to contact us if you are interested in scientific citation. Statistics on "Oil and gasoline prices - UK Brent". 1 Jan 2010 Out of one barrel of crude oil, 19.5 gallons are used to produce gasoline. Read more. Breakdown of products produced from one barrel of crude | 1,772 | 6,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-49 | latest | en | 0.921093 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-5-section-5-5-factoring-special-forms-exercise-set-page-372/78 | 1,534,515,480,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212323.62/warc/CC-MAIN-20180817123838-20180817143838-00102.warc.gz | 874,201,059 | 13,923 | ## Intermediate Algebra for College Students (7th Edition)
Published by Pearson
# Chapter 5 - Section 5.5 - Factoring Special Forms - Exercise Set: 78
#### Answer
$(x-10)(x^2+10x+100)$
#### Work Step by Step
The given binomial can be written as: $x^3-10^3$ This binomial is a difference of two cubes. RECALL: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Factor the difference of two cubes using the formula above with $a=x$ and $b=10$ to obtain: $=(x-10)(x^2+x(10)+10^2) \\=(x-10)(x^2+10x+100)$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 203 | 644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-34 | longest | en | 0.758709 |
https://cairngorms-leader.org/important-things-to-remember-when-playing-blackjack/ | 1,722,671,634,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00284.warc.gz | 124,080,599 | 15,502 | # Important Things to Remember When Playing Blackjack
A few important topics to remember when playing blackjack include the Basic Strategy, Probability of busting, Insurance, and Deck penetration. In this article, we’ll discuss all of them and more. Hopefully, you’ll be able to apply these strategies to win more games! And if you’re not sure what they all mean, take a look at our Blackjack Cheat Sheet. Then, go ahead and try them out! You might even learn something new!
## Basic strategy
If you’re unfamiliar with basic blackjack strategy, here are some key points to consider. First, it’s important to recognize that you are aiming for the closest possible total to 21. If you have a hand that contains cards of varying denominations, such as a pair of eights, then you may want to stand on the 16 against the dealer. This can greatly increase your chances of winning, but you should consider the risks associated with doubling down.
## Probability of busting
In blackjack, knowing the probability of busting is essential in the game. While the average player may have zero chance of busting, it’s far less likely if they have a hand worth twelve or less. In addition, players with totals of fifteen or more will have a higher chance of busting on anything but an Ace. However, players with 12-15 decks must also take into account the “safe” cards. The math formula used in this article can be used to determine the probability of busting in any number of decks.
## Insurance
Insurance is available on most blackjack games, and many players like to take it. However, seasoned players recommend against it, as it could threaten their bankroll. If you’re looking for ways to reduce the risk of losing your bankroll, read on to learn how to make a smart insurance decision. The best way to find an insurance provider is to read the blackjack rules. Many casinos have online blackjack rules and regulations posted right on their tables. There are a few different ways to get this type of insurance.
## Limiting the deck penetration
One important aspect of card counting is limiting the amount of deck penetration. In blackjack, penetration is the number of cards dealt before a new deck is shuffled. Many casinos instruct dealers to place a cut card at 50% penetration or less, which means that the deck is shuffled after only one hand is played. This is not a very profitable percentage for card counters, and they should look for a higher percentage.
## Double down
A good strategy for winning at blackjack is to double down on some hands, such as when the dealer shows you a weak card. The dealer will most likely receive another 10 in the next round, which means that you have a total of fifteen or sixteen. If you do not win that round, you will have to draw again, or you will go bust. You can also double down when the dealer has a five or six, because that is the only way to win a game.
## Hard hands
When forming your strategy, knowing about Blackjack’s hard hands is essential. A combination of two cards with a different value is known as a hard hand. For example, a nine and four combination is called a hard thirteen. A pair of sevens is a hard fourteen. Both of these hands have low value, but are easy to improve and carry low risk. Here are some strategies for these hands. To improve them, you should understand how to play these hands correctly. | 704 | 3,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.959728 |
https://search.r-project.org/CRAN/refmans/Compositional/html/james.html | 1,638,923,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00314.warc.gz | 549,406,352 | 2,319 | James multivariate version of the t-test {Compositional} R Documentation
## James multivariate version of the t-test
### Description
James test for testing the equality of two population mean vectors without assuming equality of the covariance matrices.
### Usage
```james(y1, y2, a = 0.05, R = 999, graph = FALSE)
```
### Arguments
`y1` A matrix containing the Euclidean data of the first group. `y2` A matrix containing the Euclidean data of the second group. `a` The significance level, set to 0.05 by default. `R` If R is 1 no bootstrap calibration is performed and the classical p-value via the F distribution is returned. If R is greater than 1, the bootstrap p-value is returned. `graph` A boolean variable which is taken into consideration only when bootstrap calibration is performed. IF TRUE the histogram of the bootstrap test statistic values is plotted.
### Details
Multivariate analysis of variance without assuming equality of the covariance matrices. The p-value can be calculated either asymptotically or via bootstrap. The James test (1954) or a modification proposed by Krishnamoorthy and Yanping (2006) is implemented. The James test uses a corrected chi-square distribution, whereas the modified version uses an F distribution.
### Value
A list including:
`note` A message informing the user about the test used. `mesoi` The two mean vectors. `info` The test statistic, the p-value, the correction factor and the corrected critical value of the chi-square distribution if the James test has been used or, the test statistic, the p-value, the critical value and the degrees of freedom (numerator and denominator) of the F distribution if the modified James test has been used. `pvalue` The bootstrap p-value if bootstrap is employed. `runtime` The runtime of the bootstrap calibration.
### Author(s)
Michail Tsagris.
R implementation and documentation: Michail Tsagris mtsagris@uoc.gr and Giorgos Athineou <gioathineou@gmail.com>.
### References
G.S. James (1954). Tests of Linear Hypothese in Univariate and Multivariate Analysis when the Ratios of the Population Variances are Unknown. Biometrika, 41(1/2): 19-43.
Krishnamoorthy K. and Yanping Xia. On Selecting Tests for Equality of Two Normal Mean Vectors (2006). Multivariate Behavioral Research 41(4): 533-548.
```hotel2T2, maovjames, el.test2, eel.test2, comp.test ```
```james( as.matrix(iris[1:25, 1:4]), as.matrix(iris[26:50, 1:4]), R = 1 ) | 593 | 2,441 | {"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-2021-49 | latest | en | 0.787241 |
http://scanraid.com/Nishio_Forcing_Chains | 1,670,158,510,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00606.warc.gz | 44,153,413 | 6,490 | Print Version Page Index
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# Nishio Forcing Chains
Nishio has a long history in the community as a semi 'trial and error' approach to very hard Sudoku bottlenecks, but in the implementation on this website it fits very nicely into the family of forcing chains and is a formal 'pattern' based strategy that uses AICs. In fact it is very close to Digit Forcing Chains since Nishio works on a single candidate as well. Where Digit Forcing Chains take a candidate and consider the consequences of the candidate being both ON and OFF, Nishio tries to find a contradiction when the candidate is merely ON. Two chains emanate from the candidate in different directions and try to join up later on another candidate. The rule is, if the start candidate is ON and this leads to another candidate being both ON and OFF it is indicating an impossible state of affairs. So the original candidate cannot be a solution.
A Nishio forcing chain is sometimes the exact reverse of a Digit Forcing Chain. If you look at the first diagram on that page you'll notice the first 'type' in the set of four ways a contradiction can be found is a destination candidiate that is ON when the start candidate is both ON and OFF.
In this example the starting candidate is 6 in J4. The shorter blue chain simply says that when 6 in J4 is ON it removes the 6s in box 7 leaving 6 in G2 as the only remaining 6 and therefore the solution. On the other hand, the longer purple chain implies 6 cannot be the solution in G2. A 6 in J4 means 6s are removed from the rest of box 8 and that turns ON the six in H9. That forces 2 in J9 which in turn removes 2 from J1 making G2 2 and not 6.
Therefore the 6 in J4 cannot be a solution and it can be removed.
Go back to Digit Forcing Chains Continue to Cell Forcing Chains
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## ... by: Leren
This Nishio chain is also a Discontinuous Nice Loop type 3 - but for some reason your solver bypasses this AIC strategy. The solver's choice of Nishio chains also seems somewhat arbitrary - you could just as easily transfer the required contradiction to any of the cells in the loop. Also the graphic is a bit confusing - the 6 in G2 should be coloured green to highlight the contradiction in that cell (2 candidates both coloured green). Are there cases where Nishio chains are not also Nice loops?
Andrew Stuart writes:
Correct on the connection with Discontinuous Nice Loop type 3. The green cell highlight I think is a carry over from previous strategies where I mark the end of the chain. Probably not necessary. The final candidate 6 in G2 is yellow as that follows the graphical convention of showing eliminations in yellow/red text. It would be green if it was part of the chain.
The solver returns the first instance of the strategy used but currently cant cycle through all instances of that strategy so I'm sure other formations of the same chain are possible as it's also a loop. I'd like to allow that in the solver and it's an upgrade I have in the job queue. There is a limit on the length of normal single chain AICs which I have to impose to stop the program taking too long to return (mainly for the grading feature which has to run through all the strategies). That means that some AICs might not be found and instead two chains in the double chain strategies like Nishio get to find the same result as each chain is under the limit. I've had several goes at optimising the AIC searchs but not to any great improvement in speed.
I haven't broadened Nishio to non loop type entities because the narrow definition I've worked with has start and end points on single candidates. But I'll be the first to admit there is overlap with other chaining strategies.
Article created on 16-July-2012. Views: 73058 | 954 | 4,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-49 | latest | en | 0.920269 |
https://sandialabs.github.io/Prove-It/packages/proveit/linear_algebra/__pv_it/axioms/d23e0d0ed950b5933d0b44fe5d690f1aac82aacc0/expr.html | 1,708,634,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00864.warc.gz | 522,293,185 | 5,875 | # from the theory of proveit.linear_algebra¶
In [1]:
import proveit
# Automation is not needed when building an expression:
proveit.defaults.automation = False # This will speed things up.
proveit.defaults.inline_pngs = False # Makes files smaller.
# import Expression classes needed to build the expression
from proveit import Function, S, T, V, W, v
from proveit.logic import Equals, Forall
In [2]:
# build up the expression from sub-expressions
sub_expr1 = [v]
expr = Forall(instance_param_or_params = [V, W], instance_expr = Forall(instance_param_or_params = [S, T], instance_expr = Forall(instance_param_or_params = sub_expr1, instance_expr = Equals(Function(LinMapAdd(S, T), sub_expr1), Add(Function(S, sub_expr1), Function(T, sub_expr1))), domain = V), domain = LinMap(V, W)))
expr:
In [3]:
# check that the built expression is the same as the stored expression
assert expr == stored_expr
assert expr._style_id == stored_expr._style_id
print("Passed sanity check: expr matches stored_expr")
Passed sanity check: expr matches stored_expr
In [4]:
# Show the LaTeX representation of the expression for convenience if you need it.
print(stored_expr.latex())
\forall_{V, W}~\left[\forall_{S, T \in \mathcal{L}\left(V, W\right)}~\left[\forall_{v \in V}~\left(\left(S + T\right)\left(v\right) = \left(S\left(v\right) + T\left(v\right)\right)\right)\right]\right]
In [5]:
stored_expr.style_options()
namedescriptiondefaultcurrent valuerelated methods
with_wrappingIf 'True', wrap the Expression after the parametersNoneNone/False('with_wrapping',)
condition_wrappingWrap 'before' or 'after' the condition (or None).NoneNone/False('with_wrap_after_condition', 'with_wrap_before_condition')
wrap_paramsIf 'True', wraps every two parameters AND wraps the Expression after the parametersNoneNone/False('with_params',)
justificationjustify to the 'left', 'center', or 'right' in the array cellscentercenter('with_justification',)
In [6]:
# display the expression information
stored_expr.expr_info()
core typesub-expressionsexpression
0Operationoperator: 9
operand: 2
1ExprTuple2
2Lambdaparameters: 27
body: 3
3Operationoperator: 9
operand: 5
4ExprTuple5
5Lambdaparameters: 36
body: 6
6Conditionalvalue: 7
condition: 8
7Operationoperator: 9
operand: 13
8Operationoperator: 11
operands: 12
9Literal
10ExprTuple13
11Literal
12ExprTuple14, 15
13Lambdaparameter: 42
body: 16
14Operationoperator: 24
operands: 17
15Operationoperator: 24
operands: 18
16Conditionalvalue: 19
condition: 20
17ExprTuple39, 21
18ExprTuple40, 21
19Operationoperator: 22
operands: 23
20Operationoperator: 24
operands: 25
21Operationoperator: 26
operands: 27
22Literal
23ExprTuple28, 29
24Literal
25ExprTuple42, 30
26Literal
27ExprTuple30, 31
28Operationoperator: 32
operand: 42
29Operationoperator: 33
operands: 34
30Variable
31Variable
32Operationoperator: 35
operands: 36
33Literal
34ExprTuple37, 38
35Literal
36ExprTuple39, 40
37Operationoperator: 39
operand: 42
38Operationoperator: 40
operand: 42
39Variable
40Variable
41ExprTuple42
42Variable | 880 | 3,021 | {"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} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.458391 |
https://discussions.unity.com/t/change-position-of-player-to-create-a-teleport-effect/192100 | 1,716,600,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00584.warc.gz | 173,392,959 | 7,178 | # Change position of player to create a teleport effect.
So I am trying to make a 2D game where the player has a teleport ability.
I thought that getting the players coordinates and then adding an x value to them is a good way to simulate teleporting.
My problem is that I don’t know how to get the players coordinates and then change them in the direction of moving.
Here is a drawing where I am trying to explain it better.
First, you’ll need to figure out the direction of the player’s movement.
If they have a rigidbody attatched, you can get a Vector3 of their velocity with Rigidbody.velocity.
If they don’t, you could find their velocity with this:
``````private Vector3 velocity;
private Vector3 previous;
void Update () {
velocity = ((transform.position - previous).magnitude) / Time.deltaTime;
previous = transform.position;
}
``````
Source: Statement and Celestium from this thread.
You could also just get their direction of movement from the same place you actually move them. For example, if you’re moving the player via keyboard input, just store the Vector3 from the keyboard before you apply it to the player, like this:
``````Vector3 movementDirection = new Vector3(Input.GetAxis("Horizontal"), 0f, Input.GetAxis("Vertical"));
``````
This is probably the best way to do it, as it won’t be affected by the player moving due to knockback, etc.
Then you could teleport the player like this:
``````public Vector3 teleportDistance = 5f;
public void Teleport () {
// We normalize the velocity vector so it has a length of 1,
// We then multiply it by how far we want to teleport.
transform.position += velocity.normalized * teleportDistance;
}
``````
I’m going to break the problem down into 2 parts.
Part 1: We need to determine which direct the player is moving
Part 2: We need to teleport the player in that direction
PART 1
The easiest way to determine player direction would be to use Unity’s built in input controls for Horizontal and Vertical movement. Simple recap on how this works: Unity tracks input using either WASD or Arrow Keys; Horizontal input gives you Left/Right or A/D, Vertical Input gives you Up/Down or W/S keys. Unity represents this input from the user with either a value of -1 or 1 respectively. So if we want to read this input from the engine it would look like this:
``````if (Input.GetAxis("Horizontal") < 0) {
//Move the player left
}
if (Input.GetAxis("Horizontal") > 0) {
//Move player right
}
``````
Now we know what direction the player is moving and we have movement controls set up all in one place.
Part 2
Now that we have the movement set up and we always know which way the player is moving (left or right) we can have them teleport on command. I’ll be using the Space Bar as my trigger for activating a teleport in the following code.
``````float teleportDistance;
if (Input.GetAxis("Horizontal") < 0 && Input.GetKeyDown(KeyCode.Space)) {
transform.position = new Vector2(transform.position.x - teleportDistance, transform.position.y);
}
if (Input.GetAxis("Horizontal") > 0 && Input.GetKeyDown(KeyCode.Space)) {
transform.position = new Vector2(transform.position.x + teleportDistance, transform.position.y);
}
``````
What this does is it determines if we are moving either left or right and if we have pressed the Space Bar. If both of these conditions are true then we set the player’s position to their current position + whatever distance we want to move them.
EDIT:
I should also note that with this code I am assuming that you have a 2D side perspective that allows the player to move left and right similar to Mario games. If you have a top down perspective game let me know and I can help you with the modified code.
Also I am using C# for the example code. | 836 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.917767 |
https://physics.icalculator.com/coefficient-of-finesse-calculator.html | 1,726,051,113,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00215.warc.gz | 422,597,960 | 6,196 | # Coefficient of Finesse Calculator
In the field of Physics, the coefficient of finesse is a parameter used to characterize the performance of optical resonators, such as Fabry-Perot interferometers and optical cavities. It quantifies the ability of a resonator to transmit and reflect light. This tutorial provides an overview of the coefficient of finesse, the associated calculations and formulas, real-life applications, key individuals in the discipline, and interesting facts. Understanding the coefficient of finesse is crucial for the analysis and design of optical resonant systems and plays a significant role in the field of optics and photonics.
🖹 Normal View 🗖 Full Page View Interface Power Reflectance
Coefficient Of Finesse =
## Example Formula
The coefficient of finesse (F) is calculated using the following formula:
F = π√R / (1 - R)
Where:
1. F: Coefficient of finesse
2. R: Reflectivity of the mirrors or surfaces in the optical resonator
## Who Wrote/Refined the Formula
The concept of the coefficient of finesse and its associated formula have been refined and developed by numerous physicists and researchers in the field of optics and photonics. While no single individual can be attributed to the creation of this specific formula, it is a result of the collective efforts of scientists who have contributed to the understanding and characterization of optical resonators.
## Real Life Application
The coefficient of finesse finds various real-life applications in the field of optics. One example is in the design and optimization of Fabry-Perot interferometers used in spectroscopy, where it determines the resolution and spectral selectivity of the instrument. It is also essential in laser cavities and optical resonators, where the coefficient of finesse affects the laser's output characteristics and the behavior of resonant optical systems.
## Key Individuals in the Discipline
Several individuals have made significant contributions to the field of optics and the study of optical resonators. Notable figures include Charles Fabry and Alfred Perot, who developed the Fabry-Perot interferometer in the late 19th century. Their work laid the foundation for the understanding and practical application of optical resonators, which are characterized by parameters such as the coefficient of finesse.
## Interesting Facts
1. The coefficient of finesse is related to the finesse parameter, which quantifies the spectral resolution and selectivity of an optical resonator.
2. The higher the coefficient of finesse, the narrower the resonant modes and spectral features of the optical resonator.
3. Coherent light sources, such as lasers, can exhibit high finesse when coupled to optical resonators with high reflectivity.
## Conclusion
The coefficient of finesse is a fundamental parameter used to characterize the performance of optical resonators. It plays a crucial role in the design and optimization of optical systems, such as Fabry-Perot interferometers and laser cavities. Understanding the calculations and formulas associated with the coefficient of finesse allows researchers and engineers to analyze and engineer resonant systems with desired spectral characteristics. The field of optics and photonics continues to advance through the exploration and application of the coefficient of finesse, contributing to various technologies and scientific discoveries.
## Physics Calculators
You may also find the following Physics calculators useful. | 672 | 3,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-38 | latest | en | 0.882604 |
https://epxx.co/artigos/qam_new.html | 1,544,756,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00597.warc.gz | 575,429,577 | 3,846 | New version of QAM modem
## New version of QAM modem
As I promised, I rewrote parts of Python QAM modem to use complex numbers, which leaves code much shorter and clearer (provided that you understand complex numbers).
Also, I put the common code in a separate file, to avoid redundancy in TX and RX parts. The FIR lowpass filter was improved too.
The files are m2co.py, m2tx.py and m2rx.py. All of them can be found at https://epxx.co/artigos/modulation.
I think I went far enough with this. I was planning to write a "version 3", but I am not sure I will. Versions 1 and 2 already make good classroom examples. FWIW I will tell what's in my mind for v3, maybe someone else picks it up.
First, the "classical" demodulation approach showed its limits in version 2; baud rate can not go above 1000 or 1100 Hz for a carrier of 1800 Hz. The FIR filter has a trade-off between frequency precision and time-domain precision.
Increasing baud rate demands increasing frequency precision, but then symbols begin to smear together after filtering. In order to reach V.29 or V.32 baud rate (which is almost the same as carrier frequency), some other detection approach must be employed.
I played with some formulas, remembered some trigonometry identities at Wikipedia, and came up with one idea.
Since the derivative of QAM signal is well-defined by a simple formula, and there are two unknown variables that shape the QAM signal — amplitude and phase — it is possible to "invert" the formula and find amplitude and phase given two derivatives taken from time-domain signal.
The signal derivative is at hand; it is just the difference between two consecutive samples. If we measure two derivatives 90 degrees apart from each other, formulas can be further simplified.
This approach was tested on spreadsheet, and should work in a Python implementation. It seems possible to determine symbol in half carrier wavelength at the most, normally a quarter wavelength is enough, which means that decoding is possible when baud rate approaches or surpasses carrier frequency.
Another advantage of this technique is more resilience to small carrier deviations. They would appear just as a very small but persistent phase delay in every sample, which can be ignored or compensated for.
I have no idea if this is the technique that advanced modems actually use, but I bet it is at least some variation of this basic approach. | 513 | 2,418 | {"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-2018-51 | longest | en | 0.953724 |
https://www.tutorialspoint.com/check-if-a-number-can-be-expressed-as-a-b-in-python | 1,685,968,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00566.warc.gz | 1,123,600,811 | 10,257 | # Check if a number can be expressed as a^b in Python
Suppose we have a number n. We have to check whether we can make express it like a^b or not.
So, if the input is like 125, then the output will be True as 125 = 5^3, so a = 5 and b = 3
To solve this, we will follow these steps −
• if num is same as 1, then:
• return true
• for initialize i := 2, when i * i <= num, update (increase i by 1), do:
• val := log(num) / log(i)
• if val - integer part of val is nearly 0, then:
• return true
• return false
Let us see the following implementation to get better understanding −
## Example
Live Demo
#include<iostream>
#include<cmath>
using namespace std;
bool solve(int num) {
if (num == 1)
return true;
for (int i = 2; i * i <= num; i++) {
double val = log(num) / log(i);
if ((val - (int)val) < 0.00000001)
return true;
}
return false;
}
int main() {
int n = 125;
cout << solve(n);
}
## Input
125
## Output
1 | 283 | 923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-23 | latest | en | 0.576582 |
https://socratic.org/questions/consider-a-solution-that-contains-80-0-r-isomer-and-20-0-s-isomer-if-the-observe | 1,558,583,683,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257002.33/warc/CC-MAIN-20190523023545-20190523045545-00434.warc.gz | 624,359,665 | 6,119 | # Consider a solution that contains 80.0% R isomer and 20.0% S isomer. If the observed specific rotation of the mixture is –43.0°, what is the specific rotation of the pure R isomer?
Feb 1, 2016
The specific rotation of the $R$ isomer is -67 °.
#### Explanation:
$R$ is in excess, so it must have a negative specific rotation; $S$ has an equal positive specific rotation.
The formula for the enantiomeric excess ($e e$) of $R$ is
ee = %R - %S = 80 % - 20 % = 60 %
Another equation for $e e$ is
ee = "observed specific rotation"/"maximum specific rotation" × 100 %
Inserting numbers, we get
60 color(red)(cancel(color(black)(%))) = "-40.3 °"/"maximum specific rotation" × 100 color(red)(cancel(color(black)(%)))
$\text{maximum specific rotation" = [α]_"D" = "-40.3 °" × 100/60 = "-67 °}$
This is the value of [α]_"D" for the $R$ isomer. | 263 | 847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-22 | latest | en | 0.790642 |
https://www.airmilescalculator.com/distance/lvo-to-xtg/ | 1,603,656,175,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00067.warc.gz | 605,817,177 | 45,254 | # Distance between Laverton (LVO) and Thargomindah (XTG)
Flight distance from Laverton to Thargomindah (Laverton Airport – Thargomindah Airport) is 1303 miles / 2096 kilometers / 1132 nautical miles. Estimated flight time is 2 hours 57 minutes.
Driving distance from Laverton (LVO) to Thargomindah (XTG) is 2234 miles / 3596 kilometers and travel time by car is about 40 hours 23 minutes.
## Map of flight path and driving directions from Laverton to Thargomindah.
Shortest flight path between Laverton Airport (LVO) and Thargomindah Airport (XTG).
## How far is Thargomindah from Laverton?
There are several ways to calculate distances between Laverton and Thargomindah. Here are two common methods:
Vincenty's formula (applied above)
• 1302.504 miles
• 2096.177 kilometers
• 1131.845 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1300.077 miles
• 2092.271 kilometers
• 1129.736 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Laverton Airport
City: Laverton
Country: Australia
IATA Code: LVO
ICAO Code: YLTN
Coordinates: 28°36′48″S, 122°25′26″E
B Thargomindah Airport
City: Thargomindah
Country: Australia
IATA Code: XTG
ICAO Code: YTGM
Coordinates: 27°59′11″S, 143°48′39″E
## Time difference and current local times
The time difference between Laverton and Thargomindah is 2 hours. Thargomindah is 2 hours ahead of Laverton.
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## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 167 kg (368 pounds).
## Frequent Flyer Miles Calculator
Laverton (LVO) → Thargomindah (XTG).
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Total frequent flyer miles:
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Round trip? | 544 | 1,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | latest | en | 0.816251 |
https://rezahajargasht.wordpress.com/page/2/ | 1,521,760,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00793.warc.gz | 716,629,289 | 15,345 | Red Back in our backyard
We were doing this experiment with my daughter but if you look closely, you can see something is resisting not to come out of the tap with the water.
It turned out to be a redback spider which I hadn’t seen before !
Lorenz Curve Slides
Here are the slides of the “Lorenz Curve” paper I presented on Friday in UWA. I will give link to the full paper soon (some slight edits on the paper is still going on).
Income Density Estimation
All measurements of poverty, inequality and so forth rely in some ways on estimation of a representation of an income (or consumption) distribution. The most commonly used representation is the PDF therefore in this post I review common ways of density estimation. The excellent handbook chapter of Cowell and Flachaire (2015) covers this topic in some details (see also Lubrano 2013). My take here is instead brief, informal, and based on personal experiences. Continue reading
A Prize for “Economic Measurement”
“Angus Deaton” has won this year’s “Nobel” prize in economics which is welcomed by this blog. He is one of leading figures in the area of “Economic Measurement”. He has made significant contributions to better measurement (and understanding) of consumption, demand functions, saving, poverty, happiness, purchasing power parities and more. We will be discussing some of these contributions in this blog in the future.
Lorenz Curves 2: Definition and Specification
Inference for Lorenz curves requires its rigorous definition and good specifications. This post briefly discusses alternative definitions and ways of specifying Lorenz curves: Continue reading
Hardy-Littlewood Collaboration
One of the most successful collaborations in the history of science has been between two great mathematicians Hardy and Littlewood. They had set the following rules Continue reading
Folded Normal Stochastic Frontier Model
What is the distribution of absolute value of a normally distributed random variable? Continue reading | 416 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-13 | longest | en | 0.914277 |
https://hiddenperspectives.org/qa/question-how-much-does-it-cost-to-have-a-home-professionally-cleaned.html | 1,606,569,841,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00298.warc.gz | 321,314,146 | 9,255 | # Question: How Much Does It Cost To Have A Home Professionally Cleaned?
## How do you quote a house cleaning job?
To give a customer an estimate, add the times for the areas given by the customer, divide those times by 60 to get how long it will take to clean and multiply that number by your hourly rate.
For example, 200/60 = 3.33 x \$30 = \$100.
The number you get will be the amount you charge your customer..
If you hire someone directly for a one time or infrequent cleaning, most etiqueete professionals say you should tip. \$10 to \$20 per cleaning is a good ballpark range. If you have a home cleaner year round, it is probably not necessary to tip each time. Instead, you can give a cash bonus or gift at the end of the year.
## How often should a house cleaner come?
If it wasn’t hard enough keeping on top of the general housekeeping chores, then you may not want to know how often you should be deep cleaning the rest of your home….How Often Should You Deep Clean Your Home?What to CleanHow often You Need to CleanCarpetsEvery WeekSkirting BoardsEvery QuarterWindows and DrapesEvery Three to Six months7 more rows
## How much does it cost to clean a 1500 sq ft house?
Deep Cleaning Costs by SizeSizeTime to cleanAverage cost<1,000 sq.ft.1.5 - 3 hours\$150 - \$1751,000 - 1,200 sq.ft.2 - 3 hours\$175 - \$1951,500 sq.ft.2.5 - 3.5 hours\$195 - \$2151,700 sq.ft.3 - 4 hours\$225 - \$2755 more rows•Mar 21, 2020
## Are house cleaners worth it?
If your time is worth \$50 per hour and the maid charges \$25 per hour, you may conclude that your time is better spent elsewhere. However, if you decide your time is worth \$15 per hour and the maid charges \$25 per hour, you probably should clean your home yourself. It’s simply more cost-effective.
## What is a reasonable price for house cleaning?
House cleaning services cost around \$30/hr. Depending on things like the size of your home and the extent of service required, the rate can be as low as \$20/hr and increase up to about \$35/hour or even up to \$50/hr for 2 person teams.
## What is the difference between standard cleaning and deep cleaning?
A regular clean is designed to help maintain a certain level of cleanliness around your house. Cleaning the bathrooms – toilet, bath, mirror, sink, etc. … A deep cleaning service will remove the deep dirt and grime in your home. It will cover the areas that aren’t typically covered in a regular cleaning service.
## What is a professional house cleaning checklist?
Dust cabinets, door panels, and baseboards. Clean and disinfect surfaces. Spot clean cabinet fronts. Clean, disinfect, and shine showers and tubs. Clean and disinfect toilets inside and out.
## How long does it take to clean a 3 bed house?
Weekly house cleaning (average: 3–5hrs) We know every house is different, but as a general rule of thumb: each bedroom you have will mean an extra hour of cleaning. For example a 2 bedroom home will require 2hrs of domestic cleaning, a 3 bedroom home will need 3 hrs and so on.
## What can a cleaner do in 2 hours?
What can be accomplished in 2 hours?Vacuuming the entire house.Cleaning the bathrooms, including toilets.Cleaning the kitchen, including quickly mopping the floor.A few assorted small tasks like wiping surfaces down.
## How many hours do I need a cleaner for?
A. For a family of 4-5 living in a typical modern four bedroom home with 2 bathrooms, Recommended weekly cleaning would typically be between 6-12 hours per week. This would not include extra services like laundry.
## How do you prepare for a cleaning lady?
7 Things to Do Before a House Cleaner ArrivesPick up household items that may be lying about. … Pick up any important documents, bills and other papers. … Put away the pets. … Identify and repair any broken items around the home. … Identify problem areas that need special cleaning attention. … Make sure they can get in.
## How do professionals clean a house?
When you have more cleaners in the home each person is responsible for different tasks.Empty all trash and replace trash bags.Pick up/straighten/make beds if needed.Remove cobwebs, dust baseboards.Dust ceiling fans.Clean window sills and wipe down doors.Dust all furniture including bottoms and sides.More items…
Generally, it is okay and within limits, to ask the following:Polishing silver or brass.Cleaning of internal windows.Washing and special care cleaning of ornaments.To fold washing and put it away.Help you de-clutter.Tidy up an area.
## What is the fastest way to clean a whole house?
Fast House Cleaning TipsClean the whole house, not one room at time. … Gather all your cleaning tools in a caddy. … Clear the clutter. … Dust and vacuum. … Wipe mirrors and glass. … Disinfect countertops and surface areas. … Focus on tubs, sinks and toilets. … Sweep, then mop.More items…•
## How do you price a cleaning job?
The average cost of house cleaning is \$90 to \$150 and the average national hourly rate is \$25 to \$90 per cleaner. A single family home should cost \$120 to \$150 to clean, according to Home Advisor. Estimating a house cleaning job isn’t as simple as quoting whatever your competitors are charging.
## What should a house cleaner clean?
Cleaning services offeredThorough cleaning of kitchens and bathrooms.Vacuuming throughout the house.Mopping.Making beds.Dusting all surfaces, including baseboards and light fixtures.
## What is included in a deep house cleaning?
We wipe clean all surfaces in the home, including tables, countertops, appliances, chairs, dressers, window sills, sink basins and faucets. We spend extra time in the bathroom, cleaning the bathtub, shower, mirrors and toilet. Light fixtures, picture frames and baseboards – we dust them all.
## Is my house too dirty for a cleaning service?
If life and work have overtaken your time and your home is suffering, consider hiring a cleaning service. Even if you believe you home is too messy for a cleaning service, rest assured the service has probably seen worse. Your home is never too dirty for a good professional residential cleaning service.
## How long should it take to clean a house?
1 1/2 Hours a living space this size warrants this amount of time to be clean by one cleaner if only its been kept up with on a weekly basis. If this was a space that wasn’t maintained, based on our experience will take 2 1\2 hours or more.
## When cleaning a house where do you start?
Our Ultimate Cleaning GuideStep 1: Dust Your House. … Step 2: Clean Furniture Fabric. … Step 3: Clean Mirrors and Glass. … Step 4: Clean Surfaces. … Step 5: Clean the Kitchen and Bathroom. … Step 6: Clean Floors. … Step 7: Vacuum the House. | 1,529 | 6,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-50 | latest | en | 0.909035 |
https://mathematica.stackexchange.com/questions/94163/help-needed-for-understanding-ndsolve-architecture-vacuum-heating-simulation | 1,709,271,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00382.warc.gz | 379,599,201 | 44,320 | # Help needed for understanding NDSolve architecture -Vacuum Heating Simulation
I am a beginner in mathematica and I am facing a problem that is way bigger than my knowledge..
I am tryng to simulate a semplificated version of Brunel Effect, wich is a Laser-Plasma interaction effect, with an electrostatic Dawson's model.
The theory is simple: the plasma, with a sharp-step density shape, is hit by a sinusoidal laser impulse, some electrons will be pulled outside of the plasma and reach the vacuum region in which they are accelerated and then they re-enter into the plasma with more energy than before.
Mathematically the problem is also quite simple: the differential equations wich governs the electrons motion are different inside and outside the plasma, but for the consistency of the fluid model the electrons trajectories should never overlap and for avoiding this i have to "switch" the label of the trajectories when overlap occur. In a few words, i have to insert some checks while integrating the equations that recognize when an overlap occur and then restart the integration with the velocities swapped from that point, i.e. i have to change the initial conditions of the 2 electrons and then continue the integration till the next overlap.
The problem is that check should act inside the NDSolve procedure and while it is doing the integration.
Here the idea of the code i had been able to develop till now: Initialitation of the Field:
ClearAll;
W = 10^6;
tau = 10*Pi/W;
c = 299792458;
Am = 0.8*W*c;
Ed[t_]:=If[t <((2Pi)/W),Sin[Pi*t/(2*tau)]*Sin[Pi*t/(2*tau)]*10*Am*Sin[W*t], Am*Sin[W*t]];
Plot[Ed[t], {t, 0, (2 Pi/W)*2}]
Define some constants and functions:
t =.
it = 0;
imax = 100;
check = 0;
check2 = 0;
aux = 0;
x0max = 4;
tmax = (2*Pi/W)*7;
V[x0_,t_]:=D[Z[x0,t],t];
x[x0_,t_]:=x0+Z[x0,t];
Initialitation of the eq.:
ndsdata = First@NDSolveProcessEquations[{D[V[x0, t], t] ==
If[x0 + Z[x0, t] > 0, -10*W^2*Z[x0, t] - Ed[t],
10*W^2*x0 - Ed[t]], V[x0, 0] == 0, Z[x0, 0] == 0}, Z[x0, t], {x0, 0, x0max},
{t, 0, tmax}, Method -> "ExplicitRungeKutta", MaxSteps -> Infinity]
Where Ed[t] is a Sinusoidal field, W is the pulse frequency, x0 the initial position of the electron, Z[x0,t] the electron displacement (the position of the electron at time t is x=x0+Z[x0,t] and V[x0,t] is the time derivate of Z[x0,t], which is the velocity of the electron
Iteration of the Solver:
For[i = 0, i < imax, i++, {it = it + 1/100000;
NDSolveIterate[ndsdata, it];
auxsol = NDSolveProcessSolutions[ndsdata];
For[j = 0, j < x0max, j++,
If[Evaluate[x[j, it] /. auxsol] >
Evaluate[x[j + 1, it] /. auxsol],
ndsdata = NDSolveReinitialize[ndsdata, {V[j, it] == V[j + 1, it], Z[j, it] ==
Z[j + 1, it]}]];If[it < ((2 Pi/W))*7, imax = imax + 1, i = imax]]
And the process the solution and Plot:
sol = NDSolveProcessSolutions[ndsdata];
this code does not work, but if i try to integrate without the swap i am able to plot the (wrong) trajectories...
the code gives me no error, but it seems that the following If cycle neither became true nor false:
If[Evaluate[x[j, it] /. auxsol]>Evaluate[x[j + 1, it] /. auxsol],
{check = 1, V[j, it] == aux,V[j, it] = V[j + 1, it], V[j + 1, it] = aux,ndsdata =NDSolveReinitialize[
ndsdata, {V[j, it] == V[j + 1, it], Z[j, it] == Z[j + 1, it]},
check2 = 1]}]
So basically the swap (or what i have tried to do as a swap) never happens.
What i obtain are simply the solution of the integration, that are physically uncorrect:
Plot[Table[Evaluate[x[x0, t] /. sol], {x0, 0, x0max}],{t,0,(2*Pi/W)*7},
PlotRange -> {{0, ((2*Pi/W))*7}, {-700, 300}}]
What i am asking is: How can work inside Ndsolve procedure and change initial conditions of the equations at a certain time, when an event (in this case the overlap) occurs?
Thank you in advance and sorry for the expositive mess..
• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Sep 7, 2015 at 17:04
• Have a look at the WhenEvent documentation. Sep 7, 2015 at 18:14
• Your code is missing several constants, which the reader will need to attempt to reproduce and solve your problem. Also, what is the equation for Z[x0, t]? Finally, it what sense doe your code not work? If you are receiving error messages, what is the first such message? Sep 7, 2015 at 19:46
• Hi, and thank you for the answer, i'll try to explain: I have reported only the code part concerned with the NDSolve Procedure, before I only Initialize the field Ed[t], some constans and then I define x[x0,t]=x0+Z[x0,t] as the spatial variation of the electron with initial position x0 and V[x0,t] as the partial derivative respect to time of Z[x0,t], which is the velocity. Z[x0,t] is the function I want to obtain with NDSolve, the differential equation which descrive its evolution is the one inside NDSolve: D[V[x0, t], t] ==If[x0 + Z[x0, t] > 0, -10*W^2*Z[x0, t] - Ed[t],10*W^2*x0-Ed[t] Sep 7, 2015 at 20:51
• Without the constants, expression for Ed[t], and equation for Z, it is unlikely that you will receive useful help. x[x0,t]=x0+Z[x0,t] does not define Z, because x also is undefined. Please add all this to your question. By the way, use @bbgodfrey in any comments you wish me to see promptly. I should add that you may be making this problem too hard for yourself. Sep 7, 2015 at 21:15
Solution with ParametricNDSolve
A fairly straightforward solution to this question can be obtained by observing that the underlying differential equation involves no derivatives with respect to x0, which therefore can be treated as a parameter. (Note that "StiffnessSwitching" should be used instead of "ExplicitRungeKutta".)
sol2 = ParametricNDSolve[{D[V[t], t] ==
If[x0 + Z[t] > 0, -10*W^2*Z[t] - Ed[t], 10*W^2*x0 - Ed[t]], V[t] == D[Z[t], t],
V[0] == 0, Z[0] == 0}, {Z, V}, {t, 0, tmax}, {x0}, Method -> "StiffnessSwitching"];
Plot[Evaluate[Table[x0 + Z[x0][t] /. sol2, {x0, 0, x0max}]],
{t, 0, tmax}, AxesLabel -> {"t", "x"}]
The question, however, requires that the lines not cross and proposes accomplishing this by swapping the values of the lines when they do. Here, the lines are sorted by
f = Table[x0 + Z[x0][t] /. sol2, {x0, 0, x0max}, {t, 0, tmax, tmax/100}];
g = Map[Sort, f // Transpose, {1}] // Transpose;
h[x0_] := Interpolation[{Table[t, {t, 0, tmax, tmax/100}], g[[x0 + 1, All]]}
//Transpose]
Plot[Evaluate[Table[h[x0][t], {x0, 0, x0max}]], {t, 0, tmax},
AxesLabel -> {"t", "x"}]
Higher resolution in x0 can be achieved in a straightforward manner, if desired.
Solution with NDSolve Components
The question proposed using NDSolve Components to obtain a solution. Doing so is not at all straightforward, in my view, but is interesting. One approach is
res = {};
ndsdata = First@NDSolveProcessEquations[{D[V[x0, t], t] ==
If[x0 + Z[x0, t] > 0, -10*W^2*Z[x0, t] - Ed[t], 10*W^2*x0 - Ed[t]],
V[x0, t] == D[Z[x0, t], t], V[x0, 0] == 0, Z[x0, 0] == 0}, {Z, V}, {x0, 0, x0max},
{t, 0, tmax}, Method -> "StiffnessSwitching"];
Do[it = i tmax/imax; NDSolveIterate[ndsdata, it];
sol = NDSolveProcessSolutions[ndsdata];
resi = Table[x[x0, it] /. sol, {x0, 0, x0max}];
res = Append[res, resi]; ord = Ordering[resi];
If[ord != Range[1, x0max + 1],
sav = Table[{Z[j, it], V[j, it]} /. sol, {j, 0, x0max}][[ord]];
zf = Interpolation[sav[[All, 1]]]; vf = Interpolation[sav[[All, 2]]];
ndsdata = First@NDSolveReinitialize[ndsdata,
{Z[x0, it] == zf[x0 + 1], V[x0, it] == vf[x0 + 1]}]], {i, imax}]
ListPlot[res // Transpose, DataRange -> {0, tmax}, Joined -> True,
AxesLabel -> {"t", "x"}]
Several points are worth making:
• This plot differs modestly from the previous plot at larger t, due to a loss of accuracy caused by 14 calls to Reinitialize with resorted initial conditions. The calculation takes about six seconds. Calling Reinitialize at every time step adds neither to the inaccuracy nor to the run-time.
• V[x0_,t_]:=D[Z[x0,t],t];, defined in the question, must be discarded, and an equivalent equation included in the argument of ProcessEquations.
• Dependent variables should be specified as {Z, V} rather than Z[x0, t]. The latter choice prevented the If in the question from executing.
• Even though the documentation might seem to suggest that
ndsdata = NDSolveReinitialize[...]
is correct, in fact
ndsdata = First@NDSolveReinitialize[...]
is necessary.
• Reinitialize followed by Iterate does not append additional solution results to existing results but instead discards the former. Hence, the earlier solution must be saved before calling Reinitialize.
• Specifying new initial conditions only at discrete values of x0 before calling Reinitialize, causes the dependent variables to be reinitialized to zero. Interpolation functions are used to here to provide continuous values to the reinitialized dependent variables.
• Consistent with the choice made in the question, x0 is sampled only at integer values during reinitialization. In practice, much higher resolution would be needed. Presumably, run-time increases linearly.
Solution with NDSolve WhenEvent
As suggested by user21, WhenEvent also might be used to trigger sorting of streamlines, although I have not attempted to do so.
• ok, I am impressed. Your answer is very deep and articulated, and will be a subject of study for me! Thank you for all your efforts, you really helped me in understaing this software! Sep 12, 2015 at 15:27
• i have a doubt: i can see that there is no change in the trajectories between the swapped and unswapped ones..but there should be some difference, in fact the swaps concerns only with the velocities of the particles when they met each other ( i mean, when x0+Z[x0,t]=x1+Z[x1,t] ,for example, i have V[x0,t]=V[x1,t] and vice versa, but since the diff. eq. depends on x0, the evolution of the particle should be different from the one started from x1. In few words, we need to swap velocities, not the entire path.. Sorry if i hadn't been clear before Sep 12, 2015 at 16:46
• @Cosimo I am no longer certain that I understand what you are seeking. It is clear that you wish to swap V, but to you also wish to swap Z, to do nothing else, or still some third action? Sep 12, 2015 at 22:40
• now that i see the results, probably i was wrong in swapping Z, instead the only important swap may be V, keeping in memory the starting x0 even after the swap, thats why with the swap the results should be physically correct.. I am very sorry for the mess, i am still studing the thoery behind this problem while trying to perform this simulations! btw you really helped me so much, thank you again! Sep 13, 2015 at 18:53
• @Cosimo I shall think about this some too, but not for a few days. I am traveling at the moment. Please let me know what you determine. Best wishes. Sep 14, 2015 at 2:27 | 3,260 | 11,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-10 | latest | en | 0.86936 |
http://www.ask.com/web?q=Is+a+Fraction+an+Integer%3F&o=2603&l=dir&qsrc=3139&gc=1 | 1,485,060,282,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281332.92/warc/CC-MAIN-20170116095121-00105-ip-10-171-10-70.ec2.internal.warc.gz | 327,848,350 | 14,641 | Web Results
## Fraction (mathematics) - Wikipedia
en.wikipedia.org/wiki/Fraction_(mathematics)
A fraction (from Latin fractus, "broken") represents a part of a whole or, more generally, any ..... Therefore, every fraction or integer except for zero has a reciprocal. The reciprocal o...
## Are fractions integers? | Reference.com
www.reference.com/math/fractions-integers-56597be72fb2c2ea
Some, but not all, fractions are integers. Fractions that can be simplified so that they have a 1 as the denominator are integers. Fractions that cannot be ...
## Rule 8: Any Integer Can Be Written as a Fraction - SOS Math
www.sosmath.com/algebra/fraction/frac3/frac31/frac318/frac318.html
You can express an integer as a fraction by simply dividing by 1, or you can express any integer as a fraction by simply choosing a numerator and denominator ...
## Integers - Fact Monster
www.factmonster.com/ipka/A0876848.html
Is It an Integer? Integers are whole numbers and their negative opposites. Therefore, these numbers can never be integers: fractions; decimals; percents ...
## terminology - Are all integers fractions? - Mathematics Stack Exchange
math.stackexchange.com/questions/318948/are-all-integers-fractions
Mar 2, 2013 ... I answered False because if an integer is written in fraction notation it is then classified as a rational number. The teacher said the answer was ...
## Whole Numbers and Integers - Math is Fun
www.mathsisfun.com/whole-numbers.html
Whole Numbers and Integers. Whole Numbers. Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on). number line positive. No Fractions!
## Math Forum: Ask Dr. Math FAQ: Integers, Rational Numbers ...
mathforum.org/dr.math/faq/faq.integers.html
Integers are the whole numbers, negative whole numbers, and zero. .... Pi is an irrational number because it cannot be expressed as a ratio (fraction) of two ...
## What Is the Difference Between Integers and Fractions?
www.quickanddirtytips.com/education/math/what-difference-between-integers-and-fractions
Jul 17, 2012 ... Learn to know the difference between integers, decimals and fractions.
## Can integers be fractions? - Quora
www.quora.com/Can-integers-be-fractions
Aug 8, 2015 ... Yes. For example: 2=6/3. 5=10/2. 6=6/1. Every integer can be expressed as a fraction. But not every fraction is an integer. More formally, the set ...
## Fractions, integers and mixed fractions - Mathinary.com
www.mathinary.com/fractions_fractions_integers_and_mixed_fractions.jsp
What is the difference between fractions, integers and mixed fractions? Learn how to convert between them by examples and descriptions. | 645 | 2,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-04 | latest | en | 0.767318 |
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Ralf Schmidt August 12, 2009 02:37
superheated wet steam (avoid condensation in a nozzle)
1 Attachment(s)
Hi!
I would like to simulate the steam flow in a flow nozzle for mass flow measurement (differential pressure method).
The aim is:
1.: to determine the minimum superheat of the steam, that condensation due to the pressure drop in the nozzle is avoided.
2.: To estimate the error in mass flow measurement, when superheat is to low and condensation in the nozzle occurs.
Has someone done something similar? Are there any experimental or numerical literature data to compare results?
The flow is NOT sonic and NOT transonic!!!
Thank You!
Ralf
Rich August 12, 2009 16:18
Ralf:
> 1.: to determine the minimum superheat of the steam, that >condensation due to the pressure drop in the nozzle is avoided.
If you know the inlet and outlet pressures, you can use the closed-form equations for quasi-one-dimensional nozzle flow. Using R and gamma for steam, you can start at the nozzle exit, using the temperature that will cause condensation at that pressure, and then work backwards to the inlet. I'd estimate that it's a half-page calculation.
>2.: To estimate the error in mass flow measurement, when superheat >is to low and condensation in the nozzle occurs.
That would probably require a multi-species simulation of quasi-one-dimensional flow using both steam and water. You would have one equation specifying the rate of evolution of water and disolution of steam. That equation would feed the two continuity equations, providing for loss of mass in the steam equation and gain of mass in the water equation. Since water will condense out and no longer be part of the flow, you could probably get away with not having a momentum equation for the water.
Regards,
Rich
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# Wiggle expression . 1 time in 1second?
FAQ • VIEW ALL
Wiggle expression . 1 time in 1second? on Oct 31, 2015 at 6:31:24 pm
Maybe it is very simple and easy subject , but i dont understand this "wiggle" expression all the time. We all know(or only me :D ) that for example
wiggle (1,15) , here "1" shows that wiggle should happen 1 time in one second.but when i try this, it happens 6 times in one second. any idea?
Re: Wiggle expression . 1 time in 1second?on Nov 1, 2015 at 5:23:25 pm
What property are you applying it to when this happens?
Re: Wiggle expression . 1 time in 1second?on Nov 1, 2015 at 8:39:56 pm
i try on rotation . i write wiggle(1,random(-2,2))
Re: Wiggle expression . 1 time in 1second?on Nov 3, 2015 at 2:05:44 pm
It's not the first part of the expression that gives you the trouble, you are giving the wiggle a new random number for every frame with the second part of the expression.
The correct wiggle expression in this case would be
wiggle (1,2)
With the value of 2 the wiggle expression itself will vary the results between -2 and 2 in relation to current value. | 340 | 1,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.864554 |
http://www.instructables.com/answers/Why-does-the-pitch-of-tapping-with-a-teaspoon-rise/ | 1,498,620,061,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00472.warc.gz | 550,010,277 | 16,454 | # Why does the pitch of tapping with a teaspoon rise in a mug of hot milk?
Need: Ceramic mug (shape doesn't matter), hot semi-skimmed or full-fat milk, metal teaspoon. When I give the milk a good stir, then begin to tap the bottom of the mug (inside) with the spoon, the pitch of tapping rises gradually. When I stir the milk thoroughly & then start tapping again, the pitch resumes its ascent from the original lowest note. If I only give a partial stir, the pitch resumes partway up from the lowest note. What's going on?
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Dr.Bill7 years ago
geeze! I thought I was the only one that noticed stuff like that.
Karred8 years ago
BeanGolem has most of it right on the money, the only thing missing is that when you heat milk up its resistance lowers (not exactly the right wording, but i've been up for thirty hours, sue me! ;D ), basically when you change the thickness/quantity of the item inside the glass your going to change how the glass resonates, with light crude oil if I remember right its +5 degrees to make a note sharp in a glass, -5 degrees to make it flat, so +10 degrees for a full note. (Farenhite (my spelling and grammar brain dieded a horrible death three hours ago) Degrees, not Celsius/Centigrade, +10 degrees C would be enough to make you wish you melted your hand when you grab the glass) Happy Milk tiem! :D
8 years ago
I was backwards, the cooler it gets, the sharper the note, because the resistance is higher, my bad! I'll go to sleep now! :D
killerjackalope8 years ago
I have an answer somewhere, I'll give the science later...
BeanGolem8 years ago
This is intriguing. I just did a little experiment at my desk. Case 1: Full mug of water. No stirring. Tap on side of glass. Result 1: A certain pitch of tappy noise Case 2: Drink half of the water in mug. No stirring. Tap on side of glass. Result 2: Pitch has gone up Case 3: Drink all of water in mug. No stirring. Tap on side of glass. Result 3: Pitch has gone up again. Conclusion: The level of water in the mug affects the system's natural frequency. The higher the water in the mug, the lower the tapping noise frequency (opposite of rubbing the rim of a wine glass). As you spin the fluid in the mug, it rises towards the outside and lowers in the middle. This causes the effective height of liquid in contact with the glass to rise, resulting in similar behavior to my experiment. Combining both of our observed results, it would appear that the volume or mass of water inside the cup has much less influence over the pitch than the height of the water in contact with the mug does. I would guess that this all has to do with the vibration of the vertical section of the mug. If you think of the mug wall as a vibrating spring, it has a certain natural frequency -- the pitch you would hear with an empty glass. When you take a certain portion of that mug wall and coat it with water, it is like adding mass to that spring, which would lower its natural frequency. Thinking of the whole mug, the more of the wall that is covered by water, the more mass there is lowering the frequency of the empty mug. That's just my guess :D | 738 | 3,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-26 | longest | en | 0.931214 |
http://www.factsabout.com/p/pu/purchasing_power_parity.html | 1,501,260,929,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00254.warc.gz | 437,582,331 | 3,591 | The theory of Purchasing power parity (PPP) asserts that in equilibrium the exchange rate that will prevail between two countries will be that which equalizes the prices of traded goods in each country. Typically, the prices of many goods would be looked at, weighted according to their importance in the economy.
Purchasing power parity exchange rates are useful for comparing living standards between countries. Actual exchange rates can give a very misleading picture of living standards. For example, if the value of the Mexican Peso falls by half compared to the dollar, the Gross Domestic Product measured in dollars will also halve. However, this doesn't necessarily mean that Mexicans are any poorer - if incomes and prices measured in pesos stay the same, they will be no worse off assuming that imported goods are not essential to the quality of life of individuals. Measuring income in different countries using purchasing power parity exchange rates helps to avoid this problem.
One issue with PPP is, however, that most sources do not state the actual content of the PPP, which is statistically deceiving e.g. Countries at Economist.
A simple and humorous example of a measure purchasing power parity is the "Big Mac index" popularised by The Economist magazine, which looks at the prices of burgers in McDonald's restaurants in different countries. If for example a Big Mac cost 4 US dollars in the US and 3 pounds in Britain, the purchasing power parity exchange rate would be 3 pounds for 4 dollars. However, if in the scenario 1 dollar could be traded for 1 pound, then the theory of PPP sudgests that over time the real exchange rate will change to match the PPP exchange rate.
## Criticism of PPP
It is wrong to assume that the prices of goods should be equal in all countries. People in different countries usually put different values on the same goods. What is a luxury goods in one country can be an ordinary daily goods in another country. PPP disregards this.
The exchange rate says how much you can buy in another country with one unit of your own currency. But purchasing power parity exchange rates has nothing to do with how much you can buy.
## Quality of Life and PPP
Even if a correct PPP is used, GDP per capita is still a measure of the economic output of the whole economy, not a direct measure of the mean or median person's quality of life. Other factors such as the quality of homes and schools, access to public services, the extent of pollution, and strength of consumer protection laws are hard to quantify and generally not fully reflected in the GDP. Thus, even a PPP-adjusted measure of GDP per capita must be used with caution, as it is only one component of quality of life.
For example, in 2002, the GDP per capita for Japan is about \$40,000, while the PPP is estimated as \$27,000, while in the US, GDP per capita is about \$27,500 and the PPP is \$36,000. The U.S. has poverty, high crime rates and slums to a greater extent than Japan, while Japan has much less physical space per person and arguably less individual freedom. Ultimately, the quality of life may depend on subjective judgement and individual preferences.
Per capita income also does not take into account inequalities in wealth distribution. | 664 | 3,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-30 | latest | en | 0.949795 |
https://www.rockpapershotgun.com/2016/11/03/yankais-triangle/ | 1,607,116,921,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00445.warc.gz | 833,888,281 | 32,225 | 7
Yankai's Triangle - A Pleasant Geometry Twiddler
This morning brought geometric gaming news in the form of Yankai’s Triangle [official site]. It’s a puzzle game of sorts, where you rotate triangles with coloured points in order to match them up. If you’ve ever played Triominoes you’ll know the sort of thing I mean.
I say “of sorts” regarding its puzzle game status because there are puzzle games which genuinely give my brain something to grind away at, exerting itself to find a solution, and there are puzzle games which don’t really involve much puzzling but you are still ultimately looking for a solution. Yankai’s Triangle is definitely in the latter camp. I’m on level 36, I think and I’ve only found a bit of grit to the puzzles a couple of times. Once where – you know when you get comfy with the way something works and the designer uses that to pull the rug from under you in a teasing way? – that happened and it made me grin and have to go back through the triangles and re-order some of them. The other time was level 34 where the game introduced a new element to the triangles which made me employ a bit more thought when finding the solution than just manipulating the shapes idly.
As per Sun’s description:
“YANKAI’S TRIANGLE is a love letter to TRIANGLES. A puzzling puzzle game about the beauty and joy of infinitely tapping on TRIANGLES for some reason.
“With an elegant innovative revolutionary TRIANGLE-first approach to interface, YANKAI’S TRIANGLE lets you tap on TRIANGLES to spin them and stuff. Colors play a part in gameplay too I think. Bring TRIANGLES to a forgotten TRIANGLE, and uncover a TRIANGLE hidden deep beneath the YANKAI.
“(YAN rhymes with KEN and KAI rhymes with EYE by the way)”
But there doesn’t really seem to be a comprehensible difficulty progression so that new mechanic has disappeared again for the following two levels and I’m back to just twiddling triangles. I’ll be honest, without that extra mechanic I mentioned but don’t want to spoil, the possibility space feels too small for the player to need to work for solutions. At most you’ll just need to go back through a few layers of previously slotted-together triangles to rotate one or two of them a few times and that’s busywork rather than difficulty. But there is something here that I like and that I think has real potential beyond what’s actually in the game so far that I’ve seen.
The other thing I like is that the triangles slot together then let you zoom out to the next layer of triangles. In and of itself that’s not noteworthy, but Sun’s previous game was Circa Infinity which was about travelling deeper into layers of concentric circles, ever expanding until you reached the boss. There’s definitely an expansion and contraction of geometry theme here which I just really like as someone’s personal game hallmark.
Wait – I got to level 41 and that was more interesting. I think it was a boss level? The shapes had teeth.
Anyway, I’m flagging it up in case that puzzle-but-not-a-puzzle-but-maybe-sometimes-slightly-a-puzzle vibe appeals, or is something you’d like to share with a kid.
It’s currently on Steam with a 1% discount so £2.26/2,96€/\$2.96 for Windows, Mac, and Linux, and also on iOS. Android is going to be a while longer.
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Philippa Warr
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Comments are now closed. Go have a lie down, Internet. | 759 | 3,382 | {"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-2020-50 | latest | en | 0.93248 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.03/jason5.html | 1,561,015,091,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00157.warc.gz | 109,922,695 | 1,790 | Quandaries and Queries Name: jason Who is asking: Other Level: All Question: 1 4 10 20 35 56 84 what is the nth term Hi Jason, The difference method will work well here: form successive rows where the first row is your original sequence, and the k-th term in the next row is always the difference between the (k+1)-th term and the k-th term in the previous row: 1-st row: 1, 4, 10, 20, 35, 56, 84, ... 2-nd row: 3, 6, 10, 15, 21, 28, ... 3-rd row: 3, 4, 5, 6, 7, ... 4-th row: 1, 1, 1, 1, ... If you can guess the 5-th term in the 4-th row, it will help you find the 6-th term in the 3-rd row, which will help you find the 7-th term in the 2-nd row, which will help you find the 8-th term in the first row. You can then start over a few times, finding the next term in each sequence; after that, you will need to try to find a general expressions for the n-th term in each row. Claude Go to Math Central | 297 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-26 | longest | en | 0.936677 |
http://lessonplanspage.com/ScienceExCanIMakeARollerCoasterMO68.htm/ | 1,369,375,828,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704234586/warc/CC-MAIN-20130516113714-00062-ip-10-60-113-184.ec2.internal.warc.gz | 154,758,334 | 12,027 | Rate this Plan:
# A Science Experiment on making a Roller Coaster
Subject:
Science
6, 7, 8
Title – ROUND AND ROUND
By – Judy Schneider
Primary Subject – Science
Secondary Subjects – Science
Grade Level – 6 – 8 (adaptable)
SCIENCE PROJECT OF THE WEEK
ROUND AND ROUND
PROBLEM: Can I construct a roller coaster that will work?
RESEARCH: Write two or three paragraphs about kinetic and potential energy.
HYPOTHESIS: What will you have to do to make your roller coaster car complete the circuit?
MATERIALS:
paper roller coaster track
flexible wire – 2 meters long, not as thick as a coat hanger wire
bead or hex nut
PROCEDURE:
Rules: 1. The roller coaster must start with a hill.
2. There must be at least two loops on the track.
3. There must be al least two hills after the last loop.
4. Every hill must be higher than the one that comes after it.
5. Every loop must be lower than the hill that comes before it.
1. Design your roller coaster. Follow the rules and use the hills and loops provided. Keep in mind what you learned from your research about kinetic and potential energy.
2. When you think your design will work, construct a model using the wire. The wire must be able to stand by itself. YOU WILL NOT BE ALLOWED TO HOLD UP THE COASTER.
3. Tape one end of the wire to the floor (hard floor, not carpet). Your “car” must travel to the end of the wire. If it doesn’t, redesign your roller coaster and try your new design.
4. Enrichment: Predict what roller coasters of the future will be like. Discuss what role friction plays in your roller coaster system.
DATA: Include your paper design. If your design didn’t work the first time, include all the designs you used.
CONCLUSION: This is not optional. You must explain what you learned by doing this activity.
Remember that you must answer the question you asked in your original problem statement.
NOTE: BE SURE TO HAVE YOUR PARENT OR GUARDIAN SIGNS YOUR WORK. PARENTS: YOUR SIGNATURE SHOWS YOUR STUDENT HAS DONE THE WORK.
TEACHER SECTION:
POSSIBLE HYPOTHESIS: Using the principles of potential and kinetic energy, the students should be able to construct the roller coaster. The potential energy at the top of the roller coaster will be enough to move the nut through all the loops and hills.
POSSIBLE CONCLUSION: Potential energy at the top of the first hill was sufficient to lift the car to the top of the first loop. The potential energy there was enough to continue moving the car through all the loops and hills. As the car moved down, potential energy changed to kinetic energy. As the car moved up, kinetic energy changed to potential energy. Friction was also a factor because it used up some of the potential energy so there was not as much kinetic energy available.
E-Mail Judy Schneider ! | 631 | 2,786 | {"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-2013-20 | latest | en | 0.891519 |
https://www.sciencelearn.org.nz/resources/1348-pedal-power | 1,726,623,308,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00503.warc.gz | 900,937,635 | 11,601 | • + Create new collection
• Competitive cyclists have power meters to carefully monitor their training and race performance. Power is a measure of how much energy is being changed every second. For cyclists, the effort they put into pedalling either makes them go faster (extra kinetic energy – related to speed) or helps them fight against drag and rolling resistance. Any improvements to aerodynamic drag need to be weighed up against the amount of power that a cyclist can still produce if they need to use a different position.
How much power can a cyclist produce?
Power is the best indicator of how well a cyclist will perform in terms of maximising their speed. An elite cyclist can produce about 5 watts (5W) of power for every kilogram of bodyweight for a 1-hour event. For example, a 70 kg cyclist who is able to maintain a power output of 350W for 1 hour would be considered to be in the elite category.
A very powerful cyclist might be able to produce 1200W or more for a few seconds. This is useful to make sure they reach maximum speed as soon as possible. In a team pursuit event, the most powerful cyclist is placed at the front for the start of a race to make sure the team reaches maximum speed quickly. For four cyclists riding in a close line, the second cyclist uses about 74% of the power of the lead cyclist. The third and fourth cyclists use about 64% of the power of the lead cyclist.
How pedalling power relates to force and speed
The power of a cyclist depends on:
• how much force the pedals are being pushed with
• the speed at which the pedals are being turned around.
The maximum power occurs when the force pushing on the pedals multiplied by the speed of the pedals is greatest.
For example, if the cyclist applies a force of 150 newtons to the pedals (150N is the force needed to lift a 15kg mass) and the speed of the pedals in a circle is 2 metres per second (2m/s), the pedalling power output of the cyclist is:
Pedalling power = force on pedals x speed of pedals
= 150N x 2m/s
= 300W
This is the same power as lifting a 30kg mass upwards a height of 1 metre every second.
Power is a measure of how quickly energy is being changed into other forms. To understand why power is force multiplied by speed. There are two main ideas:
• Change in energy is equal to the work done, which is force applied multiplied by distance moved.
• Speed equals distance moved divided by time taken.
power = change in energy/time taken = work done/time = force x distance/time = force x speed
Power can be used to calculate forwards force
The forwards force acting on the bike is not the same as the force pushing on the pedals. Assuming negligible energy losses, the forwards force on the cyclist is calculated as the pedalling power divided by the speed of the bike:
For example, if the power at the pedals is 300 watts (300W) and the speed of the bike is 40 kilometres per hour (40km/h = 11m/s), then the forwards force acting on the cyclist is:
Forwards force = pedalling power ÷ speed of bike
= 300W ÷ 11m/s
= 27N
Faster speeds require more power to counter aerodynamic drag
To calculate the force and power needed to counter aerodynamic drag and rolling resistance at different speeds, the following equations can be used:
FD = ½CD AρV2 PD = ½CD AρV3
• FD is the drag force.
• PD is the power needed to counter drag.
• CD is the drag coefficient (a number from about 0.5 to 1.1 for a cyclist, depending on bike, body position and equipment).
• A is the frontal area of the object (measured in square metres). CDA is often grouped together as a term called the effective frontal area – normally between 0.4 and 0.7m2.
• ρ is the density of air (about 1.2 kg/m3).
• V is the speed the object is travelling at (measured in metres per second – m/s).
These equations can be used to show that, if speed increases by 10%, the force of aerodynamic drag increases by 21%, but the power (effort needed by the cyclist) increases by 33%.
To calculate the force and power needed to counter rolling resistance at different speeds, the following equations can be used:
FRR = CRR x m x g PRR = FRR x V
• FRR is the rolling resistance force.
• PRR is the power needed to counter rolling resistance.
• CRR is the coefficient of rolling resistance (typically between 0.003 and 0.008 depending on tyres and the road surface). A higher number indicates more resistance.
• m is the mass of the rider plus bike (measured in kilograms – kg).
• g = 9.8 (multiplying mass by 9.8 gives the weight, which is the force downwards due to gravity
• V is the speed the object is travelling at (measured in metres per second – m/s).
Even though the force of rolling resistance doesn’t change as speed increases, the power needed to work against rolling resistance increases. As you travel faster, there is more energy being converted into heat energy each second, so this requires more effort from the rider.
Activity idea
In the Individual pursuit graphs activity, students cut out and tape different shapes, attach tiny pieces of cotton thread and use hairdryers to find which shape has the least drag. The shape that keeps airflow attached for longer reduces the low-pressure zone at the back, so it will have least drag.
Published 22 February 2011 Referencing Hub articles
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1. ## Non linear 2nd order differential equation
25 July 2009
I am not able to find the general function that integrates the following 2nd order differential equation;
y^2 y’’ + a y^3b = 0
in which 0 ≤ a ≤ 1 is a constant;
b > 0 is a constant.
The equation relates to the behaviour (orbit/motion) of a massless point subjected to a particular central motion.
I could easily find only the solutions associated with nil values for constants a and b.
I would be deeply grateful to any friend who can help me resolve the problem.
Many thanks for any help.
Leo Klem
2. Uhm, let me see if I know what you're asking.
$y'' = -ay + \frac{b}{y^2}$
$y' = -a \int {y} dy + b \int {y^{-2}} dy$
$y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2)$
$y' = -a\frac {y^2}{2} - by^{-1} + C_t$
$y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$
$y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5$
$y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T$
It's either that or I'm totally off the mark.
3. Originally Posted by Deco
Uhm, let me see if I know what you're asking.
$y'' = -ay + \frac{b}{y^2}$
$y' = -a \int {y} dy + b \int {y^{-2}} dy$
$y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2)$
$y' = -a\frac {y^2}{2} - by^{-1} + C_t$
$y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$
$y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5$
$y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T$
It's either that or I'm totally off the mark.
You're off the mark on both integrations (but close on the first)!
If you multiply your equation by $y'$ then
$y'y'' = -ay y' + \frac{b y'}{y^2}$
so
$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1
$
You can solve for $y'$ and separate but I think the resulting integral will be difficult.
4. $y' = \sqrt {- a y^2} - \sqrt {\frac{2b}{y}} + c_1$
$y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy$
$y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$
Something like that?
But since a is positive, wouldn't this result in a complex answer?
5. Originally Posted by Deco
$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$
$y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy$
$y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$
Something like that?
How on earth did you go from
$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$
to
$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?
6. Oops. Wow. Don't know what I was thinking.
7. ## Non linear 2nd order equation.Thanks, with a clarification
Many thanks to my two first interlocutors. I have to point out that the problem described by the differential equation regards a central motion. The integrating function, if it does exist, is actually useful in the explicit form given by
y = f(x; a, b, C, D)
in which x is the independent variable (it represent the "orbital" angle), a and b are the given equation constants, and C, D are the two integration constants.
Solutions in the form of an expression of functions of y do actually create difficult problems of interpretation as to the kind of motion regarded.
8. Originally Posted by Danny
How on earth did you go from
$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$
to
$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?
It is evident that the correct procedure is...
$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1 \rightarrow$
$\rightarrow y^{'}= +/- \sqrt{\frac{- a y^{3} - 2 b + 2 y c_{1}}{y}} \rightarrow$
$\rightarrow \int \sqrt{\frac{y}{-a y^{3} -2b + 2 y c_{1}}} dy = +/- x + c_{2}$ (1)
May be that the integral in (1) is a little difficult to solve ... may be ...
Kind regards
$\chi$ $\sigma$
9. ## What beyond the inverse function?
Thanks to Chisigma. You got the correct procedure, but the solution is still away. It's the point where I had actually to stop. It's clearly a "solution" in the form of an inverse function, i.e., the argument expressed as a function of the dependent variable. The integral you indicate seems however difficult to solve, but in a few particular cases, such as - for instance - assuming C1 = 0.
Could you please provide further suggestions?
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- Homework 2 (http://book.caltech.edu/bookforum/forumdisplay.php?f=131)
eychen 10-05-2013 05:17 PM
Dear all,
For Q1, I got nu_min around 0.375. Therefore, [d] would be the closet answer.
I feel the answer is reasonable, for the following consideration:
1. The probablity of getting no head outcomes in 10 trails is 1/1024.
2. For 1000 ten-times-thrown coins, the probability of getting no one with head count = 0 is [(1024-1)/1024]^1000 = 0.375.
3. In most cases, when there is no one with head count =0, the lowest head count is most likely to be 1.
4. Therefore, the expectation value of nu_min will be approximately 0*0.625 + 1*0.375 = 0.375.
Am I doing anything wrong? Or I misunderstood the question?
yaser 10-06-2013 10:11 PM
Quote:
Originally Posted by eychen (Post 11528) Dear all, For Q1, I got nu_min around 0.375. Therefore, [d] would be the closet answer. I feel the answer is reasonable, for the following consideration: 1. The probablity of getting no head outcomes in 10 trails is 1/1024. 2. For 1000 ten-times-thrown coins, the probability of getting no one with head count = 0 is [(1024-1)/1024]^1000 = 0.375. 3. In most cases, when there is no one with head count =0, the lowest head count is most likely to be 1. 4. Therefore, the expectation value of nu_min will be approximately 0*0.625 + 1*0.375 = 0.375. Am I doing anything wrong? Or I misunderstood the question?
Your reasoning is both nice and correct, and the problem is just technical that when you have 1 head, the corresponding is in fact 0.1, not 1. :(
eychen 10-07-2013 11:16 AM
Hi Yaser,
I see. nu is the fraction, not the number of the head count. Thank you! :)
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CS416 - Mathematical Modelling & Prediction Lecturers – Douglas Leith, Robert Shorten Assessment – By final exam. No marked lab or tutorial work. Tutorials – dates and times to be announced. Tutorials will be organised to run as each section of the course is completed.
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Lecturers – Douglas Leith, Robert Shorten
Assessment – By final exam. No marked lab or tutorial work.
Tutorials – dates and times to be announced. Tutorials will be organised to run as each section of the course is completed.
Web – www.hamilton.ie/cs416/
• The development and solution of a set of mathematicalequations that describe a real situation to an acceptablelevel of accuracy.
• Used in order to predict what would happen in a real situation
• Note: We are interested here in quantitative rather than purely qualitative/descriptive models.
Topic of this course: modelling for decision support
- Predictive models almost always underly decision making in high performance systems
We seek models for decision support. In this course we are not seeking “true” models, but models which are “good enough” for the task at hand.
Example. Consider modelling an ideal gas
Volume, V
Pressure, P
Temperature, T
Related by model, PV=T
Each molecule has position (x,y,z) and velocity (u,v,w)
Increasing level of detail. What level of detail should we use in our model ?
Each molecule consists of atoms. Each atom consist of other particles …
Selecting the appropriate level of abstraction/detail is a key part of modelling.
Strongly application dependent and something of an “art”. Not dealt with in detail in this course, but we do note that it is linked to what we are able to observe/measure as well as to the purpose for which the model is intended.
- because when assessing a model against the real system we cannot decide between models which generate the same predicted observations
Let Ni be the population in year i
A crude model for population growth is that
Ni=aNi-1
When a>1, the population increases (unboundedly) each year. When a<1, the population decreases each year until it reaches zero (extinction).
This is obviously an oversimplification. We need to account for overcrowding and limited resources i.e. we expect the value of parameter a to vary with the population N.
Now have
Ni=a(Ni-1)Ni-1
Lets assume that a(N)= (1-N) for N in the interval [0,1]
See www.hamilton.ie/cs416/logisticmap.htm for arguments in support of this choice.
• Ni= (1-Ni-1)Ni-1 “Logistic Map”
• Note that this is a very much simplified model. Are these simplifications justified ? That is, how “good” is this model.
• depends on what we want to use it for.
• can compare predictions of model with observed behaviour to validate model/establish accuracy. Availability of observations places fundamental limit on quality of model.
• What can we use the model for ?
• The structure of the model reflects our understanding of the structure of the system
• a model organises our experiences and observations
• We can solve the model equations to make predictions
• - decision support
We can solve the model equations to make predictions …
Graphical Solution
Ni=Ni-1
Ni
Ni= (1-Ni-1)Ni-1 with =2
Ni-1
With =2, a stationary point exists to which all solutions in interval [0,1] are eventually attracted.
NB: “stationary point” = “equilibrium point” = “steady-state solution”
This is not always the case. For example, consider when =3.75
The population never settles down to a constant value
Ni
Ni-1
Queue length, q. Service rate, B packets/s.
Packets arrive at times t0,t1,t2,…
Ignoring servicing of packets just now, when a packet arrives we have
q(tn)=min(q(tn-1)+1, qmax)
Now, interval between two packets is tn-tn-1. During this interval B(tn-tn-1) packets are serviced i.e. removed from the queue. The queue size cannot fall below zero. So, we have as our model
Q=min(q(tn-1)+1-B(tn-tn-1), qmax)
q(tn)=max(Q,0)
Now let’s model the behaviour of the source. Suppose we have one source and that it sends a new packet in response to the server signalling that it has finished servicing a packet. Also suppose that its time T for the packet to travel from the source to the queue.
• Send first packet at time t0.
• Packet arrives at queue at time to+T
• Service rate is B packets/s, so at time to+T+1/B server signals that packet has been serviced.
• Send second packet – time is now t1=t0+T+1/B
• Packet arrives at queue at time to+2T+1/B
Queue now doesn’t overflow, but server is idle for time T between packets arriving. Can we do better ?
• Additive increase/multiplicative decrease congestion control (AIMD). Define a new variable W, the number of packets sent at each acknowledgement. Suppose also that the queue signals when it’s length exceeds a threshold, qmax.
• while q<qmax, let W=W+1 when a packet is acknowledged/seviced.
• when q=q,ax, W is reset to W/2.
• Our model becomes:
• tn-1+1/B when q(tn-1)>0
• tn =
• tn-1+T+1/B when q(tn-1)=0
• Q=min(q(tn-1)+ W(tn-1) -B(tn-tn-1) , qmax)
• q(tn)=max(Q,0)
• W(tn-1)+1 when q(tn)<qmax
• W(tn)=
• W(tn-1)/2 when q(tn)=qmax
Sending rate adapts to maintain full queue (server never idle), with almost no dropped packets
queue size, q(t)
sending rate, w(t)
packets dropped
time, t
What if now have a second source ?
Source 1
Source 2
Would like first source to reduce its sending rate so as to allow second source to successfully send packets – dynamic resource allocation.
Fair sharing source 1 should roughly halve its sending rate and source 2 take up the slack …
Two sources, both using AIMD. Source 2 becomes inactive after 100s.
queue size, q(t)
sending rate, w1(t)
sending rate, w2(t)
time, t
Engine torque, TH
Shaft torque, TL
Force, F
TL=NTH, N is gearbox ratio
Force exerted by wheel is TLR, with friction coeff, R radius of wheel
Air creates drag force -v, with v the velocity of vehicle
Newton’s Law: Force = mass*acceleration
F=NTHR- v=m a
Noting that a=dv/dt, we have the following model for the speed of the vehicle.
m dv/dt= NTHR - v
Suppose the vehicle has two gears, N1 and N2. The gear used is selected by an automatic transmission. The model is then
N1THR - v in gear 1
m dv/dt=
N2THR - v in gear 2
+ a model of the the decision making process used by the automatic transmission.
Introduce map”)taxonomy of models. It turns out that most systems can be modelled using a fairly small set of model structures.
Study solutions, esp. numerical solutions/simulations
Can derive models from first principles or learn model from observations (or more usually by a combination of both approaches). We will not cover first principles modelling as very application specific. But will introduce machine learning approaches (including probabilistic reasoning ideas).
• Course Outline
• How do we derive a model for a system ?
• How do we extract information from it (esp. how do we obtain quantitative solutions and analyse their properties) ?
• this course is structured around these questions.
• Difference Equations
• Differential Equations
• Hybrid
• Linear
• Nonlinear
• Time-invariant
• Time-varying
• Other aspects of models can also be usefully classified, but not pursued here.
• Especially deterministic/stochastic models – stochastic models not covered in this course.
Can combine these two classifications e.g. linear differential equations
• Difference Equations
• Differential Equations
• Hybrid
Simple Example of a Difference Equation:
y(k) = a y(k-1), k=1,2,…
This is equivalent to the (infinite) set of equations:
y(1)=ay(0)
y(2)=ay(1)
y(3)=ay(2)
etc.
If have observations of y(0), y(1), etc, then this defines a relation between these observations. If y(1), y(2) etc are unknown, the equations can be solved to find them.
Difference Equations map”)
Logistic Map is another example of a difference equation
y(k)=(1-y(k-1))y(k-1), k=1,2,…
Can also include an external input u in the difference equation, e.g.
y(k) = a y(k-1)+bu(k-1), k=1,2,…
This is equivalent to the (infinite) set of equations:
y(1)=ay(0)+bu(0)
y(2)=ay(1)+bu(1)
y(3)=ay(2)+bu(2)
etc.
Difference Equations map”)
Definition:
Suppose there is a defined sequence of values y(k), k=0,1,2,… (e.g. representing values observed at equally-spaced time points). A difference equation is an equation relating the value y(k) to other values y(i), ik.
A difference equation is said to be causal wheny(k) is related to values y(i) with i<k.
i.e. y(k)=f(y(k-1),y(k-2),…,y(k-n), u(k-1),u(k-2),…,u(k-m))
where m,n are some constants.
NB: We write y(k)=f(y(k-1),y(k-2),…) but could equally well write this as y(k+1)=f(y(k),y(k-1),…), and this is often done.
u is an external input
Difference Equations map”)
A difference equation is said to be linear when the function f on the RHS is a linear function
i.e y(k)=a1(k)y(k-1) + a2 (k)y(k-2)+…+an (k)y(k-n)
+b1 (k)u(k-1)+b2 (k)u(k-2)+…+bm (k)u(k-m))
where a1(k), a2(k), …,an (k)and b1 (k),b2 (k), …, bm (k) are time-varying parameters. When these parameters are constants (do not vary with time), the model is said to be linear time-invariant.
• Difference Equations
• Differential Equations
• Hybrid
• Linear
• Nonlinear
• Time-invariant
• Time-varying
+
Linear Time-Invariant Difference Equations
Difference Equations map”)
A solution to a difference equation is a function y(k) that satisfies the equation. Solutions are readily derived by recursion.
e.g. for y(k)=ay(k-1) we have that
y(1)=ay(0)
y(2)=ay(1)=a2y(0)
y(3)=ay(2)=a3y(0)
etc.
i.e. a solution is y(k)=aky(0)
NB: We need to specify y(0) in order to solve this equation. This is called the initial condition for the equation. We need to specify both the equation and its initial condition in order to define a solution.
More generally,
y(k)=f(y(k-1),y(k-2),…,y(k-n), u(k-1),u(k-2),…,u(k-m))
We assume that the input values u(k) are defined beforehand. We must also specify y(0), y(1), …, y(n-1) in order to define a solution – in general the initial condition must specify n values.
• Note that a difference equation need not have any solution.
• e.g. y(k)2=-(1+y(k-1)2)
• has no solution since y(k)2 can never be negative.
• Also, even when a solution exists, it need not be unique i.e. there may exist many solutions.
• e.g. sin y(k) = y(k-1)
• Generally, however, a model of a physical system can be expected to possess a solution which is unique.
• Recall that linear time-invariant difference equations have the form
• y(k)=a1y(k-1) + a2y(k-2)+…+any(k-n)
• +b1u(k-1)+b2u(k-2)+…+bmu(k-m))
• Consider the simplest system y(k)=ay(k-1) “first-order system”
• We have that y(1)=ay(0)
• y(2)=ay(1)=a2y(0)
• y(3)=ay(2)=a3y(0)
• etc
• So the solution is y(k)=aky(0). Note that the solution behaves as anexponential – we can rewrite it as y(k)=exp(k loga)y(0)
• For a<1, y(k)0 as k - system is said to be stable
• For a>1, y(k) as k - system is said to be unstable
• For a=1, solution neither grows of decays
• - system is said to be critically stable
Also, rate of convergence/divergence varies with the value of a
a=0.95
y(k)
a=0.75
a=0.25
k
Consider now y(k)=a1y(k-1) + a2y(k-2) “second-order system”
By analogy to the first-order case, try a solution of the form y(k)=ky(0) where is some (as yet unknown) constant. Then we need,
k y(0)=a1 k-1y(0)+a2 k-2y(0)
i.e. k - a1 k-1 - a2 k-2 = 0
Dividing through by k-2 gives
2 - a1 - a2 = 0
i.e. = a1/2 (a12+4a2)/2
We can work this through to derive an explicit solution. We won’t do this though. Observe that the situation where a12+4a2 < 0 (and so is complex valued) looks like its going to be different from when a12+4a2 < 0 (and so is real valued).
Consider now y(k)=a1y(k-1) + a2y(k-2) “second-order system”
a12+4a2 < 0
“underdamped”
a12+4a2 > 0
“overdamped”
a1=1,a2=-0.2
(a12+4a2=+0.2)
y(k)
a1=1,a2=-0.5
(a12+4a2=-1)
k
a1=1,a2=-0.75
a1=1,a2=-0.5
y(k)
y(k)
k
k
a1=1,a2=-0.95
a1=1,a2=-1
y(k)
y(k)
• NOTE:
• = a1/2 (a12+4a2)/2
• Similarly to first-order case,
• For ||<1, y(k)0 as k - system is stable
• For ||>1, y(k) as k - system is unstable
• For ||=1, solution neither grows of decays
• – system is critically stable
• When is real valued, system is overdamped (with special case called critically damped when a12+4a2=0)
• solution to system is the sum of pure exponentials
• When is complex valued, system is underdamped
• -solution to system is oscillatory, with envelope that decays exponentially for stable systems, grows exponentially for unstable systems.
• For stable systems, the solution converges to a final value as k. This is called the equilibrium point of the system (also called stationary point or steady-state value).
Linear time-invariant difference equation:
y(k)=a1y(k-1) + a2y(k-2)+…+any(k-n)
+b1u(k-1)+b2u(k-2)+…+bmu(k-m))
When the input u is zero, at an equilibrium point y we must have: y=a1 y+a2 y+…+an y i.e. y=0
When input is non-zero, the equilibrium point will depend on the input. E.g. say u(k)=u, a constant value. Then
y=a1 y+a2 y+…+an y+b1u+b2u+…+bmu
i.e. y= u (b1+b2+…+bm)/(a1+a2+…+an)
For linear difference equations, in qualitative terms only a small number of types of solution can exist (stable, unstable, overdamped, underdamped etc).
For nonlinear difference equations, the situation is much richer.
e.g. depending on the value of the parameter , the Logistic Map y(k)= (1-y(k-1))y(k-1) not onlyhas solutions which are stable and unstable but also has steady oscillatory solutions and chaotic solutions (complex oscillations) - these are both types of equilibrium solution which do not have just a single value y.
Difference Equations – y(k)= Models (1-y(k-1))y(k-1)
=3.5
=1
“stable”
y(k)
y(k)
k
k
y(k)
y(k)
=3.9
“complex oscillation”
=4.1
“unstable”
k
k
So far have concentrated on difference equations. But physical systems are usually described by differential equations.
Recall Newton’s law:
F(t) = m a(t)
Force = mass x acceleration
We obtain velocity by integrating the acceleration
i.e.
Conversely, acceleration is obtained by differentiating velocity
i.e.
f(x) physical systems are usually described by differential equations.
0
xo
x
The derivative df/dx of a curve f(x) at a point xo is just the tangent to the curve at xo.
Higher-order derivatives are obtained recursively
i.e. , and more generally
f(x) physical systems are usually described by differential equations.
0
xo
x
The integral of a curve f(x) at a point xo is just the area under the curve between 0 and xo.
Differential Equations physical systems are usually described by differential equations.
• A simple example is
• dy(t)/dt = ay(t)
• Verify that the solution to this differential equation is:
• y(t)=exp(at) y(0)
• compare with the first-order difference equation y(k)=ay(k-1)
• which has solution y(k)=aky(0)=exp(k log(a)) y(0).
Differential Equations physical systems are usually described by differential equations.
Another example:
Can also include an external input u in the differential equation, e.g.
dy(t)/dt = a y(t)+bu(t)
Definition:
Suppose there is a function y(t) defined on an interval [to,t1]. A differential equation is an equation relating the value y(t) to some of its derivatives.
In general, a differential equation is of the form:
Differential Equations physical systems are usually described by differential equations.
A differential equation is said to be linear when the function f on the RHS is a linear function
i.e
where a1(t), a2(t), …,an (t)and b1 (t),b2 (t), …, bm (t) are time-varying parameters. When these parameters are constants (do not vary with time), the model is said to be linear time-invariant.
• Difference Equations
• Differential Equations
• Hybrid
• Linear
• Nonlinear
• Time-invariant
• Time-varying
Linear Time-Invariant Differential Equations
+
Differential Equations physical systems are usually described by differential equations.
A solution to a differential equation is a function y(t) that satisfies the equation.
E.g. the solution to dy(t)/dt = ay(t) is y(t)=exp(at) y(0)
Similarly to difference equations, we need to specify the initial condition y(0) in order to solve this equation i.e. we need to specify both the equation and its initial condition in order to define a solution.
E.g. the differential equation:
has general solution of the form y(t)=A+Bt where A,B are some constants. To find the values of these constants we use
y(0)=A, dy(0)/dt=B
Differential Equations physical systems are usually described by differential equations.
• Also, similarly to the difference equation y(k)=ay(k-1), the solution to dy(t)/dt = ay(t) is y(t)=exp(at) y(0)
• i.e.
• For a<0, y(t)0 as t - system is said to be stable
• For a>0, y(t) as t - system is said to be unstable
• For a=0, y(t)=y(0) - system is said to be critically stable
• Similarly to the second-order linear difference equation, the second-order linear differential equation
• exhibits overdamped, underdamped and critically damped responses depending on the values of a1 and a2.
Second-order Linear Differential Equation physical systems are usually described by differential equations.
a1=-0.7,a2=-0.75
a1=-0.29,a2=-0.93
y(t)
y(t)
time, t
time, t
a1=0,a2=-1
a1=-0.05,a2=-1
y(t)
y(t)
Compare with 2 physical systems are usually described by differential equations.nd order difference eqn y(k)=a1y(k-1) + a2y(k-2)
a1=1,a2=-0.75
a1=1,a2=-0.5
y(k)
y(k)
k
k
a1=1,a2=-0.95
a1=1,a2=-1
y(k)
y(k)
• Linear physical systems are usually described by differential equations.difference and differential equations are closely related.
• NOTE: This is generally only true for linear equations. Nonlinear equations seem to be fundamentally different,
• e.g. chaos can exist in first-order difference equations (such as the logistic map), but not in first-order differential equations (we need to go to at least third order to find a differential equation which exhibits chaos).
• Recall definition of derivative:
• This suggests that a derivative might be approximated by the finite difference:
• for small h. This is called the Euler approximation. We expect that as h is made smaller, the accuracy of the approximation improves.
Consider the first-order linear differential equation: physical systems are usually described by differential equations.
and now replace the exact derivative by its Euler approximation. We have
i.e. y(t+h)=y(t)+ha1(t)y(t)
Considering the time instants kh, k=0,1,2,… we have
y((k+1)h)=y(kh)+ha1(kh)y(kh)
which we can write as
y(k+1)=y(k)+ha1(k)y(k)
We have converted our first-order differential equation into a first-order difference equation. This conversion is only approximate, but the approximation becomes arbitrarily accurate as h is made small.
More generally, a linear differential equation: physical systems are usually described by differential equations.
can be approximated by a linear difference equation of the same order.
Example
Example – car braking (ABS) physical systems are usually described by differential equations.
Example – car braking (ABS) physical systems are usually described by differential equations.
• Inertia of car: m dv/dt = -Fx
• Inertia of wheel: J dw/dt = rFx – Tb
• m mass of car
• V velocity of car
• J rotational inertia of wheel
• w rotational velocity of wheel
• Fx tyre friction force
• Tb braking torque
• Tyre friction force Fx model
• Fx = Fz()
• () = 1.28(1-e-24 ) – 0.52
• = (v-wr)/v “wheel slip”
Fz is vertical force (weight) of car
()
Hybrid Systems physical systems are usually described by differential equations.
• Difference Equations
• Differential Equations
• Hybrid
What are hybrid systems ? Mixture of discrete event systems and differential/difference equations.
+ =
Discrete event
systems
(automata)
Differential/
Difference
Equations
Hybrid system
Physical processes
Etc.
Logic, software, hardware, switches;
Collisions;
Communication networks;
Note: is just another way of writing dq/dt, dr/dt physical systems are usually described by differential equations.
Example: Server with congestion control
(differential equation version of scheme discussed in lecture one).
Hybrid Systems physical systems are usually described by differential equations.
Example – Bouncing ball
Consider the vertical motion of a ball in a gravitational field with uniform acceleration g. Let x1 be the height of the ball and x2 its velocity. Then when x10 we have
dx1/dt=x2
dx2/dt=-g
Note: is just another way of writing dx1/dt, dx2/dt
Bouncing ball equations
Height vs time
“Switched differential equation”
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations
We obtain the solution to a switched equation by piecing together the solutions obtained in each mode. The piecing together is done such that the x1, x2 are continuous.
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations
• What can go wrong ?
• 1. Problems in continuous evolution
• Existence
• Uniqueness
• 2. Problems in the hybrid execution
• Chattering
• Zeno
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations:
Chattering
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations:
Chattering
Note that chattering solutions do not always completely “stop”. Often, the system continues to evolve in a so-called sliding mode.
-1 w0
dw/dt=
1 w<0
dv/dt=-v
v
Solution “slides” along the surface w=0
0
w
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations:
Zeno Execution
Hybrid Systems
Zeno execution = is infinite, but execution does not extend to t=+
Hybrid Systems physical systems are usually described by differential equations.
- Solutions of Hybrid Equations:
Similarly to difference and differential equations, hybrid systems can be classified as being stable, unstable etc.
Various definitions of stability are used, but all involve the solution decaying to an equilibrium solution as time .
The presence of switching in hybrid systems can sometimes lead to unexpected behaviour….
Hybrid Systems – Instability arising from switching physical systems are usually described by differential equations.
An example of decision-making and instability
a
c
b
“Car in the desert scenario”
Hybrid Systems - Instability arising from switching physical systems are usually described by differential equations.
System 1: Stable linear time-invariant differential equation:
dx1/dt=dx2/dt
dx2/dt=-2dx1/dt-0.2dx2/dt
Hybrid Systems - Instability arising from switching physical systems are usually described by differential equations.
System 2: Stable linear time-invariant differential equation:
dx1/dt=2dx2/dt
dx2/dt=-3dx1/dt-0.2dx2/dt
Hybrid Systems - Instability arising from switching physical systems are usually described by differential equations.
Switching between system 1 and system 2.
Introduce taxonomy of models. It turns out that most systems can be modelled using a fairly small set of model structures.
Study solutions, esp. numerical solutions/simulations
Can derive models from first principles or learn model from observations (or more usually by a combination of both approaches). We will not cover first principles modelling as very application specific. But will introduce machine learning approaches (including probabilistic reasoning ideas).
Course Outline | 6,662 | 24,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-51 | latest | en | 0.895057 |
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