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RE: How to compute SD of portfolio of asset classes?
1. The two approaches are mathematically equivalent, provided the variances
and covariances are estimated compatibly in the two approaches. All else
being equal, your empirical difference is likely due to Excel's COVAR()
function being misnamed--in the naming convention of the VAR(), VARP(),
STDEV(), and STDEVP() functions, COVAR() should be called COVARP(); Excel
provides no covariance analog for VAR() and STDEV().
In particular, =COVAR(data,data) gives the same value as =VARP(data), rather
than =VAR(data)
To calculate a covariance analog of VAR() and STDEV(), use either
The first approach is simpler for your application, since the correction
factor would be the same for all covariances.
2. A simpler approach would be
It doesn't avoid calculating the variance covariance matrix, but it does
avoid the other intermediates.
"nomail1983@xxxxxxxxxxx" wrote:
According to literature, the std dev of a portfolio of asset classes is
computed by:
Sqrt(Sum(Sum(w[i]*w[j]*Covar(X[i],X[j]), j=1,...,N), i=1,...,N)), where
w[i] is the allocation weight factor and X[i] is the historical
%returns for each of N asset classes.
But I believe I can also compute a std dev of the balanced porfolio by:
Stdev(Sum(w[i]*X[i,t], i=1,...,N), t=1,...,M), where X[i,t] is the
%return for each of N asset classes in each of M time periods.
The two results are different, at least empirically. Which one is the
correct one to use? Or when should I use each one for the purpose of
determining the std dev of a portfolio?
That is not really an Excel question. But I know there are a few sharp
folks in this forum who are schooled in statistics and financial
mathematics. I hope to hear from them.
And here __is__ a related Excel question: what is the best way to
formulate the first expression, namely Sqrt(Sum,(Sum(...)...))?
Here is what I did....
Assume that X[i,t] is in C5:L11. That is, C5:L5 is the 10-year
%returns for Class 1; C6:L6 for Class 2; etc for each of 7 asset
classes. Also assume that w[i] is in A5:A11 -- the allocation weight
factors for each asset class. (Of course, Sum(w[i]) = 100%.)
First, in B15:H21, I compute the matrix Covar(X[i],X[j]). Thus,
B15:B21 is Covar(X[1],X[j]), the covariance between the 1st class and
each of the classes; C15:C21 is Covar(X[2],X[j]), the covariance among
the 2nd class and all classes; etc. For example, the following
computes Covar(X[1],X[2]):
Then in B23:H23, I compute the array Sumproduct(w[j],Covar(X[i],X[j]),
j=1,...,7) for i=1,...,7. For example, the following computes
Finally, I compute
Sqrt(Sumproduct(w[i],Sumproduct(w[j],Covar(X[i],X[j], j=1,...,7),
i=1,...,7)) with the following array formula:
Is all that necessary? Or am I missing another way to perform the
computations that would obviate the need for one or more intermediate
Relevant Pages
• Re: Complex Number Covariance Matrix
... "snapshot" of each of these every time interval. ... believe that is what is usually meant by the covariance of complex quantities. ... which the above quantities are normalized using the
variances. ...
• Re: Confidence Interval from the covariance
... pmfg.pan@mail.pt (Paulo Gonçalves) wrote in message ... >> Are the variables whose covariance you're estimating jointly normal? ... >> Are their variances known, ...
• Re: mvnrnd, I need some better understanding
... I understand that 'mvnrnd' maintains the covariances across variables> within a given month, but, does 'mvnrnd' also maintain covariances from> month to month along the same variable? ...
Should I be applying the the variance multiplier to all elements of> the covariance matrix, or only to the diagonal (variances of the> covariance matirx)? ...
• Re: covariance matrix
... Can anyone tell me what the covariance matrix looks ... The covariance matrix is a 3 by 3 symmetric matrix in your case, with the diagonals being the variances and the off-diagonals being
the covariances. ... | {"url":"http://www.tech-archive.net/Archive/Excel/microsoft.public.excel.misc/2007-01/msg00057.html","timestamp":"2014-04-16T13:03:16Z","content_type":null,"content_length":"13018","record_id":"<urn:uuid:06f64383-c83e-4a50-b45d-02bf3a8fde6a>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00273-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to Shade Under a Normal Density in R
April 3, 2011
By Tony Cookson
easiest-to-find method for shading under a normal density is to use the polygon() command
. That link is to the first hit on Google for "Shading Under a Normal Curve in R." It works (like a charm), but it is not the most intuitive way to let users produce plots of normal densities.
In response to the standard polygon() approach, I wrote a function called shadenorm() that will produce a plot of a normal density with whatever area you want shaded. The code is available here:
Copy into an R script and run all of it if you want to use the shadenorm() command. To show you how to use it, I also recorded a video tutorial in R to demonstrate how to use the shadenorm() command.
Here is the code I use in the video:
There are a lot of applications where you may want to produce a plot with a normal density with some region shaded. I hope you find this function useful.
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Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 764.05077
Autor: Alavi, Yousef; Behzad, Mehdi; Erdös, Paul; Lick, Don R.
Title: Double vertex graphs. (In English)
Source: J. Comb. Inf. Syst. Sci. 16, No.1, 37-50 (1991).
Review: Let G be a (V,E) graph of order p \geq 2. The double vertex graph V[2](G) of G is the graph whose vertex set consists of all 2-subsets of V such that two distinct vertices {x,y} and {u,v} are
adjacent if and only if |{x,y}\cap{u,v}| = 1 and if x = u, then y and v are adjacent in G. For this class of graphs we develop basic properties and study regular, eulerian, bipartite graphs, as well
as general structural properties of these graphs.
Classif.: * 05C75 Structural characterization of types of graphs
05C45 Eulerian and Hamiltonian graphs
05C40 Connectivity
Keywords: regular graph; Eulerian graph; double vertex graph; bipartite graphs
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag
│Books │Problems │Set Theory │Combinatorics │Extremal Probl/Ramsey Th. │
│Graph Theory │Add.Number Theory│Mult.Number Theory│Analysis │Geometry │
│Probabability│Personalia │About Paul Erdös │Publication Year│Home Page │ | {"url":"http://www.emis.de/classics/Erdos/cit/76405077.htm","timestamp":"2014-04-20T18:51:40Z","content_type":null,"content_length":"3774","record_id":"<urn:uuid:c0807a70-963c-4a09-a690-196dbfe51529>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00330-ip-10-147-4-33.ec2.internal.warc.gz"} |
January 27th 2009, 11:52 AM
i have two questions.
so one problem is:
i don't remember if you apply the 2 to the (3x) or if it's like algebra and you just divide it by 2 to cancel out. oh and then you'd divide log 3x by log x right to combine them because they have
the same base?
the other one is:
15log base 1/3(1/27) + logbase 1/4(256)= u
so does the 15 go to the 1/3 or 1/27 or do you just divide it.
and you can't combine them because they have different bases right?
sorry if this is confusing..
January 27th 2009, 01:00 PM
i have two questions.
so one problem is:
i don't remember if you apply the 2 to the (3x) or if it's like algebra and you just divide it by 2 to cancel out. oh and then you'd divide log 3x by log x right to combine them because they have
the same base?
the other one is:
15log base 1/3(1/27) + logbase 1/4(256)= u
so does the 15 go to the 1/3 or 1/27 or do you just divide it.
and you can't combine them because they have different bases right?
sorry if this is confusing..
2 log(3x) - log x = 2
2 (log 3 + log x ) - log x = 2
can you do it from here?
$<br /> log_{10}1000 = 3<br />$
thats because 10^3 = 1000
can you work out your second equation now?
January 27th 2009, 02:55 PM
thanks for the help.
i understand what you meant for the first problem and i was able to solve it
and i understand the example you gave me, but i don't understand how it relates to the problem
January 27th 2009, 08:27 PM
$<br /> log_{(\frac{1}{3})}\frac{1}{27} = 3 <br />$
$<br /> log_{(\frac{1}{4})}256 = -4<br />$ | {"url":"http://mathhelpforum.com/algebra/70193-logs-print.html","timestamp":"2014-04-19T20:08:53Z","content_type":null,"content_length":"6527","record_id":"<urn:uuid:82bd26cf-0ef5-4afa-a047-fc56ea63ba99>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00363-ip-10-147-4-33.ec2.internal.warc.gz"} |
Model Checking in the Cloud
October 30th, 2012 | Tags: cloud computing, EDA, verification | Category: Cloud, EDA, Software
Model Checking in the Cloud
Last week I was invited in Cambridge, UK, to participate to a panel at the FMCAD conference (Formal Methods in Computer-Aided Design). The subject: “Model Checking in the Cloud”. With another four
people, we discussed the questions laid out by the panel moderator:
• How can model checking leverage the advantages of distributed and multi-core systems in the cloud?
• What are possible solutions beyond an “embarrassingly parallel” approach of running a single property per core?
• Is there a specific subset of properties that might be more suitable to this form of analysis?
• What issues need to be addressed for design houses to adopt this technology and will the current license model of EDA tools change to adapt to the new requirements?
These were my slides.
We had a lively discussion about cloud and verification. Of course I am a strong promoter of cloud-based solution for compute-intensive tasks like functional verification. In that very case though,
the conversation focused on model checking –not logic simulation or equivalence checking.
My take is that there is no “embarrassingly parallel” approach to model checking. The naïve approach that consists of running a single property per core is dubious, because (1) you cannot do load
balancing with this strategy –each property takes a very different time to be resolved, and (2) a single property can easily require years for a single core to tackle.
This means back to the white board to rethink the problem: how can we leverage a distributed computing system readily available in the cloud –a network of machines, each with its own cores, RAM, and
disk— for model checking?
Model checking is about verifying whether a sequential system satisfies some timing properties. These properties are usually expressed in some formal language like LTL (Linear Temporal Logic) or CTL
(Computation Tree Logic). For instance we can check that “if some event occurs, then some property is later always true”; or “some property eventually comes true”; or “some property holds infinitely
often in the future”.
One major component used in model checking is state space exploration: visit the states that can be reached from some initial states with a finite state machine. Another major, practical component,
is bounded model checking: instead of looking at the behavior of a system infinitely in the future, only consider a finite window in the future. This motivated my looking at three pillars of model
checking: explicit state exploration, implicit state exploration, and bounded model checking.
Explicit State Exploration
This simply consists of explicitly generating every successors state of a given state, then the successors of the successors, etc. The key point is recognizing which states have been already visited.
This problem is essentially memory limited. So in the context of cloud computing, this boils down to:
• How to efficiently partition the state space so that each node in the network can explore its state space?
• How to implement a very large distributed hash table for the visited states? We are talking Terabytes or Petabytes of data.
My conclusion: the technology for very large distributed hash table already exists. It would be interesting to see how such a large distributed system could help tackling industrial-scale model
checking problems like those encountered in chip design.
Implicit State Exploration
State exploration can also be done implicitly by using a data structure such as BDDs (Binary Decision Diagrams). In essence, we use a data structure to represent the characteristic function of a set
of states, as opposed to an explicit collection of states. This leads to very different algorithms that can dramatically outperform explicit state exploration.
My conclusion: the problem is still memory limited, but some attendees made a good case that it is rather time limited. The problem of state space partitioning remains, and there are arguably more
sophisticated solutions here. One key is to distribute the BDDs on multiple nodes of a network. This is definitely a challenge, a problem worthy of more research.
Bounded Model Checking
This consists of unrolling the system under verification k times, k being the width of the temporal widow we consider, and build a propositional formula expressing the property on that window. We
then check whether that formula is satisfiable or not (a SAT problem).
There are distributed SAT solvers, but scaling them up is non-trivial, as the nodes of the network need to share information. This is still a promising avenue of research.
“Don’t fight the exponential”
During the live exchanges between the panel and the audience, one notable objection was that model checking deals with vast state space (read: growing exponentially with the parameters of the
system), and that there was no point in “fighting the exponential”. There was no point in more computing power. Instead, as was argued by some attendees, we should be “smarter” about verification:
better models, better abstractions, better heuristics.
Of course we should look for “smarter” verification, with automated abstraction and more efficient algorithms. But we should not underestimate what we can achieve with raw computing power. What made
Deep Blue the world chess champion, beating the then reigning human champion Kasparov back in 1997, was its very large compute power, smartly pieced together. What make Google Translate useable for
basic translation are not decades of research in natural languages and semantics, but gigantic Markov chains built from Petabytes of data gathered by web crawlers. This is the same principle that
made IBM’s Watson the winner of the quiz show Jeopardy! against the two best human players in 2011. These systems are not inherently smart. They rely on huge computing power resulting from
intelligently assembling large amount of hardware, which processes very large amount of data.
To those arguing that more computing power is not a big help for problems that are exponential, I agree. However if the problem has a worst-case exponential complexity but exhibits a polynomial
behavior in many practical cases, then throwing more computing power does help. In particular, BDD and SAT solvers, and therefore model checking, can benefit from large compute networks.
The cloud makes possible building compute grids made of hundreds or thousands of nodes. This is now up to the research community to leverage these previously unseen, cheap computing capacities to
address ever more complex model checking problems. This however requires rethinking classical algorithms and recast them on very large distributed systems. And this will require being smart.
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October 30th, 2012 | Tags: cloud computing, EDA, verification | Category: Cloud, EDA, Software | {"url":"http://www.ocoudert.com/blog/2012/10/30/model-checking-in-the-cloud/","timestamp":"2014-04-18T18:12:09Z","content_type":null,"content_length":"97919","record_id":"<urn:uuid:b28e5d1b-6523-413d-9ef9-99128fd94d39>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00107-ip-10-147-4-33.ec2.internal.warc.gz"} |
University of Illinois
Mathematics Colloquium — Special Lecture
Spring 2012
Jonathan Chaika
University of Chicago
Interval exchange transformations
Interval exchange transformations are invertible, piecewise order preserving isometries of the unit interval with finitely many discontinuities. Starting from rotations of the circle, which they
generalize, this talk will present their connections to flows on flat surfaces, rational billiards and symbolic coding. Recent results on diophantine approximation for interval exchange
transformations will be presented.
Tuesday, January 24, 2012, 245 Altgeld Hall, 4:00 p.m. | {"url":"http://www.math.illinois.edu/Colloquia/12SP/chaika_jan24-12.html","timestamp":"2014-04-18T18:38:31Z","content_type":null,"content_length":"13852","record_id":"<urn:uuid:4e7c7193-f0f3-467f-9bf3-19d98a354a71>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00523-ip-10-147-4-33.ec2.internal.warc.gz"} |
A Fleet Of Nine Taxis Is To Be Dispatched To Three ... | Chegg.com
A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and 1 to airport C. IN how many distinct ways can this be accomplished?
A) If exactly one of the taxis is in need of repair what is the probability that it is dispatched to airport C?
B) If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?
Statistics and Probability | {"url":"http://www.chegg.com/homework-help/questions-and-answers/fleet-nine-taxis-dispatched-three-airports-way-three-go-airport-five-go-airport-b-1-airpor-q918404","timestamp":"2014-04-18T09:59:42Z","content_type":null,"content_length":"21240","record_id":"<urn:uuid:007a7905-ed9f-4558-86ae-17158879bda3>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00161-ip-10-147-4-33.ec2.internal.warc.gz"} |
the encyclopedic entry of stopped down
, particularly as it relates to
, the
depth of field
(DOF) is the portion of a scene that appears sharp in the image. Although a
can precisely focus at only one distance, the decrease in sharpness is gradual on either side of the focused distance, so that within the DOF, the unsharpness is imperceptible under normal viewing
For some images, such as landscapes, a large DOF may be appropriate, while for others, such as portraits, a small DOF may be more effective.
The DOF is determined by the subject distance (that is, the distance to the plane that is perfectly in focus), the lens focal length, and the lens f-number (relative aperture). Except at close-up
distances, DOF is approximately determined by the subject magnification and the lens f-number. For a given f-number, increasing the magnification, either by moving closer to the subject or using a
lens of greater focal length, decreases the DOF; decreasing magnification increases DOF. For a given subject magnification, increasing the f-number (decreasing the aperture diameter) increases the
DOF; decreasing f-number decreases DOF.
When focus is set to the hyperfocal distance, the DOF extends from half the hyperfocal distance to infinity, and is the largest DOF possible for a given f-number.
The advent of digital technology in photography has provided additional means of controlling the extent of image sharpness; some methods allow DOF that would be impossible with traditional
techniques, and some allow the DOF to be determined after the image is made.
Apparent sharpness
Precise focus is possible at only one distance; at that distance, a point object will produce a point image. At any other distance, a point object is defocused, and will produce a blur spot shaped
like the aperture, which for the purpose of analysis is usually assumed to be circular. When this circular spot is sufficiently small, it is indistinguishable from a point, and appears to be in
focus; it is rendered as “acceptably sharp”. The diameter of the circle increases with distance from the point of focus; the largest circle that is indistinguishable from a point is known as the
acceptable circle of confusion, or informally, simply as the circle of confusion. The acceptable circle of confusion is influenced by visual acuity, viewing conditions, and the amount by which the
image is enlarged. The increase of the circle diameter with defocus is gradual, so the limits of depth of field are not hard boundaries between sharp and unsharp.
Several other factors, such as subject matter, movement, and the distance of the subject from the camera, also influence when a given defocus becomes noticeable.
For a 35 mm motion picture, the image area on the negative is roughly 22 mm by 16 mm (0.87 in by 0.63 in). The limit of tolerable error is usually set at 0.05 mm (0.002 in) diameter. For 16 mm film,
where the image area is smaller, the tolerance is stricter, 0.025 mm (0.001 in). Standard depth-of-field tables are constructed on this basis, although generally 35 mm productions set it at 0.025 mm
(0.001 in). Note that the acceptable circle of confusion values for these formats are different because of the relative amount of magnification each format will need in order to be projected on a
full-sized movie screen.
(A table for 35 mm still photography would be somewhat different since more of the film is used for each image and the amount of enlargement is usually much less.)
The image format size also will affect the depth of field. The larger the format size, the longer a lens will need to be to capture the same framing as a smaller format. In motion pictures, for
example, a frame with a 12 degree horizontal field of view will require a 50 mm lens on 16 mm film, a 100 mm lens on 35 mm film, and a 250 mm lens on 65 mm film. Conversely, using the same focal
length lens with each of these formats will yield a progressively wider image as the film format gets larger: a 50 mm lens has a horizontal field of view of 12 degrees on 16 mm film, 23.6 degrees on
35 mm film, and 55.6 degrees on 65 mm film. What this all means is that because the larger formats require longer lenses than the smaller ones, they will accordingly have a smaller depth of field.
Therefore, compensations in exposure, framing, or subject distance need to be made in order to make one format look like it was filmed in another format.
Above: DOF at various apertures
Above: Selective focus
Effect of lens aperture
For a given subject framing and camera position, the DOF is controlled by the lens aperture diameter, which is usually specified as the f-number, the ratio of lens focal length to aperture diameter.
Reducing the aperture diameter (increasing the f-number) increases the DOF; however, it also reduces the amount of light transmitted, and increases diffraction, placing a practical limit on the
extent to which DOF can be increased by reducing the aperture diameter.
Motion pictures make only limited use of this control; to produce a consistent image quality from shot to shot, cinematographers usually choose a single aperture setting for interiors and another for
exteriors, and adjust exposure through the use of camera filters or light levels. Aperture settings are adjusted more frequently in still photography, where variations in depth of field are used to
produce a variety of special effects.
Camera movements and DOF
When the lens axis is perpendicular to the image plane, as is normally the case, the plane of focus (POF) is parallel to the image plane, and the DOF extends between parallel planes on either side of
the POF. When the lens axis is not perpendicular to the image plane, the POF is no longer parallel to the image plane; the ability to rotate the POF is known as the Scheimpflug principle. Rotation of
the POF is accomplished with camera movements (tilt, a rotation of the lens about a horizontal axis, or swing, a rotation about a vertical axis). Tilt and swing are available on most view cameras,
and are also available with specific lenses on some small- and medium-format cameras.
When the POF is rotated, the near and far limits of DOF are no longer parallel; the DOF becomes wedge-shaped, with the apex of the wedge nearest the camera. With tilt, the height of the DOF increases
with distance from the camera; with swing, the width of the DOF increases with distance.
Rotating the POF with tilt or swing (or both) can be used either to maximize or minimize the part of an image that is within the DOF.
Obtaining maximum DOF
Lens DOF scales
Many lenses for small- and medium-format cameras include scales that indicate the DOF for a given focus distance and f-number; the 35 mm lens in the image above is typical. That lens includes
distance scales in feet and meters; when a marked distance is set opposite the large white index mark, the focus is set to that distance. The DOF scale below the distance scales includes markings on
either side of the index that correspond to f-numbers; when the lens is set to a given f-number, the DOF extends between the distances that align with the f-number markings.
Zone focusing
When the 35 mm lens above is set to f/11 and focused at approximately 1.4 m, the DOF (a “zone” of acceptable sharpness) extends from 1 m to 2 m. Conversely, the required focus and f-number can be
determined from the desired DOF limits by locating the near and far DOF limits on the lens distance scale and setting focus so that the index mark is centered between the near and far distances; the
required f-number is determined by finding the markings on the DOF scale that are closest to the near and far distances. For the 35 mm lens above, if it were desired for the DOF to extend from 1 m to
2 m, focus would be set to approximately 1.4 m and the aperture set to f/11. The DOF limits can be determined from a scene by focusing on the farthest object to be within the DOF and noting the
distance on the lens distance scale, and repeating the process for the nearest object to be within the DOF. If the near and far distances fall outside the largest f-number markings on the DOF scale,
the desired DOF cannot be obtained; for example, with the 35 mm lens above, it is not possible to have the DOF extend from 0.7 m to infinity.
Some distance scales have markings for only a few distances; for example, the 35 mm lens above shows only 3 ft and 5 ft on its upper scale. Using other distances for DOF limits requires visual
interpolation between marked distances; because the distance scale is nonlinear, accurate interpolation can be difficult. In most cases, English and metric distance markings are not coincident, so
using both scales to note focused distances can sometimes lessen the need for interpolation. Many autofocus lenses have smaller distance and DOF scales and fewer markings than do comparable
manual-focus lenses, so that determining focus and f-number from the scales on an autofocus lens may be more difficult than with a comparable manual-focus lens. In most cases, using the lens DOF
scales on an autofocus lens requires that the lens or camera body be set to manual focus.
On a view camera, the focus and f-number can be obtained by measuring the focus spread and performing simple calculations; the procedure is described in more detail in the section Focus and f-number
from DOF limits. Some view cameras include DOF calculators that indicate focus and f-number without the need for any calculations by the photographer.
Hyperfocal distance
The hyperfocal distance is the nearest focus distance at which the DOF extends to infinity; focusing the camera at the hyperfocal distance results in the largest possible depth of field for a given
f-number. Focusing beyond the hyperfocal distance does not increase the far DOF (which already extends to infinity), but it does decrease the DOF in front of the subject, decreasing the total DOF.
Some photographers refer to this as “wasting DOF”; however, see The object field method below for a rationale for doing so. If the lens includes a DOF scale, the hyperfocal distance can be set by
aligning the infinity mark on the distance scale with the mark on the DOF scale corresponding to the f-number to which the lens is set. For example, with the 35 mm lens shown above set to f/11,
aligning the infinity mark with the ‘11’ to the left of the index mark on the DOF scale would set the focus to the hyperfocal distance. Focusing on the hyperfocal distance is a special case of zone
focusing in which the far limit of DOF is at infinity.
The object field method
Traditional depth-of-field formulas and tables assume equal circles of confusion for near and far objects. Some authors, such as Merklinger (1992), have suggested that distant objects often need to
be much sharper to be clearly recognizable, whereas closer objects, being larger on the film, do not need to be so sharp. The loss of detail in distant objects may be particularly noticeable with
extreme enlargements. Achieving this additional sharpness in distant objects usually requires focusing beyond the hyperfocal distance, sometimes almost at infinity. For example, if photographing a
cityscape with a traffic bollard in the foreground, this approach, termed the object field method by Merklinger, would recommend focusing very close to infinity, and stopping down to make the bollard
sharp enough. With this approach, foreground objects cannot always be made perfectly sharp, but the loss of sharpness in near objects may be acceptable if recognizability of distant objects is
Moritz von Rohr also used an object field method, but unlike Merklinger, he used the conventional criterion of a maximum circle of confusion diameter in the image plane, leading to unequal front and
rear depths of field.
Limited DOF: selective focus
Depth of field can be anywhere from a fraction of a millimeter to virtually infinite. In some cases, such as landscapes, it may be desirable to have the entire image in focus, and a large DOF is
appropriate. In other cases, artistic considerations may dictate that only a part of the image be in focus, emphasizing the subject while de-emphasizing the background, perhaps giving only a
suggestion of the environment (
Langford 1973
, 81). For example, a common technique in
horror films
is a closeup of a person's face, with someone just behind that person visible but out of focus. A
still photograph might use a small DOF to isolate the subject from a distracting background. The use of limited DOF to emphasize one part of an image is known as
selective focus
differential focus
Although a small DOF implies that other parts of the image will be unsharp, it does not, by itself, determine how unsharp those parts will be. The amount of background (or foreground) blur depends on
the distance from the plane of focus, so if a background is close to the subject, it may be difficult to blur sufficiently even with a small DOF. In practice, the lens f-number is usually adjusted
until the background or foreground is acceptably blurred, often without direct concern for the DOF.
Sometimes, however, it is desirable to have the entire subject sharp while ensuring that the background is sufficiently unsharp. When the distance between subject and background is fixed, as is the
case with many scenes, the DOF and the amount of background blur are not independent. Although it is not always possible to achieve both the desired subject sharpness and the desired background
unsharpness, several techniques can be used to increase the separation of subject and background.
For a given scene and subject magnification, the background blur increases with lens focal length. If it is not important that background objects be unrecognizable, background de-emphasis can be
increased by using a lens of longer focal length and increasing the subject distance to maintain the same magnification. This technique requires that sufficient space in front of the subject be
available; moreover, the perspective of the scene changes because of the different camera position, and this may or may not be acceptable.
The situation is not as simple if it is important that a background object, such as a sign, be unrecognizable. The magnification of background objects also increases with focal length, so with the
technique just described, there is little change in the recognizability of background objects. However, a lens of longer focal length may still be of some help; because of the narrower angle of view,
a slight change of camera position may suffice to eliminate the distracting object from the field of view.
Although tilt and swing are normally used to maximize the part of the image that is within the DOF, they also can be used, in combination with a small f-number, to give selective focus to a plane
that isn't perpendicular to the lens axis. With this technique, it is possible to have objects at greatly different distances from the camera in sharp focus and yet have a very shallow DOF. The
effect can be interesting because it differs from what most viewers are accustomed to seeing.
Near:far distribution
The DOF beyond the subject is always greater than the DOF in front of the subject. When the subject is at the hyperfocal distance or beyond, the far DOF is infinite; as the subject distance
decreases, near:far DOF ratio increases, approaching unity at high magnification. The oft-cited “rule” that 1/3 of the DOF is in front of the subject and 2/3 is beyond is true only when the subject
distance is 1/3 the hyperfocal distance.
Depth of field formulas
The basis of these formulas is given in the section Derivation of the DOF formulas; refer to the diagram in that section for illustration of the quantities discussed below.
Hyperfocal Distance
Let $f$ be the lens focal length, $N$ be the lens f-number, and $c$ be the circle of confusion for a given image format. The hyperfocal distance $H$ is given by
$H approx frac \left\{f^2\right\} \left\{N c\right\}$
Moderate-to-large distances
Let $s$ be the distance at which the camera is focused (the “subject distance”). When $s$ is large in comparison with the lens focal length, the distance $D_\left\{mathrm N\right\}$ from the camera
to the near limit of DOF and the distance $D_\left\{mathrm F\right\}$ from the camera to the far limit of DOF are
$D_\left\{mathrm N\right\} approx frac \left\{H s\right\} \left\{H + s\right\}$
$D_\left\{mathrm F\right\} approx frac \left\{H s\right\} \left\{H - s\right\} mbox\left\{ for \right\} s < H$
When the subject distance is the hyperfocal distance,
$D_\left\{mathrm F\right\} = infty$
$D_\left\{mathrm N\right\} = frac H 2$
The depth of field $D_\left\{mathrm F\right\} - D_\left\{mathrm N\right\}$ is
mathrm {DOF} approx frac {2 Hs^2} {H^2 - s^2} mbox{ for } s < H
For $s ge H$, the far limit of DOF is at infinity and the DOF is infinite; of course, only objects at or beyond the near limit of DOF will be recorded with acceptable sharpness.
Substituting for $H$ and rearranging, DOF can be expressed as
$mathrm \left\{DOF\right\} approx frac \left\{2 N c f^2 s^2\right\} \left\{f^4 - N^2 c^2 s^2\right\}$
Thus, for a given image format, depth of field is determined by three factors: the focal length of the lens, the f-number of the lens opening (the aperture), and the camera-to-subject distance.
When the subject distance $s$ approaches the focal length, using the formulas given above can result in significant errors. For close-up work, the hyperfocal distance has little applicability, and it
usually is more convenient to express DOF in terms of image magnification. Let $m$ be the magnification; when the subject distance is small in comparison with the hyperfocal distance,
$mathrm \left\{DOF\right\} approx 2 N c left \left(frac \left\{m + 1\right\} \left\{m^2\right\} right \right),$
so that for a given magnification, DOF is independent of focal length. Stated otherwise, for the same subject magnification, all focal lengths give approximately the same DOF. This statement is true
only when the subject distance is small in comparison with the hyperfocal distance, however.
The discussion thus far has assumed a symmetrical lens for which the entrance and exit pupils coincide with the front and rear nodal planes, and for which the pupil magnification (the ratio of exit
pupil diameter to that of the entrance pupil) is unity. Although this assumption usually is reasonable for large-format lenses, it often is invalid for medium- and small-format lenses.
When $s ll H$, the DOF for an asymmetrical lens is
$mathrm \left\{DOF\right\} approx frac \left\{2 N c \left(1 + m/P\right)\right\}\left\{m^2\right\},$
where $P$ is the pupil magnification. When the pupil magnification is unity, this equation reduces to that for a symmetrical lens.
Except for close-up and macro photography, the effect of lens asymmetry is minimal. At unity magnification, however, the errors from neglecting the pupil magnification can be significant. Consider a
telephoto lens with $P = 0.5$ and a retrofocus wide-angle lens with $P = 2$, at $m = 1.0$. The asymmetrical-lens formula gives $mathrm \left\{DOF\right\} = 6 N c$ and $mathrm \left\{DOF\right\} = 3 N
c$, respectively. The symmetrical-lens formula gives $mathrm \left\{DOF\right\} = 4 N c$ in either case. The errors are −33% and 33%, respectively.
Focus and f-number from DOF limits
Not all images require that sharpness extend to infinity; for given near and far DOF limits $D_\left\{mathrm N\right\}$ and $D_\left\{mathrm F\right\}$, the required f-number is smallest when focus
is set to
$s = frac \left\{2 D_\left\{mathrm N\right\} D_\left\{mathrm F\right\} \right\}$
{D_{mathrm N} + D_{mathrm F} }
When the subject distance is large in comparison with the lens focal length, the required f-number is
$N approx frac \left\{f^2\right\} \left\{c\right\}$
frac {D_{mathrm F} - D_{mathrm N} } {2 D_{mathrm N} D_{mathrm F} }
In practice, these settings usually are determined on the image side of the lens, using measurements on the bed or rail with a view camera, or using lens DOF scales on manual-focus lenses for small-
and medium-format cameras. If $v_\left\{mathrm N\right\}$ and $v_\left\{mathrm F\right\}$ are the image distances that correspond to the near and far limits of DOF, the required f-number is minimized
when the image distance $v$ is
$v approx frac \left\{ v_\left\{mathrm N\right\} + v_\left\{mathrm F\right\} \right\} \left\{2\right\}$
= v_{mathrm F} + frac { v_{mathrm N} - v_{mathrm F} } {2}
In practical terms, focus is set to halfway between the near and far image distances. The required f-number is
$N approx frac \left\{ v_\left\{mathrm N\right\} - v_\left\{mathrm F\right\} \right\} \left\{ 2 c \right\}$
The image distances are measured from the camera's image plane to the lens's image nodal plane, which is not always easy to locate. In most cases, focus and f-number can be determined with sufficient
accuracy using the approximate formulas above, which require only the difference between the near and far image distances; view camera users often refer to the difference $v_\left\{mathrm N\right\} ,
- , v_\left\{mathrm F\right\}$ as the focus spread. Most lens DOF scales are based on the same concept.
Foreground and background blur
If a subject is at distance $s$ and the foreground or background is at distance $D$, let the distance between the subject and the foreground or background be indicated by
$x_mathrm d = left | D - s right |$
The blur disk diameter $b$ of a detail at distance $x_mathrm d$ from the subject can be expressed as a function of the focal length $f$, subject magnification $m_mathrm\left\{s\right\}$, and f-number
$N$ according to
$b = frac \left\{fm_mathrm s\right\} N frac \left\{ x_mathrm d \right\} \left\{ s pm x_mathrm d\right\}$
The minus sign applies to a foreground object, and the plus sign applies to a background object.
The blur increases with the distance from the subject; when $b le c$, the detail is within the depth of field, and the blur is imperceptible. If the detail is only slightly outside the DOF, the blur
may be only barely perceptible.
For a given subject magnification, f-number, and distance from the subject of the foreground or background detail, the degree of detail blur varies with the lens focal length. For a background
detail, the blur increases with focal length; for a foreground detail, the blur decreases with focal length. For a given scene, the positions of the subject, foreground, and background usually are
fixed, and the distance between subject and the foreground or background remains constant regardless of the camera position; however, to maintain constant magnification, the subject distance must
vary if the focal length is changed. For small distance between the foreground or background detail, the effect of focal length is small; for large distance, the effect can be significant. For a
reasonably distant background detail, the blur disk diameter is
$b approx frac \left\{fm_mathrm s\right\} \left\{N\right\} ,$
depending only on focal length.
The blur diameter of foreground details is very large if the details are close to the lens.
The ratio $b/c$ is independent of camera format; the blur then is in terms of circles of confusion.
The magnification of the detail also varies with focal length; for a given detail, the ratio of the blur disk diameter to imaged size of the detail is independent of focal length, depending only on
the detail size and its distance from the subject. This ratio can be useful when it is important that the background be recognizable (as usually is the case in evidence or surveillance photography),
or unrecognizable (as might be the case for a pictorial photographer using selective focus to isolate the subject from a distracting background). As a general rule, an object is recognizable if the
blur disk diameter is one-tenth to one-fifth the size of the object or smaller (Williams 1990, 205), and unrecognizable when the blur disk diameter is the object size or greater.
The effect of focal length on background blur is illustrated in van Walree's article on Depth of field
Practical complications
The distance scales on most medium- and small-format lenses indicate distance from the camera's image plane. Most DOF formulas, including those in this article, use the object distance $s$ from the
lens's object nodal plane, which often is not easy to locate. Moreover, for many zoom lenses and internal-focusing non-zoom lenses, the location of the object nodal plane, as well as focal length,
changes with subject distance. When the subject distance is large in comparison with the lens focal length, the exact location of the object nodal plane is not critical; the distance is essentially
the same whether measured from the front of the lens, the image plane, or the actual nodal plane. The same is not true for close-up photography; at unity magnification, a slight error in the location
of the object nodal plane can result in a DOF error greater than the errors from any approximations in the DOF equations.
The asymmetrical lens formulas require knowledge of the pupil magnification, which usually is not specified for medium- and small-format lenses. The pupil magnification can be estimated by looking
into the front and rear of the lens and measuring the diameters of the apparent apertures, and computing the ratio (rear diameter divided by front diameter). However, for many zoom lenses and
internal-focusing non-zoom lenses, the pupil magnification changes with subject distance, and several measurements may be required.
Most DOF formulas, including those discussed in this article, employ several simplifications:
1. Paraxial (Gaussian) optics is assumed, and technically, the formulas are valid only for rays that are infinitesimally close to the lens axis. However, Gaussian optics usually is more than
adequate for determining DOF, and non-paraxial formulas are sufficiently complex that requiring their use would make determination of DOF impractical in most cases.
2. Lens aberrations are ignored. Including the effects of aberrations is nearly impossible, because doing so requires knowledge of the specific lens design. Moreover, in well-designed lenses, most
aberrations are well corrected, and at least near the optical axis, often are almost negligible when the lens is stopped down 2–3 steps from maximum aperture. Because lenses usually are stopped
down at least to this point when DOF is of interest, ignoring aberrations usually is reasonable. Not all aberrations are reduced by stopping down, however, so actual sharpness may be slightly
less than predicted by DOF formulas.
3. Diffraction is ignored. DOF formulas imply that any arbitrary DOF can be achieved by using a sufficiently large f-number. Because of diffraction, however, this isn't quite true. Once a lens is
stopped down to where most aberrations are well corrected, stopping down further will decrease sharpness in the center of the field. At the DOF limits, however, further stopping down decreases
the size of the defocus blur spot, and the overall sharpness may increase. Consequently, choosing an f-number sometimes involves a tradeoff between center and edge sharpness, although viewers
typically prefer uniform sharpness to slightly greater center sharpness. The choice, of course, is subjective, and may depend upon the particular image. Eventually, the defocus blur spot becomes
negligibly small, and further stopping down serves only to decrease sharpness even at DOF limits. Typically, diffraction at DOF limits becomes significant only at fairly large f-numbers; because
large f-numbers typically require long exposure times, motion blur often causes greater loss of sharpness than does diffraction. Combined defocus and diffraction is discussed in Hansma (1996),
Conrad's Depth of Field in Depth (PDF), and Jacobson's Photographic Lenses Tutorial
4. Post-capture manipulation of the image is ignored. Sharpening via techniques such as deconvolution or unsharp mask can increase the DOF in the final image, particularly when the original image
has a large DOF. Conversely, image noise reduction can reduce the DOF.
5. For digital capture with color filter array sensors, demosaicing is ignored. Demosaicing alone would normally reduce the DOF, but the demosaicing algorithm used might also include sharpening.
The lens designer cannot restrict analysis to Gaussian optics and cannot ignore lens aberrations. However, the requirements of practical photography are less demanding than those of lens design, and
despite the simplifications employed in development of most DOF formulas, these formulas have proven useful in determining camera settings that result in acceptably sharp pictures. It should be
recognized that DOF limits are not hard boundaries between sharp and unsharp, and that there is little point in determining DOF limits to a precision of many significant figures.
DOF vs. format size
To a first approximation, DOF is inversely proportional to format size. More precisely, if photographs with the same final-image size are taken in two different camera formats at the same subject
distance with the same field of view and f-number, the DOF is, to a first approximation, inversely proportional to the format size. Strictly speaking, this is true only when the subject distance is
large in comparison with the focal length and small in comparison with the hyperfocal distance, for both formats, but it nonetheless is generally useful for comparing results obtained from different
To maintain the same field of view, the lens focal lengths must be in proportion to the format sizes. Assuming, for purposes of comparison, that the 4×5 format is four times the size of 35 mm format,
if a 4×5 camera used a 300 mm lens, a 35 mm camera would need a 75 mm lens for the same field of view. For the same f-number, the image made with the 35 mm camera would have four times the DOF of the
image made with the 4×5 camera.
In many cases, the DOF is fixed by the requirements of the desired image. For a given DOF and field of view, the required f-number is proportional to the format size. For example, if a 35 mm camera
required 11, a 4×5 camera would require 45 to give the same DOF. For the same ISO speed, the exposure time on the 4×5 would be sixteen times as long; if the 35 camera required 1/250 second, the 4×5
camera would require 1/15 second. In windy conditions, the exposure time with the larger camera might allow motion blur. Adjusting the f-number to the camera format is equivalent to maintaining the
same absolute aperture diameter.
The greater DOF with the smaller format can be either an advantage or a disadvantage, depending on the desired effect. For the same amount of foreground and background blur, a small-format camera
requires a smaller f-number and allows a shorter exposure time than a large-format camera; however, many point-and-shoot digital cameras cannot provide a very shallow DOF. For example, a
point-and-shoot digital camera with a 1/1.8″ sensor (7.18 mm × 5.32 mm) at a normal focal length and 2.8 has the same DOF as a 35 mm camera with a normal lens at 13.
In some cases, camera movements (tilt or swing) can be used to better fit the DOF to the scene, and achieve the required sharpness at a smaller f-number.
semiconductor photolithography
applications, depth of field is extremely important as
integrated circuit
layout features must be printed with high accuracy at extremely small size. The difficulty is that the
surface is not perfectly flat, but may vary by several
. Even this small variation causes some distortion in the projected image, and results in unwanted variations in the resulting pattern. Thus photolithography engineers take extreme measures to
maximize the optical depth of field of the photolithography equipment. To minimize this distortion further, chip makers like
are forced to use
chemical mechanical polishing
machines to make the wafer surface even flatter before lithographic patterning.
Ophthalmology and optometry
A person may sometimes experience better vision in daylight than at night because of an increased depth of field due to constriction of the
Digital techniques for increasing DOF
Focus stacking
Focus stacking
is a
digital image processing
technique which combines multiple images taken at different focus distances to give a resulting image with a greater depth of field than any of the individual source images. Available programs for
multi-shot DOF enhancement include Syncroscopy AutoMontage, PhotoAcute Studio,
Helicon Focus
Getting sufficient depth of field can be particularly challenging in macro photography. The images to the right illustrate the increase in DOF that can be achieved by combining multiple exposures.
Wavefront coding
Wavefront coding
is a method that convolves rays in such a way to provide an image where fields are in focus simultaneously with all planes out of focus by a constant amount.
Plenoptic cameras
plenoptic camera
uses a microlens array to capture
4D light field
information about a scene.
Derivation of the DOF formulas
DOF limits
A symmetrical lens is illustrated at right. The subject, at distance
, is in focus at image distance
. Point objects at distances
$D_mathrm F$
$D_mathrm N$
would be in focus at image distances
$v_mathrm F$
$v_mathrm N$
, respectively; at image distance
, they are imaged as blur spots. The depth of field is controlled by the aperture stop diameter
; when the blur spot diameter is equal to the acceptable
circle of confusion $c$
, the near and far limits of DOF are at
$D_mathrm N$
$D_mathrm F$
. From similar triangles,
$frac \left\{v_mathrm N - v\right\} \left\{v_mathrm N\right\} = frac c d$
$frac \left\{v- v_mathrm F\right\} \left\{v_mathrm F\right\} = frac c d$
It usually is more convenient to work with the lens f-number than the aperture diameter; the f-number $N$ is related to the lens focal length $f$ and the aperture diameter $d$ by
$N = frac f d,;$
substituting into the previous equations and rearranging gives
$v_mathrm N = frac \left\{fv\right\} \left\{f - Nc\right\}$
$v_mathrm F = frac \left\{fv\right\} \left\{f + Nc\right\}$
The image distance $v$ is related to an object distance $s$ by the thin lens equation
$frac 1 s + frac 1 v = frac 1 f,;$
Substituting into the two previous equations and rearranging gives the near and far limits of DOF:
$D_\left\{mathrm N\right\} = frac \left\{s f^2\right\} \left\{f^2 + N c \left(s - f \right) \right\}$
$D_\left\{mathrm F\right\} = frac \left\{s f^2\right\} \left\{f^2 - N c \left(s - f \right) \right\}$
Hyperfocal distance
Setting the far limit of DOF
$D_\left\{mathrm F\right\}$
to infinity and solving for the focus distance
$s = H = frac \left\{f^2\right\} \left\{N c\right\} + f,$
where $H$ is the hyperfocal distance. Setting the subject distance to the hyperfocal distance and solving for the near limit of DOF gives
$D_\left\{mathrm N\right\} = frac \left\{f^2 / \left(N c \right) + f\right\} \left\{2\right\} = frac \left\{H\right\}\left\{2\right\}$
For any practical value of $H$, the focal length is negligible in comparison, so that
$H approx frac \left\{f^2\right\} \left\{N c\right\}$
Substituting the approximate expression for hyperfocal distance into the formulas for the near and far limits of DOF gives
$D_\left\{mathrm N\right\} = frac \left\{H s\right\}\left\{H + \left(s - f \right)\right\}$
$D_\left\{mathrm F\right\} = frac \left\{H s\right\}\left\{H - \left(s - f \right)\right\}$
Combining, the depth of field $D_\left\{mathrm F\right\} - D_\left\{mathrm N\right\}$ is
mathrm {DOF} = frac {2 H s (s - f )} {H^2 - (s - f )^2} mbox{ for } s < H
Moderate-to-large distances
When the subject distance is large in comparison with the lens focal length,
$D_\left\{mathrm N\right\} approx frac \left\{H s\right\} \left\{H + s\right\}$
$D_\left\{mathrm F\right\} approx frac \left\{H s\right\} \left\{H - s\right\} mbox\left\{ for \right\} s < H$
mathrm {DOF} approx frac {2 H s^2} {H^2 - s^2} mbox{ for } s < H
For $s ge H$, the far limit of DOF is at infinity and the DOF is infinite; of course, only objects at or beyond the near limit of DOF will be recorded with acceptable sharpness.
When the subject distance $s$ approaches the lens focal length, the focal length no longer is negligible, and the approximate formulas above cannot be used without introducing significant error. At
close distances, the hyperfocal distance has little applicability, and it usually is more convenient to express DOF in terms of magnification. Substituting
$s = frac \left\{m + 1\right\} \left\{m\right\} f$
$s - f = frac \left\{f\right\} \left\{m\right\}$
into the formula for DOF and rearranging gives
mathrm {DOF} = frac {2 f (m + 1 ) / m } { (f m ) / (N c ) - (N c ) / (f m ) }
At the hyperfocal distance, the terms in the denominator are equal, and the DOF is infinite. As the subject distance decreases, so does the second term in the denominator; when $s ll H$, the second
term becomes small in comparison with the first, and
$mathrm \left\{DOF\right\} approx 2 N c left \left(frac \left\{m + 1\right\} \left\{m^2\right\} right \right),$
so that for a given magnification, DOF is independent of focal length. Stated otherwise, for the same subject magnification, all focal lengths for a given image format give approximately the same
DOF. This statement is true only when the subject distance is small in comparison with the hyperfocal distance, however. Multiplying the numerator and denominator of the exact formula by
$frac \left\{N c m\right\} \left\{f\right\}$
$mathrm \left\{DOF\right\} = frac$
{2 N c left (m + 1 right )} {m^2 - left (frac {N c} {f} right )^2}
Decreasing the focal length $f$ increases the second term in the denominator, decreasing the denominator and increasing the value of the right-hand side, so that a shorter focal length gives greater
DOF. The effect of focal length is greatest near the hyperfocal distance, and decreases as subject distance is decreased. However, the near/far perspective will differ for different focal lengths, so
the difference in DOF may not be readily apparent. When the subject distance is small in comparison with the hyperfocal distance, the effect of focal length is negligible, and, as noted above, the
DOF essentially is independent of focal length.
Near:far DOF ratio
From the “exact” equations for near and far limits of DOF, the DOF in front of the subject is
$s - D_\left\{mathrm N\right\} = frac \left\{Ncs\left(s - f\right)\right\} \left\{f^2 + Nc\left(s - f\right)\right\},,$
and the DOF beyond the subject is
$D_\left\{mathrm F\right\} - s = frac \left\{Ncs\left(s - f\right)\right\} \left\{f^2 - Nc\left(s - f\right)\right\}$
The near:far DOF ratio is
$frac \left\{s - D_\left\{mathrm N\right\}\right\} \left\{D_\left\{mathrm F\right\} - s\right\}$
= frac {f^2 - Nc(s - f)} {f^2 + Nc(s - f)}
This ratio is always less than unity; at moderate-to-large subject distances, $f ll s$, and
$frac \left\{s - D_\left\{mathrm N\right\}\right\} \left\{D_\left\{mathrm F\right\} - s\right\}$
approx frac {f^2 - Ncs} {f^2 + Ncs} = frac {H - s} {H + s}
When the subject is at the hyperfocal distance or beyond, the far DOF is infinite, and the near:far ratio is zero. It's commonly stated that approximately 1/3 of the DOF is in front of the subject
and approximately 2/3 is beyond; however, this is true only when $s approx H/3$.
At closer subject distances, it's often more convenient to express the DOF ratio in terms of the magnification
$m = frac f \left\{s - f\right\}$
Substitution into the “exact” equation for DOF ratio gives
$frac \left\{s - D_\left\{mathrm N\right\}\right\} \left\{D_\left\{mathrm F\right\} - s\right\}$
= frac {m - Nc/f} {m + Nc/f}
As magnification increases, the near:far ratio approaches a limiting value of unity.
Focus and f-number from DOF limits
Not all images require that sharpness extend to infinity; the equations for the DOF limits can be combined to eliminate $Nc$ and solve for the subject distance. For given near and far DOF limits $D_\
left\{mathrm N\right\}$ and $D_\left\{mathrm F\right\}$, the subject distance is
$s = frac \left\{2 D_\left\{mathrm N\right\} D_\left\{mathrm F\right\} \right\}$
{D_{mathrm N} + D_{mathrm F} }
The equations for DOF limits also can be combined to eliminate $s$ and solve for the required f-number, giving
$N = frac \left\{f^2\right\} \left\{c\right\}$
frac {D_{mathrm F} - D_{mathrm N} } {D_{mathrm F} (D_{mathrm N} - f ) + D_{mathrm N} (D_{mathrm F} - f ) }
When the subject distance is large in comparison with the lens focal length, this simplifies to
$N approx frac \left\{f^2\right\} \left\{c\right\}$
frac {D_{mathrm F} - D_{mathrm N} } {2 D_{mathrm N} D_{mathrm F} }
Most discussions of DOF concentrate on the object side of the lens, but the formulas are simpler and the measurements usually easier to make on the image side. If $v_\left\{mathrm N\right\}$ and $v_\
left\{mathrm F\right\}$ are the image distances that correspond to the near and far limits of DOF, the required f-number is minimum when the image distance $v$ is
$v = frac \left\{2 v_\left\{mathrm N\right\} v_\left\{mathrm F\right\} \right\}$
{v_{mathrm N} + v_{mathrm F} }
The required f-number is
$N = frac \left\{f\right\} \left\{c\right\}$
frac { v_{mathrm N} - v_{mathrm F} } {v_{mathrm N} + v_{mathrm F} }
The image distances are measured from the camera's image plane to the lens's image nodal plane, which is not always easy to locate. In most cases, focus and f-number can be determined with sufficient
accuracy using the approximate formulas
$v approx frac \left\{ v_\left\{mathrm N\right\} + v_\left\{mathrm F\right\} \right\} \left\{2\right\}$
= v_{mathrm F} + frac { v_{mathrm N} - v_{mathrm F} } {2}
$N approx frac \left\{ v_\left\{mathrm N\right\} - v_\left\{mathrm F\right\} \right\} \left\{ 2 c \right\},$
which require only the difference between the near and far image distances; focus is simply set to halfway between the near and far distances. View camera users often refer to the difference $v_\left
\{mathrm N\right\} , - , v_\left\{mathrm F\right\}$ as the focus spread; it usually is measured on the bed or focusing rail. On manual-focus small- and medium-format lenses, the focus and f-number
usually are determined using the lens DOF scales, which often are based on the two equations above.
For close-up photography, the f-number is more accurately determined using
$N approx frac \left\{1\right\} \left\{ 1 + m \right\} frac \left\{ v_\left\{mathrm N\right\} - v_\left\{mathrm F\right\} \right\} \left\{ 2 c \right\},$
where $m$ is the magnification.
Foreground and background blur
If the equation for the far limit of DOF is solved for
, and the far distance replaced by an arbitrary distance
, the blur disk diameter
at that distance is
$b = frac \left\{fm_mathrm s\right\} \left\{N\right\} frac \left\{ D - s \right\} \left\{ D \right\}$
When the background is at the far limit of DOF, the blur disk diameter is equal to the circle of confusion $c$, and the blur is just imperceptible. The diameter of the background blur disk increases
with the distance to the background. A similar relationship holds for the foreground; the general expression for a defocused object at distance $D$ is
$b = frac \left\{fm_mathrm s\right\} \left\{N\right\} frac \left\{ left| D - s right | \right\} \left\{ D \right\}$
For a given scene, the distance between the subject and a foreground or background object is usually fixed; let that distance be represented by
$x_mathrm d = left | D - s right | ;$
$b = frac \left\{fm_mathrm s\right\} \left\{N\right\} frac \left\{ x_mathrm d \right\} \left\{ D \right\}$
or, in terms of subject distance,
$b = frac \left\{fm_mathrm s\right\} \left\{N\right\} frac \left\{ x_mathrm d \right\} \left\{ s pm x_mathrm d \right\} ,$
with the minus sign used for foreground objects and the plus sign used for background objects. For a relatively distant background object,
$b approx frac \left\{fm_mathrm s\right\} N$
In terms of subject magnification, the subject distance is
$s = frac \left\{ m_mathrm s + 1 \right\} \left\{ m_mathrm s \right\} f ,$
so that, for a given f-number and subject magnification,
$b = frac \left\{fm_mathrm s\right\} \left\{N\right\} frac \left\{ x_mathrm d \right\} \left\{ frac \left\{ m_mathrm s + 1\right\} \left\{m_mathrm s\right\} f pm x_mathrm d \right\}$
= frac {fm_mathrm s ^2} {N} frac { x_mathrm d } { left (m_mathrm s + 1 right ) f pm m_mathrm s x_mathrm d }
Differentiating $b$ with respect to $f$ gives
$frac \left\{mathrm d b\right\} \left\{mathrm d f\right\}$
= frac {pm m_mathrm s ^3 x_mathrm d ^2} {N left [left (m_mathrm s + 1 right ) f pm m_mathrm s x_mathrm d right ]^2 }
With the plus sign, the derivative is everywhere positive, so that for a background object, the blur disk size increases with focal length. With the minus sign, the derivative is everywhere negative,
so that for a foreground object, the blur disk size decreases with focal length.
The magnification of the defocused object also varies with focal length; the magnification of the defocused object is
$m_mathrm d = frac \left\{v_mathrm s\right\} \left\{D\right\} = frac \left\{ left \left(m_mathrm s + 1 right \right) f \right\} \left\{ D \right\},$
where $v_mathrm s$ is the image distance of the subject. For a defocused object with some characteristic dimension $y$, the imaged size of that object is
$m_mathrm d y = frac \left\{ left \left(m_mathrm s + 1 right \right) f y \right\} \left\{ D \right\}$
The ratio of the blur disk size to the imaged size of that object then is
$frac b \left\{ m_mathrm d y \right\} = frac \left\{m_mathrm s\right\} \left\{ m_mathrm s + 1 \right\} frac \left\{x_mathrm d \right\} \left\{ Ny \right\},$
so for a given defocused object, the ratio of the blur disk diameter to object size is independent of focal length, and depends only on the object size and its distance from the subject.
Asymmetrical lenses
The discussion thus far has assumed a symmetrical lens for which the entrance and exit pupils coincide with the object and image nodal planes, and for which the pupil magnification is unity. Although
this assumption usually is reasonable for large-format lenses, it often is invalid for medium- and small-format lenses.
For an asymmetrical lens, the DOF ahead of the subject distance and the DOF beyond the subject distance are given by
$mathrm \left\{DOF_N\right\} = frac$
{N c (1 + m/P)} {m^2 [1 + (N c ) / (f m ) ] }
$mathrm \left\{DOF_F\right\} = frac$
{N c (1 + m/P)} {m^2 [1 - (N c )/ (f m ) ] },
where $P$ is the pupil magnification.
Combining gives the total DOF:
$mathrm \left\{DOF\right\} = frac \left\{2 f \left(1/m + 1/P \right) \right\}$
{ (f m ) / (N c ) - (N c ) / (f m ) }
When $s ll H$, the second term in the denominator becomes small in comparison with the first, and
$mathrm \left\{DOF\right\} approx frac \left\{2 N c \left(1 + m/P\right)\right\}\left\{m^2\right\}$
When the pupil magnification is unity, the equations for asymmetrical lenses reduce to those given earlier for symmetrical lenses.
Effect of lens asymmetry
Except for close-up and macro photography, the effect of lens asymmetry is minimal. A slight rearrangement of the last equation gives
$mathrm \left\{DOF\right\} approx frac \left\{2 N c\right\} \left\{m\right\}$
left (frac 1 m + frac 1 P right )
As magnification decreases, the $1/P$ term becomes smaller in comparison with the $1/m$ term, and eventually the effect of pupil magnification becomes negligible.
• Hansma, Paul K. 1996. View Camera Focusing in Practice. Photo Techniques, March/April 1996, 54–57. Available as GIF images on the Large Format page
• Langford, Michael J. 1973. Basic Photography. 3rd ed. Garden City, NY: Amphoto. ISBN 0-8174-0640-9
• Larmore, Lewis. 1965. Introduction to Photographic Principles. 2nd ed. New York: Dover Publications, Inc.
• Merklinger, Harold M. 1992. The INs and OUTs of FOCUS. v. 1.0.3. Bedford, Nova Scotia: Seaboard Printing Limited. Version 1.03e available in PDF at http://www.trenholm.org/hmmerk/. ISBN
• Ray, Sidney F. 2002. Applied Photographic Optics 3rd ed. Oxford: Focal Press. ISBN 0-240-51540-4
• Shipman, Carl. 1977. SLR Photographers Handbook. Tucson: H.P. Books. ISBN 0-912656-59-X
• Williams, John B. 1990. Image Clarity: High-Resolution Photography. Boston: Focal Press. ISBN 0-240-80033-8
Further reading
• Hummel, Rob (editor). 2001. American Cinematographer Manual. 8th ed. Hollywood: ASC Press. ISBN 0-935578-15-3
See also
External links | {"url":"http://www.reference.com/browse/stopped+down","timestamp":"2014-04-19T03:29:02Z","content_type":null,"content_length":"138979","record_id":"<urn:uuid:a8eb342d-3af8-4bb9-92fd-32fb1c904afc>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00444-ip-10-147-4-33.ec2.internal.warc.gz"} |
CPI movements over the past 60 years
Why use index numbers?
3.1 Deriving useful price measures for single, specific items such as Granny Smith apples is a relatively straightforward exercise. An estimate of the average price per kilogram in each period is
sufficient for all applications. Price change between any two periods would simply be calculated by direct reference to the respective average prices.
3.2 However, if the requirement is for a price measure that covers a number of diverse items, the calculation of a 'true' average price is both complicated and of little real meaning. For example,
consider the problem of calculating and interpreting an average price for two commodities as diverse as apples and motor vehicles. Because of this, price measures such as the CPI are typically
presented in index number form.
Description of a price index
3.3 Price indexes provide a convenient and consistent way of presenting price information that overcomes problems associated with averaging across diverse items. The index number for a particular
period represents the average price in that period relative to the average price in some base period for which, by convention, the average price has been set to equal 100.0.
3.4 A price index number on its own has little meaning. For example, the CPI All groups index number of 179.4 in the September quarter 2011 says nothing more than the average price in the September
quarter 2011 was 79.4% higher than the average price in the base year 1989–90 (when the index was set to 100.0). The value of index numbers stems from the fact that index numbers for any two periods
can be used to directly calculate price change between the two periods.
Percentage change is different to a change in index points
3.5 Movements in indexes from one period to any other period can be expressed either as changes in index points or as percentage changes. The following example illustrates these calculations for the
All groups CPI (weighted average of the eight capital cities) between the September quarter 2010 and the September quarter 2011. The same procedure is applicable for any two periods.
│ │ Index number│
│September quarter 2011 │ 179.4│
│less September Quarter 2010 │ 173.3│
│Change in index points │ 6.1│
│Percentage change │ 6.1/173.3 x 100 = 3.5%│
Movements in the CPI best measured using percentage changes 3.6
For most applications, movements in price indexes are best calculated and presented in terms of percentage change. Percentage change allows comparisons in movements that are independent of the level
of the index. For example, a change of 2 index points when the index number is 120 is equivalent to a percentage change of 1.7%, but if the index number was 80 a change of 2 index points would be
equivalent to a percentage change of 2.5% – a significantly different rate of price change. Only when evaluating change from the base period of the index will the points change be numerically
identical to the percentage change.
Percentage changes are not additive 3.7
The percentage change between any two periods
be calculated, as in the example above, by direct reference to the index numbers for the two periods. Adding the individual quarterly percentage changes will not result in the correct measure of
longer–term percentage change. That is, the percentage change between say the June quarter one year and the June quarter of the following year typically will not equal the sum of the four quarterly
percentage changes. The error becomes more noticeable the longer the period covered and the greater the rate of change in the index. This can readily be verified by starting with an index of 100 and
increasing it by 10% (multiplying by 1.1) each period. After four periods, the index will equal 146.4 delivering an annual percentage change of 46.4%, not the 40% given by adding the four quarterly
changes of 10%.
Calculating index numbers for periods longer than quarters 3.8
Although the CPI is compiled and published as a series of quarterly index numbers, its use is not restricted to the measurement of price change between particular quarters. Because a quarterly index
number can be interpreted as representing the average price during the quarter, index numbers for periods spanning more than one quarter can be calculated as the simple (arithmetic) average of the
relevant quarterly indexes. For example, an index number for the year 2010 would be calculated as the arithmetic average of the index numbers for the March, June, September and December quarters of
│ │ Index number│
│March quarter 2010 │ 171.0│
│June quarter 2010 │ 172.1│
│September quarter 2010 │ 173.3│
│December quarter 2010 │ 174.0│
│Divided by │ 4.0│
│equals CPI Index 2010 │ 172.6│
This characteristic of index numbers is particularly useful. It allows for comparison of average prices in one year (calendar or financial) with those in any other year. It also enables prices in say
the current quarter to be compared with the average prevailing in some prior year.
ANALYSING THE CPI 3.9
The quarterly change in the All groups CPI represents the weighted average price change of all the items included in the CPI. While publication of index numbers and percentage changes for components
of the CPI are useful in their own right, these data are often not sufficient to enable important contributors to overall price change to be reliably identified. What is required is some measure that
encapsulates both an item’s price change and its relative importance in the index.
Points contribution and points contribution change 3.10
If the All groups index number is thought of as being derived as the weighted average of indexes for all its component items, the index number for a component multiplied by its weight to the All
groups index results in what is called its ‘points contribution’. It follows that the change in a component item’s points contribution from one period to the next provides a direct measure of the
contribution to the change in the All groups index resulting from the change in that component’s price.
Information on points contribution and points contribution change (or points change) is of immense value when analysing sources of price change and for answering ‘what if’ type questions. Consider
the following data extracted from the September quarter 2011 CPI publication:
│ │ Index number │ Percentage change│ Points contribution │Points change│
│Item │June qtr 2011│Sept qtr 2011│ │June qtr 2011│Sept qtr 2011│ │
│All groups │ 178.3│ 179.4│ 0.6│ 178.3│ 179.4│ 1.1│
│Electricity│ 248.4│ 267.7│ 7.8│ 3.55│ 3.83│ 0.28│
Using points contributions 3.12
Using only the index numbers themselves, the most that can be said is that between the June and September quarters 2011, the price of Electricity increased by more than the overall CPI (by 7.8%
compared with an increase in the All groups of 0.6%). The additional information on points contribution and points change can be used to:
Calculate the effective weight for Electricity in the June and the September quarters
(given by the points contribution for Electricity divided by the All groups index)
For the June quarter, the weight is calculated as 3.55/178.3 x 100 = 2.0% and for the September quarter is 3.83/179.4 x 100 = 2.1%. Although the underlying quantities are held fixed, the effective
weight in expenditure terms has increased due to the price of Electricity increasing by more than the prices of all other items in the CPI basket (on average).
Calculate the percentage increase that would have been observed in the CPI if all prices other than those for Electricity had remained unchanged
(given by the points change for Electricity divided by the All groups index number in the previous period). For the September quarter 2011 this is calculated as 0.28/178.3 x 100 = 0.16%. In other
words, a 7.8% increase in Electricity prices in September quarter 2011 would have resulted in an increase in the overall CPI of 0.2%.
Calculate the average percentage change in all other items excluding Electricity
(given by subtracting the points contribution for Electricity from the All groups index in both quarters and then calculating the percentage change between the resulting numbers which represents the
points contribution of the ‘other’ items). For the above example, the numbers for All groups excluding Electricity are: June quarter 2011, 178.3–3.55 = 174.8; September quarter 2011, 179.4–3.83 =
175.6; and the percentage change (175.6–174.8)/174.8 x 100 = 0.5%. In other words, prices of all items other than Electricity increased by 0.5% on average between the June and September quarters
Estimate the effect on the All groups CPI of a forecast change in the prices of one of the items
(given by applying the forecast percentage change to the item's points contribution and expressing the result as a percentage of the All groups index number). For example, if the price of Electricity
was forecast to increase by 25% in the December quarter 2011, then the points change for Electricity would be 3.83 x 0.25 = 1.0, which would deliver an increase in the All groups index of 1.0/179.4 x
100 = 0.6% . In other words, a 25% increase in the Electricity price in the December quarter 2011 would have the effect of increasing the CPI by 0.6%. Another way commonly used to express this impact
is ‘Electricity’ would contribute 0.6 percentage points to the change in the CPI.
ABS rounding conventions 3.13
To ensure consistency in the data produced from the CPI, it is necessary for the ABS to adopt a set of consistent rounding conventions or rules for the calculation and presentation of data. The
conventions strike a balance between maximising the usefulness of the data for analytical purposes and retaining a sense of the underlying precision of the estimates. These conventions need to be
taken into account when using CPI data for analytical or other special purposes.
Index numbers are always published to a reference base of 100.0. Index numbers and percentage changes are always published to one decimal place, with the percentage changes being calculated from the
rounded index numbers. Points contributions are published to two decimal places, with points contributions change being calculated from the rounded points contributions. Index numbers for periods
longer than a single quarter (e.g. for financial years) are calculated as the simple arithmetic average of the relevant rounded quarterly index numbers.
The following questions and answers illustrate the uses that can be made of the CPI. For more information about which index to use in contracts see 'Use of Price Indexes in Contracts' at
CPI can be used to compare money values over time 3.15 Question 1:
What would $200 in 2005 be worth in the September quarter 2011?
3.16 Response 1:
This question is best interpreted as asking ‘How much would need to be spent in the September quarter 2011 to purchase what could be purchased in 2005 for $200?’ As no specific commodity is
mentioned, what is required is a measure comparing the general level of prices in the September quarter 2011 with the general level of prices in the calendar year 2005. The All groups CPI would be an
appropriate choice.
Because CPI index numbers are not published for calendar years, two steps are required to answer this question. The first is to derive an index for the calendar year 2005. The second is to multiply
the initial dollar amount by the ratio of the index for September quarter 2011 to the index for 2005.
The index for the calendar year 2005 is obtained as the simple arithmetic average of the quarterly indexes for March (147.5), June (148.4), September (149.8) and December (150.6) 2005 giving 149.1
rounded to one decimal place. The index for the September quarter 2011 is 179.4.
The answer is then given by:
$200 x 179.4/149.1 = $240.64
Forecasting impact of price changes on the CPI 3.19 Question 2
: What would be the impact of a 10% increase in vegetable prices on the All groups CPI in the December quarter 2011?
3.20 Response 2:
Two pieces of information are required to answer this question; the All groups index number for the September quarter 2011 (179.4), and the September quarter 2011 points contribution for Vegetables
An increase in vegetable prices of 10% would increase vegetables points contribution by 2.33 x 10/100 = 0.23 index points which would result in an All groups index number of 179.6 for the December
quarter 2011, an increase of 0.2%.
Indexes used should be representative of specific items 3.22 Question 3
: How does the CPI reflect changes in electricity prices?
3.23 Response 3:
The All groups CPI measures price change for all goods and services. In Table 4 of
Consumer Price Index, Australia
(cat. no. 6401.0) there is a range of component indexes by capital city which can be used. The example below sets out the price change for electricity compared to the All groups CPI over the last 10
years. This shows that the price of electricity has increased faster than the headline number.
│ │All groups CPI index number│ Electricity index number│
│September quarter 2001 │ 134.2│ 133.8│
│September quarter 2010 │ 173.3│ 237.9│
│September quarter 2011 │ 179.4│ 267.7│
│Percentage change – 1 year ago │(179.4 – 173.3)/173.3 x 100│(267.7 – 237.9)/237.9 x 100│
│ │ = 3.5%│ = 12.5%│
│Percentage change – 10 years ago │(179.4 – 134.2)/134.2 x 100│(267.7 – 133.8)/133.8 x 100│
│ │ = 33.7%│ = 100.1%│
Price indexes can be used to estimate changes in volumes 3.24 Question 4
: Household Expenditure Survey data show that average weekly expenditure per household on Food and non–alcoholic beverages increased from $158.58 in 2003–04 to $216.40 in 2009–10 (i.e. an increase of
36.5%). Does this mean that households, on average, purchased 36.5% more Food and non–alcoholic beverages in 2009–10 than they did in 2003–04?
3.25 Response 4:
This is an example of one of the most valuable uses that can be made of price indexes. Often the only viable method of collecting and presenting information about economic activity is in the form of
expenditure or income in monetary units (e.g. dollars). While monetary aggregates are useful in their own right, economists and other analysts are frequently concerned with questions related to
volumes, for example, whether more goods and services have been produced in one period compared with another period. Comparing monetary aggregates alone is not sufficient for this purpose as dollar
values can change from one period to another due to either changes in quantities or changes in prices (most often a combination).
To illustrate this, consider a simple example of expenditure on oranges in two periods. The product of the quantity and the price gives the expenditure in any period. Suppose that in the first period
10 oranges were purchased at a price of $1.00 each and in the second period 15 oranges were purchased at a price of $1.50 each. Expenditure in period one would be $10.00 and in period two $22.50.
Expenditure has increased by 125%, yet the volume (number of oranges) has only increased by 50% with the difference being accounted for by a price increase of 50%. In this example all the price and
quantity data are known, so volumes can be compared directly. Similarly, if prices and expenditures are known, quantities can be derived.
But what if the actual prices and quantities are not known? If expenditures are known and a price index for oranges is available, the index numbers for the two periods can be used as if they were
prices to adjust the expenditure for one period to remove the effect of price change. If the price index for oranges was equal to 100.0 in the first period, the index for the second period would
equal 150.0. Dividing expenditure in the second period by the index number for the second period and multiplying this result by the index number for the first period provides an estimate of the
expenditure that would have been observed in the second period had the prices remained as they were in the first period. This can easily be demonstrated by reference to the oranges example:
$22.50/150.0 x 100.0 = $15.00 = 15 x $1.00
So, without ever knowing the actual volumes (quantities) in the two periods, the adjusted second period expenditure ($15.00), can be compared with the expenditure in the first period ($10.00) to
derive a measure of the proportional change in volumes $15/$10 = 1.50, which equals the ratio obtained directly from the comparison of the known quantities.
We now return to the question on expenditure on Food and non–alcoholic beverages recorded in the HES in 2003–04 and 2009–10. As the HES data relates to the average expenditure of Australian
households, the ideal price index would be one that covers the retail prices of Food and non–alcoholic beverages for Australia as a whole. The price index which comes closest to meeting this ideal is
the index for the Food and non–alcoholic beverages of the CPI for the weighted average of the eight capital cities. The Food and non–alcoholic beverages index number for 2003–04 is
(149.3+152.0+154.7+153.3)/4 = 152.3 and for 2009–10 it is (186.6+189.9+191.3+190.7)/4 = 189.6. Using these index numbers, recorded expenditure in 2009–10 ($216.40) can be adjusted to 2003–04 prices
as follows:
$216.40/189.6 x 152.3 = $173.82
│Food and non–alcoholic beverages │2003–04│2009–10│
│HES expenditure │$158.58│$216.40│
│Food and non–alcoholic beverages price index number │ 152.3│ 189.6│
│revalued 2009–10 quantities at 2003–04 prices │$158.58│$173.83│
│Volume change ($173.83 – $158.58)/$158.58 x 100 │ │ 9.6%│
The revalued 2009–10 quantities at 2003–04 prices of $173.82 can then be compared to the expenditure recorded in 2003–04 ($158.58) to deliver an estimate of the change in volumes. This indicates a
volume increase of 9.6% between 2003–04 and 2009–10. Over the same period food prices increased
(189.6 – 152.3)/152.3 x100 = 24.5% and total expenditure (($216.40 – $158.58)/$158.58 x 100) increased 36.5%,
This page last updated 6 September 2012 | {"url":"http://www.abs.gov.au/AUSSTATS/abs@.nsf/Latestproducts/6440.0Main%20Features302011?opendocument&tabname=Summary&prodno=6440.0&issue=2011&num=&view=","timestamp":"2014-04-21T10:58:41Z","content_type":null,"content_length":"47686","record_id":"<urn:uuid:b279e28f-7b6e-493c-af4e-46c103fbeffe>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00329-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Beta distribution is a special case of the Dirichlet distribution, and is related to the Gamma distribution. It has the probability distribution function
It is often seen in Bayesian inference and order statistics.
a : float
Alpha, non-negative.
b : float
Beta, non-negative.
size : tuple of ints, optional
The number of samples to draw. The ouput is packed according to the size given.
out : ndarray
Array of the given shape, containing values drawn from a Beta distribution.
The number of samples to draw. The ouput is packed according to the size given.
Array of the given shape, containing values drawn from a Beta distribution. | {"url":"http://docs.scipy.org/doc/numpy-1.4.x/reference/generated/numpy.random.mtrand.RandomState.beta.html","timestamp":"2014-04-21T12:12:20Z","content_type":null,"content_length":"9096","record_id":"<urn:uuid:0d26577e-bec3-406c-81fd-a872cda8e2de>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00547-ip-10-147-4-33.ec2.internal.warc.gz"} |
SCU Dept. of Mathematics and Computer Science -- Sample Math
[Return to Department of Mathematics and Computer Science homepage]
[Return to Mathematics Major page]
Sample Curriculum
Class of 2012
This chart is based on the university and college requirements ("core curriculum") in effect in Fall 2008 for the class of 2012. Some university and college requirements will change beginning with
students beginning as freshmen in Fall 2009.
│Quarter│Freshman Year │Sophomore Year │Junior Year │Senior Year │
│ │ │ │ │Math 103 (Adv. Lin. Alg.)│
│ │Math 11 (Calc I) │Math 21 (14 after Fall '09) (Calc IV) │Math 102 (Adv. Calc.) │or Math 111 (Abs. Alg.) │
│Fall │CSCI 10 (Intro CS) (A.2) │U.S. │Math Up. Div. (A.1) │or Math 176 (Comb.) │
│ │English 1 │Foreign Lang. I │Math 100 (or Engl. Writing)│Free Elective │
│ │Rel. Studies (Intro) │Math 51 (Discr. Math)* │Free Elective │Free Elective │
│ │ │ │ │Free Elective │
│ │Math 12 (Calc II) │Math 52 (Abst. Alg.) │Math Up. Div. │Math Up. Div. │
│Winter │Physics 31 (formerly 4) │Foreign Lang. II │Rel. Studies (Adv.) │Free Elective │
│ │West. Culture I │Math 22 (Diff. Eq.)** │World Cultures II │Free Elective │
│ │English 2 │Rel. Studies (Interm) │Free Elective │Free Elective │
│ │Math 13 (Calc. III) │Math 53 (Lin. Alg) │Math Up. Div. │Math Up. Div. │
│Spring │Physics 32 & 32L (formerly 5)│Soc. Science │Fine Arts │Free Elective │
│ │West. Culture II │World Cultures I │Ethnic/Women's St. │Free Elective │
│ │Free Elective │Ethics │Free Elective │Free Elective │
*Required for Class of 2012 and afterwards. **Required for Class of 2009 and afterwards.
A. General Comments
1. Students planning to go on to graduate school in pure mathematics should take Math 105 (Complex), 111 (Abst. Alg. I), 112 (Abst. Alg. II), 113 (Topology), 153 (Interm. Analysis I), 154 (Interm.
Analysis II).
2. Students who show proficiency in a high-level programming language may substitute another course to fulfill the technology requirement in lieu of Math 10.
3. Students are referred to the Bulletin for details concerning other recommendations and substitutions.
B. Pre-Secondary Teaching Options and Recommendations:
B-I: Special Recommendations for Those Preparing for Admission to California Teacher Training Credential Programs
The State of California requires that students seeking a credential to teach mathematics or computer science in California secondary schools must pass the California Subject Examination for Teachers
(CSET), a subject area competency examination. The secondary teaching credential additionally requires the completions of an approved credential program, which can be completed as a fifth year of
study and student teaching.
B-II: Students may consider completing the Emphasis in Mathematics Education. In addition to the general requirement for majoring in Mathematics,
1. Students must complete Math 101 (Geom.), 102, 111, 122 (Prob. & Stats. I), 123 (Prob. & Stats. II) or 8 (Stats), 170 (Devel. Math.), either 175 (Numb. Th.) or 178 (Cryptography).
2. Students must complete Educ 198B (Second. School Teach.).
3. Physics 11 and 12 (formerly 20 and 21) may be substituted for Physics 31 and 32 (formerly 4 and 5).
4. Students are strongly recommended to complete the Urban Education minor.
C. Special Recommendations for the Emphasis in Applied Mathematics
1. Students must complete Math 22, 102, 122, 123 (Prob. & Stats. II), 166 (Num. Analysis), and 176.
2. Students must also complete two courses from Math 144 (PDE), 155 (ODE), 164 (Comp. Simul.), 165 (Lin. Prog.), 178 (Cryptography), or an approved alternative (but not from courses designated as
the computer science sequence [see Computer Science Sample Curriculum]).
D. Special Recommendations for the Emphasis in Financial Mathematics
In addition to the general requirement for majoring in Mathematics,
1. Students must complete Math 102, 122, 123, 125, 144, 166.
2. Students must also complete Business 70, Accounting 11, 12, and Finance 121, 124.
E. Special Recommendations for Students interested in Computer Applications
Students are referred to the requirements for C.S. minors and the upper division tracks recommended for C.S. majors.
At least 12 upper division courses (60 units) are required for graduation. Thus, if the English Composition course is not an upper division course, at least three of the free electives must be upper
divsion courses.
The information presented on this webpage is not intended as the official statement of graduation requirements. The student is referred to the current University Bulletin.
Last Updated: 14 April 2009 | {"url":"http://math.scu.edu/samplecurrma.shtml","timestamp":"2014-04-18T16:00:56Z","content_type":null,"content_length":"8130","record_id":"<urn:uuid:80229e3b-06a3-4cfe-bd04-6f9fff5a17c6>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00463-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Master Methode
August 6th, 2011, 01:50 PM #1
Junior Member
Join Date
Aug 2011
In master method they talk about functions being asymptotically larger and polynomialy larger.
But I don't feel much difference between them. What are their definitions ?
And in one case they say nlog2(n) is asymptotically larger than n but is not polynomialy larger than n . The reason they have given is that nlog(n)/n = log(n) is asymptotically less than n^e for
any positive constant e. But for me the last part seems incorrect. Because for 0<e< 0.2 n^e is asymptotically smaller than log(n).
Am I correct?
Please help me.
Re: Master Methode
the whole asymtoticaly and polynominal-thing is much more show than really substance. There are a few Functions between log and polonoms and exp, but almost none of them has some real impact.
(ok... maybe the iterated logarithm is sometimes important)
(thats my opinion)
For your question x^0.1 grows much faster than log(x). Just type very big numbers in your calculator. x^0.01 also grows faster than log(x). This is not a proof, just try around to get a feeling.
The proof for this goes like this:
If you paint x^n and exp(x) you will see, that exp(x) grows much faster than every x^n for each possible n in the natural Numbers. This is very easy to show, if you write exp(x)=1+x+x^2/2+x^2/6+
.... . Then you mirror the diagramms at the line x=y, then exp grows faster than every x^n is transformed to log(x) grows slower than every x^(1/n). 1/n can be smaller than every positive e. -->
voila :-)
If you really want a proof with every details, you will get a few pages written with complicated formulas, but I hope the above gave some impression about the main points.
August 7th, 2011, 05:49 AM #2
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Answer on Mechanics | Kinematics | Dynamics Question for Colton
Question #1060
A boy drags a box across the room with a strap at an angle of 25 degrees above the horizontal. The boy exerts a force of 10 N. If the mass of the box is 3 kg and the coefficient of kinetic friction
is 0.1, what is the acceleration of the box?
Expert's answer
II Newton's law for the box:
ma = F cos (25) - μ mg;
a = F/m cos(25) - μ g = 10/3 *cos(25) - 0.1*10 = 2.02 m/s
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Useful octave commands relating to strings
To read a string in from a file, see part 2 below.
Part 1
I found this information by typing
help -i string
in octave.
char (CELL_ARRAY)
- Built-in Function: char (S1, S2, ...)
Create a string array from a numeric matrix, cell array, or list of
If the argument is a numeric matrix, each element of the matrix is
converted to the corresponding ASCII character. For example,
char ([97, 98, 99])
=> "abc"
- Function File: int2str (N)
- Function File: num2str (X, PRECISION)
- Function File: num2str (X, FORMAT)
Convert a number to a string. These functions are not very
flexible, but are provided for compatibility with MATLAB. For
better control over the results, use `sprintf' (*note Formatted
- Function File: strcat (S1, S2, ...)
Return a string containing all the arguments concatenated. For
s = [ "ab"; "cde" ];
strcat (s, s, s)
=> "ab ab ab "
Searching and Replacing
- Function File: findstr (S, T, OVERLAP)
Return the vector of all positions in the longer of the two strings
S and T where an occurrence of the shorter of the two starts. If
the optional argument OVERLAP is nonzero, the returned vector can
include overlapping positions (this is the default). For example,
findstr ("ababab", "a")
=> [ 1, 3, 5 ]
findstr ("abababa", "aba", 0)
=> [ 1, 5 ]
- Function File: index (S, T)
Return the position of the first occurrence of the string T in the
string S, or 0 if no occurrence is found. For example,
index ("Teststring", "t")
=> 4
*Caution:* This function does not work for arrays of strings.
- Function File: strcmp (S1, S2)
Compares two strings, returning 1 if they are the same, and 0
*Caution:* For compatibility with MATLAB, Octave's strcmp
function returns 1 if the strings are equal, and 0 otherwise.
This is just the opposite of the corresponding C library function.
- Function File: strrep (S, X, Y)
Replaces all occurrences of the substring X of the string S with
the string Y. For example,
strrep ("This is a test string", "is", "&%$")
=> "Th&%$ &%$ a test string"
- Function File: substr (S, BEG, LEN)
Return the substring of S which starts at character number BEG and
is LEN characters long.
If OFFSET is negative, extraction starts that far from the end of
the string. If LEN is omitted, the substring extends to the end
of S.
For example,
substr ("This is a test string", 6, 9)
=> "is a test"
This function is patterned after AWK. You can get the same
result by `S (BEG : (BEG + LEN - 1))'.
String Conversions
- Function File: hex2dec (S)
Return the decimal number corresponding to the binary number stored
in the string S. For example,
hex2dec ("1110")
=> 14
If S is a string matrix, returns a column vector of converted
numbers, one per row of S. Invalid rows evaluate to NaN.
- Function File: dec2bin (N, LEN)
Return a binary number corresponding the nonnegative decimal number
N, as a string of ones and zeros. For example,
dec2bin (14)
=> "1110"
If N is a vector, returns a string matrix, one row per value,
padded with leading zeros to the width of the largest value.
- Function File: dec2base (N, B, LEN)
Return a string of symbols in base B corresponding to the the
nonnegative integer N.
dec2base (123, 3)
=> "11120"
If N is a vector, return a string matrix with one row per value,
padded with leading zeros to the width of the largest value.
If B is a string then the characters of B are used as the symbols
for the digits of N. Space (' ') may not be used as a symbol.
dec2base (123, "aei")
dec2base (123, "aei")
=> "eeeia"
The optional third argument, LEN, specifies the minimum number of
digits in the result.
- Function File: base2dec (S, B)
Convert S from a string of digits of base B into an integer.
base2dec ("11120", 3)
=> 123
If S is a matrix, returns a column vector with one value per row
of S. If a row contains invalid symbols then the corresponding
value will be NaN. Rows are right-justified before converting so
that trailing spaces are ignored.
If B is a string, the characters of B are used as the symbols for
the digits of S. Space (' ') may not be used as a symbol.
base2dec ("yyyzx", "xyz")
=> 123
- Function File: str2num (S)
Convert the string S to a number.
- Mapping Function: toascii (S)
Return ASCII representation of S in a matrix. For example,
toascii ("ASCII")
=> [ 65, 83, 67, 73, 73 ]
- Mapping Function: tolower (S)
Return a copy of the string S, with each upper-case character
replaced by the corresponding lower-case one; nonalphabetic
characters are left unchanged. For example,
tolower ("MiXeD cAsE 123")
=> "mixed case 123"
- Mapping Function: isdigit (S)
Return 1 for characters that are decimal digits.
Part 2: LOADING STRINGS
If the string is just ascii and one character per line then we can just use:
load filename.dat
and it will strip the input into a number-array, which can then be processed however we like.
If the string is ascii but all concatenated then we can just prepend an octave friendly header. Eg:
# name: name_we_want_to_use_in_octave_workspace
# type: string
# elements: 1
# length: 71
The hashed comments tell octave what it should expect to read. (If the length is not the full length of the string, that's fine it just reads as many characters as we tell it to. (Note that the
spacing in the comment structure seems to be important. Eg
# type: string
works, but
#type: string
seems not to.
The string is read in as a character array. We can access substrings by doing
Also, we can convert the elements to decimal (not characters) by doing using, say
name_we_want_to_use_in_octave_workspace - '0'
this substracts the ascii value of zero from each of the character array elements and magically turns everything into proper integers. David MacKay and Simon Osindero | {"url":"http://users-deprecated.aims.ac.za/~mackay/octave/strings.html","timestamp":"2014-04-19T19:36:36Z","content_type":null,"content_length":"7323","record_id":"<urn:uuid:09c00310-3e74-43e3-9088-94e957e9181e>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00578-ip-10-147-4-33.ec2.internal.warc.gz"} |
Morse equations and unstable manifolds of isolated invariant sets
Rodríguez Sanjurjo, José Manuel (2003) Morse equations and unstable manifolds of isolated invariant sets. Nonlinearity, 16 (4). pp. 1435-1448. ISSN 0951-7715
Restricted to Repository staff only until 31 December 2020.
Official URL: http://iopscience.iop.org/0951-7715/16/4/314
We describe a new way of obtaining the Morse equations of a Morse decomposition of an isolated invariant set. This is achieved through a filtration of truncated unstable manifolds associated with the
decomposition. The results in the paper make it possible to calculate the Morse equations (and also the Conley index) in many interesting situations without using index pairs. We also study the
intrinsic topology of the unstable manifold and obtain new duality properties of the cohomological Conley index.
Item Type: Article
Uncontrolled Keywords: Index theory, Morse-Conley indices; Strange attractors, chaotic dynamics; Stability theory
Subjects: Sciences > Mathematics > Geometry
Sciences > Mathematics > Topology
ID Code: 16824
References: Bhatia N P and Szeg¨o G P 1970 Stability Theory of Dynamical Systems (Grundlehren der Mat. Wiss. 161) (Berlin: Springer)
Borsuk K 1975 Theory of Shape, Monografie Matematyczne, Tom 59 (Mathematical Monographs) vol 59 (Warsaw: PWN-Polish Scientific Publishers)
Conley C 1978 Isolated invariant sets and the Morse index CBMS Regional Conference Series in Mathematics vol 38 (Providence, RI: American Mathematical Society)
Conley C and Zehnder E 1984 Morse-type index theory for flows and periodic solutions for Hamiltonian equations Commun. Pure Appl. Math. 37 207–53
Dydak J and Segal J 1978 Shape Theory. An Introduction (Lecture Notes in Mathematics vol 688) (Berlin: Springer)
Giraldo A and Sanjurjo J M R 1999 On the global structure of invariant regions of flows with asymptotically stable attractors Mathematische Zeitschrift 232 739-746
Guckenheimer J and Holmes P 1983 Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields (Berlin: Springer)
Gunther B and Segal J 1993 Every attractor of a flow on a manifold has the shape of a finite polyhedron Proc. AMS 119 321–9
Kapitanski L and Rodnianski I 2000 Shape and Morse theory of attractors Comm. Pure Appl. Math. 53 218–42
Lorenz E N 1963 Deterministic non-period flows J. Atmos. Sci. 20 130–41
Mardesic S 1974 Pairs of compacta and trivial shape Trans. AMS 189 329–36
Mardesic S and Segal J 1982 Shape Theory. The Inverse System Approach (North-Holland Mathematical Library) vol 26 (Amsterdam: North-Holland)
Mawhin J and Willem M 1989 Critical Point Theory and Hamiltonian Systems (New York: Springer)
McCord C 1988 Mappings and homological properties in the Conley index theory Ergodic Theory Dynam. Syst. 8 (Charles Conley Memorial Volume) 175–98
McCord C 1992 Poincare–Lefschetz duality for the homology Conley index Trans. AMS 329 233–52
Robbin J W and Salamon D 1988 Dynamical systems, shape theory and the Conley index Ergodic Theory Dynam. Syst. 8 (Charles Conley Memorial Volume) 375–93
Rybakowski K P 1987 The Homotopy Index and Partial Differential Equations, Universitext (Berlin: Springer)
Rybakowski K P and Zehnder E 1985 AMorse equation in Conley’s index theory for semiflows on metric spaces Ergodic Theory Dynam. Syst. 5 123–43
Salamon D 1985 Connected simple systems and the Conley index of isolated invariant sets Trans. AMS 291 1–41
Sanjurjo J M R 1994 Multihomotopy, Cech spaces of loops and shape groups Proc. London Math. Soc. 69 330–44
Sanjurjo J M R 1995 On the structure of uniform attractors J. Math. Anal. Appl. 192 519–28
Spanier E H 1966 Algebraic Topology (New York: McGraw-Hill)
Sparrow C 1982 The Lorenz Equations: Bifurcations, Chaos, and Strange Attractors (Berlin: Springer)
Tucker W 1999 The Lorenz attractor exists C. R. Acad. Sci. Paris t. 328, Serie I 1197–202
Deposited On: 23 Oct 2012 09:04
Last Modified: 07 Feb 2014 09:36
Repository Staff Only: item control page | {"url":"http://eprints.ucm.es/16824/","timestamp":"2014-04-19T07:24:48Z","content_type":null,"content_length":"32599","record_id":"<urn:uuid:073c5cb1-6b06-4388-a89a-d7704307ce44>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00656-ip-10-147-4-33.ec2.internal.warc.gz"} |
Fountain Hills ACT Math Tutors
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Hamtramck Calculus Tutor
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21 Subjects: including calculus, geometry, statistics, algebra 1
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Using Two Irrationals to Generate All Positive Integers
Date: 10/03/2003 at 00:59:55
From: Sarah
Subject: two irrational numbers a, b generating all positive integers
If a and b are positive irrational numbers such that 1/a + 1/b = 1,
then every positive integer can be uniquely expressed as either floor
(ka) or floor(kb), where k is a positive integer.
I don't know how to approach this kind of problem. It's asking me to
show that somehow a and b can generate every possible positive
integer. How do I do that?
Date: 10/03/2003 at 05:21:26
From: Doctor Jacques
Subject: Re: two irrational numbers a, b generating positive integers
Hi Sarah,
This is a very interesting question indeed.
We may first record some useful facts:
* As neither a or b is 1/2, if, for example, a < b, we have:
1 < a < 2
2 < b
* As neither a or b is rational, no number of the form ka, kb, k/a,
k/b, (for integer k), is an integer.
We show first that every integer N >= 1 can be expressed at least in
one way as floor(ka) or floor(mb).
Assume that this is not the case. This means that there are integers
k, m, and N such that
1) ka < N
2) N+1 < (k+1)a
3) mb < N
4) N+1 < (m+1)b
where all the inequalities are strict because a and b are irrational,
as noted above.
With a little algebra, we can deduce from this:
1) (k/N) < (1/a) < (k+1)/(N+1)
2) (m/N) < (1/b) < (m+1)/(N+1)
and, adding these together:
(k+m)/N < (1/a) + (1/b) = 1
1 = (1/a) + (1/b) < (k+m+2)/(N+1)
which leads to:
k+m < N < k + m + 1
This says that there is an integer N strictly between consecutive
integers (k+m) and (k+m+1), an obvious contradiction.
Let us now consider the unicity question. As a and b are > 1, there
is at most one integer k and one integer m such that N = floor(ka) or
N = floor(mb). We must show that it is not possible to have both,
i.e. that we cannot have:
N = floor(ka) = floor(mb)
We use a similar technique: assume, by contradiction, that the above
equality holds for some k, m, N. We have:
1) N < ka < N+1
2) N < mb < N+1
Can you transform the above relations to get a contradiction similar
to the previous one (an integer strictly between two consecutive
Please feel free to write back if you require further assistance.
- Doctor Jacques, The Math Forum | {"url":"http://mathforum.org/library/drmath/view/64448.html","timestamp":"2014-04-19T03:09:44Z","content_type":null,"content_length":"7412","record_id":"<urn:uuid:15142ac2-eaf3-4b62-8bad-f8e4d4159e83>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00661-ip-10-147-4-33.ec2.internal.warc.gz"} |
To read ARRNGE files, apply absorption corrections and calculate mean structure factors
Adds calculated structure factors to a list of hkl and observed F's
To group together equivalent reflections for structures with non-zero propagation vectors
To group together equivalent reflections and give some statistics on their degree of equivalence, indices may be submultiples of integers.
To read ARRNGE files and prepare for LSQ with extinction
Calculation of bond lengths and angles. Can also write cards for geometric slack constraints
To calculate the magnetic structure factors for a given list of indices.
To calculate the magnetic interaction vectors for each domain for a given list of indices. The results are given on polarisation axes.
Least squares refinement from magnetic structure factor data (paramagnetic version)
Analysis of raw D3 data 1999 version
Calculates extinction coefficients and writes a data file for least squares
Calculates form factors and radial electron densities from Slater type radial wave-functions
General fourier inversion program showing atom positions
General fourier inversion program
To generate reflections and create a D3 measuring file
To generate magnetic and fundamental reflection indices and calculate the corresponding stucture factors.
To generate reflection indices and calculate the corresponding stucture factors.
To calculate the mean structure factors of satellite reflections from data arranged by ARRINC
To generate D cards for the UB matrix read from a record stored on the ILL data base
To draw magnetic structures in 3D
Least squares refinement from magnetic structure factor data
To generate indices of reflections in a magnetic powder diagram
Interactive calculation of weighted means
Prepares the data for an animated 3d display of a magnetic structure
Magnetic least squares with multipole description of form factors
Structure factor least squares using multipole form factors
To draw magnetic structures in 3D in SGI inventor format
Least squares refinement from neutron polarimetry data
To calculate the positions and intensities of the lines in a powder diagram
Least squares refinement from measured structure factors
Least squares refinement from structure factors measured on twinned crystals
Calculation of gamma and magnetic structure factors from flipping ratios
P. Jane Brown e-mail: brown@ill.fr
Institut Laue Langevin,
Grenoble, FRANCE | {"url":"http://www.ccp14.ac.uk/ccp/web-mirrors/ill-hewat/dif/ccsl/appenx/maialfa.html","timestamp":"2014-04-19T06:59:09Z","content_type":null,"content_length":"6365","record_id":"<urn:uuid:3ae6ad85-b789-44f9-9c3f-153546d99439>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00144-ip-10-147-4-33.ec2.internal.warc.gz"} |
Poiseuille's Equation
Copyright © University of Cambridge. All rights reserved.
'Poiseuille's Equation' printed from http://nrich.maths.org/
Water is pumped at a steady rate through a straight circular pipe of length $L$ and radius $R$.
I make the assumption that if the flow is steady then the flow rate $Q$ of water through the pipe can only depend on $L$, $R$, the constant difference in pressure $\Delta P$ between the two ends of
the pipe and the viscosity $\mu$ of the fluid. How strongly do you agree with the validity of this assumption from a practical point of view? Under what circumstances are they most likely to be
The standard equation governing flow along a circular pipe is called the Poiseille-equation:
$$Q = k\frac{R^4 \Delta P}{\mu L}\, \mbox{for a constant } k$$
Show that this makes sense from the point of view of the physical units of the various variables. Would any other combinations of variables combine to give a flow rate?
How might you devise an experiment to determine the numerical value of $k$?
Suppose that you push water along a horizontal pipe into the air. If you have a fixed amount of force at your disposal, are narrow pipes or wide pipes best to get the largest flow rate? | {"url":"http://nrich.maths.org/6512/index?nomenu=1","timestamp":"2014-04-17T12:43:05Z","content_type":null,"content_length":"4524","record_id":"<urn:uuid:29e4ef3a-3ff0-44dc-9474-23816ae6807d>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00309-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mentoring the Cherokee Creek Checkers Club #AlgPoW with the Noyce Scholars
Last Friday, I was invited to speak about the Problems of the Week to the Northeast region Noyce Scholars (early career and future STEM educators, and a really fun group to do math with!).
Since I’d been thinking a lot about the AlgPoW that just wrapped up, Cherokee Creek Checkers Club, we looked at that problem and some students’ work on the problem.
We began with this scenario:
Cherokee Creek Middle School has a checkers club with an enthusiastic group of players. Unfortunately, they started this semester in a bit of a slump, winning only 40 of their first 90 games in
various inter-school competitions. That gave them a winning percentage of only about 44%.
Then they found a new coach, who inspired them and significantly improved their overall level of play. This month they went on a hot streak, winning 3 out of every 5 games they played.
We used a scenario to loosen our thinking up and try to focus on three big preparation themes: prerequisite knowledge, big math ideas, and multiple approaches.
Looking at the scenario, the Noyce Scholars noticed and wondered about cool math like:
• Quantities are represented as percentages, fractions, and ratios in this problem.
• How could you determine which coach was better? How do you know for sure?
• What’s the overall winning percentage, based on the number of games played, before and after the new coach.
• If they won 40 out of 90 games at first, how many games might they play over a year? A week?
• How come winning 3 out of 5 is an improvement over winning 40 out of 90?
• They increased their winning percentage by 16 percent
All of these noticings and wonderings gave us insight into two of our three foci.
1. What do students need to know/understand to work on a problem about this story?
□ Fractions, ratios, and percents and how they are related
□ How winning percentages are calculated
□ That two coaches are being compared and one might be better than the other; the team improved over time
2. What big math ideas are at work here?
□ Comparing two quantities using fractions, ratios, and percentages
□ Change in one quantity as a function of another e.g. overall winning percentage as a function of games played with the new coach
Our third focus is multiple approaches, and for that we needed a problem to solve. I revealed that after playing some more games, the team’s overall winning percentage climbed to exactly 52% and we
wondered, “how many games did they play? How many did they win?” Recalling that they won 3 out of every 5 games, we set to work.
The multiple methods that we used were Make a Table, Guess and Check, and Make a Mathematical Model (an algebraic model, in this case).
Then we took a look at the work of an Algebra I student, M.:
They played 85 games that month
For this problem I did guess and check. First I tried 40/90 and 30/50 because it says they won 3/5 games so 70/140 was 50% which was too little so I tried 33+40/55+90 and got 50.3% so I realized
I needed to go up a lot more. Then I tried 60/100 + 40/90 and 100/190 and got 52.6% so I went down a bit and did 94/180 and got 52.22222% so I went down more and did 91/175 and 52% evenly. so
they had played 90 games and now they played a total of 175, so they played 85 games that month
Check- 40/90 + 51/90 is 91/175 which is 52%
Here are some of the initial things we noticed and wondered about M’s work:
• She checks her work at the bottom
• In the check, she adds 40/90 + 51/90 = 91/175 and we wondered how she was using that calculation to check her work
• We also wondered in the check if she meant 51/85 instead of 51/90 because 51/85 was used earlier in the problem and would represent winning 3 out of every 5 games (17 sets of 5 games)
• We wondered how she thought of her first guess, 30/50
• We noticed that sometimes she used “and” and sometimes she used the “+” sign
• We noticed she used fraction notation and wondered if she was using it informally, using “+” to mean “and” and using “/” to mean “out of”
• We noticed that she got the correct answer, and it didn’t take her very many guesses at all!
And finally, we thought of some questions we could ask M. to help her revise her work and keep reflecting on the problem such as:
• This time, it took you 5 guesses to get the answer. What are some ways you could find an answer with fewer guesses next time you solve a similar problem?
• How did you decide what guess to start at? What did you notice in the problem that led to your first guess?
• How would you explain your steps and calculations to another student who is stuck and doesn’t understand how to make a guess and check it?
• Each time you did a calculation you wrote it a different way. If you pick one of the ways (e.g. 33+40/55+90) and translate all the other calculations into that format, what patterns do you notice
in the calculations?
The final step, which we ran out of time for, is to get feedback on the questions we generated. Which of them are engaging? Which inspire reflection and meta-cognition? What do you think of the tone
of each question?
By the way, if you’re interested in the process of looking at student work and finding the right question to ask, we are focusing a new set of Professional Development courses on it this year. We’d
love to have you join us!
If you teach with the PoWs, or work with pre-service teachers, you might also be interested in our free mentoring, which pairs volunteer pre-service teachers with students working on the PoWs. The
mentors learn to ask good questions and the students get feedback on their mathematical thinking.
Some “Cherokee Creek Checkers Club” links in case you are interested:
• Information about accessing “Finding a Supplement” (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36
student logins as well! | {"url":"http://mathforum.org/blogs/pows/mentoring-the-cherokee-creek-checkers-club-algpow-with-the-noyce-scholars/","timestamp":"2014-04-19T09:31:00Z","content_type":null,"content_length":"23425","record_id":"<urn:uuid:f1382438-d394-48f4-b4aa-ce5906f00ca5>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00426-ip-10-147-4-33.ec2.internal.warc.gz"} |
CoinToss help
Join Date
Mar 2013
Rep Power
hi all, i am coding a coin flipping program that i am having a little trouble with. the program has to toss the coin until it has 3 of the same as well as return the number of try's it took
to get it. this is what i came up with so far
import java.util.*;
public class CoinFlip2 {
public static void main(String[] args) {
Random rand = new Random();
int H = 1;
int T = 0;
int tries = 0;
int result = rand.nextInt(1) + 0;
while(H != 3 || T !=3)
if (result == H){
H = H + 1;
else {
if (result == T){
T = T + 1;
i came across the
num1 = new random
num2 = new random
num3 = new random
while(num1 num2 and num3 are not equal) {
num1 = new random
num2 = new random
num3 = new random
println num1 and num2 and num3
i thought i understood what they was saying, but not so much
thank you for any help in the right directions on this
Join Date
Oct 2010
Rep Power
Hi Rawdogz, welcome to the forums.
I dont understand the comments below the code but you seem to be confusing counters with the values used in the conditions.
If for example the first random number is one which matches 'H'. 'H' is then increased by one to two which will never be generated by 'rand' and so the code will never be reached.
Instead of comparing 'result' to 'H' / 'T', you can call rand.nextInt(1) directly. The second if statement is redundent as if the result is not a head then its tails.
Your while condition is causing issues. It continues to loop until both 'H' and 'T' equal three which is unlikely.
It would also be advisable to follow naming conventions for variable names. All variables should begin with a lower case letter unless its a constant in which case all letters are usually
upper case.
And please enclose you code within [code] tags.
Join Date
Mar 2013
Rep Power
Hi Rawdogz, welcome to the forums.
I dont understand the comments below the code but you seem to be confusing counters with the values used in the conditions.
If for example the first random number is one which matches 'H'. 'H' is then increased by one to two which will never be generated by 'rand' and so the code will never be reached.
Instead of comparing 'result' to 'H' / 'T', you can call rand.nextInt(1) directly. The second if statement is redundent as if the result is not a head then its tails.
Your while condition is causing issues. It continues to loop until both 'H' and 'T' equal three which is unlikely.
It would also be advisable to follow naming conventions for variable names. All variables should begin with a lower case letter unless its a constant in which case all letters are usually
upper case.
And please enclose you code within [code] tags.
thank you for the reply Ronin attracted is the actual assignment.
Write a complete program that plays a coin flipping game, displays the individual flips, reports when a game is LOST or WON and shows the number of flips needed to complete the game.
The algorithm is as follows:
Simulate the flip of a coin using a JAVA random number generator.
Flip the coin once to initialize the flip value and print out the flip
Inside of a loop, repeatedly flip the coin until 3 consecutive flips have the same value (3 heads or 3 tails)
a. Display the flip results after each flip
When the game ends, report the total number of flips
Note: you can use 0 for heads and 1 for tails, but if you have time, add another method that will convert the integers 0 and 1 to the characters H and T
in my research on how to write this program i came across this model on how to write the program but i really didn't understand it
num1 = new random
num2 = new random
num3 = new random
while(num1 num2 and num3 are not equal) {
num1 = new random
num2 = new random
num3 = new random
println num1 and num2 and num3
Join Date
Oct 2010
Rep Power
Ah, I understand where you were going with the code. 'H' and 'T' are character values and not the name of the variables to use.
I see what the snippet of code in #3 is doing but this doesn't apply directly to your problem. The problem with that code is it generates three random numbers and loops if these are not
identical. An example would help explain the problem:
First go generates 0, 0, 1 which causes the code to loop. Second go generates 1, 1, 0 which causes the code to loop. Based on your requirements though the program should have terminated as
there were three consecutive 1s.
Forgetting the 'H' and 'T', in additon to the counter for the number of tries, you could use a simple counter to count the number of instances of one side (heads or tails). If the side
changes then the counter is reset and it starts again. You could use a boolean value to determine if you are currently counting heads or tails.
Once the counter reaches three then print out the number of tries required. | {"url":"http://www.java-forums.org/new-java/70020-cointoss-help.html","timestamp":"2014-04-17T11:10:53Z","content_type":null,"content_length":"77065","record_id":"<urn:uuid:471ff676-bdf8-4159-af57-904295ce51da>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00211-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cardinal sequences
István Juhász, William Weiss
In this article we characterize all those sequences of cardinals of length
J. Bagaria, Locally generic Boolean algebras and cardinal sequences, Alg. Univ. 47 (2002), pp. 283 - 302.
R. La Grange, Concerning the cardinal sequence of a Boolean algebra, Alg. Univ. 7 (1977), pp. 307 - 313.
J. C. Martinez, A consistency result on thin-tall superatomic Boolean algebras, Proc. AMS 115 (1992), pp. 473 - 477.
Downloading the paper | {"url":"http://www.renyi.hu/pub/setop/bill1/bill1_abs.html","timestamp":"2014-04-21T14:41:18Z","content_type":null,"content_length":"2821","record_id":"<urn:uuid:59754b5d-5de9-42c8-9994-897e2e0663e5>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00142-ip-10-147-4-33.ec2.internal.warc.gz"} |
Group theory
Date: Tue, 22 Nov 1994 08:21:15 -0800
From: (Matthew Baker)
Subject: Question
Could you please answer this??
The four rotational symmetries of the square satisfy the four requirements
for a group, and so they are called a subgroup of the full symmetry group.
(Notice that the identity is one of these rotational symmetries and that
the product of two rotations is another rotation in the subgroup.)
a. Do the four line symmetries of the square form a subgroup?
b. Does the symmetry group of the equilateral triangle have a subgroup?
Date: Mon, 28 Nov 1994 11:22:49 -0500 (EST)
From: Dr. Sydney
Subject: Re: Question
Hello there! Thanks for writing Dr. Math! What a good question you are
When you ask about the four line symmetries, I assume you are talking about
reflections about the 4 axes of symmetry, right? Well, it turns out these 4
reflections will not be a subgroup of the total symmetry group because the
set of all reflections is not closed. Can you see why? If you draw a
picture and label axes and vertices, and try and perform two of these
reflections, you'll find you will not get another reflection, but instead
you'll get a rotation. Try it out! Another reason they don't form a
subgroup is because the identity element is not an element of the set.
However, if you take the set that contains the identity element and any ONE
reflection, you will get a subgroup. Can you see why?
On to your next question...the symmetry group of the equilateral triangle
does have subgroups; in fact it has 4 subgroups in addition to the trivial
subgroups. Can you guess what they may be? (Hint: very similar to the
symmetry group of the square--consider rotations and reflections.) See if
you can figure this one out--if you want you can write back with your
thoughts on the problem--I'd love to hear what you think.
I'm quite impressed you are doing group theory in high school--I am just
learning group theory this semester in an abstract algebra class. Please do
write back with any more questions you might have.
--Sydney, a math doctor | {"url":"http://mathforum.org/library/drmath/view/54741.html","timestamp":"2014-04-18T09:01:24Z","content_type":null,"content_length":"7321","record_id":"<urn:uuid:0b3d94c8-be7d-417a-bdc3-5f298a5f8b6a>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00378-ip-10-147-4-33.ec2.internal.warc.gz"} |
Feldman, Joel - Department of Mathematics, University of British Columbia
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• The Behaviour of sin z In these notes we look at the values that sin z takes as z runs over C. We recall that | {"url":"http://www.osti.gov/eprints/topicpages/documents/starturl/08/007.html","timestamp":"2014-04-19T09:59:49Z","content_type":null,"content_length":"86858","record_id":"<urn:uuid:a2b3236a-679f-484b-b352-214c48796995>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00046-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculating Divorce Rates
Unreviewed Activity submitted in preparation for the NNN Writing with Numbers Workshop
This assignment acquaints students with a variety of ways to calculate the likelihood of divorce. Often the mass media and other reports tend to report divorce statistics in less useful and
occasionally useless ways. Students use Excel to calculate and observe trends in divorce rates for national data from 1900 - 2004. From the data, students discuss which of three rates is the best way
to measure divorce and write a short 2-3 page paper on their selection and the usefulness of three media related articles that discuss divorce rates.
Learning Goals
To complete this assignment, students must utilize rate calculation, graphing, and critical thinking skills along with the ability to concisely summarize divergent views on a given topic.
Context for Use
This activity was designed for a sophomore level class in family sociology. The activity needs to be introduced prior to students working with the data and should require at least part of a class for
discussion when the assignment is due.
Description and Teaching Materials
This assignment acquaints students with a variety of ways to calculate the likelihood of divorce. Often the mass media and other reports tend to report divorce statistics in less useful and
occasionally useless ways. There are four ways to calculate divorce rates. The first, often called the "apples and oranges" method involves calculating a ratio of the number of divorces to marriages
occurring in a given year. This involves comparing two groups which have little in common as only very few divorces involve marriages formed during the same year. The second, called the crude divorce
rate, involves looking at the number of divorces taking place within the entire population whether married, widowed, too young to marry or not. These are the two most commonly used statistics in the
media and in official governmental data.
The final two provide a more precise measure of divorce but are used less often. The refined divorce rate is based on the number of divorces within all marriages existing at a given time and is a
useful way to look at the chances of divorce occurring. The last approach follows the amount of divorce occurring among a cohort of people who marry during the same year over time. This is a useful
way to track when divorce occurs after marriage and to compare the likelihood of divorce within different cohorts.
In this exercise students calculate and graph the "apples and oranges," crude, and refined divorce rates from raw data from 1900 to the present in five and ten year intervals. They will be asked to
discuss the advantages and disadvantages of each method. Finally, they will read examples of how all four rates are used in the media.
Teaching Notes and Tips
Students have a tendency to identify the correct divorce rate from the spreadsheet data (refined divorce rate = number of divorces/number of existing marriages), but to select a different rate when
responding to the articles provided. On the other hand, students used the articles to identify factors that are associated with divorce such as educational level, etc.
See attached pre-test. The post-test needs work. | {"url":"http://serc.carleton.edu/nnn/quantitative_writing/examples/23804.html","timestamp":"2014-04-17T12:35:37Z","content_type":null,"content_length":"25839","record_id":"<urn:uuid:6c87fdfb-5898-4818-b7be-71f39d563a89>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00057-ip-10-147-4-33.ec2.internal.warc.gz"} |
method of characteristics
There are essentially two ways for the method of characteristics (I recognise your name from a number of MOC posts, did I post my notes on the subject?)
From one of your calculations you have:
Integrating this equation shows that:
\frac{1}{u}=\frac{x^{2}}{2}+F(\xi )
We now paramatrise the initial data, so take [itex](\xi ,1)[/itex] as the point which the characteristic passes through, this will give the initial values as [itex]u(\xi ,1)=\xi[/itex], evaluating
the characteristic at this point yields [itex]1=k\xi[/itex], giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate [itex]u[/itex] at the point
[itex](\xi ,1)[/itex] to obtain
\frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )
From this you can compute [itex]F(\xi )[/itex] and then from there you can substitute for [itex]\xi[/itex] by using the equation of the characteristic.
Now to find u you have the solution | {"url":"http://www.physicsforums.com/showthread.php?t=515925","timestamp":"2014-04-20T08:35:31Z","content_type":null,"content_length":"31475","record_id":"<urn:uuid:5e4f7564-d997-4f00-a63d-ec9c730f3825>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00183-ip-10-147-4-33.ec2.internal.warc.gz"} |
EE 8215 – Nonlinear Systems
Course description
Introduction. Examples of nonlinear systems. State-space models. Equilibrium points. Linearization. Range of nonlinear phenomena: finite escape time, multiple isolated equilibria, limit cycles,
chaos. Bifurcations. Phase portraits. Bendixson and Poincare-Bendixson criteria. Mathematical background: existence and uniqueness of solutions, continuous dependence on initial conditions and
parameters, normed linear spaces, comparison principle, Bellman-Gronwall Lemma. Lyapunov stability. Lyapunov's direct method. Lyapunov functions. LaSalle's invariance principle. Estimating region of
attraction. Center manifold theory. Stability of time-varying systems. Input-output and input-to-state stability. Small gain theorem. Passivity. Circle and Popov criteria for absolute stability.
Perturbation theory and averaging. Singular perturbations. Feedback and input-output linearization. Zero dynamics. Backstepping design. Control Lyapunov functions.
Class schedule
TuTh, 2:30pm - 3:45pm, MechE 102; Jan 21 - May 9, 2014
Instructor and Teaching Assistant
• Teaching Assistant
Xiaofan Wu
Office: Keller Hall 2-276
Regular office hours: Wednesday, 4:45pm - 5:45pm
Extra office hours: Monday, 4:45pm - 5:45pm (only when HW is due Tuesday)
Textbook and software
Grading policy
• Homework policy
Homework is intended as a vehicle for learning, not as a test. Moderate collaboration with your classmates is allowed. However, I urge you to invest enough time alone to understand each homework
problem, and independently write the solutions that you turn in. Homework is generally handed out every Thursday, and it is due at the beginning of the class a week later. Late homework will not
be accepted. Start early!
• Even though I plan to cover everything from scratch, the students would benefit from an exposure to linear systems (EE/AEM 5231 or an equivalent course). Those interested should contact the | {"url":"http://www.ece.umn.edu/users/mihailo/courses/s14/index.html","timestamp":"2014-04-21T00:09:06Z","content_type":null,"content_length":"4862","record_id":"<urn:uuid:42073010-951a-4014-9c93-d84226bc2d9b>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00158-ip-10-147-4-33.ec2.internal.warc.gz"} |
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st: re: labeling question
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st: re: labeling question
From David Airey <david.airey@vanderbilt.edu>
To statalist@hsphsun2.harvard.edu
Subject st: re: labeling question
Date Sun, 28 Dec 2008 19:06:08 -0600
I should say also that graph_by doesn't respect the variable format command, although I think it should.
I have a variable that is numeric, like:
but about 20 levels to the variable. I want a to use graph_by where the by variable is the above, but I don't want the labels to show the float values. I could _string_ the numeric variable and
graph_by that, but then the order is not numeric on the graph.
How do I rapidly change a float numeric variable to a labeled numeric variable such that my graph_by is labeled more neatly, and in numeric order?
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2008-12/msg00993.html","timestamp":"2014-04-17T12:47:08Z","content_type":null,"content_length":"5914","record_id":"<urn:uuid:2f5fd277-4140-44a8-899b-55c22283564b>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00320-ip-10-147-4-33.ec2.internal.warc.gz"} |
Wolfram Demonstrations Project
Stratification as a Device for Variance Reduction
Stratification is a commonly used device for variance reduction in statistical estimation problems. A judicious choice of the strata may lead to substantial reductions in the variance of the
estimator. In this Demonstration, the estimation of the mean of a truncated gamma distribution with parameters [2, 5], adequately normalized, is considered.
Once the strata limits are established, the variance of the estimator depends also on the sample allocation, that is, the sample size allocated to each stratum. This Demonstration affords a simple
way to look at the behavior of the variance as a function of strata definition and sample allocation. This variance is compared with the variance in simple random sampling (simple r.s.), that is,
the sampling scheme with the same total sample size and no stratification. Moving the sliders for strata definition and sample sizes yields the change in the value of the variance depicted on the
scale placed at the right of the distribution plot.
Contrary to common belief, an inadequate stratification may lead to variances greater than those obtained in simple random sampling. | {"url":"http://demonstrations.wolfram.com/StratificationAsADeviceForVarianceReduction/","timestamp":"2014-04-17T09:39:47Z","content_type":null,"content_length":"43089","record_id":"<urn:uuid:39cdbaf2-857c-47d9-bed8-6b936363ed4d>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00236-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Forums - View Single Post - The Should I Become a Mathematician? Thread
Would it be advantageous to obtain two bachelors degrees in mathematics (one in pure math and the other in applied math). Would this combination open up more career options (i.e. a quant/financial
engineer, operations research, etc..) than just majoring in pure or applied math alone. I am leaning towards becoming a quant/financial engineer in some company in the future. Also, the two bachelor
degrees in pure and applied math would be obtained in 5 years versus 8 years. So I figure that this is a good deal. What are your opinions? | {"url":"http://www.physicsforums.com/showpost.php?p=1056535&postcount=191","timestamp":"2014-04-16T16:17:13Z","content_type":null,"content_length":"7231","record_id":"<urn:uuid:3e1386a6-200e-4b03-9aa6-41628adac013>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00298-ip-10-147-4-33.ec2.internal.warc.gz"} |
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8.1.2 Solar Cell Current-Voltage Characteristics and Equivalent Circuit
It is important to look a bit more closely at the IV-characteristics of a silicon pn-junction solar cell. The proper equation for that was already introduced before
In a kind of short-hand notation, and because it is what electrical engineers always do, we could symbolize that with the normal diode symbol:
However, it is customary to use a "two diode" model, where the first diode describes just the simple pn-junction behavior without carrirer generation and recombination in the space charge region
(SCR), and the second diode, which then must be switched in parallel, describes the "non-ideality" of the real pn-junction. The equivalent circuit diagram then would consist of two diodes in
Since our junctions area is > 100 cm^2, there might be regions where we have a local "shunt" in the junction. In the worst case this is a local short-circuit, caused for example because a metal
particle was embedded in the junction (we obviously need to be clean if we make those junctions). The best we can do to model globally the many possibilties for shunts is to add a shunt resistor
R[Sh] in parallel to our diodes. "Global" in contrast to "local" means that we only care for whatever we can measure for the total solar cell, i.e. whatever we can get using only the two wires
attached to it.
If we now add the internal series resistance that is always there in series to what we already have, and consider that the photocurrent flowing across the junctions(s) is simply a constant current
generator in electrical engineering terms, we obtain the final (and most simple) equivalent cuircuit diagram of a solar cell and a somewhat modified master equation for the equivalent circuit
┃ æ ö æ eU[eff] nkT ö ┃
┃ I = I[1]^ · ç exp eU[eff] kT^ 1 ÷ + I[2]^ · ç exp ^ 1 ÷ + U[eff] R[SH] I[Ph] ┃
┃ è ø è ø ┃
┃ ┃
┃ "Ideal" Diode "Bad" Diode ┃
┃ ┃
┃ U[eff] = U I · R[SE] ┃
It is easy to see how the master equation was changed.
First we switched from current densities j to currents I and simply used the terms I[1] and I[2] as abbreviations for the pre-exponential factors. I[1] and I[2] belong to the first (= ideal) and
second (non-ideal) diode in the equivalent circuit diagram.
Next, we considered that the junction voltage U[eff] is not identical to the externally measured voltage U but smaller by the voltage drop in the series resistor R[SE]; i.e. U[eff] = U I · R[SE].
Then we generalized the non-ideal diode a bit by not just having a factor of two in the denominator of the exponent, but an ideality factor n. For n = 2 we would have the old equation, but since we
arrived at that part by rather hand-waving approximations, real (non-ideal) diodes may be better described by some number other than 2. Our ideal diode has n = 1.
Finally we add the current lost in the shunt resistor, which is given by U[eff] / R[SH]. We have a positive sign because this current must be opposed to the photo current, which has a negative sign
in the conventions chosen.
Everything is quite clear - except that we can not solve this equation anymore analytically and plot U(I) curves with R[SH], R[SE], and n as parameters.
We have reached a point where we need to do some exercise:
It is absolutely essential that you do this exercise - at least ponder the question and look carefully at the solutions provided.
You will learn that the series and shunt resistance does not only matter very much but that we need to have extremely low values for the series resistance in the order of at most a few mW per (10 ×
10) cm^2 solar cell.
Class Exercise: Cu has a specific resistivity of about 2µWcm. Given a solid Cu wire with 1 mm^2 cross section, what length will give you already 10 mW?
OK - coming back from the exercise (or just the solution) you learned that:
The series resistance really matters for high efficiency cells; i.e. for all solar cells with let's say h > 10 %. In fact, optimizing the series resistance of a standard commercial solar cells
(now, in 2007, with h » 15 %) is one of the major tasks in solar cell R&D. We will look at some number in another exercise coming up in the next module.
The shunt resistance, if not extremely small, is less obnoxious for the efficieny h, but generates high leakage currents in reverse direction, increasing with the voltage. This is deadly in modules
and thus cannot be tolerated either.
As far as the ideality factor is concerned, we do not notice major problems. Here you have to look at details to see that the impact of the second diode is ther and detrimental.
The way to lump everything together globally is to describe the global IV-characteristics with 3 numbers:
1. The fill factor FF, defined as the relation between the area of the large yellow rectangle to the more orange area that is centered at the optimal working point.. That implies that the inner
rectangle has the largest area of any rectangle (Area = U · I = power delivered by the solar cell). The fill factor measures the degree of "rectangularity" of the characteristics.
2. The open-circuit voltage U[OC], i.e. the maximum voltage the solar cell will produce for the load resistor
3. The short-circuit current I[SC].
The efficiency h then is directly proportional to U[OC], I[SC], and FF, i.e.
┃ h = const · U[OC] · I[SC] · FF ┃
How do we measure the IV-characteristics of a real solar cell coming out of a production line?
Easy, you might think: Apply a voltage, measure the current, change the voltage, measure the current again... Do it automatically by using a voltage ramp and keeping track of the current. Yes -
that would be perfectly OK except that solar cells from a real production come off the line at the rate of about 1 solar cell per second! Your measurement scheme is too slow for this because
whenever the current changes the capacity of the SCR must be recharged and this takes some time.
The way it is done is simple but powerful: You impress a constant current on your solar cell by some external power source. At a given illumination intensity this takes a certain voltage as
determined by the IV-characteristics. If you change the illumination intensity, the voltage needed to drive the constant current changes, too. If you change your illumination intensity very quickly
be just using a flash light, your voltage can follow just as quickly because you do not change the charge in the SCR capacitor.
The U(t) curve for a fixed current contains all the information you want - but now you can get it with flashing speed!
You will actually do this in a Lab course; the link provides the details.
We are not yet done. Even if you have understood and committed to memory everything pointed out above with all its implications, at least two topics must be still be addressed:
1. How about other types of solar cells?
2. Connection between local and global parameters.
The first question can be dealt with rather easily - in principle:
Whatever type of solar cell you have - Si bulk, µ-crystalline Si thin film type, amorphous Si, CIGS or CdTe thin films, dye-based TiO[2] electrolytic cells - to name just a
few, they must have some characteristics similar to a diode, and you can always find a suitable equivalent circuit diagram for modelling its behaviour.
This equivalent circuit diagram might be somewhat different from the one given above, but you always have the problems associated with series and shunt resitances after you
solved the basic junction problems.
Quite often your choices for contacts are severely restricted and your series resistances get so large that you only can get decent efficiencies by switching them in series
in relatively small units. More to that will be found in the link.
The second point is the difficult one. What he have discussed so far is the global behaviour of a solar cell - what you measure for the whole "global" 100 mm x 100 mm cell, being uniformly
Now take your virtual knife and cut your solar cell into 10 000 (1 x 1) mm^2 cells (without any damage and so on), an measure the IV-characteristics of those 10 000 local
solar cells. We assume that you know how to do this - it is possible albeit not easy.
If your global solar cell was not very uniform (rather likely for many solar cell types), your local solar cells may show wildly different behaviour. The IV characteristics
of the, let's say, 8 local cells that contain serious shunts (= short circuits) will be far more affected by this than your global cell (where you have another 9 992
shunt-free cells switched in parallel).
You get the point: Your major global parameters like short circuit current I[SC], open circuit voltage U[OC], fill factor FF, series and shunt resistances R[SE] and R[SH], diode ideality factor n,
are somehow determined by the local values, but it is not immediately obvious how.
Besides the global short circuit current I[SC], which obviously is just the sum of the local short circuit currents, for all other parameters the relationship between local
and global parameters is non-trivial or very complex. In the case of the global ideality factor n, the number you extract from your global IV curves may not even have much
to do with the junction properties.
Do we need to know the local parameters? After all, what counts is the real = global solar cell. If it is as good as it should be, the local parameters might be of no interest.
Well - yes and no. Even if everything is OK globally, some small local defects may limit the life time of the solar cell, i.e. be a risk to its expected 20+ years of
harvesting energy.
More important, if you don't know your local parameters, you have very little guidance in making your solar cell better. If, for example, the global short circuit current is
too low, that may indicate that it is just too low everywhere or that it might be allright on most parts of the cell but really lousy in some small areas. If you don't know
what is the reason, you can't fix the problem.
To give an idea what we are talking about, look at the pictures of the "multi-crystalline" solar cell below:
A perfect solar cell would show only one color - the optimal value. The colorful pictures thus demonstrate that there is room for improvement and that you have to have the information contained in
the pictures to be able to make progress
How are maps like these obtained? With the so-called CELLO-Plus technique, developed in Kiel. This is a rather sophisticated characterization technique that is described in some more detail in the
© H. Föll (Semiconductor Technology - Script) | {"url":"http://www.tf.uni-kiel.de/matwis/amat/semitech_en/kap_8/backbone/r8_1_2.html","timestamp":"2014-04-20T08:14:33Z","content_type":null,"content_length":"33851","record_id":"<urn:uuid:4409c58d-ca21-45c2-b61c-57eb7cb4dd1d>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00113-ip-10-147-4-33.ec2.internal.warc.gz"} |
NYJM Abstract - 18-43 - Joel H. Shapiro
Joel H. Shapiro
Strongly compact algebras associated with composition operators
view print
Published: October 20, 2012
Keywords: Composition operator, multiplication operator, strongly compact algebra
Subject: Primary 47B33, 47B35; Secondary 30H10
An algebra of bounded linear operators on a Hilbert space is called strongly compact whenever each of its bounded subsets is relatively compact in the strong operator topology. The concept is
most commonly studied for two algebras associated with a single operator T: the algebra alg(T) generated by the operator, and the operator's commutant com(T). This paper focuses on the strong
compactness of these two algebras when T is a composition operator induced on the Hardy space H^2 by a linear fractional self-map of the unit disc. In this setting, strong compactness is
completely characterized for alg(T), and "almost'' characterized for com(T), thus extending an investigation begun by Fernández-Valles and Lacruz [A spectral condition for strong compactness, J.
Adv. Res. Pure Math. 3 (4) 2011, 50-60]. Along the way it becomes necessary to consider strong compactness for algebras associated with multipliers, adjoint composition operators, and even the
Cesàro operator.
Author information
Fariborz Maseeh Department of Mathematics and Statistics, Portland State University, Portland OR 97207 | {"url":"http://www.kurims.kyoto-u.ac.jp/EMIS/journals/NYJM/j/2012/18-43.html","timestamp":"2014-04-21T02:05:19Z","content_type":null,"content_length":"7986","record_id":"<urn:uuid:cfe14237-3d75-491d-83c2-6cacaa2896e7>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00354-ip-10-147-4-33.ec2.internal.warc.gz"} |
Gauss's Day of Reckoning
Gauss's Day of Reckoning
A famous story about the boy wonder of mathematics has taken on a life of its own
Let me tell you a story, although it's such a well-worn nugget of mathematical lore that you've probably heard it already:
In the 1780s a provincial German schoolmaster gave his class the tedious assignment of summing the first 100 integers. The teacher's aim was to keep the kids quiet for half an hour, but one young
pupil almost immediately produced an answer: 1 + 2 + 3 + ... + 98 + 99 + 100 = 5,050. The smart aleck was Carl Friedrich Gauss, who would go on to join the short list of candidates for greatest
mathematician ever. Gauss was not a calculating prodigy who added up all those numbers in his head. He had a deeper insight: If you "fold" the series of numbers in the middle and add them in
pairs—1 + 100, 2 + 99, 3 + 98, and so on—all the pairs sum to 101. There are 50 such pairs, and so the grand total is simply 50×101. The more general formula, for a list of consecutive numbers
from 1 through n, is n(n + 1)/2.
The paragraph above is my own rendition of this anecdote, written a few months ago for another project. I say it's my own, and yet I make no claim of originality. The same tale has been told in much
the same way by hundreds of others before me. I've been hearing about Gauss's schoolboy triumph since I was a schoolboy myself.
The story was familiar, but until I wrote it out in my own words, I had never thought carefully about the events in that long-ago classroom. Now doubts and questions began to nag at me. For example:
How did the teacher verify that Gauss's answer was correct? If the schoolmaster already knew the formula for summing an arithmetic series, that would somewhat diminish the drama of the moment. If the
teacher didn't know, wouldn't he be spending his interlude of peace and quiet doing the same mindless exercise as his pupils?
There are other ways to answer this question, but there are other questions too, and soon I was wondering about the provenance and authenticity of the whole story. Where did it come from, and how was
it handed down to us? Do scholars take this anecdote seriously as an event in the life of the mathematician? Or does it belong to the same genre as those stories about Newton and the apple or
Archimedes in the bathtub, where literal truth is not the main issue? If we treat the episode as a myth or fable, then what is the moral of the story?
To satisfy my curiosity I began searching libraries and online resources for versions of the Gauss anecdote. By now I have over a hundred exemplars, in eight languages. (The collection of versions is
available here.) The sources range from scholarly histories and biographies to textbooks and encyclopedias, and on through children's literature, Web sites, lesson plans, student papers, Usenet
newsgroup postings and even a novel. All of the retellings describe what is recognizably the same incident—indeed, I believe they all derive ultimately from a single source—and yet they also exhibit
marvelous diversity and creativity, as authors have struggled to fill in gaps, explain motivations and construct a coherent narrative. (I soon realized that I had done a bit of ad lib embroidery
After reading all those variations on the story, I still can't answer the fundamental factual question, "Did it really happen that way?" I have nothing new to add to our knowledge of Gauss. But I
think I have learned something about the evolution and transmission of such stories, and about their place in the culture of science and mathematics. Finally, I also have some thoughts about how the
rest of the kids in the class might have approached their task. This is a subject that's not much discussed in the literature, but for those of us whose talents fall short of Gaussian genius, it may
be the most pertinent issue. | {"url":"http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning","timestamp":"2014-04-18T18:25:48Z","content_type":null,"content_length":"131548","record_id":"<urn:uuid:d147ad0e-47f1-4174-8017-759d2ecc06e4>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00562-ip-10-147-4-33.ec2.internal.warc.gz"} |
Derivative of an integral
December 2nd 2007, 03:54 PM #1
Derivative of an integral
$h(x)=\int_{6}^{x^{2}} \sqrt{5+r^{3}}dr$
Find h'(x)
Note: I'm having a tough time with these, so explain it like you would to an idiot.
This is what I've come up with so far:
$h'(x)=\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)\frac{du}{dx}$
$h'(x)=\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)2x$
But I can't figure out how to find $\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)$
Last edited by angel.white; December 2nd 2007 at 04:10 PM.
Always remember:
Thanks, I got
$\sqrt{5+x^{6}}\cdot 2x - \sqrt{5+6^{3}}\cdot 0$
$\sqrt{5+x^{6}}\cdot 2x$
And that was correct. Appreciate it.
yes. but note that since the derivative of a constant is zero, you can essentially neglect to write the sqrt(5 + 6^3) part. when you see a constant as a limit, just ignore it.
note that the results here follow by the second fundamental theorem of calculus (as it is sometimes called), galactus' result follows by applying the chain rule to the theorem
yes. but note that since the derivative of a constant is zero, you can essentially neglect to write the sqrt(5 + 6^3) part. when you see a constant as a limit, just ignore it.
note that the results here follow by the second fundamental theorem of calculus (as it is sometimes called), galactus' result follows by applying the chain rule to the theorem
Man, integrals are killing me right now O.o
Got done with all the homework, but I really need to study before the test, but due to my other class schedules and my working, I don't think I'll get another opportunity :/ I might not do as
well on this one. Right now I have to just use galactus' method without understanding it, simply b/c of time restraints, but I bought some calc books off amazon.com that I'll try to go through
during Christmas break, before Calc2. Then hopefully I will be able to really wrap my head around the integrals.
Man, integrals are killing me right now O.o
Got done with all the homework, but I really need to study before the test, but due to my other class schedules and my working, I don't think I'll get another opportunity :/ I might not do as
well on this one. Right now I have to just use galactus' method without understanding it, simply b/c of time restraints, but I bought some calc books off amazon.com that I'll try to go through
during Christmas break, before Calc2. Then hopefully I will be able to really wrap my head around the integrals.
look up the second fundamental theorem of calculus. if you don't get it, come back here and ask questions
December 2nd 2007, 03:59 PM #2
December 2nd 2007, 04:18 PM #3
December 2nd 2007, 07:28 PM #4
December 2nd 2007, 08:41 PM #5
December 2nd 2007, 08:43 PM #6 | {"url":"http://mathhelpforum.com/calculus/23998-derivative-integral.html","timestamp":"2014-04-18T22:25:58Z","content_type":null,"content_length":"52098","record_id":"<urn:uuid:490eb339-8572-4507-bba1-0ddbdd4275ea>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00109-ip-10-147-4-33.ec2.internal.warc.gz"} |
Addition Games
This list of addition games and worksheets will help you learn addition facts and increase your addition skills.
Repair the pipes by adding numbers. Select the pairs of numbers that will get the given sum.
Balancing Act tests your ability to add quickly. The object of the game is to make the Play math lines game and shoot numbered balls to the advancing ball line to make the number 10. Stars and
two sides of the scale balance. Click on a weight, and it will hop to the other side. coins may appear so try to shoot them.
The problems begin with three weights (easy) and get to six weights (hard).
Clear the playfield making a necessary sum of numbers on neighboring tiles. Train your Collect the required sum, rotating the digits in the falling figure. Train your addition and logical thinking
addition and logical thinking skills. Play online with other players. skills. Compete with other players online.
Complete the sum by swapping the digits and change the cells to white. Train your Play Dress Up Addition. Dress Up the girl with tops, bottoms, and shoes. Spend exactly the Goal amount to
addition and logical thinking skills. Play online with other players. advance to the next level
Find the chosen sum by diagonally, horizontally, or vertically circling a group of Destroy UFOs with your addition skills. In the math space game, Addition Attack, kids must add
numbers. Beat the clock and score as many points possible! single-and-double-digit numbers together while shooting spaceships trying to invade Earth.
Shoot down the aliens and impove your addition skills. Single player spaceship additon (Froggy Hops) Fill in the missing number in the number sentence. Finding 10 more or 1 more than a given
game. number, (counting on). Click on the tops of the lilly pads!
The whale is stuck in the lake, she wants to go out to the sea. Click on the number Recall number facts up to 10+10. See the Funky Mummy dance when you get it right! Get 10 answers correct
bond (add to 4, 5, 6, 7, 8, 9 or 10) that would complete the water pipe. without clicking on a wrong sarcophacus and you can write in Egyptian hieroglyphs.
Recall bonds of 20 (add to 20). See the Funky Mummy dance when you get it right! Get 10 Learn and practice single digit addition facts by turning the tumblers on a safe. If you get four correct you
answers correct without clicking on a wrong sarcophacus and you can write in Egyptian can enter the safe and play a game.
Collect fuel by flying into it. Shoot the fuel that you don't want. The faster you
shoot the fuel that you don't want, the faster you see new fuel! Tests your addition Add 5 units of fuel to each ship that lands at your refuelling base.
Choose the beavers that add up to 10.In a similar way to the games at the seaside,
imagine your mouse pointer is a hammer, you must tap all the animals on the head with a Use base ten blocks to model grouping in addition.
total of 10.
Marble Math is a fun and educational activity for children learning addition. There are five different levels
Place the digits in the right places in the addition problem in order to create a of addition and a fun bonus activity after each level. Children must score 70% or greater to advance to the
correct solution! next level. Audio helps children understand how to count and play the activity. Have fun with this addition
Arithmetiles Beat the clock and advance to the next level. Discover the secrets of Arithmetiles. Create as
many Arithmetile combinations as possible before time runs out. You must add, subtract, multiply or divide to Bee Smart Math Addition
get the given number before time runs out. Beat the clock and advance to an entirely different puzzle. Power (Spider Attack)
Players can choose between Timed and Untimed games and play over 100 puzzle levels.
Balancing Act Balancing Act tests your ability to add quickly. The object of the game is
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We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"http://www.onlinemathlearning.com/addition-games.html","timestamp":"2014-04-18T00:35:45Z","content_type":null,"content_length":"67261","record_id":"<urn:uuid:1cbfa81c-604e-4296-ab48-50301cfd2ac9>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00422-ip-10-147-4-33.ec2.internal.warc.gz"} |
Superscript and Subscript Unicode Range
We modeled a fraction feature from things we saw in other fonts and even got some pretty cool code from this very fabulous build forum. When there is no glyph for a particular fraction the glyphs
used in the substitution are encoded in the Private Use area.
Now, however, we’re wondering if they would be better in the superscript and subscript range (2070-209F). The main use of the font thus far is typesetting in InDesign. The client wants to use this
same font for their web product though we’re not sure what that will be just yet. So, our thought is that if all the numerator and denominator glyphs are encoded in the superscript and subscript
range we’ll have a little more of a direct “conversion” from the InDesign file using the fraction feature to the superscript and subscript Unicode range that should (I think?) be pretty easy to
display on the web.
We’re hesitant, however, as very few fonts seem to use the superscript and subscript range in any significant way.
Can anyone comment on what the intent of this range is and/or why the numerous fonts we checked out all use private use range or unencoded glyphs for the numerator and denominator glyphs?
Thanks very much for slogging through this far too long post.
Login or register to post comments
…why the numerous fonts we checked out all use private use range or unencoded glyphs for the numerator and denominator glyphs?
I don’t know which fonts you checked, but I generally use differently sized and positioned glyphs for superiors/inferiors and numerator/denominator, with unencoded “.numr” and “.dnom” glyphs, e.g:
• Login or register to post comments | {"url":"http://typophile.com/node/95084","timestamp":"2014-04-16T08:38:26Z","content_type":null,"content_length":"11038","record_id":"<urn:uuid:73ae59ae-5ecc-43c0-bf3e-d202a24cb696>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00522-ip-10-147-4-33.ec2.internal.warc.gz"} |
U James - r73rtf
Principal Investigator Matthew R James Project r73
Systems Engineering, Machine CM
Research School of Physical Sciences and Engineering
Co-Investigator Sonny Yuliar, Peter Dowell
Systems Engineering,
Research School of Physical Sciences and Engineering
H-infinity control
Our research activities have been in the field of robust H-infinity control for nonlinear dynamical systems. This field addresses the problem of achieving high performance systems in the presence of
uncertainties. This problem arises in many engineering problems ranging from flight control, chemical process regulation to automated washing machines. The term "H-infinity" here is a conventional
symbol to indicate that in the design process we minimize the maximum or worst effects of the uncertainties on the systems.
Our general goal is to develop a control synthesis theory as well as efficient numerical methods for computing the required control policies. The fundamental equations that we deal with are of the
nature of dynamic programming ones. In particular, they are in the form of first order nonlinear partial differential equations (PDE's). We employ a finite difference technique to discretize these
PDE's and use viscosity solution methods to analyse the convergence properties of the discretized equations. The development of the basic numerical methods was initiated in 1993. Our work since the
beginning of 1994 has been focussing on obtaining accelerated versions of the basic methods.
What are the basic questions addressed?
The basic questions are: how to discretize the PDE
infu in U supw in W ( DS(x) . f(x,u,w) + L(x,u,w) ) = 0 in X,
in which DS(x) denotes the gradient of S(x) with respect to x, f, L are determined from the problem data, U, W, X are spaces of u, w, x respectively, and how to compute a solution to the discretized
equation efficiently.
What are the results to date and the future of the work?
We employ a finite difference method to discretize the PDE and a viscosity solution technique to obtain convergence properties. The discretized PDE is computed using value space and policy space
iteration methods. These constitute our basic numerical schemes. We have also obtained accelerated versions of the basic schemes. The future of the work is to utilize the numerical method in a real
control design problem.
What computational techniques are used and why is a supercomputer required?
The computational techniques involve iteratively computing S^k(x), k = 1,2,...., for all x in (X^n)^d, via
S^k(x) = infu in U^d supw in W^d ( [[Sigma]]z in N^d(x) p(x,z;u,w) S^k-1(z) + [[lambda]](x,u,w) L(x,u,w) ),
in which d is the grid size, p and c are scalar functions, and the notation U^d indicates the discretized U space, and so on. As k ->[[infinity]], S^k(x) ->S(x). Thus computation of S^k at x requires
S^k-1(z) for all z in the neighbourhood of x, N^d(x). This computation can be efficiently carried out using a parallel machine by assigning a single processor to each point x in (X^n)^d.
Numerical Approximation of the H[[infinity]] Norm for Nonlinear Systems, M R James, and S Yuliar, to appear in Automatica, June, 1995.
Robust Output Feedback Control for Bilinear Systems, C A Teolis, S Yuliar, M R James, and J Baras, in Proc. 33rd IEEE Conference on Decision and Control (CDC), USA, December 1994. | {"url":"http://anusf.anu.edu.au/annual_reports/annual_report94/UJames-r73.html","timestamp":"2014-04-16T13:23:58Z","content_type":null,"content_length":"4483","record_id":"<urn:uuid:f3a9ce01-057d-49ee-82f9-da9c506eeecf>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00178-ip-10-147-4-33.ec2.internal.warc.gz"} |
rate of change
August 22nd 2007, 08:46 AM
rate of change
A machine is rolling metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume V is 128 $\pi$ cubic inches. At what rate
is the length H changing when the radius r is 2.5 inches? $[Hint: V = \pi r^2 h]$
August 22nd 2007, 09:32 AM
A machine is rolling metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume V is 128 $\pi$ cubic inches. At what rate
is the length H changing when the radius r is 2.5 inches? $[Hint: V = \pi r^2 h]$
Use your hint.
Note: It would help if you could post in a less general category and profide a more descriptive title. It might give a clue as to your level or what class you are in or what tools should be used.
In this case, it's a nice differential calculus problem. If you're in a different class, that will be a problem.
r and h are functions of time.
V is constant, so dV = 0
dr is given as -0.05"/sec
Find dh when r = 2.5"
Complete derivative:
$V = \pi r^2 h$
$dV = \pi \left(r^{2}\frac{dh}{dt} + h*\left(2r\frac{dr}{dt}\right)\right)$
dV is known.
r is given.
dh/dt is what we need. We'll have to solve for it after everything else is found.
h - You will have to solve for this from the Hint.
dr/dt is given.
Substitute and solve.
Additional Note: Please show your work. It is a great place to learn where you are in your mathematics understanding.
August 22nd 2007, 09:42 AM
It's calc...I'mnot sure what to do =(
August 22nd 2007, 01:08 PM
Samantha, Samantha...I did nearly the entire problem for you. It is time for you to start showing a little more gumption. Please go read the previous post and pick out the statements that contain
instructions for you to follow.
Really, there remains only a little algebra. If you cannot do it, you should go have a really frank conversation with your academic advisor. | {"url":"http://mathhelpforum.com/calculus/17963-rate-change-print.html","timestamp":"2014-04-17T07:22:28Z","content_type":null,"content_length":"7636","record_id":"<urn:uuid:77a9296b-fc4c-4901-b884-de937b6b0d80>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00233-ip-10-147-4-33.ec2.internal.warc.gz"} |
Peaches in General
Copyright © University of Cambridge. All rights reserved.
'Peaches in General' printed from http://nrich.maths.org/
There's a problem which goes like this :
A monkey had some peaches.
He ate half of them plus one more.
On the second day, he ate half of the rest plus one more.
On the third day, he ate half of the rest plus one more again.
On the fourth day, he found there was only one left.
How many did he have at the beginning?
By the time you reach Stage 4 you will start to feel that problems like this are examples or instances of something more general.
Try this problem as it is and then generalise your solution as far as you think you can.
Using a spreadsheet may be a help, but you can judge that for yourself. | {"url":"http://nrich.maths.org/5957/index?nomenu=1","timestamp":"2014-04-20T05:55:40Z","content_type":null,"content_length":"3960","record_id":"<urn:uuid:3f813946-2755-4fa8-b307-7428093889a7>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00535-ip-10-147-4-33.ec2.internal.warc.gz"} |
User user26756
bio website
visits member for 1 year, 6 months
seen Apr 26 '13 at 12:15
stats profile views 81
2 awarded Critic
Apr Galois descent for semilinear endomorphisms
2 revised edited tags; edited tags
Mar Galois descent for semilinear endomorphisms
27 comment I edited my Question to give mor information.
Mar Galois descent for semilinear endomorphisms
27 comment @Keerthi Madapusi Pera: Yes, see my EDIT.
Mar Galois descent for semilinear endomorphisms
27 revised added 939 characters in body; edited body
26 asked Galois descent for semilinear endomorphisms
Mar Two different definitions of $\sigma$-L-spaces in Kottwitz I and II
23 comment Thanks for the comments. I was not aware that $F^{nr}$ is the compositum of $F$ and $K$ in this case. Can you perhaps give a reference? Kottwitz assumes in his paper that $k$ is
algebraically closed. But actually I'm interested in the more general situation of $k$ just a perfect field of char. $p$. Do the definitions agree in this more general situation?
Mar Two different definitions of $\sigma$-L-spaces in Kottwitz I and II
23 revised added 230 characters in body
Mar Two different definitions of $\sigma$-L-spaces in Kottwitz I and II
23 revised deleted 215 characters in body
Mar Two different definitions of $\sigma$-L-spaces in Kottwitz I and II
22 revised edited body; deleted 1 characters in body
22 asked Two different definitions of $\sigma$-L-spaces in Kottwitz I and II
Jan Constructible topology on schemes
25 comment @Ricky I think you are right. Huber uses spectral spaces in some proofs, but I think he doesn't do anything with them that could not be done without them. They are afaik not crucial to
the theory of adic spaces.
25 awarded Teacher
Jan Constructible topology on schemes
25 revised deleted 48 characters in body
25 answered Constructible topology on schemes
16 awarded Scholar
Jan Structure of f.g. modules over a non-commutative ring
16 comment Thanks. That's what I'm looking for.
16 accepted Structure of f.g. modules over a non-commutative ring
Jan Structure of f.g. modules over a non-commutative ring
16 revised edited body
Jan asked Structure of f.g. modules over a non-commutative ring | {"url":"http://mathoverflow.net/users/26756/user26756?tab=activity","timestamp":"2014-04-19T15:23:55Z","content_type":null,"content_length":"44496","record_id":"<urn:uuid:f60d9019-92a9-4012-8ee3-ced571ef15bf>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00460-ip-10-147-4-33.ec2.internal.warc.gz"} |
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What's the derivative of (1/3x^2)?
July 25th 2007, 07:35 PM #1
Jul 2007
What's the derivative of (1/3x^2)?
What's the derivative of (1/3x^2)? I dont really understand how to use the power rule with this one. I know its Y'=Nx^N-1 but how do I put this fraction into something like that? Is it like 3x^
-3? then taking the derivative of that becomes -9^-4???? The entire quiestion is (1/3x^-2) - (5/2x) I dont understand how to find the derivative of the other side either really. Help GREATLY
What's the derivative of (1/3x^2)? I dont really understand how to use the power rule with this one. I know its Y'=Nx^N-1 but how do I put this fraction into something like that? Is it like 3x^
-3? then taking the derivative of that becomes -9^-4???? The entire quiestion is (1/3x^-2) - (5/2x) I dont understand how to find the derivative of the other side either really. Help GREATLY
$<br /> \frac{d}{dx} \left[\frac{1}{3x^2}\right]= \frac{1}{3}~\frac{d}{dx}\left[ \frac{1}{x^2}\right]=$$\frac{1}{3}~\frac{d}{dx} \left[ x^{-2} \right]<br />$
Then apply the power rule to $x^{-2}$ with $N=-2$
What's the derivative of (1/3x^2)? I dont really understand how to use the power rule with this one. I know its Y'=Nx^N-1 but how do I put this fraction into something like that? Is it like 3x^
-3? then taking the derivative of that becomes -9^-4???? The entire quiestion is (1/3x^-2) - (5/2x) I dont understand how to find the derivative of the other side either really. Help GREATLY
$y=\frac{1}{3x^2} = \frac{1}{3}x^{-2}$
$\implies y' = \frac{1}{3} \cdot (-2) \cdot x^{-2-1}$
Thanks guys!
July 25th 2007, 07:50 PM #2
Grand Panjandrum
Nov 2005
July 25th 2007, 07:51 PM #3
Global Moderator
Nov 2005
New York City
July 25th 2007, 08:01 PM #4
Jul 2007 | {"url":"http://mathhelpforum.com/calculus/17213-what-s-derivative-1-3x-2-a.html","timestamp":"2014-04-21T07:20:41Z","content_type":null,"content_length":"41307","record_id":"<urn:uuid:f6f9d970-ca60-4951-a9fc-0c8f4b63bf5e>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00109-ip-10-147-4-33.ec2.internal.warc.gz"} |
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: The 'Half-Twin' Puzzle
Relativity and Cosmology
This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (http://www.relativity-4-engineers.com), in the eBook
or in the snippets I post here.
Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).
A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"
Regards, Jorrie
Previous in Blog: Cosmic Ballistics Anyone? Next in Blog: Amazing New Einstein Test
The 'Half-Twin' Puzzle
The 'Twin Paradox' of relativity is around a century old, but it still gets a lot of attention on many forums. I have also written on it a few times in this Relativity and Cosmology Blog, e.g. here
and here. An interesting variant of the classical twin-paradox is to consider the situation at the halfway stage, which may be labeled the 'Half-Twin Puzzle' of relativity.
On the right is a graphic of what I call the 'RGB scenario', an all-inertial variant of the classical twin paradox. It differs only in that the 'away-twin' (Red) does not make a quick turn-around to
come home again, but a third inertial observer (Blue) does a 'R/B flyby' in the opposite direction and performs the return leg. Both speeds are 0.866c relative to Green, giving a Lorentz factor γ = 1
/√(1-0.866^2) = 2 in both directions. This means that the 'moving' clock appears to tick at half the rate of the 'stationary' clock.
At the R/B flyby, Blue sets her clock to read the same as Red's clock (2 units) and since they are both inertial, presumably their clocks will tick at the same rate relative to Green. When Blue later
passes Green, they compare clocks first-hand and conclude that Green has aged 8 units, double the sum of Red's and Blue's aging during the two halves of the test (a sum of 4 units). This is standard,
verified relativity and not disputable.
The 'half twin puzzle' is where we simply ask the question: at the halfway (R/B) flyby event, which twin (Green or Red) has aged less since the original (G/R) flyby event? Here are some issues,
lurking in very subtle, perhaps confusing logic. To understand them, we must look at the full GRB scenario, after which Green, Red and Blue can hold a teleconference and compare results. They will
conclude that the proper time that has elapsed for Red until the R/B flyby (from 0 to 2) equals the proper time that has elapsed for Blue from the R/B to the B/G flybys (from 2 to 4). This is
inevitable, since Red and Blue flew the same distance at the same speed relative to Green.
Green, Red and Blue will further agree that Green's elapsed proper time between the G/R and G/B flybys was 8 units, as measured by Green's own clock. It seems logical that between the G/R and R/B
flybys, Green "really aged" 4 units while Red "really aged" only 2 units (Green's dotted line of simultaneity). After all, Red and Blue each recorded half of the 4 units, so how can Green not record
half of the 8 units at the halfway point?
The first apparent paradox is this: since twins Green and Red were both stationary in their own inertial frames for the duration of the test, they should be equivalent and the one cannot age more (or
less) rapidly than the other over the duration of this test. To argue that they age differently will mean giving preference to one inertial frame over another. Since the 2 units for Red is a given,
it means that Green should also have aged only 2 units.
Secondly, according to special relativity, Red has the 'right' to view Green as moving at v = -0.866c, with a Lorentz factor of γ = 2. So per Red, Green should have aged only 2/γ = 1 unit between the
G/R and R/B flybys (Red's dotted line of simultaneity). Thirdly, to make matters even worse, according to Blue's dotted line of simultaneity, Green's clock should have read 7 units when the R/B flyby
occurred. After all, Green flew at 0.866c relative to Blue and while Blue aged 2 units, Green should have aged only 2/γ = 1 unit.
How do we reconcile these apparently paradoxical conclusions, i.e., did Green age 1, 2, 4 or 7 units during this 'Half-Twin test'? We will let interested readers puzzle a little, before offering
attempted resolutions of the paradoxes.
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
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passingtongreen #
Guru 1
Join Date: Aug 01/12/2010 2:13 PM
Location: Glen
Mills, PA. My "knowledge" of Relativity is gleaned from popular books, that is why it is in quotation marks, it could be faulty but I think I understand the principles.
Posts: 2188 My first observation is that the velocities are as measured by Green, Red and Blue will not measure each other as moving at 1.732c, further, each will see the other with a different
clock speed than Green will see.
Good Answers:
104 Lewis Carroll Epstein said we need a myth, the myth is that everybody moves through spacetime at the speed of light. It seemed to work well, particularly if seconds and light seconds
are graphed as equal lengths. If drawn that way, the red and blue lines would be at 30^o to the horizontal and the relative velocity problem more obvious. I have to play with that to
see if I can figure their view of things.
Any action we take to protect liberty is itself, a threat to liberty.
Jorrie #
Guru 3
In reply to #
Join Date: May
2006 01/13/2010 9:12 AM
Location: Hi passingtongreen,
34.03S, 23.03E
The relative speeds Green_Red and Green_Blue are 0.866c and Red_Blue = 2 x 0.866/(1+0.866^2) ~ 0.99c.
Posts: 3457
In my diagram, the x- and the t-scales are meant to be the same. It is however a Minkowski spacetime diagram, not an Epstein space-propertime diagram, as you suggested (with the 30^o
Good Answers: to the horizontal). In the Minkowski, the angle to the vertical (time axis) is atan(v/c). The Epstein diagram is not very useful if you have to consider more than two inertial frames.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Anonymous #
01/13/2010 7:57 AM
It is true that Red really ages 2 units between the GR and RB coincidences. But beyond that, there's never a coincident event involving Red, and therefore it is incorrect to conclude that
he "would have" aged the same way as Blue did. In fact, it is even absurd to say that. For Red to really compare, he himself must've made the return leg, in which case he has suffered an
infinite acceleration at the RB point, thereby making any analysis meaningless.
To say Green's clock "must have read" one thing or another "when" RB happened is meaningless unless some standard of simultaneity is set up. Red and Blue can judge the apparent rate of
Green's clock by, say, receiving light signals that Green sent every second of his clock, and then figuring out how long ago in their own frame the signal must've been sent (at any point
of their journey, they know how far they are from Green). It is from this apparent rate that, say, Red concludes that "I calculate Green's clock to be half as fast as mine, which means
that now when I'm meeting Blue, Green must be 1 unit old". It is in a similar manner that Blue concludes "I calculate Green's clock to be half as fast as mine, and I am now meeting him
when his clock shows 8, which means that when I met Red 2 units ago, Green must've been 7 units old."
Of course, these "must've been"s and "when"s don't make any real sense, because there is no coincident event to compare.
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Jorrie #
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Location: Hi Guest, you wrote: "It is true that Red really ages 2 units between the GR and RB coincidences. But beyond that, there's never a coincident event involving Red, and therefore it
34.03S, 23.03E is incorrect to conclude that he "would have" aged the same way as Blue did. In fact, it is even absurd to say that."
Posts: 3457 But, won't you agree that since they are both true inertial observers, we are allowed to say that they should age the same? On what basis shall we give preference to one or the
Good Answers: 42
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Anonymous #
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01/13/2010 11:52 AM
Yes, in fact all three are inertial and their proper times move at the same rate. Indeed there is no preference of one over the other. But one cannot take what Blue observes to conclude
that Red "would have" aged the same amount in the latter half. In this phase, Red would have moved somewhere further and he no longer shares any spacetime event with either of Blue and
Green, so that they never get a chance to compare their time readings. The only way they could compare elapsed proper times in their respective frames is if their trajectories began and
ended at the same two spacetime points.
One way is that whatever clock Red himself has on his ship, he instantaneously hands over to Blue at their meeting, and then to say that that clock records the total proper time. But then,
the trajectory in spacetime followed by that clock would involve a bit of acceleration, and so, although both Red and Blue are inertial, the clock they exchange would be noninertial while
being exchanged. It would then no longer record the proper time along a spacetime geodesic.
The trajectories of Red, Blue and Green are geodesics, but because it's a flat (Minkowskian) world, they don't intersect at more than one point so as to enable comparison of the proper
times along different geodesics. We can compare the proper time along the two contours --- one being Green's trajectory and the other the clock's --- both trajectories going between the
same two events GR and GB. They should be the same if both trajectories are inertial. But the clock's trajectory is noninertial, and therefore it is no surprise if it records a different
Jorrie #
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Location: Hi Guest, you wrote: "Yes, in fact all three are inertial and their proper times move at the same rate. Indeed there is no preference of one over the other. But one cannot take what
34.03S, 23.03E Blue observes to conclude that Red "would have" aged the same amount in the latter half."
Posts: 3457 But, we have a clear condition that Blue and Red are moving at the same speed, just in opposite directions relative to Green. We are not really interested in Red's future aging, only
that Blue will age at the same amount between the two 'inbound' events than what Red aged between the two 'outbound' events.
Good Answers:
42 I do not think you can dispute that Blue will age 2 units during this interval. It is just a "Minkowski triangle" with two equal, time-like sides. AFAIK, one can analyze it in any
inertial frame and get the same numerical result.
"One way is that whatever clock Red himself has on his ship, he instantaneously hands over to Blue at their meeting, and then to say that that clock records the total proper time."
Why can't B just "hand over" the time, not the physical clock, as stated in the puzzle? Blue then sets her own clock according to this time and carries on, with no acceleration of
any clock.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Jorrie #
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Location: Hi again Guest, just to get the 'solution' closer to back on track, I fully agree with what you wrote before (I gave a GA for it):
23.03E "Red and Blue can judge the apparent rate of Green's clock by, say, receiving light signals that Green sent every second of his clock, and then figuring out how long ago in their own
frame the signal must've been sent (at any point of their journey, they know how far they are from Green). It is from this apparent rate that, say, Red concludes that "I calculate Green's
Posts: clock to be half as fast as mine, which means that now when I'm meeting Blue, Green must be 1 unit old". It is in a similar manner that Blue concludes "I calculate Green's clock to be
3457 half as fast as mine, and I am now meeting him when his clock shows 8, which means that when I met Red 2 units ago, Green must've been 7 units old.""
Good That said, what do you reckon is the age of A at the time of the R/B flyby?
42 -J
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
varun_achar #
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Hi again Jorrie,
The thing with this situation is that the first leg of Red's journey and the second leg of Blue's journey are sections of flat-space geodesics, but gluing them together at RB produces a
broken line which is NOT a geodesic, and therefore time measured along it does not yield the proper time for inertial observers. You're right, they needn't exchange the clock itself, but
the fact remains that while Red ages by 2 between GR and RB, and Blue ages by the same between RB and GB, we have no specific event for Blue before and for Red after until when to
consider as the "equivalent" leg to the inner journey.
As for the question "what do you reckon is the age of A (Green) at the time of the R/B flyby?", it is nothing absolutely defined, because simultaneity is relative. So while for Red (from
his back-calculations from the light signals etc.), that moment seems simultaneous with the 1-unit point for Green, for Green it appears simultaneous with the 4-unit point, and for Blue,
with Green's 7-unit point. There is no universal standard to it.
I think what takes a little time to sink in is that although Red and Blue are at the same event, their current estimates of Green's time are different, because they aren't moving
We can take another example in terms of flat-space geodesics. If the times along all piecewise-geodesic contours between the same two end points were the same, then it would mean that
time for even noninertial observers, who can be thought of as moving along paths whose infinitesimal sections are geodesics, must experience the same time. But this is not true. We
cannot just add the times along the outward and inward sections (which are sections of different geodesics) and get any meaningful measure of time.
Had the metric been curved, say like that of a sphere, then two geodesics could have the same end points and still be distinct, and could have the same length. But in the Minkowski case,
this never happens. So the only way two observers meeting at two different events can agree on the time in between is if they move together all along. Otherwise, they can't both be
inertial and also meet at two points.
Does this convince you?
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Jorrie #
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Location: Hi varun_achar, if these are really your first posts on CR4, then very welcome here!
23.03E You are making very good points (worth your first official GA) and I'm glad to have you onboard. I don't necessarily need convincing, but I'm trying to poke holes in whatever
resolutions others come up with to see how robust they are. I also throw arguments on the table and see if/how they can be shot down. In the end we should be able to agree on a
Posts: 3457 'best-buy' solution...
Good What if we say that since Green and Red are the main actors in the 'half-twin test', the most democratic solution is to assume they are completely equivalent. For convenience, we set up
Answers: 42 an inertial frame (A) where the situation is drawn symmetrical. Green and Red now moves at 0.577c in opposite directions relative to frame A, still giving their relative speed of
0.866c, as before.
In the A-frame, both Green and Red have aged 2 units at the time of the R/B flyby (Blue is left out for less clutter). Assume Green and Red have agreed to each send the other a
time-stamped radio signal when their respective clocks read 2 units (the black arrows). They will each receive the others time signal at 7.46 units on their respective clocks.^[1]
Doesn't this 'prove' that Green actually aged the same 2 units than what Red aged at the R/B flyby?
I've had the counter-argument that I 'cheated' by selecting a sort-off 'preferred frame' and hence get this symmetrical result.The answer is that for the given inputs, we can choose any
frame as reference and we will get exactly the same 7.46 units for Red and Green. This is a frame independent result...
[1] For the general readership, I've used the relativistic Doppler shift to calculate this time, because for me it's the easiest: T2 = (1+0.866)/√[1-0.866^2] T1. T1 = 2, the period at
the transmitter, giving T2 = 7.46, the period at the receiver.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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varun_achar #
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Thank you Jorrie for the warm welcome and for your reception of my answers. Indeed, these have been my first comments here on CR4. I am no engineer, therefore I haven't been a very
frequent visitor. My father, who is one, occasionally shares with me posts that interest him. That is how I came upon this your page.
I agree that your suggested frame makes things much more evidently understandable. Just a small remark on my second comment: While there is no requirement for R and B to actually
exchange an object, envisaging them doing so serves to highlight the fact that the broken line GR-RB-GB, which in this imagination would be the clock's world line, is not an inertial
line, making it analogous to the "Twin Paradox" in the way we usually hear of it.
Thank you for the diagrams, which are extremely helpful in understanding such situations.
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Hi Jorrie,
A very interesting variant of the Twins paradox, you have there! Takes away all the unnecessary acceleration business, that does nothing but cloud the real solution.
The problem lies in asking questions that do not have any frame invariant meaning. Like "Doesn't this 'prove' that Green actually aged the same 2 units than what Red aged at the R/B
The notion of simultaneity is relative and depends on the frame. So which birthday of G was "at with the R/B flyby" event, depends on the frame. In the A-frame it is 2, in the G-frame it
is 4, in the R-frame it is 1 and in the B-frame it is 7. This can be seen from your diagrams clearly.
There is no meaning to the question, what is his "real" age when the R/B flyby takes place. It depends on the frame of reference.
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Location: Hi guest, you wrote: "There is no meaning to the question, what is his "real" age when the R/B flyby takes place. It depends on the frame of reference."
23.03E I agree that this is the 'standard view'. But, as I have shown in reply #9, there are some frame-independent quantities available in the 'half-twin' case. My quest is to find out if one
of these quantities cannot perhaps be converted to a frame-independent age of Green when some distant event happens. After all, since they are both inertial, one would suspect Green and
Posts: 3457 Red to age at exactly the same rate - the only problem seems to be one of measurement...
Good I know it's perhaps a futile quest, but I have one more 'ace up the sleeve' for a little later. :)
Answers: 42
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varun_achar #
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How about some fourth guy, say Yellow, does with Green what Blue does with Red? Then there'd be a corresponding YG crossing to compare with the RB, and that would give a definite
meaning to Green's age "at" the RB crossing....
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Location: 34.03S, Hi varun_achar, "How about some fourth guy, ..."
As a matter of fact, I've already done such a drawing - it is the "ace up the sleeve" that I was talking about earlier. I'm not 100% sure how to interpret it, but I'll post it
Posts: 3457 anyway for all to think along these lines.
Good Answers: 42 I used Lilac for the "fourth guy", L, who first has a flyby with G at 2 units, which time he transfers to his clock. Then he has a flyby with B when they each record 2.4 units on
their own clocks. I think this means that we can say with some certainty:
1. B retrospectively knows that G's clock read 2 units when R's clock (and his own) was at 2 units. Or not?
2. This result is coordinate independent, because we get the same values irrespective of what reference frame we use (the A-frame is just a very convenient one).
3. This does not negate the coordinate dependent results, like "per G's definition of simultaneity, G's clock read 4 units when R's clock read 2 units".
4. It is compatible with the notion that all inertial frames should be treated equal, including ticking at the same clock rates.
5. We have replaced the "Minkowski triangle" with a "Minkowski kite". I do not fully understand the implications if this, but maybe it defines a kind of "absolute simultaneity"?
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
varun_achar #
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Hi Jorrie,
Nice, so my hunch was right. Maybe we don't really need a very profound interpretation of this; to me it seems like merely setting more constraints and thereby defining more
features. What do you think?
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Location: Hi Varun,
34.03S, 23.03E
I'm still tying to work out what we can and cannot say about such a scenario. Returning to my prior 5-pointer:
Posts: 3457
"1. B retrospectively knows that G's clock read 2 units when R's clock (and his own) was at 2 units. Or not?"
Good Answers:
42 I think it is clear that the L/G flyby had to be at t[G]=t[L]=2, otherwise the identical readings t[G]=t[L]=2.4 at the L/B flyby could not be true.
"3. This does not negate the coordinate dependent results, like "per G's definition of simultaneity, G's clock read 4 units when R's clock read 2 units"."
The definition of simultaneity is conventional (Einstein's), which is a good convention, because it simplify physics considerably. It is however not the only convention that will
satisfy the Lorentz transformations (LTs) and so predict the same outcomes for experiments.
"5. We have replaced the "Minkowski triangle" with a "Minkowski kite". I do not fully understand the implications if this, but maybe it defines a kind of "absolute simultaneity"?"
It looks like for every event in Minkowski spacetime, there is at least one other 'complimentary event' that can be viewed as simultaneous without using Einstein's simultaneity
convention. By this I mean doing the experiment 'BGRL' above, which relies on Minkowski space (and hence on the LTs), but not on Einstein's simultaneity convention.
IMO, this is the same as saying that for every space-like^[1] pair of events in Minkowski spacetime, there exists one inertial frame in which the events are simultaneous as per
Einstein's simultaneity convention.
There may also be time-like complimentary events, where there is one inertial frame in which the events are co-located.
Any more features?
[1] Space-like means nothing, not even light, can travel between the events, while time-like means material objects can.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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You are right. The terms "spacelike" and "timelike" are in fact motivated by the fact that spacelike (timelike) -separated events always have some inertial frame in which their
separation is purely in space (time), and their time (space) coordinate(s) is (are) the same. In other words, spacelike-separated events don't have a "proper time interval" between them
but what I'd call a "proper space interval", whereas timelike-separated events have a "proper time interval" between them, which is the time-separation between them in a frame where they
have the same spatial coordinates.
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Location: Yup!
23.03E I suppose we can now take the terminology further by saying that like we have 'proper distance' versus 'coordinate distance' and 'proper time' versus 'coordinate time', we can have '
proper simultaneity' versus 'coordinate simultaneity'.
Posts: 3457
Interestingly, the lines of 'proper simultaneity' is nothing more than the hyperbolas of constant time-like interval in Minkowski spacetime, as shown on the right.^[1] Each event on the
Good ct-axis has only on hyperbola (i.e., one definition of 'proper simultaneity'), irrespective of who observes the event. Contrast this to 'coordinate simultaneity', a straight line with a
Answers: 42 different slope for each inertial observer of the event. Galilean simultaneity obviously has a single, zero-slope line of simultaneity for each event, so 'proper simultaneity' must not
be confused with 'absolute simultaneity'.
A single time-like hyperbola represents the set of events with the same proper time separation from the origin, i.e. as observed by various observers co-located with the origin and with
those events. This is what I call a 'line of proper simultaneity'.
A single space-like hyperbola represents the set of events with the same proper spatial separation from the origin, i.e. as observed by various observers for whom those events are
apparently simultaneous with the origin. May we call this a 'line of proper separation', or is there a better term?
[1] For interested parties, the figure depicted is Fig. 2.7 of the eBook Relativity-4-Engineers. More discussion there...
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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This is a nice construction. I have one question to ask, whose answer doesn't seem trivial and I must think about: What is the extent of "objectiveness" of these lines with respect to a
change of the origin (translation), and to a boost? For now we can ignore spatial rotations since we're visualizing it in a plane.
My hunch is that the whole scene would diffeomorphically deform into one which looks identical. For instance, if we wanted to draw the same diagram with the ct'-axis as the chosen time
axis (which amounts to a boost) and ct'=2 as the origin (translation), then the proper-time-1 hyperbola would deform into proper-time-zero asymptotes (+ and - pi/4 lines), while the rest
of them would all deform in accordance with these new asymptotes. The boost part would deform them to a new parametrization, but would keep them congruent to themselves before the
deformation. The interesting part is the translation, which would deform each hyperbola to the shape of its immediate neighbour. If, for instance, we shifted the origin to the ct=1
point, then as mentioned before, the corresponding hyperbola would deform into the V-shaped asymptotes. At this point, if we apply a small translation further (to some ct=1+delta), the V
would suddenly get reflected into the inverted V and thereafter become a hyperbola on the bottom half-plane.
Similar things would happen to the spacelike hyperbolas on applying spatial translations.
It would be great if we had some sort of a tool where we could graphically choose the origin and axes and then see the diffeomorphism evolve the diagram into the new frame.
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I forgot to mention: Since the points sharing a hyperbola continue to do so under these diffeomorphisms, and ones on different hyperbolas remain on different ones, this is a
Lorentz-invariant definition of simultaneity.
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Good point about the Lorentz invariance.
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BTW, I did not notice your #19 and #20 before I posted my #21 (they did a 'flyby' in cyberspace :)
Good Answers: 42
I'll also have to think a bit about translations and boosts.
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Location: Hi Varun,
23.03E Before I get too carried away and I take others with me, it may be wise to point out that AFAIK, "proper simultaneity" is not an accepted mainstream term and there may not even exist such
a concept in modern science. It may well be that it has been treated in some paper that I do not know of - the closest I've got to a good summary of the Conventionality of Simultaneity is
Posts: in a document by Prof. Allen I. Janis, formerly of the University of Pittsburgh. He referenced a large number of papers that I have not worked through yet. Einstein's simultaneity is
3457 usually called "standard synchrony", while non-Einstein definitions of simultaneity are usually called "nonstandard synchrony".
Good Be that as it may, it is scientifically valid to define a 'new' simultaneity, different from Einstein's and then analyze the effects of such a definition, whatever its official
Answers: terminology may be. Prof. Janis mentioned that "all of special relativity has been reformulated (in an unfamiliar form) in terms of nonstandard synchronies (Winnie, 1970a and 1970b)".
Returning to my definition of "proper simultaneity" (or maybe I should have called it "hyperbolic simultaneity"). In order to make it science and not philosophy, we should answer some
technical questions:
1. How would clocks be synchronized in such a definition of simultaneity?
2. What would the observable effects be, if any?
3. What will the "unfamiliar laws" of physics look like?
The answers are not all clear to me, but for 1 and 2, I can mention the following:
A1. 'Fast move' a number of clocks, all set to zero as they pass the origin simultaneously, at different constant speeds until their co-located observers read time t on the clock faces.
Do not resynchronize them (the Einstein way) and they will sit on a "hyperbola of (proper) simultaneity". This is somewhat loosely described, but I think you get the gist. Any
A2. The full 'twin-paradox scenario' will yield the same result as Einstein's, because no clocks are ever resynchronized there. The Michelson-Morley experiment would not be influenced,
because it too does not use clock synchronization. Any one-way result will however differ from Einstein's, including the one-way speed of light. Light propagation will not be isotropic -
the one-way speed of light depends on the simultaneity convention used.
A3. Other than the anisotropy of light propagation, I have no idea. I can however imagine that things may be rather ugly. Clever Einstein...
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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Hi Jorrie,
These are issues to be addressed indeed. My suggestions are as follows: a light pulse will travel along the hyperbola containing the source event. So for synchronized clocks, we could
just use a source of light stationary in some accepted frame which emits a spherically-symmetric (in the 3D case) pulse every second, say. The zeroth pulse would propagate along the
rectangular asymptotes, the first one would propagate along the ct=1 hyperbola, and so on. Observers in other frames would measure time based on the count of pulses they have received.
This way, when they compare time readings of events with other observers, they would all agree on your definition, i.e. the "proper simultaneity". One could perhaps also call it "null
The only problem is that such a system could be established only between nearby locations, since the light intensity would wane with distance. A possible solution is to have relay
points, where the light pulse is detected and a new pulse with appropriate calculations is broadcast, with prior understanding between stations about the calculations on how to get their
readings to match.
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Nope, I was wrong about all that. I treated it as though the hyperbola defined by a constant distance from the origin contains points which are null-separated, which is not true! Now my
description of the diffeomorphism and all that falls apart. Light obviously travels along null geodesics and not along hyperbolas.
Now it is a matter to consider what happens to the diagram under Lorentz transformations. By your definition, points having the same Minkowski separation from the origin are
simultaneous. However, with a translation, the origin would change and so would the sets of points which have the same separation from it. The hyperbola containing the new origin
wouldn't "turn into" the light cone, alas!
It seems to me that the only definition of simultaneity which would result in the same sets of simultaneous points in all inertial frames is to say that null-separated points are
simultaneous. What do you think?
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Location: Because of the 'standard configuration' Minkowski, with the various frame clocks synchronized at the origin, I think we need not consider translations, but only spacetime rotations
34.03S, 23.03E (Lorentz transformation). I think my definition does result in one set of simultaneous points for all inertial frames in the standard configuration (figure on the right).
Posts: 3457 In any case, my interest in Lorentz invariant simultaneity stems from the quest to determine if there are any technical grounds for a view that Green and Red have the same proper age
at the R/B flyby (figure below). It seems to be so from this discussion, but I'm sure there will be a lot of resistance as well. Anyway, there may still be a fatal flaw in the whole
Good Answers: argument.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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Location: A further thought on the problem: you wrote:
34.03S, 23.03E
"It seems to me that the only definition of simultaneity which would result in the same sets of simultaneous points in all inertial frames is to say that null-separated points are
Posts: 3457 simultaneous. What do you think?"
Good Answers: The null-separated events do indeed form the 'x-axis' for all frames in this scheme, but then there are hyperbolas above it that form the lines of simultaneity for time-like
42 separated events.
One problem is that with a null x-axis, distance coordinates are undetermined and so are all one-way speeds, including the one-way speed of light. However, two-way speed is
determined and hence we can have a conventional one-way distance, I think. Does that mean we can still have a conventional one-way speed determination?
Any ideas?
PS: I've found this interesting article on The Conventionality of Simultaneity by Vesselin Petkov.
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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Location: Further to my prior post, where I wrote: "However, two-way speed is determined and hence we can have a conventional one-way distance, I think. Does that mean we can still have a
34.03S, conventional one-way speed determination?".
A light signal (dotted red) from the origin will propagate up the light cone and if reflected back by a mirror colocated with the ct' = 2 event, the return signal will arrive at ct=
Posts: 3457 1.74. This can be used to fix a coordinate distance of x=0.87 for the event, irrespective of which synchrony convention is followed (only one clock involved). This gives the two-way
speed of light at c.
Good Answers:
42 One can also postulate that the one-way speed of light must then also be c, but it is not directly measurable. The fact that the light-cone is defined as the 'real x-axis' means that
the one-way light travel time was zero - hence infinite one-way speed for light. This is a consequence of the 'new' clock synchrony model - a clock present at the reflection event must
read zero.
Interestingly, while the relative speed of the two frames (ct and ct') was set as v = 0.4c in standard Einstein synchrony, their relative speed in the new synchrony is v' = 0.87/2=
0.435c. This does mean that we have a 'mixed' method of distance measurement and clock synchrony. For distance we still use half the two-way light travel time, but for clock synchrony
we don't.
Ouch... This is becoming very complex now! The question is: since clock synchrony is a convention, is this type of clock synchrony valid?
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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01/18/2010 9:37 AM
Interesting questions raised. I will have to look through in detail.....
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Location: I wrote: "... their relative speed in the new synchrony is v' = 0.87/2=0.435c. This does mean that we have a 'mixed' method of distance measurement and clock synchrony. "
23.03E This bears some resemblance to the "Epstein space-propertime" diagram^[1], where we plot proper time against coordinate space (or 'your time' against 'my space'), as pictured on the
right. The diagram is drawn for v=0.8c, giving the relative time dilation factor as 1/γ = √(1-0.8^2) = 0.6.
Posts: 3457
The Epstein diagram shows the light-cone (the "real" x-axis) at an angle perpendicular to the time axis. In essence, the hyperbolas are just replaced by circles. The angle of the ct'
Good axis is now v/c = sin(-Φ), where in the Minkowski diagram it is v/c = tan(-Φ). Here the scale of the ct' axis is the same as that of the ct axis, directly indicating that there is no
Answers: 42 desynchronization of clocks (the lines of simultaneity are circles).
An interesting connection, maybe, but not quite the same thing, I think...
[1] Relativity Visualized, Lewis Carrol Epstein, 1997
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
varun_achar #
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Posts: 14 01/19/2010 3:25 PM
This is another interesting construction. I read the paper on the conventionality of simultaneity. It seems to me that a definition which would conform to our intuition would suffice for
everyday (nonrelativistic) life, while the proper time definition works very well in the context of, say, cosmology, where the universe we see "now" is more or less a light cone (except
that the spacetime isn't globally Minkowski).
Jorrie #
Guru 31
In reply to #
Join Date:
May 2006 01/20/2010 3:07 PM
Location: After all these 'diversions', it may be time to try and answer the final question of the OP: "How do we reconcile these apparently paradoxical conclusions, i.e., did Green age 1, 2, 4 or
34.03S, 7 units during this 'Half-Twin test'?"
The modern answer is that because the definition of simultaneity is a convention, devoid of any "real physical meaning", we cannot really tell. Einstein adopted a convention for his
Posts: 3457 Special Relativity that makes the speed of light isotropic, i.e., the one-way speed of light will be observed as the same in all directions. This gives every inertial observer a
different view of what is simultaneous and what not.
Answers: 42 However, there is nothing that prevent us from adopting a different method (convention) for synchronizing clocks, e.g., like the "proper simultaneity" that I suggested in #18 above. The
price that we will pay for such a 'non-standard' synchronization procedure is that the laws of physics will be much more difficult to express mathematically. Sonego & Pin wrote in
Foundations of anisotropic relativistic mechanics, 2009: Misner, Thorne and Wheeler concisely and effectively wrote: "Time is defined so that motion looks simple". We could paraphrase
them saying: "Clocks are synchronized so that physics looks simple."
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Jorrie #
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Join Date:
May 2006 01/21/2010 2:10 PM
Location: I wrote: "Sonego & Pin wrote in Foundations of anisotropic relativistic mechanics, 2009: Misner, Thorne and Wheeler concisely and effectively wrote: "Time is defined so that motion looks
34.03S, simple". We could paraphrase them saying: "Clocks are synchronized so that physics looks simple.""
Just to illustrate how things can become more complex than necessary when we deviate from Einstein's convention for clock synchronization, here is a (more or less) complete graphical
Posts: representation of "proper simultaneity" (as roughly suggested in #18 above).
According to the literature that I've referenced so far, this is a perfectly valid convention for defining simultaneity - in fact it even has some technical appeal to it as well.^[1]
Good However, having a non-linear coordinate system will never 'look simple' and it may be difficult to calculate (or even plot) certain characteristics.
42 In a nutshell, the bottom part (below the blue/black/red ±X,X,X axes, the light-cone), may be ignored completely, because it is the non-causal area of Minkowski spacetime, where neither
material objects, nor electromagnetic effects can go. The light cone is the new x-axis for all inertial frames. The time- and spatial axes are still linear, but the spacetime 'slices of
constant time' are hyperbolic and all inertial observers share the same definition of simultaneity. Hence, all inertial frames agree on the x, cT coordinates of any event that happens
inside the light-cone.
Distances are still measured by the Einstein (or radar) method - time the two-way propagation of light, using only one clock. Take half the time and divide it by the speed of light and
you have the spatial separation. We set up a grid of distances in this way, with non-synchronized clocks, static in the inertial frame, at every grid position of interest. To achieve
'proper simultaneity' (in principle), we 'shoot' a set of identical clocks, synchronized at the origin, in every direction in such a way that speeds (or magnitude of momenta) in the
original coordinates are all identical. A perfectly symmetrical 'explosion' should be able to achieve this.
As such clocks pass the static clocks of the grid, each grid clock is set in accordance with the passing clock's reading, with no correction for travel time or other offsets. The moving
clocks all register proper time and hence the synchronization is according to proper time. While such clocks are traveling, they are assumed to be perfectly inertial, with the initial
acceleration either happening before they pass the origin, or alternatively, is of negligible duration.
If this does not make any sense, don't worry - we will not use such a scheme - it is just to illustrate an alternative definition of simultaneity, however impractical. If it makes sense
and you spot a flaw in the argument (in principle, not in practice), please let me know.
[1] One of the problems with most alternative, non-Einstein conventions of simultaneity is that they tend to make the speed of light (and hence the Newtonian laws of motion) anisotropic
for moving frames. The described convention seems to retain the isotropy of light propagation and hence the same laws of motions in different directions for all inertial frames.
I could not find a mention of this specific convention in the literature, but I will keep looking...
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Jorrie #
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In reply to #
Join Date:
May 2006 01/22/2010 10:31 AM
Location: Oops! In the footnote of my prior post, there is a subtle problem. I wrote: "The described convention seems to retain the isotropy of light propagation and hence the same laws of
34.03S, motions in different directions for all inertial frames."
This "proper simultaneity" convention has the rather nasty characteristic that the measured one-way speed of light, although isotropic, will be infinite! The reason is that all clocks
Posts: 3457 'crossing' the x-axis read zero at that time. If the light-sphere is emitted when the clock at the origin reads zero, it will arrive at the reflection event just as the clock there also
reads zero. The light travel time will hence appear to be zero...
Answers: 42 -J
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Jorrie #
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Join Date:
May 2006 02/01/2010 2:47 AM
Location: I kept looking and I found this type of synchrony inside the paper that I referenced earlier: "Sonego & Pin wrote in Foundations of anisotropic relativistic mechanics, 2009"
23.03E It is in fact a peer reviewed paper, with full citation and (partial) synopsis:
Posts: 3457 "Foundations of anisotropic relativistic mechanics, Sonego, Sebastiano; Pin, Massimo (Universit`a di Udine, Via delle Scienze 208, 33100 Udine, Italy); Journal of Mathematical Physics,
Volume 50, Issue 4, pp. 042902-042902-28 (2009).
Answers: 42 "We lay down the foundations of particle dynamics in mechanical theories that satisfy the relativity principle and whose kinematics can be formulated employing reference frames of the
type usually adopted in special relativity. Such mechanics allow for the presence of anisotropy, both conventional (due to non-standard synchronisation proto-cols) and real (leading to
detectable chronogeometrical effects, independent of the choice of synchronisation)."
The relevant equations are in section 2.1.1, page 7 of the pdf. If I understand the paper correctly, the non-standard synchrony described in this Blog is valid as one of the
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
NeilB #
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Join Date: 03/02/2010 4:59 PM
Mar 2010
G aged 4 units and R aged 2 units. In R's frame (and B's), the distance between G and the point where R and B cross is half the distance as measured by G and his clock ticks half as many
Posts: 2 times as G until they cross. This remains true, and becomes more apparent, in the diagram where R considers his own frame as stationary and G as receding, because the relative velocity
of R and B to each other is less than twice their velocities with respect to G.
Jorrie #
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Join Date: May
2006 03/02/2010 11:09 PM
Location: 34.03S, Hi Neil,
You are right, but only by the convention you adopt for simultaneity, as I explained in my Answer above. One has to complete the other half of the 'half-twin paradox' before you
Posts: 3457 can really tell. One-way aging remains a thing defined by convention, not physics...
Good Answers: 42 -J
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
NeilB #
Participant 37
Join Date: Mar 03/03/2010 12:01 PM
Ah, this was a puzzle on synchronizing distant clocks. I didn't pick up on that.
Posts: 2
I've ordered your book. I'm no engineer but my math isn't too bad, so hopefully I can follow some of it. OT, but I am rereading Epstein's Relativity Visualized. Whatever happened to
him? His geometric explanations in this book are truly ingenious.
Reply Off Topic (Score 5)
Jorrie #
Guru 38
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Join Date: May
2006 04/30/2010 7:54 AM
Location: Hi Neil,
34.03S, 23.03E
Sorry, I missed this reply of yours.
Posts: 3457
You can ask questions on Relativity 4 Engineers here, if you like (or perhaps in the FAQ section that I've opened in this Blog - see header above).
Good Answers:
42 Yes, I also enjoyed Epstein's visualizations of relativity very much. His work did not get much credit in the scientific circles, because it is a little narrow-minded, although very
useful for the novice. One cannot do much more than that with most of it - e.g., his 'Epstein space-propertime diagram' has very limited utility, but it is a very nice
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
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Show Members | {"url":"http://cr4.globalspec.com/blogentry/11156/The-Half-Twin-Puzzle","timestamp":"2014-04-16T21:53:41Z","content_type":null,"content_length":"220915","record_id":"<urn:uuid:f0aca8f4-bbe6-41dd-9aa2-ddf83ca25af8>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00213-ip-10-147-4-33.ec2.internal.warc.gz"} |
How to demonstrate bigO cost of algorithms?
Mitja nun at meyl.com
Wed Jun 2 23:25:32 CEST 2004
Rusty Shackleford <rs at overlook.homelinux.net>
(news:slrncbs3d0.9l3.rs at frank.overlook.homelinux.net) wrote:
> Thanks for that explanation -- it really helped. I forgot that O(n)
> can translate into dn + c where d and c are constants.
You'd NEVER write, e.g., O(n)=3n+7. What you'd say would be simply O(n)=n.
As already pointed out, this means that complexity (and, with that,
execution time) is PROPORTIONAL to n. It doesn't give you the number of
seconds or something.
Consider the following:
def foo(n):
for i in range(n):
for j in range(i):
print i, j
The "print" statement is executed n*n/2 times. However, O(n)=n^2: if you
increase n seven times, the complexity increases 49 times.
> I think I'm going to keep trying to figure out ways to demonstrate
> big O stuff, because I just don't understand it until I see it in a
> non-abstract sense.
> Thanks again.
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HARMOKO, EKO TRI (2009) ANALISIS PENGARUH ROA DAN LEVERAGE FACTOR TERHADAP RENTABILITAS MODAL SENDIRI PADA PT.BANK RAKYAT INDONESIA (PERSERO) TBK. Other thesis, University of Muhammadiyah Malang.
ANALISIS_PENGARUH_ROA_DAN_LEVERAGE_FACTORTERHADAP_RENTABILITAS_MODAL_SENDIRI_PADA_PT.pdf - Published Version
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TITLE : “Analysis Influence ROA and Leverage Factor on Self-Capital Rentability for PT. BANK RAKYAT INDONESIA (Persero) Tbk.” (Eko Tri Harmoko, Dr. Ahmad Juanda, MM., Ak, Ihyaul Ulum, SE., M.si)
Objective of this research is to know, did ROA and Leverage Factor have influence also which factor is more dominant to influences the development of Self-Capital Rentability for PT. BANK RAKYAT
INDONESIA (Persero) Tbk. Analysis device that was used is statistic test non-parametric Spearman Rho Rank Order Correlation. Spearman Rho Rank Order Correlation uses to analyze relation that data is
ordinal shaped or in order. Meanwhile, correlation analysis Spearman used to describes relation between two variables. The Analysis Result shows that correlation between ROA with ROE is 0,395 so it
has been proved that r count < r table makes ROA had non-real influence to ROE. Correlation between Leverage factor with ROE that has 0,361 this makes Leverage Factor had non-real influence to ROE,
this included in weak category because it placed on level interval 0,2 – 0,4. If ROA higher , so ROE is higher too as well, and vice versa if ROA lower so ROE is lower, it has Quod Erat Demonstrandum
in this research.
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Viscous Fluids - Lesson - www.TeachEngineering.org
Lesson: Viscous Fluids
Students are introduced to the similarities and differences in the behaviors of elastic solids and viscous fluids. Several types of fluid behaviors are described—Bingham plastic, Newtonian, shear
thinning and shear thickening—along with their respective shear stress vs. rate of shearing strain diagrams. In addition, fluid material properties such as viscosity are introduced, along with the
methods that engineers use to determine those physical properties.
Grade Level: 11 (10-12) Lessons in this Unit: 12345
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coterminal angle
Pretty simple concept I learned in my Honors
Algebra II
, but the term "coterminal
" doesn't seem to catch on...
Two angles are coterminal when they are in the same position in the
Cartesian coordinate plane
. For example, the angles 0 and 2π (in
) are coterminal since they both represent the same position.
, if you have the angle
, then any coterminal angles to
can be expressed as:
θ[coterminal] = θ + 2πk
is any
real integer
). (If you're
measuring angles in degrees
instead of radians, then the term would be "360°
" instead of "2π
One of the properties (I don't think there are many more...) of coterminal angles is that the value of a
trig function
for angles that are coterminal will be the same for
angles (e.g. since π/6 and 13π/6 are coterminal angles, sin(π/6) = sin(13π/6) = 1/2).
A concept similar to coterminal angles is that of the
reference angle | {"url":"http://everything2.com/title/coterminal+angle","timestamp":"2014-04-21T08:37:33Z","content_type":null,"content_length":"18773","record_id":"<urn:uuid:77622c71-90da-4d58-8f70-feb6cd8dd023>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00125-ip-10-147-4-33.ec2.internal.warc.gz"} |
If tan(cot θ) = cot(tan θ) then sinθ.cosθ =
π denotes pi. - Homework Help - eNotes.com
If tan(cot θ) = cot(tan θ) then sinθ.cosθ =
π denotes pi.
xangle sen (ctg x)/ cos (ctg x) = cos(tg x)/ sen(tg x) developing: sen (ctg x) sen(tg x) = cos(tg x) cos(ctg x) cos(tg x)cos(ctg x) - sen(tg x)sen(ctg x) = 0 This is the formula for the cosine
addition of angles, where the angles are tg x and ctg x cos(tg x + cotg x)=0 cos( sen x /cos x + cos x /sen x) = 0 now: cosine is zero for x= pigrego/2 or x = 3/2 pigreco Therefore: cos( 1 /senx cosx
) =0 1/senx cosx = pigreco /2 1/senx cosx = 3/2 pigreco senx cosx = 2/pigreco senx cosx = 2/3 pigreco We can also write: sen 2x = 4/pigreco and sen 2x = 4/3 pigreco
Set x as angle
sen (ctg x)/ cos (ctg x) = cos(tg x)/ sen(tg x)\\ developing: sen (ctg x) sen(tg x) = cos(tg x) cos(ctg x)
x)cos(ctg x) - sen(tg x)sen(ctg x) = 0
is the formula for the cosine addition of angles, where the angles are tg x and ctg x
x + cotg x)=0 \\ cos( sen x /cos x + cos x /sen x) = 0 \\ now: cosine is zero for x= pigrego/2 or x = 3/2 pigreco Therefore:\\ cos( 1 /senx cosx ) =0
cosx = pigreco /2 \\ 1/senx cosx = 3/2 pigreco\\ senx cosx = 2/pigreco\\ senx cosx = 2/3 pigreco\\ We can also write: sen 2x = 4/pigreco and sen 2x = 4/3
d. `2/pi`
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the encyclopedic entry of Hendecagon
In geometry, a hendecagon (also undecagon) is an 11-sided polygon. The name "undecagon" is often seen as incorrect, but the matter is up for debate. The prefix should be the Greek 'hen', not the
Latin 'un' (which is also of Greek origin) . A hendecagon has 11 sides. A regular hendecagon has internal angles of 147.272727... degrees. The area of a regular hendecagon with side length a is given
$A = frac\left\{11\right\}\left\{4\right\}a^2 cot frac\left\{pi\right\}\left\{11\right\} simeq 9.36564,a^2.$
A regular hendecagon is not
compass and straightedge
Use in coinage
The Canadian dollar coin, the loonie, is patterned on a regular hendecagonal prism, as is the Indian two-rupee coin.
It was also patterned on the Susan B. Anthony dollar of the United States from 1979-1981 and again in 1999.
See also
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FitDay Discussion Boards - View Single Post - Weight Loss for Math Geeks
Weight Loss for Math Geeks
Calculating your recommended intake of calories is easy with this formula:
655 + (4.3 x weight in pounds) + (4.7 x height in inches) - (4.7 x age in years)
66 + (6.3 x weight in pounds) + (12.9 x height in inches) - (6.8 x age in years)
This will give you your basal metabolic rate (BMR)
Take this number
x 0.2 if you are sedentary
x 0.3 if you are lightly active
x 0.4 if you are moderately active
x 0.5 if you are extremely active
Add this number to your BMR
This will give you your recommended calorie intake per day if you want to maintain your current weight
Subtract 500 calories per day to result in an approximate 1 lb/wk weight loss.
The important thing to remember is that you should recalculate your BMR as you lose weight. This formula should be used with a three to five times a week moderate to heavy exercise regimen of mixed
cardio and weights.
I've found that the above formula used with a ratio of 50% carbs to 25% fat and 25% protein and a moderate workout regimen results in around a one pound weight loss per week for me. Fit Day does an
awesome job of helping me keep up with my fitness and weight goals
Hope this will help someone else as well | {"url":"http://www.fitday.com/fitness/forums/59156-post1.html","timestamp":"2014-04-21T07:22:40Z","content_type":null,"content_length":"15851","record_id":"<urn:uuid:a1c0c5f2-d1ed-4441-90b6-f40830cd6387>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00364-ip-10-147-4-33.ec2.internal.warc.gz"} |
the first resource for mathematics
Predictor-corrector methods for periodic second-order initial-value problems.
(English) Zbl 0631.65074
This paper deals with the numerical solution of initial value problems for the special second order equation ${y}^{"\text{'}}=f\left(t,y\right)$, by means of predictor-corrector methods. The authors
define the phase lag with respect to the homogeneous solution component as the phase error introduced by the numerical scheme when applied to the test equation ${y}^{"\text{'}}=-{k}^{2}y$. Thus some
predictor-corrector methods introduced by the authors [ibid. 3, 417-437 (1983; Zbl 0533.65045)] for first order differential equations are modified to make them applicable to special second order
equations and the coefficients are chosen satisfying the requirement that both the algebraic and phase lag orders be as large as possible.
In this way several numerical predictor-corrector schemes with orders 4 and 6 and phase lag orders up to 10 are proposed. The paper ends with some numerical experiments comparing the behaviour of
these methods and the classical fourth order Runge-Kutta-Nyström (RKN) method. It is concluded that for problems with periodic solutions the proposed schemes are more efficient than the RKN method.
65L05 Initial value problems for ODE (numerical methods)
34A34 Nonlinear ODE and systems, general
34C25 Periodic solutions of ODE | {"url":"http://zbmath.org/?q=an:0631.65074","timestamp":"2014-04-17T15:33:40Z","content_type":null,"content_length":"22262","record_id":"<urn:uuid:9e25189d-e610-4952-b921-75f39ef90aed>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00622-ip-10-147-4-33.ec2.internal.warc.gz"} |
Solution to the Shopkeeper Problem
I don't think I need to post the solution as Christophe has already solved all of them in no time.
Problem 1: One Side Only (Simple)
This is simply the numbers 2^0,2^1,2^2 ….. that is 1,2,4,8,16 ……….
So for making 1000 kg we need
up to
1, 2, 4, 8, 16, 32, 64, 128, 512.
Problem 2: Both Sides (Medium)
For this answer is 3^0,3^1,3^2 …. That is 1,3,9,27,81,243,729
Problem 3: Incremental (Hard)
This is exactly a problem solved by Gray code.
Gray codes are named after the Frank Gray who patented their use for shaft encoders in 1953
A Gray code represents each number in the sequence of integers {0...2^N-1} as a binary string of length N in an order such that adjacent integers have Gray code representations that differ in only
one bit position. Marching through the integer sequence therefore requires flipping just one bit at a time.
Example (N=3): The binary coding of {0...7} is {000, 001, 010, 011,
100, 101, 110, 111}, while one Gray coding is {000, 001, 011, 010,
110, 111, 101, 100}.
For this answer we need as many blocks as per Solution to Problem 1.
For easy understanding let me describe the case where the packets range from 1 to 7 which can be easily extended to 1 - 125 range.
Now if we want to make packets of all weights from 1 to & we will do the following
001 We measure 1kg,using 1kg block.
011 We measure 3kg by placing 2 kg block also
010 We remove 1kg block and measure 2 kg.
110 We add 4kg weight and measure 6kg weight …
Now we can see answer to our problem is Gray code of 7 bits. Now our range is 1 to 125 and not 1 to 127.This can be solved by using appropriate Gray code making the following numbers falling to the
end of the sequence you are starting with
2 comments:
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In the first answer, 256 is missing in the list:
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Math Help
April 20th 2009, 09:24 PM #1
MHF Contributor
Jul 2008
Taxi Ride
The cost of a taxi ride is calculated by adding a fixed amount to a charge for each 1/4 of the trip. If one mile costs $4 and 3/2 miles costs $5.50, what is the cost of a three-mile trip?
I made a table:
1 $4
3/2 $5.50
3 $x
This question came from the chapter titled linear equations in two variables.
After doing the math, I got $11 for 3 miles but the math book tells me that the correct answer is $10 for 3 miles.
How do I set up a system of linear equations for this question? How do I find $10?
a little help
This does not require a system of equations, that's a little overkill.
So, when you are setting up linear equations, you should know that a general linear equation looks like this: y=mx+b, where m is the slope and b is the y-intercept. So, you need to find the slope
here, which you find using the equation $m=(y_2 - y_1)/(x_2 - x_1)$
So, what points will you use for this little formula above?
The linear equation in 2 variables they are talking about is the above, as opposed to the x=4 or y=10 linear equations.
Once you find out the slope here, you can use the point-slope formula to find your equation $y - y_1 = m(x - x_1)$
Ok, so you have your equation. Now just plus in 3 for the variable you used for miles and simplify.
Let me know if that helped
yes but...
This does not require a system of equations, that's a little overkill.
So, when you are setting up linear equations, you should know that a general linear equation looks like this: y=mx+b, where m is the slope and b is the y-intercept. So, you need to find the slope
here, which you find using the equation $m=(y_2 - y_1)/(x_2 - x_1)$
So, what points will you use for this little formula above?
The linear equation in 2 variables they are talking about is the above, as opposed to the x=4 or y=10 linear equations.
Once you find out the slope here, you can use the point-slope formula to find your equation $y - y_1 = m(x - x_1)$
Ok, so you have your equation. Now just plus in 3 for the variable you used for miles and simplify.
Let me know if that helped
I do not see the connection between my question and the slope of lines.
Can you show me your method?
The slope is basically defined as the change that is going on. You said "The cost of a taxi ride is calculated by adding a fixed amount to a charge for each 1/4 of the trip" which means that this
is the slope. It is going up a fixed amount for every quarter mile. So, basically, just use the chart that you made and use the 'y' as the cost and the 'x' as the distance traveled. This will
give you the slope. Try it out.
Not sure how to do it the previously mentioned way. But, I've come up with $10, quite easily my way.
I thought of it this way.
1 mile is $4.00
1 and 1/2 mile is $5.50.
The problem says it is a FIXED amount and a charge for every 1/4 mile.
Divide the $4.00 by 4, that's $0.25 cents every 1/4 mile.
But you'll notice the increase from 1 mile to 1 and a half is a $1.50 increase, which can't be possible at $0.25 every 1/4mile. So the initial FIXED amount has to be $4.00 for the FIRST MILE,
then $0.75 for every 1/4 mile after that.
So 1 mile = $4.00
2 miles contains 8 one-fourth miles. So $0.75 * 8 = $6
$4.00 + $6.00 = $10.00 = 3 mile trip.
that's how I achieved the answer. Not sure about the above way.
The cost of a taxi ride is calculated by adding a fixed amount to a charge for each 1/4 of the trip. If one mile costs $4 and 3/2 miles costs $5.50, what is the cost of a three-mile trip?
I made a table:
1 $4
3/2 $5.50
3 $x
This question came from the chapter titled linear equations in two variables.
After doing the math, I got $11 for 3 miles but the math book tells me that the correct answer is $10 for 3 miles.
How do I set up a system of linear equations for this question? How do I find $10?
Since this comes from the linear equations in two variables, you want two equations with two unknowns. The 1/4 isn't needed though....
Let x=the fixed charge, and y=price per mile. Then the two equations become 4=x+y and 5.50=x+1.5y. That can be solve by substitution, or by subtraction. These two equations are your system, both
of the form
cost=fixed amount + miles*(charge per mile).
Solving, you get y=3, and x =1. Thus, for 3 miles, cost =x+3y=1+3(3)=10.
if you really want the 1/4 mile in there, you measure miles in 1/4's, and you get
4=x+4y and 5.50=x+6y.... cost=fixed amount + (number of quarter miles) *(charge per mile). Your y then should be (3/4).
um.... seriously?
Ok, first off, it's completely unnecessary to use 2 equations.
Use the formulas I posted and get it that way.
if you wanted to do it without them, then you can see that the slope is for every half mile is $1.50, or every mile is $3.
Hence, to then figure out the base cost, is going to be $4 - $3 = $1 (which is the y-intercept).
Then, you use the idea of the slope is $3/mile (our slope, m)
So, when we put this stuff back into the equation (which is what you are supposed to learn from these exercises) is that y=mx+b or by substitution is y = 3x + 1; where y is the cost of the trip
and x is the mileage.
So, when you use the formula y=3x+1, and plug in for 3 miles, you will get $10, or to use the units, it'll be $y = ($3/mile) \times X miles + $1
which means the miles cancels out and you are left with dollars, or $y = $3 \times X + $1
Not sure how to do it the previously mentioned way. But, I've come up with $10, quite easily my way.
I thought of it this way.
1 mile is $4.00
1 and 1/2 mile is $5.50.
The problem says it is a FIXED amount and a charge for every 1/4 mile.
Divide the $4.00 by 4, that's $0.25 cents every 1/4 mile.
But you'll notice the increase from 1 mile to 1 and a half is a $1.50 increase, which can't be possible at $0.25 every 1/4mile. So the initial FIXED amount has to be $4.00 for the FIRST MILE,
then $0.75 for every 1/4 mile after that.
So 1 mile = $4.00
2 miles contains 8 one-fourth miles. So $0.75 * 8 = $6
$4.00 + $6.00 = $10.00 = 3 mile trip.
that's how I achieved the answer. Not sure about the above way.
Thank you for breaking this question down for me.
nice work...
Since this comes from the linear equations in two variables, you want two equations with two unknowns. The 1/4 isn't needed though....
Let x=the fixed charge, and y=price per mile. Then the two equations become 4=x+y and 5.50=x+1.5y. That can be solve by substitution, or by subtraction. These two equations are your system, both
of the form
cost=fixed amount + miles*(charge per mile).
Solving, you get y=3, and x =1. Thus, for 3 miles, cost =x+3y=1+3(3)=10.
if you really want the 1/4 mile in there, you measure miles in 1/4's, and you get
4=x+4y and 5.50=x+6y.... cost=fixed amount + (number of quarter miles) *(charge per mile). Your y then should be (3/4).
Thank you for breaking down this question.
April 20th 2009, 09:52 PM #2
April 21st 2009, 08:59 PM #3
MHF Contributor
Jul 2008
April 21st 2009, 09:12 PM #4
April 21st 2009, 11:18 PM #5
Apr 2009
April 22nd 2009, 01:31 AM #6
April 22nd 2009, 10:27 AM #7
April 22nd 2009, 03:35 PM #8
MHF Contributor
Jul 2008
April 22nd 2009, 03:36 PM #9
MHF Contributor
Jul 2008 | {"url":"http://mathhelpforum.com/algebra/84765-taxi-ride.html","timestamp":"2014-04-20T13:03:47Z","content_type":null,"content_length":"56241","record_id":"<urn:uuid:d177f6be-d174-4d83-8518-4d372605789c>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00641-ip-10-147-4-33.ec2.internal.warc.gz"} |
binomial theorem for all powers
August 31st 2012, 02:12 AM
binomial theorem for all powers
My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?
Thanks in advance.
August 31st 2012, 02:49 AM
Prove It
Re: binomial theorem for all powers
Because it's possible that infinite series might diverge (go to infinity). You need to check the interval for which the series is convergent.
August 31st 2012, 02:58 AM
Re: binomial theorem for all powers
In that case, shouldn't the restriction be "x =/= +infinity or -infinity (when n is just a small number)"?
If -1<x<1, will the value of (1+x)^n be very small?
I really don't understand..
August 31st 2012, 03:11 AM
Prove It
Re: binomial theorem for all powers
The series depends on x. But the series itself is only convergent if it follows specific conditions, which means that it will only be convergent for particular values of x (the ones which will
satisfy the conditions). I suggest you research convergence tests for infinite series.
August 31st 2012, 04:10 AM
Re: binomial theorem for all powers
I think that you are missing the point.
${\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^n {{\binom{n}{k} x^k}}$ for any $x\in\mathbb{R}$.
So it would be true for $-1<x<1.$
August 31st 2012, 04:21 AM
Prove It
Re: binomial theorem for all powers
The OP was actually referring to the GENERALISED Binomial Theorem, which is an infinite series. | {"url":"http://mathhelpforum.com/algebra/202748-binomial-theorem-all-powers-print.html","timestamp":"2014-04-16T06:51:50Z","content_type":null,"content_length":"9050","record_id":"<urn:uuid:8e03a0c9-e6f9-4b88-96f8-170a42bd30b8>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00498-ip-10-147-4-33.ec2.internal.warc.gz"} |
Originally posted on 05/01/2011
Ambient occlusion is an approximation of the amount by which a point on a surface is occluded by the surrounding geometry, which affects the accessibility of that point by incoming light. In effect,
ambient occlusion techniques allow the simulation of proximity shadows - the soft shadows that you see in the corners of rooms and the narrow spaces between objects. Ambien occlusion is often subtle,
but will dramatically improve the visual realism of a computer-generated scene:
The basic idea is to compute an
occlusion factor
for each point on a surface and incorporate this into the lighting model, usually by modulating the ambient term such that more occlusion = less light, less occlusion = more light. Computing the
occlusion factor can be expensive; offline renderers typically do it by casting a large number of rays in a normal-oriented hemisphere to sample the occluding geometry around a point. In general this
isn't practical for realtime rendering.
To achieve interactive frame rates, computing the occlusion factor needs to be optimized as far as possible. One option is to pre-calculate it, but this limits how dynamic a scene can be (the lights
can move around, but the geometry can't).
Way back in 2007, Crytek implemented a realtime solution for
which quickly became the yardstick for game graphics. The idea is simple: use per-fragment depth information as an approximation of the scene geometry and calculate the occlusion factor in
screen space
. This means that the whole process can be done on the GPU, is 100% dynamic and completely independent of scene complexity. Here we'll take a quick look at how the
method works, then look at some enhancements.
Crysis Method
Rather than cast rays in a hemisphere, Crysis samples the depth buffer at points derived from samples in a sphere:
This works in the following way:
• project each sample point into screen space to get the coordinates into the depth buffer
• sample the depth buffer
• if the sample position is behind the sampled depth (i.e. inside geometry), it contributes to the occlusion factor
Clearly the quality of the result is directly proportional to the number of samples, which needs to be minimized in order to achieve decent performance. Reducing the number of samples, however,
produces ugly 'banding' artifacts in the result. This problem is remedied by randomly rotating the sample kernel at each pixel, trading banding for high frequency noise which can be removed by
blurring the result.
method produces occlusion factors with a particular 'look' - because the sample kernel is a sphere, flat walls end up looking grey because ~50% of the samples end up being inside the surrounding
geometry. Concave corners darken as expected, but convex ones appear lighter since fewer samples fall inside geometry. Although these artifacts are visually acceptable, they produce a stylistic
effect which strays somewhat from photorealism.
Normal-oriented Hemisphere
Rather than sample a spherical kernel at each pixel, we can sample within a hemisphere, oriented along the surface normal at that pixel. This improves the look of the effect with the penalty of
requiring per-fragment normal data. For a deferred renderer, however, this is probably already available, so the cost is minimal (especially when compared with the improved quality of the result).
Generating the Sample Kernel
The first step is to generate the sample kernel itself. The requirements are that
• sample positions fall within the unit hemisphere
• sample positions are more densely clustered towards the origin. This effectively attenuates the occlusion contribution according to distance from the kernel centre - samples closer to a point
occlude it more than samples further away
Generating the hemisphere is easy:
for (int i = 0; i < kernelSize; ++i) {
kernel[i] = vec3(
random(-1.0f, 1.0f),
random(-1.0f, 1.0f),
random(0.0f, 1.0f)
This creates sample points on the surface of a hemisphere oriented along the z axis. The choice of orientation is arbitrary - it will only affect the way we reorient the kernel in the shader. The
next step is to scale each of the sample positions to distribute them within the hemisphere. This is most simply done as:
kernel[i] *= random(0.0f, 1.0f);
which will produce an evenly distributed set of points. What we actually want is for the distance from the origin to falloff as we generate more points, according to a curve like this:
We can use an accelerating interpolation function to achieve this:
float scale = float(i) / float(kernelSize);
scale = lerp(0.1f, 1.0f, scale * scale);
kernel[i] *= scale;
Generating the Noise Texture
Next we need to generate a set of random values used to rotate the sample kernel, which will effectively increase the sample count and minimize the 'banding' artefacts mentioned previously.
for (int i = 0; i < noiseSize; ++i) {
noise[i] = vec3(
random(-1.0f, 1.0f),
random(-1.0f, 1.0f),
Note that the
component is zero; since our kernel is oriented along the
-axis, we want the random rotation to occur around that axis.
These random values are stored in a texture and tiled over the screen. The tiling of the texture causes the orientation of the kernel to be repeated and introduces regularity into the result. By
keeping the texture size small we can make this regularity occur at a high frequency, which can then be removed with a blur step that preserves the low-frequency detail of the image. Using a 4x4
texture and blur kernel produces excellent results at minimal cost. This is the same approach as used in
The SSAO Shader
With all the prep work done, we come to the meat of the implementation: the shader itself. There are actually two passes: calculating the occlusion factor, then blurring the result.
Calculating the occlusion factor requires first obtaining the fragment's view space position and normal:
vec3 origin = vViewRay * texture(uTexLinearDepth, vTexcoord).r;
I reconstruct the view space position by combining the fragment's linear depth with the interpolated
. See
Matt Pettineo's blog
for a discussion of other methods for reconstructing position from depth. The important thing is that
ends up being the fragment's view space position.
Retrieving the fragment's normal is a little more straightforward; the scale/bias and normalization steps are necessary unless you're using some high precision format to store the normals:
vec3 normal = texture(uTexNormals, vTexcoord).xyz * 2.0 - 1.0;
normal = normalize(normal);
Next we need to construct a change-of-basis matrix to reorient our sample kernel along the origin's normal. We can cunningly incorporate the random rotation here, as well:
vec3 rvec = texture(uTexRandom, vTexcoord * uNoiseScale).xyz * 2.0 - 1.0;
vec3 tangent = normalize(rvec - normal * dot(rvec, normal));
vec3 bitangent = cross(normal, tangent);
mat3 tbn = mat3(tangent, bitangent, normal);
The first line retrieves a random vector
from our noise texture.
is a
which scales
to tile the noise texture. So if our render target is 1024x768 and our noise texture is 4x4,
would be (1024 / 4, 768 / 4). (This can just be calculated once when initialising the noise texture and passed in as a uniform).
The next three lines use the
Gram-Schmidt process
to compute an orthogonal basis, incorporating our random rotation vector
The last line constructs the transformation matrix from our
vectors. The
vector fills the
component of our matrix because that is the axis along which the base kernel is oriented.
Next we loop through the sample kernel (passed in as an array of
), sample the depth buffer and accumulate the occlusion factor:
float occlusion = 0.0;
for (int i = 0; i < uSampleKernelSize; ++i) {
// get sample position:
vec3 sample = tbn * uSampleKernel[i];
sample = sample * uRadius + origin;
// project sample position:
vec4 offset = vec4(sample, 1.0);
offset = uProjectionMat * offset;
offset.xy /= offset.w;
offset.xy = offset.xy * 0.5 + 0.5;
// get sample depth:
float sampleDepth = texture(uTexLinearDepth, offset.xy).r;
// range check & accumulate:
float rangeCheck= abs(origin.z - sampleDepth) < uRadius ? 1.0 : 0.0;
occlusion += (sampleDepth <= sample.z ? 1.0 : 0.0) * rangeCheck;
Getting the view space sample position is simple; we multiply by our orientation matrix
, then scale the sample by
(a nice artist-adjustable factor, passed in as a uniform) then add the fragment's view space position
We now need to project
(which is in view space) back into screen space to get the texture coordinates with which we sample the depth buffer. This step follows the usual process - multiply by the current projection matrix (
), perform
-divide then scale and bias to get our texture coordinate:
Next we read
out of the depth buffer (
). If this is in front of the sample position, the sample is 'inside' geometry and contributes to occlusion. If
is behind the sample position, the sample doesn't contribute to the occlusion factor. Introducing a
helps to prevent erroneous occlusion between large depth discontinuities:
As you can see,
works by zeroing any contribution from outside the sampling radius.
The final step is to normalize the occlusion factor and invert it, in order to produce a value that can be used to directly scale the light contribution.
occlusion = 1.0 - (occlusion / uSampleKernelSize);
The Blur Shader
The blur shader is very simple: all we want to do is average a 4x4 rectangle around each pixel to remove the 4x4 noise pattern:
uniform sampler2D uTexInput;
uniform int uBlurSize = 4; // use size of noise texture
noperspective in vec2 vTexcoord; // input from vertex shader
out float fResult;
void main() {
vec2 texelSize = 1.0 / vec2(textureSize(uInputTex, 0));
float result = 0.0;
vec2 hlim = vec2(float(-uBlurSize) * 0.5 + 0.5);
for (int i = 0; i < uBlurSize; ++i) {
for (int j = 0; j < uBlurSize; ++j) {
vec2 offset = (hlim + vec2(float(x), float(y))) * texelSize;
result += texture(uTexInput, vTexcoord + offset).r;
fResult = result / float(uBlurSize * uBlurSize);
The only thing to note in this shader is
, which allows us to accurately sample texel centres based on the resolution of the AO render target.
The normal-oriented hemisphere method produces a more realistic-looking than the basic
method, without much extra cost, especially when implemented as part of a deferred renderer where the extra per-fragment data is readily available. It's pretty scalable, too - the main performance
bottleneck is the size of the sample kernel, so you can either go for fewer samples or have a lower resolution AO target.
A demo implementation is available
Wikipedia article on SSAO
has a good set of external links and references for information on other techniques for achieving real time ambient occlusion.
22 comments:
1. Thank you so much for this tutorial. It's a pretty hard effect to implement, I used your tutorial for to add ssao to my own python molecular viewer https://github.com/chemlab/chemlab. What I've
obtained so far is this: http://troll.ws/image/d9ab2364 (I have to add the blur step) it looks right to me but there are many mistakes that I could have made.
I would have never been able to implement such a cool looking effect without your tutorial and your code. I sincerely thank you.
1. I've added blur and have a very small problem.
In this picture I've rendered with 128 samples a set of procedurally-generated sphere imposters, the problem I'm having is that around each sphere there is a thin halo of non-occlusion:
Do you have any idea/suggestion about what causes this issue and how to solve this problem?
The shader I'm using are in this directory: https://github.com/chemlab/chemlab/tree/master/chemlab/graphics/postprocessing/shaders
2. This is the main issue with indiscriminately blurring the AO result. Areas of occlusion/non-occlusion will tend to 'leak' - most noticeably where there are sharp discontinuities in the depth
The solution is to use a more complex blur which samples the depth buffer and only blurs pixels which are at a similar depth.
Another option might be to simply dilate the AO result slightly (after applying the blur). I'm not sure how well this will work, though.
2. Hi,
Thanks for the great tutorial! I have a question for you:
What do I need in order to compute the vViewRay vector used for the unprojection?
The article you linked to says that vViewRay is a vector pointing towards the far-clipping plane - how would I obtain it? Would this be done in the vertex shader of the SSAO fullscreen quad pass
or via some other means? Maybe you can share your method of obtaining it :)?
1. You can compute the view ray from the normalized device coordinates of the fragment in question and the field of view angle and aspect ratio of the camera, like this:
float thfov = tan(fov / 2.0); // can do this on the CPU
viewray = vec3(
ndc.x * thfov * aspect,
ndc.y * thfov,
You can do this either in the vertex shader (and interpolate the view ray), or directly in the fragment shader (compute ndc as texcoords * 2.0 - 1.0).
Matt actually has another, more in-depth blog post on this topic, which may help clarify things better.
3. Hey John,
First off, thanks for these great tutorials.
I have been trying to implement your ssao for a few day and I'm stumped. I took a few images. Some of them are averages across 8 samples.
Origin: https://dl.dropboxusercontent.com/u/11216481/SSAO/origin.png
Depth: https://dl.dropboxusercontent.com/u/11216481/SSAO/depth.png
SamplePosition: https://dl.dropboxusercontent.com/u/11216481/SSAO/sampleP.png
SampleDepth: https://dl.dropboxusercontent.com/u/11216481/SSAO/sampleDepth.png
Offset.xy: https://dl.dropboxusercontent.com/u/11216481/SSAO/offset.png
Could you please let me know if these look right. I have a feeling the sample depth is not right.
Any help would be greatly appreciated. Thanks!
1. The sample depth does look a bit odd, but it's difficult to tell what's wrong with it. What does the end result look like?
2. I made some progress. When I make uRadius linear (* 1.0/(far-near)) the final result seems a lot better. However everything is flipped around. I tried negating axis of the sample position but
it produces strange results. The ssao seems to be calculated corrected, it's just not in the right place lol.
Result: https://dl.dropboxusercontent.com/u/11216481/SSAO/final.png
3. Is uRadius not linear anyway? It should simply a scale value for the sample kernel. Also, remember that uRadius should be appropriate to the scale of the scene.
Are you using a left or right handed system; is +z into, or out of the screen?
4. That was the initial problem. uRadius was much to large and it was throwing off the samplePosition and therefore the offset for the sample depth lookup.
I am using a left handed system (thanks, I wasn't thinking about it). I negated the z component of the viewRay (I calculate it as per the comments above). This properly oriented the ssao and
produces great results... mostly.
looking down z: https://dl.dropboxusercontent.com/u/11216481/SSAO/final-oriented-downz.png
The above screen has the camera facing in the -z direction. When I turn the camera around 180 to face +z I get a strange artifact. The artifact is gradual when turning the camera and gets the
worst when facing +z dead on.
looking up z: https://dl.dropboxusercontent.com/u/11216481/SSAO/final-oriented-upz.png
The artifacts move slightly when with animation in the scene; some objects are moving but they are away from the sphere pyramid. It's like the sample depth lookup is wrapping around. However,
I have my depth texture clamped to border so that should happen.
For more info, I am using a 32x32 kernel and 4x4 noise texture.
Have you seen anything like this?
5. Looks like the kernel isn't being properly oriented - check that your normals are correct.
6. You were totally right. The problem was with my normal texture. It wasn't in view space... **facepalm** Multiplied by the good ol' inverse transpose modelview and voila!
Thanks for all your help man!
4. This is what it looks like now:
I moved in closer to the cube to take this picture.
I'm not sure if this is what it's supposed to look like.
1. Yes, that looks correct - remember that you're in view space, so you'd expect to see red along the X axis, Y green, Z blue.
As you're working with right-handed coordinates you may need to negate the kernel sample.
2. I negated the sample kernel, which didn't fix my problem.
I looked at the range check value, and found out it only reached 1 when I either got very close to the object, or made the radius very large (~10 instead of 1). I doubt this is correct, and I
have no idea what could be causing it.
The depth is being linearized when fetching the sample depth from the texture.
3. Well we know that the view space position is being calculated correctly now. Are you getting any occlusion results at all? Take the range check out and see if you get any results.
4. No, I'm not getting anything, just a white screen.
5. Well all I can really say is go back to the beginning and make sure you're getting sensible results at every step. You can use the sample implementation to visualize the outputs from a
working version and check that what you're getting matches.
6. Well, I finally got it working. Thanks for all the help.
I had two things missing to make this work with a right handed system:
First, I had to negate the sample depth after sampling and linearization to make it increase and decrease with the z axis.
Second, when accumulating the occlusion, the operator comparing the sample depth and sample z needs to be '=>', not '<=' (or, if using the step function, switch the arguments).
Finally, I found a way to reduce the hallow due to blurring: instead of iterating on [0, noiseSize[, iterate on [-noiseSize / 2, noiseSize / 2[. Iterating on [-noiseSize / 2, noiseSize / 2]
and dividing by (noiseSize + 1)^2 might even be better (haven't tested it yet).
7. Good to hear - I am going to add some notes on the differences between left/right handed implementations of this as it seems to be the main source of problems for people trying to implement
it themselves.
I was going to say that you were wrong about reducing the halo, but then I realized that the simple blur code in the tutorial is actually different to the code in the demo implementation! The
demo does what you suggest - it centres the halo on the object border, which reduces the halo. This is about as good as it gets without dilating the SSAO result.
5. hi..Im student from Informatics engineering, this article is very informative, thanks for sharing :)
6. Hi there, I was learning from this tutorial, and I loved the camera movements, would you tell how you implemented the smoothness/lerping? Thanks. | {"url":"http://john-chapman-graphics.blogspot.co.uk/2013/01/ssao-tutorial.html","timestamp":"2014-04-21T12:08:24Z","content_type":null,"content_length":"135993","record_id":"<urn:uuid:3b933029-9cb7-4f35-9374-ef9fd5c270d3>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00257-ip-10-147-4-33.ec2.internal.warc.gz"} |
Lisa Jones
Lisa Jones
a question. They will receive an automated email and will return to answer you as soon as possible. Please
to ask your question.
Amy Ballou
(TpT Seller) re: Calculus Derivatives Test
Is #9 suppose to be negative??
Yes, I copied it quickly and goofed. It's definitely supposed to be negative. Sorry about that!
Amy Ballou
(TpT Seller) re: Calculus Derivatives Test
Is there any way to edit this file (ever so slightly)??
I don't think so when it's in pdf - but I'm not absolutely sure.
I'm sorry that it didn't meet your needs.
(TpT Seller) re: Form A Graphing, Continuity, and Limits with Rational Functions
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Use the footer to put the copyright symbol, your name and date of creation on EVERY page. Items, especially free ones, have been "stolen" from TPT and posted on other sites to be sold. You have to
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I think it is important that we see what others are doing; so, feel free to look at any or all of my freebies listed below for ideas.
I am your very first follower, and if you would like, you are welcome to follow my store as well: http://www.teacherspayteachers.com/Store/SciPi/Products
Keep checking the Sellers Forum under Communication for ideas and helpful suggestions given by TPT veterans. The best of luck as you sell on TPT!
Wow!! Thanks so much for all of the helpful information! I really appreciate your welcoming response to my work and look forward to following you as well. Any constructive criticisms would also be
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I've been teaching mathematics for 20+ years. I currently teach Algebra 2 and Calculus, but I have taught Algebra I, Geometry, Math Analysis and Trigonometry (PreCalc) throughout my career.
I try to incorporate practices supported by cognitive science, specifically cognitive load theory. Most of my activities are designed to promote student engagement during guided instruction. I
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Yet to be added
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I have two children and one grandchild. Prior to teaching, I worked as a research assistant on a project designed to familiarize high school mathematics teachers with applications of mathematics. I
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I've participated in a wide variety of PD throughout my career, but I cannot say enough in support of the Professional Learning Community (PLC) model. Our district's PLCs engage in content rich
lesson development throughout the year. I love working with colleagues to develop lessons and identify best practices. | {"url":"http://www.teacherspayteachers.com/Store/Lisa-Jones-43?breadcrumb=2&subject=22","timestamp":"2014-04-16T11:08:04Z","content_type":null,"content_length":"312956","record_id":"<urn:uuid:875e8cf6-16ca-4c93-a201-bacafd519a77>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00637-ip-10-147-4-33.ec2.internal.warc.gz"} |
David Wetzel on 08 Nov 11
Math journals provide advantages for students to develop a better understanding of mathematics, with teachers learning their student's views and beliefs regarding math.
shared by Erica C on 15 Oct 11 - No Cached
Online Math activities and worksheets from basic numbers through algebra and geometry
David Wetzel on 31 Jul 11
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Math - 1 views
Math test activities for students and teachers of all grade levels
quizzes for a lot of math concepts
interactives on math, including fractions, decimals, and percents and recognizing and comparing
links to math activities to teach ratios, multiplication, scaling, and numbe lines
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Excel - Formulas - Dates
1. If you copy the formula =C1*$D$9 from cell C1 to cell C2, what will the formula be in cell C5?
2. In Excel 2007 you can write simple formulas that cannot automatically update their results when values change
3. $A1 is an absolute reference to column A and a relative reference to row 1. As a mixed reference is copied from one cell to another, the absolute reference stays the same but the relative
reference changes
4. The two CDs purchased in February cost $12.99 and $16.99. The total of these two values is the CD expense for the month. You can add these values in Excel by typing a simple formula in a cell:
7. In the practice, you learned that if you misspell SUM in this formula =SUME(B4:B7), you'll get an error value of #NAME? To fix the formula, you must delete it and start over again.
9. What would happen if you used a formula to do math with dates?
10. A formula result is in cell C6. You wonder how you got the result. To see the formula, you:
11. The button shown below is use for ________ in Excel
14. In Excel if you write =TODAY() in the formula bar (Fx) | {"url":"http://www.proprofs.com/quiz-school/story.php?title=excel-formulas_1","timestamp":"2014-04-19T11:57:50Z","content_type":null,"content_length":"210029","record_id":"<urn:uuid:473aa1f4-3dc3-4707-852a-8cf722995fbe>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00308-ip-10-147-4-33.ec2.internal.warc.gz"} |
Further Results on Colombeau Product of Distributions
International Journal of Mathematics and Mathematical Sciences
Volume 2013 (2013), Article ID 918905, 5 pages
Research Article
Further Results on Colombeau Product of Distributions
^1Faculty of Electrical Engineering and Informational Technologies, Saints Cyril and Methodius University of Skopje, Karpos II bb, 1000 Skopje, Macedonia
^2Faculty of Computer Sciences, Goce Delčev University of Štip, Krste Misirkov, Br 10A, 2000 Štip, Macedonia
Received 10 August 2012; Revised 28 November 2012; Accepted 10 December 2012
Academic Editor: Shyam Kalla
Copyright © 2013 Biljana Jolevska-Tuneska and Tatjana Atanasova-Pacemska. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use,
distribution, and reproduction in any medium, provided the original work is properly cited.
Results on Colombeau product of distributions and are derived. They are obtained in Colombeau differential algebra of generalized functions that contains the space of Schwartz distributions as a
subspace and has a notion of “association” that is a faithful generalization of the weak equality in .
1. Introduction
In quantum physics one finds the need to evaluate , when calculating the transition rates of certain particle interactions; see [1]. The problem of defining products of distributions is also closely
connected with the problem of renormalization in quantum field theory. Due to the large use of distributions in the natural sciences and other mathematical fields, the problem of the product of
distributions [2] is an objective of many research studies. Starting with the historically first construction of distributional multiplication by König [3], and the sequential approach developed in [
4] by Antosik et al., there have been numerous attempts to define products of distributions, see [5–7], or rather to enlarge the number of existing products.
Having a look at the theory of distributions [8, 9] we realize that there are two complementary points of view.(1)A distribution is a continuous functional on the space of space functions (compactly
supported smooth functions). Here we have a linear action of on a test function .(2)Letting be a sequence of smooth functions converging to the Dirac measure, a family of regularization can be
produced by convolution, which converges weakly to the original distribution . Identifying two sequences and if they have the same limit, we obtain a sequential representation of the space of
distributions. Other authors use the equivalence classes of nets of regularization. The delta-net is defined by
When we work with regularization the nonlinear structure is lost by identifying sequences (nets) with the same limit. So we have to get nonlinear theory of generalized functions that will work with
regularization, but identify less. The actual construction of such algebras enjoying these optimal properties is due to Colombeau [10]. His theory of algebras of generalized functions offers the
possibility of applying large classes of nonlinear operations to distributional objects. Some of them are used on solving differential equations with nonstandard coefficients [11]. The “association
process” in Colombeau algebra providing a faithful generalization of the equality of distributions in enables us to obtain results in terms of distributions, the so-called Colombeau products. In
fact, we evaluate particular product of distributions with coinciding singularities, as embedded in Colombeau algebra, in terms of associated distributions; see [12–14]. Therefore, the results
obtained can be reformulated as regularized model products in the classical distribution theory.
In 1966, Mikusinski published his famous result [15]: where neither of the products on the left-hand side here exists, but their difference still has a correct meaning in distribution space . Formula
including balanced products of distributions with coinciding singularities can be found in mathematical and physical literature. In this paper results on product of distributions and are derived.
2. Colombeau Algebra
The basic idea underlying Colombeau’s theory in its simplest form is that of embedding the space of distributions into a factor algebra of , with regularization by convolution with a fixed
“mollifier” .
The space of test function is Functional act on test functions and points is where is required to belong to with respect to the second variable . Now let for. The sequence converges to in .
Taking this sequence as a representative of we obtain an embedding of into the algebra . However, embedding into this algebra via convolution as above will not yield a subalgebra since of course in
general. The idea, therefore, is to factor out an ideal such that this difference vanishes in the resulting quotient. In order to construct it is obviously sufficient to find an ideal containing all
differences . Taylor expansion of shows that this term will vanish faster than any power of , uniformly on compact sets, in all derivatives. We use the following space: moderate functionals, denoted
by , are defined by the property: Null functionals, denoted by , are defined by the property: In other words, moderate functionals satisfy a locally uniform polynomial estimate as when acting on ,
together with all derivatives, while null functionals vanish faster than any power of in the same situation. The null functionals form a differential ideal in the collection of moderate functionals.
We define space of functions and denote by . It is a subalgebra in in sense of natural identification.
Moderate functionals, denoted by , are defined by the property: Null functionals, denoted by , are defined by the property:
Definition 1. Space of generalized functions , generalized complex numbers, and generalized real numbers is the factor algebra defined as
The space of distributions is imbedded by convolution: Equivalence classes of sequences in will be denoted by . If is a generalized function with compact support and is a representative of , then its
integral is defined by Let , . Then(i)they are equal in the distribution sense, if (ii)they are associated if there exist a representative and of and , respectively, such that
3. Results on Some Products of Distributions
It was proved in [12] that for any the product of the generalized functions and in admits associated distributions and it holds with the particular cases for and Here we will make a generalization of
(19). In order to prove the main theorem we need the following lemmas, easily proved by induction.
Lemma 2. For one has
Lemma 3. For one has and stands for .
Theorem 4. The product of the generalized functions and for in admits associated distributions and it holds
Proof. For given we suppose that , without loss of generality. Then using the embedding rule and the substitution we have the representatives of the distribution in Colombeau algebra:
Similar, using the embedding rule and the substitution we have the representatives of the distribution in Colombeau algebra:
Then, for any we have using the substitution .
By the Taylor theorem we have that for . Using this for (25) we have where for and we have changed the order of integration.
Putting we have Thus
Next, suppose that is even, less than and that is less than . It follows from Lemma 2 that is an even or odd function accordingly as is odd or even. We thus have that is an odd function and . If then
and by changing the order of integration we can prove that again . If we suppose that is odd and less than we can prove in a similar manner that .
For the case if again we have . For and using Lemma 3 and changing the order of integration we have Further, and . So, for and
Finally we have Therefore passing to the limit, as , we obtain (22) proving the theorem.
1. S. Gasiorawics, Elementary Particle Physics, John Wiley & Sons, New York, NY, USA, 1966.
2. M. Oberguggenberger, Multiplication of Distributions and Application to Partial Differential Equations, vol. 259 of Pitman Research Notes in Mathematics Series, Longman, 1992.
3. H. König, “Neue Begründung der theorie der distributions,” Mathematische Nachrichten, vol. 9, pp. 129–148, 1953. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at
4. P. Antosik, J. Mikusinski, and R. Sigorski, theory of Distributions. The Sequential Approach, Elsevier, Amsterdam, The Netherlands, 1973.
5. B. Fisher, “The product of distributions,” The Quarterly Journal of Mathematics, vol. 22, pp. 291–298, 1971. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at
6. B. Fisher, “A noncommutative neutrix product of distributions,” Mathematische Nachrichten, vol. 108, pp. 117–127, 1982. View at Publisher · View at Google Scholar · View at MathSciNet
7. L. Z. Cheng and B. Fisher, “Several products of distributions on ${\mathbf{R}}^{m}$,” Proceedings of the Royal Society London A, vol. 426, no. 1871, pp. 425–439, 1989. View at Publisher · View at
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8. A. H. Zemanian, Distribution Theory and Transform Analysis, Dover, 1965.
9. I. M. Gel'fand and G. E. Shilov, Generalized Functions, vol. 1, chapter 1, Academic Press, New York, NY, USA, 1964. View at MathSciNet
10. J.-F. Colombeau, New Generalized Functions and Multiplication of Distributions, vol. 84 of North-Holland Mathematics Studies, North-Holland, Amsterdam, The Netherlands, 1984. View at Zentralblatt
MATH · View at MathSciNet
11. B. Jolevska-Tuneska, A. Takači, and E. Ozcag, “On differential equations with nonstandard coefficients,” Applicable Analysis and Discrete Mathematics, vol. 1, no. 1, pp. 276–283, 2007. View at
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13. B. Damyanov, “Some distributional products of Mikusiński type in the Colombeau algebra $G\left({R}^{m}\right)$,” Journal of Analysis and its Applications, vol. 20, no. 3, pp. 777–785, 2001. View
at Zentralblatt MATH · View at MathSciNet
14. B. Damyanov, “Modelling and products of singularities in Colombeau's algebra $G\left(R\right)$,” Journal of Applied Analysis, vol. 14, no. 1, pp. 89–102, 2008. View at Publisher · View at Google
Scholar · View at Zentralblatt MATH · View at MathSciNet
15. J. Mikusiński, “On the square of the Dirac delta-distribution,” Bulletin de l'Académie Polonaise des Sciences, vol. 14, pp. 511–513, 1966. View at Zentralblatt MATH · View at MathSciNet | {"url":"http://www.hindawi.com/journals/ijmms/2013/918905/","timestamp":"2014-04-16T14:08:27Z","content_type":null,"content_length":"390927","record_id":"<urn:uuid:a323d105-c70f-492c-a470-d5a8155c45a2>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00338-ip-10-147-4-33.ec2.internal.warc.gz"} |
The theory of exponential differential equations
Kirby, P. J. (2006) The theory of exponential differential equations. PhD thesis, University of Oxford.
This thesis is a model-theoretic study of exponential differential equations in the context of differential algebra. I define the theory of a set of differential equations and give an axiomatization
for the theory of the exponential differential equations of split semiabelian varieties. In particular, this includes the theory of the equations satisfied by the usual complex exponential function
and the Weierstrass p-functions.
The theory consists of a description of the algebraic structure on the solution sets together with necessary and sufficient conditions for a system of equations to have solutions. These conditions
are stated in terms of a dimension theory; their necessity generalizes Ax’s differential field version of Schanuel’s conjecture and their sufficiency generalizes recent work of Crampin. They are
shown to apply to the solving of systems of equations in holomorphic functions away from singularities, as well as in the abstract setting.
The theory can also be obtained by means of a Hrushovski-style amalgamation construction, and I give a category-theoretic account of the method.
Restricting to the usual exponential differential equation, I show that a “blurring” of Zilber’s pseudo-exponentiation satisfies the same theory. I conjecture that this theory also holds for a
suitable blurring of the complex exponential maps and partially resolve the question, proving the necessity but not the sufficiency of the aforementioned conditions.
As an algebraic application, I prove a weak form of Zilber’s conjecture on intersections with subgroups (known as CIT) for semiabelian varieties. This in turn is used to show that the necessary and
sufficient conditions are expressible in the appropriate first order language.
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Magnus hierarchy of 1-related groups
up vote 7 down vote favorite
By Magnus-Moldavansky theorem (see Wilhelm Magnus, Abraham Karrass, Donald Solitar, Combinatorial group theory. Presentations of groups in terms of generators and relations, Reprint of the 1976
second edition, Dover Publications, Inc., Mineola, NY, 2004 and http://arxiv.org/PS_cache/math/pdf/0608/0608635v3.pdf), every 1-related group is an HNN extensions of a "smaller" 1-related group. Say,
if one letter occurs in the relator with total exponent 0, one can take this letter as the free letter in the HNN extension.
Question. Is it true that in this HNN extension one of the associated subgroups is always undistorted? Is it possible that one can always represent the group as an HNN extension satisfying that
property (one of the associated subgroups is undistorted)?
For example, in some sense the "worst" 1-related group is the Baumslag group $$\langle a,t\mid a^{a^t}=a^2\rangle=\langle a,b,t\mid a^b=a^2, a^t=b\rangle.$$ In that case $t$ is a free letter, the
base group is $\langle a,b \mid a^b=a^2\rangle$, the associated subgroups are $\langle a\rangle$ (exponentially distorted in the base group) and $\langle b \rangle$ (undistorted in the base group).
Can you define what you mean by "associated subgroup"? – Ian Agol Sep 6 '11 at 0:11
@Ian: If $G$ is an HNN extension of $H$ and $t$ is the free letter conjugating subgroup $A<H$ to subgroup $B<H$ (so that $G=\langle H, t\mid A^t=B\rangle$, then $A,B$ are called the associated
subgroups of the HNN extension. – Mark Sapir Sep 6 '11 at 1:21
en.wikipedia.org/wiki/HNN_extension – Mark Sapir Sep 6 '11 at 1:25
add comment
1 Answer
active oldest votes
You can reverse engineer distortion. Consider the group $\langle a_0, a_1, a_2 |a_1^{a_2a_0}= a_1^2\rangle$. By the Freiheitssatz, the subgroups generated by $\langle a_0, a_1\
rangle$ and $\langle a_1, a_2\rangle$ are free. But both are distorted, since $a_1$ is conjugate to $a_1^2$ (if I understand the notion of distortion correctly).
up vote 6 down Now, take the HNN extension of these two free subgroups to get a 1-relator group $$\langle a_0, a_1, a_2, t | a_1^{a_2a_0}= a_1^2, a_0^t = a_1, a_1^t=a_2\rangle=\langle a, t | (a^t)
vote accepted ^{a^{t^2}a} = (a^t)^2\rangle.$$
@Ian: Thank you! – Mark Sapir Sep 6 '11 at 2:18
add comment
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magnetic moment
Definitions for magnetic moment
This page provides all possible meanings and translations of the word magnetic moment
Random House Webster's College Dictionary
magnet′ic mo′ment(n.)
1. the product of the strength of either magnetic pole of a dipole and the distance between the poles.
Category: Electricity and Magnetism
Origin of magnetic moment:
Princeton's WordNet
1. magnetic moment, moment of a magnet(noun)
the torque exerted on a magnet or dipole when it is placed in a magnetic field
1. magnetic moment(Noun)
The torque exerted on a magnet within a magnetic field; a vector, being the product of the strength of the magnet and the distance between its poles.
1. Magnetic moment
The magnetic moment of a magnet is a quantity that determines the force that the magnet can exert on electric currents and the torque that a magnetic field will exert on it. A loop of electric
current, a bar magnet, an electron, a molecule, and a planet all have magnetic moments. Both the magnetic moment and magnetic field may be considered to be vectors having a magnitude and
direction. The direction of the magnetic moment points from the south to north pole of a magnet. The magnetic field produced by a magnet is proportional to its magnetic moment as well. More
precisely, the term magnetic moment normally refers to a system's magnetic dipole moment, which produces the first term in the multipole expansion of a general magnetic field. The dipole
component of an object's magnetic field is symmetric about the direction of its magnetic dipole moment, and decreases as the inverse cube of the distance from the object.
The Standard Electrical Dictionary
1. Magnetic Moment
The statical couple with which a magnet would be acted on by a uniform magnetic field of unit intensity if placed with its magnetic axis at right angles to the lines of force of the field.
(Emtage.) A uniformly and longitudinally magnetized bar has a magnetic moment equal to the product of its length by the strength of its positive pole.
Find a translation for the magnetic moment definition in other languages:
Use the citation below to add this definition to your bibliography:
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Language Processing Systems Laboratory
Next: Distributed Parallel Processing Up: Department of Computer Previous: Mathematical Foundation of
This year was the second year for the Language Processing Systems Laboratory. As such, the achievement consisted in continuing existing lines of research by each of the lab members (as exemplified in
the publication and achievement list) and also continuing two group projects, namely Towards a Programming Language Designer's Workbench based on Logic Programming and Action Semantics, and
Granularity Optimization and Scheduling on Massively Parallel Computer Systems, which are stated in what follows.
Towards a Programming Language Designer's Workbench based on Logic Programming and Action Semantics
Language development usually proceeds in the following stages: design, specification, prototyping, and implementation (with no strict ordering). We call it a programming language life cycle. Many
programming languages have been developed based on the life cycle. However, their specifications are usually written in English: such specifications can be very misleading for both implementors and
users since the designers' original intention may be misunderstood. Even when a programming language is formally specified (normally based on denotational, operational, or axiomatic semantics
frameworks), it is still difficult to pass on the designer's intention to others because it rarely gives hints on how to implement the language. With this in mind, we are developing a Programming
Language Designer's Workbench based on a recently developed framework, action semantics. The workbench allows a language designer to design and implement a language in one setting. Thus, the language
designers can actually confirm that the designed language is in fact the one he intended. This consistency check is indeed important because it eliminates potential design errors and possible
misunderstanding among designers and implementors.
The project has been carried out in three differrent directions: (1) developing theories necessary to realize programming language designer's workbench; (2) designing methodology related to actually
building up the general-purpose compiler generator based on action semantics framework; (3) studying semantics of many constructs of programming languages (imperative, functional, logic,
object-oriented, etc.) to see how action semantics specifications affect the understanding of semantics of programming languages as well as the design of new languages.
Granularity Optimization and Scheduling on Massively Parallel Computer Systems
Despite recent advances in massively parallel processing (MPP), resource management remains a critical issue for achieving high performance on MPP architectures. Resource management involves code and
data partitioning, scheduling and control of program execution. How to manage these activities to achieve the fastest parallel execution for a single application is an important and challenging
problem that so far has not been satisfactorily resolved. CASS, which stands for Clustering And Scheduling System, is a compiler optimization tool that provides facilities for automatic partitioning
and scheduling of parallel programs on distributed memory architectures. Central to the design of CASS is grain-size optimization. The high communication overhead in current MPP architectures imposes
a minimum grain-size (i.e., computation-to-communication ratio) below which performance degrades significantly. Using task clustering techniques, CASS restructures a parallel program to one whose
grain-size matches that of the target architecture, while minimizing overall parallel execution time.
The completed work of this research in the past fiscal year includes:
• Developed two efficient clustering and scheduling algorithms, namely CASS1 (without task duplication) and CASS2 (with task duplication). Experimental results have been obtained, showing that both
of our algorithms are superior to existing approaches.
• Extended CASS to dynamic task graphs.
Refereed Journal Papers
1. S. Rajasekaran and David S. L. Wei. Selection, routing and sorting on the star graph. Journal of Parallel and Distributed Computing, Accepted to appear, 1995.
Abstract: We consider the problems of selection, routing and sorting on an $n$-star graph (with $n!$ nodes), an interconnection network which has been proven to possess many special properties.
We identify a tree like subgraph (which we call as a `$(k,1,k)$ chain network') of the star graph which enables us to design efficient algorithms for the above mentioned problems. We present an
algorithm that performs a sequence of $n$ prefix computations in $O(n^2)$ time. This algorithm is used as a subroutine in our other algorithms. We also show that sorting can be performed on the
$n$-star graph in time $O(n^{3})$ and that selection of a set of uniformly distributed $n$ keys can be performed in $O(n^2)$ time with high probability. Finally, we also present a deterministic
(non oblivious) routing algorithm that realizes any permutation in $O(n^{3})$ steps on the $n$-star graph. There exists an algorithm in the literature that can perform a single prefix computation
in $O(n\lg n)$ time. The best known previous algorithm for sorting has a run time of $O(n^3\lg n)$ and is deterministic. To our knowledge, the problem of selection has not been considered before
on the star graph.
2. M. A. Palis, J. C. Liou, and David S. L. Wei. Task clustering and scheduling for distributed memory parallel architectures. IEEE Trans. on Parallel and Distributed Systems, Accepted to appear,
Abstract: This paper addresses the problem of scheduling parallel programs represented as directed acyclic task graphs for execution on distributed memory parallel architectures. Because of the
high communication overhead in existing parallel machines, a crucial step in scheduling is task clustering, the process of coalescing fine grain tasks into single coarser ones so that the overall
execution time is minimized. The task clusting problem is NP-hard, even when the number of processors is unbounded and task duplication is allowed. A simple greedy algorithm is presented for this
problem which, for a task graph with arbitrary granularity, produces a schedule whose makespan is at most twice optimal. Indeed, the quality of the schedule improves as the granularity of the
task graph becomes larger. For example, if the granularity is at least 1/2, the makespan of the schedule is at most 5/3 times optimal. For a task graph with $n$ tasks and $e$ inter-task
communication constraints, the algorithm runs in $O(n(n \lg n + e))$ time, which is $n$ times faster than the currently best known algorithm for this problem. Similar algorithms are develpoed
that produce: (1) optimal schedlues for coarse grain graphs; (2) 2-optimal schedules for trees with no task duplication; and (3) optimal schedules for coarse grain trees with no task duplication.
Refereed Proceeding Papers
1. David S. L. Wei. Quicksort and permutation routing on the hypercube and deBruijn networks. In International Symposium on Parallel Architectures, Algorithms, and Networks, pages 238--245. IEEE
Computer Society, IEEE Computer Society Press, December 1994.
Abstract: We consider the problems of sorting and routing on some unbounded interconnection networks, namely hypercube and de Bruijn network. We first present two efficient implementations of
quicksort on the hypercube. The first algorithm sorts $N$ items on an $N$-node hypercube, one item per node, in $O(\frac{\log^{2}N}{\log\log N})$ time with high probability, while the other one
sorts N items on an $(N/\log N)$-node hypercube, $\log N$ items per node, in $O(\log^{2}N)$ time with high probability, which achieves optimal speedup in the sense of $PT$ product. Both
algorithms beat the previously best known parallel quicksort that runs in $O(\log^{2} N)$ expected time on a butterfly of $N$ nodes. We also present a deterministic (nonoblivious) permutation
routing algorithm which runs in $O(d \cdot n^{2})$ time on a $d$-ary de Bruijn network of $N=d^{n}$ nodes. To the best of our knowledge, this routing algorithm is so far the fastest deterministic
one for the de Bruijn network of arbitrary degree. The previously best known one runs in $O((\log d) \cdot d \cdot n^{2})$ time. All algorithms presented are simple, the constants hidden behind
the big Oh being small.
2. M. Palis, J. C. Liou, and David S. L. Wei. A greedy task clustering heuristic that is provably good. In International Symposium on Parallel Architectures, Algorithms, and Networks, pages
398--405. IEEE Computer Society, IEEE Computer Society Press, December 1994.
Abstract: A simple greedy algorithm is presented for task clustering with duplication (or recomputation) which, for a task graph with arbitrary granularity, produces a schedule whose makespan is
at most twice optimal. For a task graph with $n$ tasks and $e$ inter-task communication constraints, the algorithm runs in $O(n(n \lg n + e))$ time, which is $n$ times faster than the currently
best known algorithm for this problem. Similar algorithms are develpoed that produce: (1) optimal schedlues for coarse grain graphs; (2) 2-optimal schedules for trees with no task duplication;
and (3) optimal schedules for coarse grain trees with no task duplication.
3. D. F. Hsu and David Wei. Permutation routing and sorting on directed de bruijn network. In International Conference on Parallel Processing, Wisconsin, U.S.A, August 1995.
Abstract: We show that any deterministic oblivious routing scheme for the permutation routing on a $d$-ary de Bruijn network of $N=d^{n}$ nodes, in the worst case, will take $\Omega(\sqrt{N})$
steps under the {\it single-port} model. This lower bound improves the results shown in \cite{BH82} and \cite{KKT90} provided $d$ is not a constant. A matching upper bound is also given. We also
present an efficient general sorting algorithm which runs in $O((\log d) \cdot d \cdot n^{2})$ time for directed de Bruijn network with $d^{n}$ nodes, degree $d$ and diameter $n$.
4. Harvey Abramson. Why Johnny Can't Read the Screwiest Writing System in the World and How to Help Him Learn: On the Necessity of Japanese--English Hyperdictionaries. In Proceedings - PACLING '93,
April 1994.
5. Harvey Abramson. Logic Programming and Grammars. In The Encyclopedia of Language and Linguistics, 1994.
6. Harvey Abramson. Prolog Based Aids for the Study of Japanese. In Proceedings - Practical Applications of Prolog Conference, pages 1--13, April 1994.
7. Harvey Abramson. Breaking Out of the Web of Kanji or What is a Dictionary? . In Proceedings - Language Engineering on the Information Highway Conference, September 1994.
8. Harvey Abramson and Veronica Dahl. Extending Logic Grammars with ID-LP Specifications. In Proceedings - Iberamia 94, IV Congreso Iberoamerican de Inteligencia Artificial, pages 230--246, October
9. K.-G. Doh and D. A. Schmidt. 1st international workshop on action semantics. In Peter D. Mosses, editor, The facets of action semantics: some principles and applications, pages 1--15. BRICS,
April 1994.
Abstract: A distinguishing characteristic of action semantics is its facet system, which defines the variety of information flows in a language definition. The facet system can be analyzed to
validate the well-formedness of a language definition, to infer the typings of its inputs and outputs, and to calculate the operational semantics of programs. We present a single framework for
doing all of the above. The framework exploits the internal subsorting structure of the facets so that sort checking, static analysis, and operational semantics are related, sound instances of
the same underlying analysis. The framework also suggests that action semantics's extensibility can be understood as a kind of ``weakening rule'' in a ``logic'' of actions. In this paper, the
framework is used to perform type inference on specific programs, to justify meaning-preserving code transformations, and to ``stage'' an action semantics definition of a programming language
into a static semantics stage and a dynamic semantics stage.
10. K.-G. Doh. Acm-sigplan symposium on partial evaluation and semantics-based program manipulation. In Action transformation by partial evaluation, pages 230--240. SIGPLAN, ACM, June 1995.
Abstract: The compiler automatically generated from action-semantics directed compiler generators first translates a source program to its action denotation, and then to a target code. To improve
the efficiency of the target code, it is essential to transform the action denotation into the proper one leading to a better target code. In this paper, an automatic action transformation based
on partial evaluation is presented, along with methodology to analyze action semantics equations, allowing necessary analyses to be performed at compiler-generation time before source programs
are given.
Doctoral Dissertations Advised
1. Jing-Chiou Liou, 1994.
Grain-Size Optimization and Task Scheduling on Distributed Memory Parallel Architectures, New Jersey Institute of Technology. Thesis advisor: David S. L. Wei (with Dr. Micheal Palis).
Academic Activities
1. Harvey Abramson, Natural Language Understanding and Logic Programming, April 1994.
Member Program Committee 'Natural Language Understanding and Logic Programming 5, Lisbon.
1. David S. L. Wei, 1994. Invited speaker: National Tsing Hua University, Taiwan.
2. David S. L. Wei, 1995. Invited speaker: The New York Academy of Sciences.
3. David S. L. Wei, 1995. Invited speaker: Laboratory for Computer Science, Massachusetts Institute of Technology.
4. David S. L. Wei, 1994. Advisor of Student Karate Club, University of Aizu.
5. David S. L. Wei, 1994. Referee of IEEE Trans. on Computers, and Journal of Parallel and Distributed Computing.
6. David S. L. Wei. Invited speaker: National Taiwan University.
7. Kyung-Goo Doh, 1994. Referee for Journal of Functional Programming.
8. Kyung-Goo Doh, 1994.
Refereed for ACM-SIGPLAN Symposium on Partial Evaluation and Semantics-based Program Manipulation.
9. Kyung-Goo Doh, 1994. Talk at the Department of Computer Science, Aarhus University, Denmark.
Next: Distributed Parallel Processing Up: Department of Computer Previous: Mathematical Foundation of | {"url":"http://web-ext.u-aizu.ac.jp/official/researchact/annual-review/1994/sections/sectionstar2_1_3.html","timestamp":"2014-04-19T22:17:03Z","content_type":null,"content_length":"17189","record_id":"<urn:uuid:54b404b2-3999-46f2-8468-8047372e5030>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00519-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts by
Total # Posts: 36
how do you convert mg to moles?
The dissolved Ca++ concentration in a river is approximately 30 mg/L. This concentration is equivalent to ___ ppm or ___ mol/L, or ___ cmol/L. It would provide ___ cmolc+/L. (c+ is a subscript)
math - please help
What is the rotational speed (rpm) of the 64 in diameter rear tires of a tractor traveling at 6.5 mph if there is no slippage? How do I work this? Thank you.
Estimate the gallons of paint needed to cover the outside of a cylindrical silo which is 30 feet in diameter and 70 feet high. Do not include any roof area. The paint covers approximately 150 square
convert: 6000 kg to ___ tonnes (metric tons) 6000 kg to ___tons 1 acre-ft of water to ___gallons 32 psi to ___kpa 10 psfito ___ oz/in^2
physics - please help
I forgot to mention that I got 6.417E29 for part A and 2.688E34 for part B. Both answers were wrong. Please help. Thank you.
physics - please help
Assume the earth is a uniform sphere and that its path around the sun is curcular. Calculate the kinetic energy that the earth has because of its rotation about its own axis. For comparison, the
total energy used in the US in one year is about 9.33 x 10^9 J b) Calculate the ki...
Physics - still confused -Please help
In 9.7 s a fisherman winds 2.1 m of fishing line onto a reel whose radius is 30 cm (assumed to be constant as an approximation). The line is reeled in at constant speed. Determine the angular speed
of the reel in rad/s. Thank you.
Physics Please help
I'm still confused. to find angular speed = tnngenital speed/radius, would the radii cancel out, leaving only the tangential speed? still confused as to set up
In 9.7 s a fisherman winds 2.1 m of fishing line onto a reel whose radius is 30 cm (assumed to be constant as an approximation). The line is reeled in at constant speed. Determine the angular speed
of the reel in rad/s. Thank you.
A box is being with a velocity v by a force P (parallel to to v) along a level horizontal floor. the normal force is Fn, the kinetic frictional force if fk, and the weight of the box is mg. Decide
which forces do positive, zero or negative work and why?
Physics - position time graph
The graph shows position as a function of time for 2 trains running on parallel tracks. The graph shows line A at a 45 degree angle,beginning at origin. Curve B also begins at origin, continues up
and to the right (all in quadrant 1, like a sq root function) It crosses Line A ...
Two forces, F1 and F2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 69.0 N and F2 = 24.5 N. What is the horizontal acceleration (magnitude and direction) of
the block? F1 is pointing at the top left corner of the box forming a 65 degree...
Physics Please Help
Force 1 is pushing on the top left corner of the box. The arrow forms a 65 degree angle with the verticle (as if the box were extended straight left, from 9:00 to the arrow = 65 degrees) I found F1
horizontal component to be 69 cos 65 = 29.16 N F2 is pushing to the left. F2 = ...
Physics Please Help
Two forces, F1 and F2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 69.0 N and F2 = 24.5 N. What is the horizontal acceleration (magnitude and direction) of
the block? F1 is pointing at the top left corner of the box forming a 65 degree ...
algebra - exponentials
My question is how do I know that N increases by a factor of 2.9 each day. (see below) Let N(t) be number of bacteria after t days. N(t) = Pa^t for some constants P and a. N(2) = 5400 and N(6)=
382,000. a)before working problem, estimate value of N(5). b) write down 2 equations...
algebra - please help
Let N(t) be number of bacteria after t days. N(t) = Pa^t for some constants P and a. N(2) = 5400 and N(6)=382,000. a)before working problem, estimate value of N(5). b) write down 2 equations for P
and a, one when t=2 and one when t=6, and use these two equations to compute a a...
If log (base b) 3 = 1.099 and log (base b) 6 = 1.792, find log (base b) 18 = ???
I won't have access to this on a test. How can I work it on a calculator.
Find all positive solutions to: x^6 + 7x = 4x^4 - x^2 + 8
height of a projectile thrown straight up from top of 480 ft hill t seconds after bing thrown up with initial velocity vo ft/s is h = -16t^2 +vot + 480 If vo = 96, find times when the projectile will
be at a) height of 592 ft above the ground and b) crash on the ground.
algebra - Please help
Use the equation y = ax^2 + bx + c where x is the horizontal distance a baseball has traveled and y is the height of the ball at the given distance. 1)Are some values of y unreasonable? 2)How is this
information important for choosing a window for the graph? 3) With respect to...
Use the equation y = ax^2 + bx + c where x is the horizontal distance a baseball has traveled and y is the height of the ball at the given distance. 1)Are some values of y unreasonable? 2)How is this
information important for choosing a window for the graph? 3) With respect to...
Produce a third degree polynomial that has exactly the roots -3 and 5 with y-intercept 1350. I understand I could start with (x+3)(x-5)^2, but i don't know how to get the y-intercept 1350 with the
same conditions. Please help. thank you.
Produce a third degree polynomial that has exactly the roots -3 and 5 with y-intercept 1350. I understand I could start with (x+3)(x-5)^2, but i don't know how to get the y-intercept 1350 with the
same conditions. Please help. thank you.
physics - need help
Resistors of 30 ohms and 60 ohms are connected in parallel and joined in series to a 10 ohm resistor. The circuit voltage is 180 volts. Find: the voltage of the parallel circuit the voltage across
the parallel circuit the current through the 10 ohm resistor
what current will flow through a resistance of 11 ohms when the emf impressed acrosss the resistance is 2.8 volts?
Thank you
If an ammeter records 8 amperes and a voltmeter records 1.1 volts, the resistance of the conductor in the circuit is ___ ohms
If an auto weighing 13,345 N causes a bridge to sag 0.0254 m, a truck weighing 26,690 N would cause the bridge to sag how many meters, assuming the truck was on the same spot as the auto had been?
How do I even set this up?
So F = 22500 Am I right in thinking it is POSITIVE?
a bullet with a mass of 5 g and a speed of 600 m/s penetrates a tree to a depth of 4 cm. Assume that a constant frictional force stops the bullet. Calculate the magnitude of this frictional force. I
used KE = 1/2 mv^2 = 900 J then F (friction ) = W/d = 22500 N BUT I don't ...
a bullet with a mass of 5 g and a speed of 600 m/s penetrates a tree to a depth of 4 cm. Assume that a constant frictional force stops the bullet. Calculate the magnitude of this frictional force.
*Also, I'm not sure if the answer would be positive or negative.
a spring with spring constant of 30 N/m is stretched 0.2 m from its equilibrium position. How much work must be done tostretch it an additional 0.1 m?
the kilowatt-hour is a unit of: force, work, power, or time?
a 2575 kg van runs into the back of a 825 kg compact car at rest. They move off together at 8.5 m/s. Assuming no friction with the ground, find the initial speed of the van. | {"url":"http://www.jiskha.com/members/profile/posts.cgi?name=Nilan","timestamp":"2014-04-17T22:12:53Z","content_type":null,"content_length":"15212","record_id":"<urn:uuid:b17b9a56-3eba-41d0-a14f-8f15e76f0012>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00489-ip-10-147-4-33.ec2.internal.warc.gz"} |
COMSOL Multiphysics
COMSOL Multiphysics Version 4.1 introduces many new ideas and concepts to make it easier for you to create models. The following are the most important updates to the graphical user interface:
New layout:
In version 4 it is easier to organize and design models with the intuitive structure of the COMSOL Desktop. The Model Builder displays all the features of your model in one place.
Personalized desktop:
Control how you organize the COMSOL Desktop layout—your preferences are saved for the next time you open COMSOL Multiphysics.
Sequences of operations:
You can now build custom model sequences to generate your geometry, physics, mesh, studies, and results. The sequences can be edited and changes are automatically updated across the model. Operations
in the sequences can be modified by the parametric solver.
Geometric parameter sweeps:
With geometry sequences COMSOL can perform geometric parameter sweeps with full associativity from the user interface.
Material settings:
Materials are now administrated in one node in the Model Builder. You can select a material and its properties for each domain and for all physics in that domain.
Multiple solutions and meshes:
With the new layout, you can save and view multiple solutions and meshes, and compare and contrast the results in the Results branch.
Use probes to visualize scalar quantities during computations. The quantities can be defined as integrals, max/min, the average of a field quantity, or the value at a point. This now works for time
stepping and parameter sweeps.
Dynamic help:
The new context-dependent help enables easy browsing with extended search functionality. The Help Desk provides access to the complete documentation set in PDF and HTML formats.
Improved graphics:
Faster, better looking graphics.
Model settings:
Access to the model settings is easier and more intuitive. When you select a node in the Model Builder, a docked window containing associated settings displays at the same time.
Predefined selections:
Define selections of domains, boundaries, edges, and points. These predefined selections are available in the Settings windows for the physics interfaces, meshing, and when studying the results.
LiveLink™ family of products for integrating with CAD and MATLAB®
The new LiveLink™ for SolidWorks®, LiveLink™ for Inventor®, and LiveLink™ for Pro/ENGINEER® products connect COMSOL Multiphysics directly with these CAD programs for interactively linking parameters
specified in a CAD system with simulation geometry. In addition, LiveLink for MATLAB is available for incorporating COMSOL Multiphysics models in the MATLAB technical computing and programming
Cluster support
A floating network license for COMSOL Multiphysics can be extended at no additional cost to computational nodes for clusters on the Windows Cluster Server 2003, Windows HPC Server 2008, and Linux
Version 4 also introduces:
• New and enhanced physics interfaces and predefined multiphysics interfaces
• New parallel solvers and higher solver performance:
• A modal solver for frequency response and time domain (for structural and acoustics simulations, for example)
• The MUMPS and SPOOLES direct solvers for parallel and cluster computing
• A fast-frequency sweep solver using AWE (asymptotic waveform evaluation) for electromagnetic waves simulations, for example
New Functionality in Version 4.1
General COMSOL Desktop Functionality
• Undo/redo of many operations in the Model Builder and the Settings window.
• Copy-paste and duplication of selected nodes in the Model Tree, such as functions, selections, study steps, plot groups and plots, and images and data for export.
• Automatic save and recovery of models during solver operations.
• The dtang operator is a complement to the d and pd differentiation operators. dtang(f,x) computes the derivative of f with respect to x, in the tangential direction along a boundary, when f is
defined only on the boundary and therefore cannot be differentiated using the d operator.
• License borrowing is now available directly in the user interface. This functionality allows floating network license users to borrow a license and run COMSOL, for example, on a laptop or a home
computer that is not connected to a network.
• 3D Helix predefined geometric primitive.
• Parametric curves in 2D and 3D. Curves can now be described as functions of a parameters.
• Geometry sweep, in 3D, based on parametric curves.
• Polygon nodes for easily creating polygons from coordinate data.
• Rendering of work planes in 3D.
• COMSOL geometries can now be converted to the Parasolid format in the CAD Import Module. This allows for the parameterization of geometries that combine imported parts from CAD files and parts
created using COMSOL geometry operations.
• Export to the Parasolid file format using the CAD Import Module is now supported for most COMSOL geometries, including objects created by extrude and revolve. You can then access geometries
created in COMSOL in your CAD software.
• Physics-controlled meshing, which automatically creates meshes that are adapted to the physics in the model. This is for example utilized by the Fluid Flow interfaces that create a somewhat finer
mesh than the default and in addition create boundary layer meshes on no-slip boundaries.
For more information on how to use physics-controlled meshing for fluid flows, see “Physics-Controlled Meshing” in the COMSOL Multiphysics User’s Guide.
• Improved and more robust boundary layer meshing. The boundary-layer mesher can now handle cases where interior boundaries intercept the boundary layer.
• A deformed mesh can now be used as a geometry. Using the Moving Mesh interface, you are able to manually remesh in order to keep a high quality mesh as the geometry deforms.
Physics Interfaces
• Display of equations in physics interfaces, which gives a compact and precise description of your physics settings.
• Support for additional element types in the PDE interfaces: Argyris, bubble, curl, divergence, discontinuous, Hermite, and Gauss point data elements.
• The PDE interfaces now support setting the field name and dependent variables independently. You can create a vector field, under one name, that encompasses several dependent variables.
• Additional functionality in the physics interfaces for each module, see the module sections in these release notes.
Studies and Solvers
Improved usability in the Study branch:
• Settings in the study steps and the corresponding settings in the solvers are synchronized by default.
• The relative tolerance for the time-dependent solver can now be set in the study step.
• The Time discrete solver is now enabled by adding a Time Discrete study step. This solver gives you full control of the time discretization and can be used, for example, in the projection method
for fluid flow. To use this method for fluid flow, you also have to select the memory efficient form option in a Fluid Flow interface.
• A computation is easily resumed by right-clicking on the study or solver nodes and selecting Continue.
• Computations can also be started from any solver step by selecting Compute From Selected in the solver sequence.
• Time-dependent simulations can be segregated for one-way coupled multiphysics models. The solver picks the value of the variable not solved at every time step during the solution of a second
variable. This is available by selecting the All option for the variables not solved for in the Dependent Variables settings.
• Show default solver creates a new active solver if the current solver has been edited. The old solver is kept in the tree but it is no longer active. If the study step or the physics have been
changed, but the current solver has not been edited, the current solver is refreshed with the settings from the study step and physics. If a solver setting has been changed manually, an edit
overlay is displayed on the corresponding solver.
• Scaling of eigenvectors can be done using mass scaling and maximum values, in addition to the root mean square (RMS). The mass scaling gives eigenvectors that can exchange with other software in
a standard fashion. The scaling with maximum values visualize eigenmodes using the same range (max = 1).
• Mass participation factors are computed for eigenvalue/eigenfrequency studies. A mass participation factor for an eigenmode gives an estimate of how much vibrational energy is dissipated through
an eigenmode in each direction.
• Linearized frequency-domain problems can be solved about a general stationary bias solution. This bias solution can be the solution to a nonlinear problem. A perturbation operator allows you to
specify the load that should be applied in the frequency domain. In Results, you are able to visualize the load and the response to the load in the frequency domain as well as the bias solution
and combinations of the bias solution and the frequency response.
• The AMG preconditioner now support complex-valued problems. This allows for the use of memory efficient iterative solvers for problems solved in the frequency domain.
• The choice of damping factor in the automatic damped Newton method is improved. This yields a more robust solution process for nonlinear problems.
• Automatic and manual scaling can now be mixed for more robust solution of multiphysics problems.
• Explicit time stepping solver using the Runge-Kutta method is now available. This solution method may give a higher performance for problems that require very small time steps.
• Labels have been added to contour plots.
• You can now display several plot windows that visualize a plot group. (The Plot In option in 4.0a only allowed several plot windows with snapshots of the graphics window.)
• Polar plots that displays graphs in polar coordinates specified by radius and angle.
• Frequency-domain transformation using FFT for computing frequency spectra from time-dependent simulations.
• Data sets for visualization of results on surfaces and edges.
• Data series operations for computing the maximum, minimum, average, or integral of, for example, the values of a quantity in a point over time for a time-dependent model.
• Export of table data to file.
• Syntax for special characters (Greek letters, mathematical symbols, and others) are now enabled in plots. Unicode characters can also be used and will then be printed with the current font.
Special characters always use our symbol font.
• Transparency has been improved to correctly show several overlapping faces. This works for OpenGL rendering and software rendering.
Backward Compatibility vs. Version 3.5a
Deformed Geometry Interface
The Parameterized Geometry application mode in versions 3.5a, which is limited to 2D, is replaced with the Deformed Geometry interface in version 4.1. This interface is available in 2D and 3D. The
Deformed Geometry interface deforms the mesh using an arbitrary Lagrangian-Eulerian (ALE) method and is not the parameterized geometry using geometric parameter sweeps (see above), which is new
functionality in version 4.1.
In the version 4.1 interface, the Linear Displacement and Similarity Transform boundary conditions are not yet available as preset conditions. Those boundary conditions are planned for version 4.2.
In version 4.1, you can create the corresponding conditions by manually entering variables.
Pairs Boundary Conditions
Pairs are used to connect boundaries between domains that are separated by an assembly boundary (boundary between different parts in an assembly). To create an assembly, you have to either import it
from a CAD package or actively form an assembly as a final step in the COMSOL geometry sequence.
Pair boundary conditions are available as general conditions for all application modes in versions 3.5a. Most pair boundary conditions that were available in COMSOL 3.5a can be found in the physics
interfaces in version 4.1. Exceptions from this are listed under backward compatibility for each product.
Periodic PAIR Boundary Conditions
Periodic boundary conditions are used to model repetitive structures, where one boundary in a domain is identical to another boundary in the same domain.
Periodic boundary conditions are available as general conditions for all application modes in version 3.5a. In version 4.1, this functionality is a tailored condition for each physics interface.
However, some physics interfaces may lack periodic boundary conditions in version 4.1; see the Backwards Compatibility section for the modules below.
Note that all physics interfaces will include periodic boundary conditions in the future.
Axisymmetric Models
In version 3.5a equations in axisymmetric application modes use the independent variable for the radius, r, to account for axisymmetry. In version 4.1, the equations are compensated by using the
factor 2πr.
If you have manually multiplied expressions by 2π for a version 3.5a model, please note that these may be incorrect when the model is opened in version 4.1.
Report Generator
The report generator is not yet implemented in version 4.1. A report generator is planned for 4.2.
Backward Compatibility for pre-3.5a models
COMSOL 4.1 can load models saved from version 3.5a. For loading models from earlier COMSOL versions than 3.5a you need to load them in COMSOL 3.5a and then save them. For simplifying this task a
utility is available where you can convert all files in a directory from versions 3.0–3.5 to version 3.5a. See the section “COMSOL Convertpre35a Command” on page 51 for Windows, section “COMSOL
Convertpre35a Command” on page 85 for Linux, section “COMSOL Convertpre35a Command” on page 112 for the Mac in the COMSOL Installation and Operations Guide for more information. | {"url":"http://www.comsol.fi/support/releasenotes/4.1/multiphysics/","timestamp":"2014-04-20T18:35:46Z","content_type":null,"content_length":"51196","record_id":"<urn:uuid:ee1ee404-79e7-4443-9ae2-2428f62f77e3>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00241-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two identical guitar strings are stretched with the same tension...
Get your Question Solved Now!!
Two identical guitar strings are stretched with the same tension...
Introduction: Find the length of low-pitched string considering beat frequency
More Details: Two identical guitar strings are stretched with the same tension between supports that are not the same distance apart. The fundamental frequency of the higher-pitched
string is 340Hz, and the speed of transverse waves in both wires is 150 m/s. How much longer is the lower-pitched string if the beat frequency is 5Hz?
I tried using the equation L=v/(2*fundamental freq) then doing L=v/(2*(highfreq-lowfreq)) then solving (answer 2- answer 1) but my answer is wrong.
Please log in or register to answer this question.
0 Answers
Related questions | {"url":"http://www.thephysics.org/16208/two-identical-guitar-strings-are-stretched-with-same-tension","timestamp":"2014-04-16T20:08:18Z","content_type":null,"content_length":"105666","record_id":"<urn:uuid:089c6c7f-872f-4786-8e23-afc0c78730e5>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00120-ip-10-147-4-33.ec2.internal.warc.gz"} |
Roosevelt, NJ
Find a Roosevelt, NJ Math Tutor
My name is Maor, and I'm currently a third-year PhD candidate in the chemistry department at Princeton University. I obtained my Bachelor's degree in chemistry (minor in mathematics) from the
University of Massachusetts Lowell, and I obtained a master's degree in chemistry from Princeton University. Simply put: I love teaching and wish to continue doing so!
6 Subjects: including calculus, precalculus, algebra 2, algebra 1
...Algebra is the collection and categorization of many different rules, formulas and properties. A typical Algebra 1 course reinforces the very basics of solving, graphing, and writing linear
equations and inequalities. The next step introduces powers and exponents, quadratic equations along with polynomials and factoring.
10 Subjects: including ACT Math, algebra 1, algebra 2, geometry
...I love watching children succeed and at the same time enjoy their success. That is why every summer since my freshman year I have taught at the summer camp offered by my school for children
ages 5-12. I was offered the job and was requested to come back every summer if possible.
10 Subjects: including algebra 1, geometry, prealgebra, Arabic
...I was always an excellent teacher. I have always kept up with technology. The subjects that I have taught are Honors Algebra II, Honors Geometry, Algebra II, Geometry, Algebra I, General Math,
SAT prep, Basic Computer Programming, C++ Programming, Visual Basic, and GED.
7 Subjects: including SAT math, PSAT, algebra 1, algebra 2
...Through this approach, my students are much better equipped to begin their individual journey to fluency. I am qualified to teach Praxis because in addition to passing the Praxis general exam
with distinction and thus gaining my NJ teaching license (top 10% of test takers), I have tutored Praxis...
37 Subjects: including algebra 1, English, ACT Math, geometry | {"url":"http://www.purplemath.com/Roosevelt_NJ_Math_tutors.php","timestamp":"2014-04-18T23:21:07Z","content_type":null,"content_length":"23941","record_id":"<urn:uuid:539d2667-57c1-45df-977c-0deca06e9124>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00615-ip-10-147-4-33.ec2.internal.warc.gz"} |
Each year the internal revenue service adjusts the value of an exemption based on inflation and rounded to the nearest $50 In a recent year if the exemption was worth $3,100 and inflation was 4.7 percent what would be the amount of the exemption for the upcoming tax year?
Each year the internal revenue service adjusts the value of an exemption based on inflation and rounded to the nearest $50 In a recent year if the exemption was worth $3,100 and inflation was 4.7
percent what would be the amount of the exemption for the upcoming tax year?
$3,100 X 1.047 = $3,245.70 rounded to $3,250
Not a good answer? Get an answer now. (FREE)
There are no new answers.
True. Take-home pay is a person's earnings after deductions for taxes and other items.
Added 7/30/2012 10:47:23 PM | {"url":"http://www.weegy.com/?ConversationId=305950B6","timestamp":"2014-04-17T01:00:48Z","content_type":null,"content_length":"37671","record_id":"<urn:uuid:f043121b-1885-480f-859c-4124c6aebd0b>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00363-ip-10-147-4-33.ec2.internal.warc.gz"} |
Biomedical Transport Processes
Biomedical Transport Processes (BTP)
Emphasis Area Advisor: Bob Tranquillo
BTP involves three fundamental processes: momentum transfer, mass transfer, and heat transfer. They share similar biophysical and mathematical descriptions and so are commonly taught within a single
course, as we do in BMEn 3101 and its sequel, BMEn 5311.
• Momentum transfer is what underlies flow fluid in the subject known as fluid mechanics. Applications of fluid mechanics in BME range from predicting blood flow in vessels, to flow of samples in
"lab-on-chip" microfluidic systems, to flow of cell culture medium through tissue-engineered cartilage in bioreactors.
• Mass and heat transfer refer to the ability to deliver molecules and energy, respectively, from a source to a target. Applications of mass and heat transfer range from predicting blood
oxygenation rates in capillaries from oxygen in lung alveoli and in hollow fibers from pure oxygen gas in "heart-lung machines," to movement of mRNA generated in the cell nucleus to cytoplasmic
While appropriate and accurate experimentation is also key on this subject, BTP is highly mathematical and computational in nature, since the basis of making such predictions is formulating and
solving the equations that govern momentum, mass, and energy balances. This is reflected in the number of mathematical and computational ESE courses listed for this Emphasis Area.
As suggested in the above applications, BTP is relevant in almost every physiological and cellular process and almost all medical devices. Thus, this Emphasis Area is relevant for students interested
in pursuing both employment and advanced studies upon graduation.
Electives Course List
Please remember that all courses listed are merely suggestions. You may take any cohesive set of classes that meet the Engineering and Science Elective requirements and are approved by the Emphasis
Area Advisor and the Director of Undergraduate Studies.
Course Credits Engineering (E) or Science (S)
3 E
3 E
3 E
3 E
3 E
3 E
3 E
4 E
3 E
2 E
3 E
3 E
3 E
3 S
4 S
3 S
4 S
4 S
4 S
3 E
4 E
4 E
4 E
4 E
4 E
4 E | {"url":"http://bme.umn.edu/undergrad/handbook/emphasis/transport.html","timestamp":"2014-04-17T13:28:06Z","content_type":null,"content_length":"26474","record_id":"<urn:uuid:df984470-e908-41ba-8f92-f71e54643fb2>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
Please help with Rationals
April 20th 2008, 05:17 PM
Please help with Rationals
Hi, I have this function, g(x)= (x^3-x)/(x^3-4x).
Now I need help with the following...
- How to find the zeros for f?
- How to find f(-x)?
- How to find point of discontinuity?
And also, would finding the asymptotes, slant asymptote, and point of discontinuity be easier if I simplify this rational first?:
I can simplify the denominator as (x+1)(x-6) but that's it...
Is the first step to simplifying the numerator... Factoring out x?
I'd appreciate any help, thanks so much!
April 20th 2008, 05:24 PM
Hi, I have this function, g(x)= (x^3-x)/(x^3-4x).
Now I need help with the following...
- How to find the zeros for f?
- How to find f(-x)?
- How to find point of discontinuity?
And also, would finding the asymptotes, slant asymptote, and point of discontinuity be easier if I simplify this rational first?:
I can simplify the denominator as (x+1)(x-6) but that's it...
Is the first step to simplifying the numerator... Factoring out x?
I'd appreciate any help, thanks so much!
A rational functions zeros are only in the numerator...so you need to solve $x^3-x=0\Rightarrow{x(x^2-1)=0}$..so x=0,x=-1,x=1
since this is odd $f(-x)=-f(x)$
solve the denominator for 0 we get $x(x^2-4)=0$ so x=0,x=2, x=-2 all points of discontinuity
just polynomial divide...DO NOT simplify first to find slant asymptotes | {"url":"http://mathhelpforum.com/algebra/35285-please-help-rationals-print.html","timestamp":"2014-04-18T00:33:03Z","content_type":null,"content_length":"5700","record_id":"<urn:uuid:3a209160-a856-47f9-87fb-6816d2be2129>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00177-ip-10-147-4-33.ec2.internal.warc.gz"} |
Exponents and Radicals
By M Bourne
√(a + b)^10
Exponents are a very important part of algebra. An exponent is just a convenient way of writing repeated multiplications of the same number.
Radicals involve the use of the radical sign, `sqrt(\ )`. Sometimes these are called surds.
If you learn the rules for exponents and radicals, then your enjoyment of mathematics will surely increase! (Yeah, right... :-)
In this Chapter
1. Simplifying Expressions with Integral Exponents - defines exponents and shows how to use them when multiplying or dividing in algebra.
Related Sections in
"Interactive Mathematics"
The Binomial Theorem, which shows how to expand expressions like (x + y)^6.
Exponential and Logarithmic Functions, which talks about applications of expressions like `10^x` and log x.
Differentiation of polynomials, where one of the steps involves exponents.
Roots and radicals, from the basic algebra section.
2. Fractional Exponents - shows how an fractional exponent means a root of a number
3. Simplest Radical Form - this technique can be useful when simplifying algebra
4. Addition and Subtraction of Radicals - shows how to add or subtract expressions involving the square root sign
5. Multiplication and Division of Radicals (Rationalizing the Denominator) - while this method is not as important as it was in the pre-calculator days, it is still a useful technique
6. Equations with Radicals - the fun starts!
The chapter begins with 1. Simplifying Expressions with Integral Exponents »
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Algebra Lessons on DVD
Easy to understand algebra lessons on DVD. See samples before you commit.
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st: help- overlapped sample splittings
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st: help- overlapped sample splittings
From Albert Robinson <robinson9837@hotmail.com>
To statalist@hsphsun2.harvard.edu
Subject st: help- overlapped sample splittings
Date Mon, 22 Oct 2007 13:57:15 +0400
Fellow listers
I am running a panel data model with Stata and my database is an incomplete panel of 997 firms and some variables, max range: 1980 - 2000 (but some firms have a lower range).
I need your help to make some calcutation on a variable, hereafter called "mvalue" (it's a ratio). First of all, for each year, we can find negative or positive or zero values of "mvalue". What I need to build:
* by considering the distribution of this variable, I would identify as "LOW mvalue" those firms that are above the 25th percentile of the distribution of the negative values. Note that this selection should produce a dummy-column where - e.g. - the same firm could have zero and one randomly distributed or only zero or one. Vice versa, HIGH mvalue firms are above the 25th percentile of the distribution of firms experiencing positive values of the values.
** Finally, created these two dummies (two columns), for a firm to be identifies as a PERSISTENTLY LOW (HIGH) mvalue, I require that it be identified as a LOW (HIGH) mvalue firm for at least two consecutive periods. This would produce two other columns, for PERSISTENTLY LOW and PERSISTENTLY HIGH mvalue.
I am new to the list, so I apologize in advance if the answer to these questions was easy. I would appreciate your help and I would be very, very grateful I somebody could write me the Stata commands (with or without an example) useful for sample splittings of this kind.
Albert Robinson
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Proffs - various integration proffs
May 8th 2010, 12:10 PM #1
May 2010
Proffs - various integration proffs
Problem 1 :
Let f:[1,∞[ →R be a positive continuous decreasing function. Show if m,n ∈N and m<n that the integral
∫ f(x) dk from (m+1) to (n+1) ≤∑ f(k) (sum from k=1 to n) -∑ (sum from k=1 to m) ≤f(m)+∫f(x)dk (with integral over the interval from m to n)
Problem 2 :
From problem 1 prove if ∫ f(x) dk on the interval from 1 to infinity is convergent that the integral ∫ f(x) dk from (m+1) to (infinity) ≤∑ f(k) (sum from k=1 to infinity) -∑ (sum from k=1 to m)
≤f(m)+∫f(x)dk (with integral over the interval from m to infinity)
Problem 3 :
Let snm = =∑ 1/k (sum from k=1 to nm) - (sum from k=1 to n times m)
Show from problem 1 if sn=∑ 1/k (sum from k=1 to n) for all n ∈N that the sum difference limit : lim (snm - sm) = ln(n)
Problem 4 :
Proff that the functional serie fn given by fn : R→ R for n=1,2,3 ..... given by
fn(x) = ____1________
1+ exp((n(2-x))
is pointwise convergent and find the limiting function.
Is the functional serie also uniform convergent on R ? Argue for your result.
Help proving these four problems is urgently needed.
Problem 1 :
Let f:[1,∞[ →R be a positive continuous decreasing function. Show if m,n ∈N and m<n that the integral
∫ f(x) dk from (m+1) to (n+1) ≤∑ f(k) (sum from k=1 to n) -∑ (sum from k=1 to m) ≤f(m)+∫f(x)dk (with integral over the interval from m to n)
Problem 2 :
From problem 1 prove if ∫ f(x) dk on the interval from 1 to infinity is convergent that the integral ∫ f(x) dk from (m+1) to (infinity) ≤∑ f(k) (sum from k=1 to infinity) -∑ (sum from k=1 to m)
≤f(m)+∫f(x)dk (with integral over the interval from m to infinity)
Problem 3 :
Let snm = =∑ 1/k (sum from k=1 to nm) - (sum from k=1 to n times m)
Show from problem 1 if sn=∑ 1/k (sum from k=1 to n) for all n ∈N that the sum difference limit : lim (snm - sm) = ln(n)
Problem 4 :
Proff that the functional serie fn given by fn : R→ R for n=1,2,3 ..... given by
fn(x) = ____1________
1+ exp((n(2-x))
is pointwise convergent and find the limiting function.
Is the functional serie also uniform convergent on R ? Argue for your result.
Help proving these four problems is urgently needed.
Too many questions, too little self work shown: what have you done so far? Where are you stuck? These are questions at least from advanced calculus I or calculus II, so you must know something
before attacking this stuff: show us!
May 8th 2010, 05:53 PM #2
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Birch-Tate conjecture
From Encyclopedia of Mathematics
Let field Algebraic number). The Milnor Abelian group of finite order.
Let Dedekind zeta-function of
Specifically, let Galois group of the cyclotomic extension over ). Then
A numerical example is as follows. For
What is known for totally real number fields
By work on the main conjecture of Iwasawa theory [a6], the Birch–Tate conjecture was confirmed up to
Subsequently, [a7], the Birch–Tate conjecture was confirmed up to
Moreover, [a7] (see the footnote on page 499) together with [a4], also the
By the above, all that is left to be considered is the [a3].
In addition, explicit examples of families of non-Abelian extensions [a1], [a2].
The Birch–Tate conjecture is related to the Lichtenbaum conjectures [a5] for totally real number fields
[a1] P.E. Conner, J. Hurrelbrink, "Class number parity" , Pure Math. , 8 , World Sci. (1988)
[a2] J. Hurrelbrink, "Class numbers, units, and Algebraic , NATO ASI Ser. C , 279 , Kluwer Acad. Publ. (1989) pp. 87–102
[a3] M. Kolster, "The structure of the Comment. Math. Helv. , 61 (1986) pp. 376–388
[a4] M. Kolster, "A relation between the Canad. Math. Bull. , 32 : 2 (1989) pp. 248–251
[a5] S. Lichtenbaum, "Values of zeta functions, étale cohomology, and algebraic Algebraic , Lecture Notes in Mathematics , 342 , Springer (1973) pp. 489–501
[a6] B. Mazur, A. Wiles, "Class fields of abelian extensions of Invent. Math. , 76 (1984) pp. 179–330
[a7] A. Wiles, "The Iwasawa conjecture for totally real fields" Ann. of Math. , 131 (1990) pp. 493–540
How to Cite This Entry:
Birch–Tate conjecture. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Birch%E2%80%93Tate_conjecture&oldid=22123 | {"url":"http://www.encyclopediaofmath.org/index.php/Birch%e2%80%93Tate_conjecture","timestamp":"2014-04-20T15:50:45Z","content_type":null,"content_length":"24655","record_id":"<urn:uuid:b3f51c8f-6966-40c5-a37b-971ee8ff8ce0>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00234-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: Kronecker symbol
Justin C. Walker on Mon, 25 Aug 2003 20:11:06 -0700
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I'm not sure that your assumptions are correct.
Starting with Legendre, the symbol is defined for a, p, with p prime (and (a|p)=+/- 1); Jacobi extends this to (a|n) for general odd n, making it multiplicative in n. Kronecker then extends Jacobi to
n=2. I'd think that you'd end up with something like (3|4)=(3|2)^2=1.
Once you extend Legendre, I think you are out of the "a is a square mod b" realm; for example, (-7|15)=1, but -7 (== 8) is not a square mod 15.
I can't answer the second question, and a check of the source offers no hints. FWIW, I believe that quadratic (== real?) characters all have the form (a|b) (due to Dirichlet, I think).
On Monday, August 25, 2003, at 12:06 PM, Jack Fearnley wrote:
I have been having some trouble with kronecker(x,y) which is described in the
documentation as a generalization of the Legendre symbol (x|y). I was
expecting kronecker to behave like a Dirichlet character. In particular, I
expected kronecker(3,4) to equal -1 whereas it computes as 1.
I have two questions:
1) What is the precise definition of kronecker in Pari?
2) Is there a quick and easy way to generate quadratic Dirichlet characters in
2a) What about higher order Dirichlet characters?
Best Regards,
Jack Fearnley
Justin C. Walker, Curmudgeon-At-Large *
Institute for General Semantics | It's not whether you win or lose...
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Rack and Pinion calculation help
May 7th 2012, 03:14 AM #1
May 2012
My maths is rusty, but I know the basics of gear ratios, enough to say a rack and pinion works of a 1:1 ratio.
See attachment for the overview of what I am hoping to achieve.
I need the rack to move a total of 24 mm over three 1/2 turns of the black pinion gear.
The black pinion can be no smaller than 10 mm radius. It can increase in size.
The red gear can be no smaller than a 5 mm radius (due to machining tolerance), but can increase in size.
Number of teeth is unknown, but ideally a smooth operation would exist. So far, I have been basing my calculations on having 20 teeth.
This should be simple, but can't get my head around it. If anyone knows, your help will be highly appreciated! I'll keep at it. Cheers
Re: Rack and Pinion calculation help
are you sure you have the right design here?
i think you need the red gear and the black one on the same shaft, with the rack touching the red gear.
Re: Rack and Pinion calculation help
If the red and black were on the same shaft, they would both rotate in the same direction, forcing the rack down. Would that serve any purpose when the black could be the direct drive? Basically
the black pinion has a key hole inserted inside, so the red could not mechanically fit onto the same shaft as the black. Sorry, I should have mentioned that before in my query.
Re: Rack and Pinion calculation help
the black cant be the direct drive because you cant make it small enough to get 8mm per half turn. your design in post #1 wont work because the diameter of the red wheel has no impact on how far
the rack moves per half turn of the black wheel.
If you need the thing to go up at the required speed then do what i said in #2 and insert another gear between the red one and the rack.
Last edited by SpringFan25; May 7th 2012 at 04:05 AM.
Re: Rack and Pinion calculation help
Okay, I'll try that. Cheers
May 7th 2012, 03:42 AM #2
MHF Contributor
May 2010
May 7th 2012, 03:49 AM #3
May 2012
May 7th 2012, 04:03 AM #4
MHF Contributor
May 2010
May 7th 2012, 04:06 AM #5
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Math Forum - Problems Library - Discrete Math, Patterns and Recursion
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Discrete Math: Patterns and Recursion
Working with a simpler example and then observing patterns is key to solving many of the following problems. Some of these problems, like the Forever Green Nursery problem, are based on a
recursive pattern. Recursive patterns occur when the outcome of a step is dependent on the outcome of a previous step.
For background information elsewhere on our site, explore the High School Discrete Math area of the Ask Dr. Math archives. To find relevant sites on the Web, browse and search Discrete
Mathematics in our Internet Mathematics Library.
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Advogato: Blog for Bram
Today I'll be writing about computer strategies for the classic game Roshambo, also known as paper scissors stone.
But ... but ... There's no strategy in Roshambo!
While it is true that playing random can reliably do dead average, in any tournament everyone will be trying to win, hence everyone will have bias, and strategies which do better than average against
a wide range of opponents will rise to the top.
I've spent far more time than is reasonable reading about the computer Roshambo tournaments, and will present a distilled explanation of the strategies here.
Henny exemplifies the simplest strategy which works well in practice. It picks a random move which the opponent played previously, and plays the response which beats it. If the opponent has any bias
toward playing the same move over time, Henny will win. Henny is also very defensive - optimal play against it will only get an edge very slowly over time, and even that might get swallowed by noise.
Markov Chains
Another winning in practice strategy is to use markov chains - look at the last n moves your opponent made, and predict that they'll continue playing as they have most frequently in the past.
Considerable variation is possible in length of chains, which brings us to...
Iocaine Powder
The competitor Iocaine Powder magically combines the best qualities of Henny and markov chaining by looking for the longest match between the moves your opponent just played and any sequence they've
played in the past, and assuming they'll play the same next move again. If the opponent has a bias based on the last n moves for an n, this strategy will essentially pick a random time they played
the same last n moves they just did, which is the defensive strategy used by Henny.
Ironically, this algorithm is completely deterministic.
Note that it's slightly better when there are equal length matches to go with the first matching sequence rather than the last matching sequence, to perform well when an opponent plays a string of
paper, then a scissors, then a string of paper again.
It's possible to implement this technique very efficiently using trees, which strangely Iocaine Powder doesn't do (MemBot does, though).
But this strategy still has a major weakness, and Iocaine Powder has another trick up its sleeve...
Sicilian Reasoning
A braindead, fixed sequence entry named DeBruijnize nearly cleaned up in the first tournament. DeBruijnize sequences are biased against sequences they've done in the past, rather than in favor of
them. A meta-strategy called sicilian reasoning beats it nicely. Compute how well you would have done to date predicting the opponents move straightforwardly, then always shifting it up one (paper->
scissors, scissors->rock, rock->paper), then down one, then the same three predicting instead your own side. Play whichever one would have the highest score so far.
Sicilian reasoning cleanly defeats not only DeBruijnize, but all manner of cheap variants on it.
Sword and Shield
This is my own idea, which is untested, but seems reasonable.
Generate predictions via whatever method (pre-sicilian reasoning) for both your move and the opponent's move, then play a move based on the following table. The top row is your predicted move, the
left row is the opponent's (t = stone) -
p s t
p | s p s
s | t t s
t | t p p
This algorithm plays defensively by predicting both sides simultaneously, and hence may beat any opponent which bases each specific move on a prediction for only one side. It has nine sicilian
reasoning variants, found by offseting the prediction for either side one of the three possible ways.
Wimping Out
A tempting strategy is to make your bot 'wimp out' and start playing randomly if it isn't doing well. Tournaments play two programs against each other many times with no persistent information
between runs to keep this strategy from being effective.
Optimizing for Score
The winning entry, Greenberg, looks at historical performance if its history buffer was various sizes, and has exponential falloff of their performance. This makes it score very well against weaker
opponents, but makes it more vulnerable to better opponents. Score, match wins, and worst-case lossage are all subtly different things to optimize for.
The Future
Sadly, there appear to be no plans for future computer Roshambo tournaments. It would be interesting to see how strategies continue to evolve. | {"url":"http://www.advogato.org/person/Bram/diary/10.html","timestamp":"2014-04-19T22:11:43Z","content_type":null,"content_length":"8342","record_id":"<urn:uuid:9d3b9e5a-7d06-4d55-8069-6e403000caa0>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00181-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: discretizing laplacian
Replies: 2 Last Post: Mar 1, 2013 6:02 AM
Messages: [ Previous | Next ]
Re: discretizing laplacian
Posted: Feb 26, 2013 8:59 AM
Sandeep Kumar <searchsandy1712@gmail.com> writes:
>Can anybody please tell me how to discretize Laplacian. I have -
>laplacian(Un^2.Un+1), which is laplacian of product of Un^2 and Un+1.
>where Un is U(j,n) and Un+1 is U(j,n+1).
> Thanks in advance.
this makes no sense for me: with U already discretized, what should
laplacian of this mean?
If I remember right you one had laplacian(u^3) plus Crank-Nicholson,
that means you need
where of course laplacian is w.r.t x and -fortunately- x is onedimensional.
method 1:
(d/dx)^2(u(x,t_n)^3)(x=x_i) = 3u(i,n)^2(u(i+1,n)-2u(i,n)+u(i-1,n))/h^2
(d/dx)^2(u(x,t_n)^3) =
(1/h)^2 ( ((u(i+1,n)^2+u(i,n)^2)/2)*(u(i+1,n)-u(i,n))
- ((u(i,n)^2+u(i-1,n)^2)/2)*(u(i,n)-u(i-1,n)) )
Date Subject Author
2/26/13 Re: discretizing laplacian Peter Spellucci
3/1/13 Re: discretizing laplacian AMX | {"url":"http://mathforum.org/kb/message.jspa?messageID=8420201","timestamp":"2014-04-19T23:02:06Z","content_type":null,"content_length":"17907","record_id":"<urn:uuid:aeb073f0-9c91-4e89-bf5b-1ab280f04d26>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00455-ip-10-147-4-33.ec2.internal.warc.gz"} |
Sequential Diagnosis by Abstraction
S. A. Siddiqi, J. Huang
When a system behaves abnormally, sequential diagnosis takes a sequence of measurements of the system until the faults causing the abnormality are identified, and the goal is to reduce the diagnostic
cost, defined here as the number of measurements. To propose measurement points, previous work employs a heuristic based on reducing the entropy over a computed set of diagnoses. This approach
generally has good performance in terms of diagnostic cost, but can fail to diagnose large systems when the set of diagnoses is too large. Focusing on a smaller set of probable diagnoses scales the
approach but generally leads to increased average diagnostic costs. In this paper, we propose a new diagnostic framework employing four new techniques, which scales to much larger systems with good
performance in terms of diagnostic cost. First, we propose a new heuristic for measurement point selection that can be computed efficiently, without requiring the set of diagnoses, once the system is
modeled as a Bayesian network and compiled into a logical form known as d-DNNF. Second, we extend hierarchical diagnosis, a technique based on system abstraction from our previous work, to handle
probabilities so that it can be applied to sequential diagnosis to allow larger systems to be diagnosed. Third, for the largest systems where even hierarchical diagnosis fails, we propose a novel
method that converts the system into one that has a smaller abstraction and whose diagnoses form a superset of those of the original system; the new system can then be diagnosed and the result mapped
back to the original system. Finally, we propose a novel cost estimation function which can be used to choose an abstraction of the system that is more likely to provide optimal average cost.
Experiments with ISCAS-85 benchmark circuits indicate that our approach scales to all circuits in the suite except one that has a flat structure not susceptible to useful abstraction.
This page is copyrighted by AAAI. All rights reserved. Your use of this site constitutes acceptance of all of AAAI's terms and conditions and privacy policy. | {"url":"http://aaai.org/Library/JAIR/Vol41/jair41-010.php","timestamp":"2014-04-19T13:25:04Z","content_type":null,"content_length":"3820","record_id":"<urn:uuid:219df883-5f6b-4ed7-aca5-460dc0f8580d>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00083-ip-10-147-4-33.ec2.internal.warc.gz"} |
[R] Testing the equality of correlations
[R] Testing the equality of correlations
Viechtbauer Wolfgang (STAT) Wolfgang.Viechtbauer at STAT.unimaas.nl
Thu Sep 28 10:50:21 CEST 2006
It's more complicated than that, since Phi(X1,X2), Phi(X1,X3), and Phi(X1,X4) are dependent. Take a look at:
Olkin, I., & Finn, J. D. (1990). Testing correlated correlations. Psychological Bulletin, 108(2), 330-333.
Meng, X., Rosenthal, R., & Rubin, D. B. (1992). Comparing correlated correlation coefficients. Psychological Bulletin, 111(1), 172-175.
You will probably have to implement these tests yourself.
Wolfgang Viechtbauer
Department of Methodology and Statistics
University of Maastricht, The Netherlands
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-
> bounces at stat.math.ethz.ch] On Behalf Of Paul Hewson
> Sent: Wednesday, September 27, 2006 17:40
> To: Marc Bernard; r-help at stat.math.ethz.ch
> Subject: Re: [R] Testing the equality of correlations
> Off the top my head (i.e. this could all be horribly wrong), I think
> Anderson gave an asymptotic version for such a test, whereby under the
> null hypothesis, the difference between Fisher's z for each sample, z1 -
> z2, is normal with zero mean.
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Marc Bernard
> Sent: 27 September 2006 14:42
> To: r-help at stat.math.ethz.ch
> Subject: [R] Testing the equality of correlations
> Dear All,
> I wonder if there is any implemented statistical test in R to test
> the equality between many correlations. As an example, let X1, X2, X3
> X4 be four random variables. let
> Phi(X1,X2) , Phi(X1,X3) and Phi(X1,X4) be the corresponding
> correlations.
> How to test Phi(X1,X2) = Phi(X1,X3) = P(X1,X4)?
> Many thanks in advance,
> Bernard
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the Golden Ratio
Spirals and the Golden Ratio
Golden Spirals and Fibonacci Spirals
Fibonacci numbers and Phi are related to spiral growth in nature
If you sum the squares of any series of Fibonacci numbers, they will equal the last Fibonacci number used in the series times the next Fibonacci number. This property results in the Fibonacci
spiral, based on the following progression and properties of the Fibonacci series:
1^^2 + 1^^2 + 2^^2 + 3^^2 + 5^^2 = 5 x 8
1^^2 + 1^^2 + . . . + F(n)^^2 = F(n) x F(n+1)
A Golden spiral is very similar to the Fibonacci spiral but is based on a series of identically proportioned golden rectangles, each having a golden ratio of 1.618 of the length of the long side to
that of the short side of the rectangle:
The Fibonacci spiral gets closer and closer to a Golden Spiral as it increases in size because of the ratio of each number in the Fibonacci series to the one before it converges on Phi, 1.618, as the
series progresses (e.g., 1, 1, 2, 3, 5, 8 and 13 produce ratios of 1, 2, 1.5, 1.67, 1.6 and 1.625, respectively)
Fibonacci spirals and Golden spirals appear in nature, but not every spiral in nature is related to Fibonacci numbers or Phi. Most spirals in nature are equiangular spirals, and Fibonacci and Golden
spirals are special cases of the broader class of Equiangular spirals. An Equiangular spiral itself is a special type of spiral with unique mathematical properties in which the size of the spiral
increases but its shape remains the same with each successive rotation of its curve. The curve of an equiangular spiral has a constant angle between a line from origin to any point on the curve and
the tangent at that point, hence its name. In nature, equiangular spirals occur simply because they result in the forces that create the spiral are in equilibrium, and are often seen in non-living
examples such as spiral arms of galaxies and the spirals of hurricanes. Fibonacci spirals, Golden spirals and golden ratio-based spirals generally appear in living organisms, as illustrated below:
Golden spiral in human ear
Golden ratio proportions in successive spirals of a sea shell
The Nautilus shell spiral is not a Golden spiral but often still has Golden Ratio proportions
The nautilus shell is often shown as an illustration of the golden ratio in nature, but the spiral of a nautilus shell is NOT a golden spiral, as illustrated below. The golden spiral overlay is
provided by PhiMatrix golden ratio software:
Nautilus shell spiral compared to a Golden Spiral
The Nautilus spiral, however, while not a Golden spiral, often displays proportions its dimensions that are close to a golden ratio, appearing in successive rotations of the shell as the Nautilus
grows. As with all living organisms, there is variation in the dimensions of individuals, so the appearance of the golden ratio is not universal.
See also other examples and explanations of the golden ratio in the nautilus spiral.
Alternate spirals in plants occur in Fibonacci numbers
Plants illustrate the Fibonacci series in the numbers of leaves, the arrangement of leaves around the stem and in the positioning of leaves, sections and seeds.
Here a sunflower seed illustrates this principal as the number of clockwise spirals is 55 (marked in red, with every tenth one in white) and the number of counterclockwise spirals is 89 (marked in
green, with every tenth one in white.)
Pinecones and pineapples illustrate similar spirals of successive Fibonacci numbers, with the example below showing the alternating pattern of 8 and 13 spirals in a pine cone.
Leave a Comment
{ 6 comments… read them below or add one } | {"url":"http://www.goldennumber.net/spirals/","timestamp":"2014-04-18T15:39:05Z","content_type":null,"content_length":"59101","record_id":"<urn:uuid:70faba95-f5b3-4892-b51d-478684e14a45>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00147-ip-10-147-4-33.ec2.internal.warc.gz"} |
Post a reply
well anyways back to my original question, I got some help from a friend and they said to start by making the (3/2)(x-1) into one term, and then multiply the sqrt by something that will allow you to
combine it with the expanded fraction, then go from there. And they said that the final solution will end up with a square root inside of a square root, and that there wasnt a way that they knew to
get around it. | {"url":"http://www.mathisfunforum.com/post.php?tid=588&qid=6260","timestamp":"2014-04-20T16:34:21Z","content_type":null,"content_length":"21306","record_id":"<urn:uuid:332b732b-c313-4eb6-b7e3-3573d3a63017>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00619-ip-10-147-4-33.ec2.internal.warc.gz"} |
Permutations and Combinations
August 5th 2013, 10:12 PM #1
Jan 2013
Permutations and Combinations
Q1: The number of ways 8 people can be arranged in a straight line if two people insist on being separated is:
A) 15,120
B) 3,600
C) 30,240
D) 7,200
E) 5,400
No matter what I try I can't obtain any of these possible answers. I'm thinking it's something to do with the wording of the question, but I'm not sure what exactly.
Q2: A cricket team takes 11 players to a game and there are 14 players from which to choose; 1 wicket keeper, 7 batsmen and 6 bowlers.
a) How many different teams are possible if there are no restrictions?
I did: ^14C[11 ]= 364, but I'm unsure.
b) How many different teams are possible if the wicket keeper must be included?
I did ^13C[11 ]= 78, but I'm unsure.
c) How many different teams are possible if the team contains the wicket keeper, 6 batsmen and 4 bowlers?
Thanks in advance.
Re: Permutations and Combinations
Hello, Fratricide!
Q1: the number of ways 8 people can be arranged in a straight line
. . . if two people insist on being separated is:
. . (A) 15,120 . . (B) 3,600 . . (C) 30,240 . . (D) 7,200 . . (E) 5,400
$\text{The eight people are: }\:\{A,B,C,D,E,F,G,H\}$
$\text{Let }A\text{ and }B\text{ be the two unfriendly people.}$
$\text{With no restrictions, there are }\,8! = 40,\!320\text{ arrangements.}$
$\text{Now consider the arrangements in which }A\text{ and }B\:are\text{ adjacent.}$
$\text{Duct-tape }A\text{ and }B\text{ together.}$
$\text{Then we have 7 "people" to arrange: }\,\left\{\boxed{AB}, C, D, E, F, G, H\right\}$
$\text{There are: }\,7!\text{ arrangements.}$
$\text{But }A\text{ and }B\text{ could be taped like this: }\,\boxed{BA}.$
$\text{Hence, there are: }\,2\cdot7! \,=\,10,\!080\text{ ways in which }A\text{ and }B\text{ are adjacent.}$
$\text{Therefore, the answer is: }\:40,\!320 - 10,\!080 \:=\:30,\!240\:\text{ . . . answer (C)}$
Re: Permutations and Combinations
I think you are right in Q2(a)
(b) 13C10( because the wicket keeper must be include)
(c) 7C6x6C4
Re: Permutations and Combinations
Re: Permutations and Combinations
well I really not good on explanation, because my English is sucked, but I'll try..
you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers(
from 6). Therefore, 7C6x6C4
Re: Permutations and Combinations
well I really not good on explanation, because my English is sucked, but I'll try..
you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers(
from 6). Therefore, 7C6x6C4
Perfect, thanks.
August 5th 2013, 10:52 PM #2
Super Member
May 2006
Lexington, MA (USA)
August 6th 2013, 02:58 AM #3
Junior Member
Apr 2013
August 6th 2013, 09:52 PM #4
Jan 2013
August 7th 2013, 01:18 AM #5
Junior Member
Apr 2013
August 7th 2013, 02:57 AM #6
Jan 2013 | {"url":"http://mathhelpforum.com/math-topics/221056-permutations-combinations.html","timestamp":"2014-04-18T03:17:09Z","content_type":null,"content_length":"50933","record_id":"<urn:uuid:87e40c2a-eba3-42c3-b10e-78c7303de265>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00655-ip-10-147-4-33.ec2.internal.warc.gz"} |
expectation-propagation and ABC
August 23, 2011
By xi'an
“It seems quite absurd to reject an EP-based approach, if the only alternative is an ABC approach based on summary statistics, which introduces a bias which seems both larger (according to our
numerical examples) and more arbitrary, in the sense that in real-world applications one has little intuition and even less mathematical guidance on to why p(θ|s(y)) should be close to p(θ|y) for
a given set of summary statistics s.”
Simon Barthelmé and Nicolas Chopin posted a recent arXiv paper on Expectation-Propagation for Summary-Less, Likelihood-Free Inference. They sell expectation-propagation as quick and dirty version of
ABC, avoiding the selection of summary statistics by using the constraint
$||y_i-y^\star_i||\le \epsilon$
on each component of the simulated pseudo-data vector y* being the actual data. Expectation-propagation is a variational technique [Simon and Nicolas are quite fond of!] and it consists in replacing
the target with the “closest” member from an exponential family, like the Gaussian distribution. The expectation-propagation approximation is found by including a single “observation” at a time,
using the other approximations for the prior, and finding the best Gaussian in this pseudo-model. In addition, expectation-propagation provides an approximation of the evidence. In the “
likelihood-free” setting (I do not like this term because we are dealing with a specific well-defined likelihood, we simply cannot compute it!), this means computing empirical mean and empirical
variance, one observation at a time, under the above tolerance constraint.
Unless I am confused, the expectation-propagation approximation to the posterior distribution is a [sequentially updated] Gaussian distribution, which means that it will only be appropriate in
cases where the posterior distribution is approximately Gaussian. Since the three examples processed in the paper are of this kind, e.g. the above reproduction, I wonder at the performances of
the expectation-propagation method in less smooth cases, such as ridge-like or multimodal posteriors. The authors mention two limitations: “First, it [EP] assumes a Gaussian prior; and second, it
relies on a particular factorisation of the likelihood, which makes it possible to simulate sequentially the datapoints“, but those seem negligible wrt my above comment. I thus remain unconvinced by
the concluding sentence quoted above. (The current approach to ABC is to consider p(θ|s(y)) as a target per se, not as an approximation to p(θ|y).) Nonetheless, expectation-propagation constitutes a
quick approximation method that can always used as a reference against other approximations.
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Meeting Details
For more information about this meeting, contact Victor Nistor, Stephanie Zerby, Mark Levi, Jinchao Xu, Ludmil Zikatanov.
Title: Computing with Singular and Nearly Singular Integrals
Seminar: Computational and Applied Mathematics Colloquium
Speaker: Tom Beale, Mathematics, Duke University
We will describe a relatively simple, direct approach to computing a singular or nearly singular integral, such as a harmonic function given by a single or double layer potential on a smooth, closed
curve in 2D or a surface in 3D. Integral formulations are used especially for Stokes flow (viscosity-dominated fluid flow)and in electromagnetics. The present approach can be useful for moving
interfaces since the representation of the interface requires less work than boundary elements. The value of the integral is found by a standard quadrature, with the singularity replaced by a
regularized version. Correction terms are then added for the errors due to regularization and discretization. These corrections are found by local analysis near the singularity. The accurate
evaluation of a layer potential at a point near the curve or surface on which it is defined is not routine, since the integral is nearly singular. For a surface in 3D, integrals are computed in
overlapping coordinate grids on the surface. A quadrature technique of J. Wilson allows this to be done without explicit knowledge of the coordinate charts, thus making the approach more practical.
For a boundary value problem, an integral equation can be solved for the density of the needed potential. For a fluid interface, e.g. in Stokes flow, the velocity or pressure can be written in terms
of layer potentials. Recent work with W. Ying in 2D shows that the method can accurately handle boundaries which are close to each other, such as two drops merging, and work in 3D is in progress.
Collaborators include M.-C. Lai, A. Layton, S. Tlupova, J. Wilson, and W. Ying.
Room Reservation Information
Room Number: MB106
Date: 03 / 15 / 2013
Time: 03:35pm - 04:25pm | {"url":"http://www.math.psu.edu/seminars/meeting.php?id=15341","timestamp":"2014-04-17T16:04:50Z","content_type":null,"content_length":"4939","record_id":"<urn:uuid:28fd4946-8593-43c6-8cec-d63637966d07>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00422-ip-10-147-4-33.ec2.internal.warc.gz"} |
Parallel In-Place Merge
In my article on parallel merge, I developed and optimized a generic parallel merge algorithm. It utilized multiple CPU cores and scaled well in performance. The algorithm was stable, leading
directly to Parallel Merge Sort. However, it was not an in-place implementation. In this article, I develop an in-place parallel merge, which enables in-place parallel merge sort.
The STL implements sequential merge and in-place merge algorithms, as merge() and inplace_merge(), which run on a single CPU core. The merge algorithm is O(n), leading to O(nlgn) merge sort, while
the in-place merge is O(nlgn), leading to O(n(lgn)^2) in-place merge sort. Because merge is faster than in-place merge, but requires extra memory, STL favors using merge whenever memory is available,
copying the result back to the input array (under the hood).
Revisit Not-In-Place
The divide-and-conquer merge algorithm described in Parallel Merge, which is not-in-place, is illustrated in Figure 1. At each step, this algorithm moves a single element X from the source array T to
the destination array A, as follows.
Figure 1. A merge algorithm that is not in place.
The two input sub-arrays of T are from [p[1] to r[1]] and from [p[2] to r[2]]. The output is a single sub-array of A from [p3 to r3]. The divide step is done by choosing the middle element within the
larger of the two input sub-arrays—at index q[1]. The value at this index is then used as the partition element to split the other input sub-array into two sections - less than X and greater than or
equal to X. The partition value X (at index q[1] of T) is copied to array A at index q3. The conquer step recursively merges the two portions that are smaller than or equal to X— indicated by light
gray boxes and light arrows. It also recursively merges the two portions that are larger than or equal to X—indicated by darker gray boxes and dark arrows.
The algorithm proceeds recursively until the termination condition —when the shortest of the two input sub-arrays has no elements. At each step, the algorithm reduces the size of the array by one
element. One merge is split into two smaller merges, with the output element placed in between. Each of the two smaller merges will contain at least N/4 elements, since the left input array is split
in half.
This algorithm is O(n) and is stable. It was shown not to be performance competitive in its pure form. However, performance became equivalent to other merges in a hybrid form, which utilized a simple
merge to terminate recursion early. The hybrid version enabled parallelism by using of divide-and-conquer method, scaled well across multiple cores, outperforming STL merge by over 5X on quad-core
CPU, being limited by memory bandwidth. It was also used as the core of Parallel Merge Sort, which scaled well across multiple cores, utilizing a further hybrid approach to gain more speed, and
providing a high degree of parallelism, Θ(n/lg^2n). | {"url":"http://www.drdobbs.com/parallel/parallel-in-place-merge/240008783?cid=SBX_ddj_related_mostpopular_default_tools&itc=SBX_ddj_related_mostpopular_default_tools","timestamp":"2014-04-19T18:12:37Z","content_type":null,"content_length":"93948","record_id":"<urn:uuid:babc5396-ea51-45b7-ac8d-378c09c13f04>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00052-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Hunt for Addi(c)tive Monster
This is another spurious post about mathematics that happened instead of something more useful. Let's talk about one of the most common mathematical objects: a
that maps
real numbers
additive iff
for any real
This is a natural and simple condition. Well-known examples of additive functions are provided by
is called
a slope
. Are there other, non-linear additive functions? And if so, how do they look like? A bit of thinking convinces one that a non-linear additive function is not trivial to find. In fact, as we shall
show below, should such a function exist, it would exhibit extremely weird features. Hence, we shall call a non-linear additive function
a monster
. In the following, some properties that a monster function has to possess are investigated, until a monster is cornered either out of existence or into an example. Out of misplaced spite we shall
in some proofs.
First, some trivial remarks about the collection of all additive functions (which will be denoted
vector space
over field
zero vector
. This vector space is a sub-space of a vector space of all functions from
Unfortunately, composition is not compatible with scalars (
algebra over a field
Looking at an individual additive function
For any real number
and for any natural number
that is, restriction of
to any subset
is linear when restricted to the natural numbers. Specifically,
Combining these results, for any rational number
That is,
is linear when restricted to any subset of the form
Notice that we just proved that composition of linear functions is compatible with multiplication on rational scalars (see above), so that
Having briefly looked over the landscape of
, it's even... linear. Which of these it enjoys together with a monster? It's intuitively very unlikely that an additive, but non-linear function might happen to be differentiable, so let's start
with checking continuity.
Statement 0. If an additive function
, then
That is, an additive function is linear everywhere it is continuous. This means that a monster must be discontinuous at least in one point. Note that linear functions are precisely everywhere
continuous additive functions. Can a monster function be discontinuous in a single point? Or, even, can it be continuous
? It is easy to show that property of additivity constrains structure of sets on which function is continuous severely:
Statement 1. If an additive function
, it is linear.
Take an arbitrary non-zero point y. By statement 0, x,
For any natural n, take
and by the sandwich theorem, yx' above and
Taking the (obviously existing) limits of both sides of this inequality, one gets
The case of y being 0 is trivial.
Oh. A monster function cannot be continuous even at a single point—it is discontinuous everywhere. Additive functions are divided into two separated classes: nice, everywhere continuous linear
functions and unseemly, everywhere discontinuous monsters. (We still don't know whether the latter class is populated, though.) Our expectations of capturing a monster and placing a trophy in a hall
must be adjusted: even if we prove that a monster exists and construct it, an idea of drawing its
must be abandoned—it's too ugly to be depicted.
Let's think for a moment how a monster might look like. Every additive function is linear on any
Dirichlet function
is continuous at 0 and thus disqualified from monstrousness by statement 1. This shot into darkness missed. At at least we now see that not only monster must have different slopes at different
Let's return to monster properties. A monster function is discontinuous everywhere, but how badly is it discontinuous?
, a function is
locally bounded
at every point where it is continuous. A monster is continuous nowhere, but is it still locally bounded anywhere or somewhere? In a way similar to statement 1 it's easy to prove the
Statement 2. If an additive function
a < b
, then it is linear.
First, by using that
Let's prove that f is continuous at 0, then it will be linear by the statement 1. For arbitrary x from the q, such that q we have:
on the other hand, we have:
establishing that f is continuous at 0.
This indicates that a monster is not just a bad function, it's a very bad function, that takes arbitrarily large absolute values in arbitrarily small segments. Too bad. As a byproduct, a monster
cannot be monotonic on any segment, because a function monotonic on [
] is bounded there: | {"url":"http://www.cofault.com/2010/01/hunt-for-addictive-monster.html","timestamp":"2014-04-17T00:59:02Z","content_type":null,"content_length":"41903","record_id":"<urn:uuid:612ac89c-1089-476b-969b-0afe7a85cba5>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00185-ip-10-147-4-33.ec2.internal.warc.gz"} |
Taper Tricks
Last month I mentioned that I had more graphs to show you, so let me jump right in. In a previous column [May ‘08], I mentioned that audio taper pots aren’t really logarithmic in a true sense, but
that they employ a trick that allows them to mimic a log taper. Of course, I just breezed right on past that statement with no helpful explanation, so allow me to expound.
There are audiophile-grade pots that I understand provide a very close approximation of a logarithmic curve, but they’re large and expensive (and sometimes ganged – two pots in one – for stereo
rather than mono applications), and aren’t suited to the guitar. Instead, guitar manufacturers almost universally use inexpensive pots that come from a handful of companies, such as CTS.
So as to clarify the difference between logarithmic data and non-logarithmic data, let me show you a Bode plot again (a logarithmic graph – this is the same graph shown last month), along with a
second graph that shows some data in a non-logarithmic format.
You can see that the Bode plot is logarithmic in nature, because the scales multiply by powers of ten. For example, look at the distance shown between 100 and 1000 on the vertical scale. This
distance is the same as the distance between 1000 and 10,000, and the same as the distance between 500 and 5000. Why? Because the scale multiplies by powers of ten:
100 = 1 x 10
1000 = 1 x 10
10,000 = 1 x 10
500 = 5 x 10
5000 = 5 x 10
Each major division line represents a tenfold increase in value. The
of 500 to 5000, or of 100 to 1000, or of 1374 to 13,740, is the same: it’s a tenfold increase. So the distances between these sets of numbers are the same; this graph is all about relationships.
Now look at this graph:
This graph is not logarithmic, it’s linear. That is, it shows a direct, linear correlation between the vertical scale and the horizontal scale. Obviously, you’ve seen tons of graphs that use this
By the way, in case you hadn’t noticed, this second graph illustrates the trick employed by cheap guitar pots to mimic a true log curve. The graph’s vertical scale represents the output of a pot,
from 0 to 100 percent. The pot’s value is irrelevant – this scale shows percentage values, not absolute values. The horizontal scale represents the 300 degrees of rotation for a typical CTS pot as
used in the guitar industry.
The black line shows the plot that a linear taper pot would make. It’s perfectly straight: if you turn the pot 10 percent (30°) then you get 10 percent output. Turn it 70 percent (210°) and you get
70 percent output.
Now look at the blue line. This is what a true logarithmic plot would look like. You can see that you have to turn the pot pretty far before the output starts to rise appreciably. If you had a pot
whose output truly followed this plot, then it would produce a perfect volume increase or decrease. That complaint you may have heard about some pots that act more like a switch than a pot, reducing
volume suddenly and dramatically in a way that makes it difficult to control, wouldn’t apply to this pot at all. It would work
But don’t hold your breath. The gray line is what you actually get in a guitar pot. The trick? Determine the pot’s midpoint value (typically 10 percent – remember last month’s column?) and then print
two separate carbon sections – one on each side of center – to make up the overall trace. Oh, and make the two batches of carbon material that are used for the two sections have different resistive
values. Clever, huh? You can see that the result, which is really two linear sections, does a moderately effective job of following the logarithmic plot that it’s designed to mimic. Of course, it’s
far from perfect, but at $5 it’s the best you’re going to get, and it’s arguably good enough. I say arguably because it works and most people don’t notice that it’s not perfect. If you do, if you
want better and aren’t afraid to pay for it, then it may be time to start rummaging through that audiophile parts bin down at the Wal-Mart.
Happy hunting!
George Ellison
Founder, Acme Guitar Works
acmeguitarworks.com george@acmeguitarworks.com | {"url":"http://www.premierguitar.com/articles/taper-tricks-1","timestamp":"2014-04-18T08:39:19Z","content_type":null,"content_length":"58874","record_id":"<urn:uuid:191efa8f-4440-4aeb-988c-333a5ad38f2c>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00517-ip-10-147-4-33.ec2.internal.warc.gz"} |
Taper Tricks
Last month I mentioned that I had more graphs to show you, so let me jump right in. In a previous column [May ‘08], I mentioned that audio taper pots aren’t really logarithmic in a true sense, but
that they employ a trick that allows them to mimic a log taper. Of course, I just breezed right on past that statement with no helpful explanation, so allow me to expound.
There are audiophile-grade pots that I understand provide a very close approximation of a logarithmic curve, but they’re large and expensive (and sometimes ganged – two pots in one – for stereo
rather than mono applications), and aren’t suited to the guitar. Instead, guitar manufacturers almost universally use inexpensive pots that come from a handful of companies, such as CTS.
So as to clarify the difference between logarithmic data and non-logarithmic data, let me show you a Bode plot again (a logarithmic graph – this is the same graph shown last month), along with a
second graph that shows some data in a non-logarithmic format.
You can see that the Bode plot is logarithmic in nature, because the scales multiply by powers of ten. For example, look at the distance shown between 100 and 1000 on the vertical scale. This
distance is the same as the distance between 1000 and 10,000, and the same as the distance between 500 and 5000. Why? Because the scale multiplies by powers of ten:
100 = 1 x 10
1000 = 1 x 10
10,000 = 1 x 10
500 = 5 x 10
5000 = 5 x 10
Each major division line represents a tenfold increase in value. The
of 500 to 5000, or of 100 to 1000, or of 1374 to 13,740, is the same: it’s a tenfold increase. So the distances between these sets of numbers are the same; this graph is all about relationships.
Now look at this graph:
This graph is not logarithmic, it’s linear. That is, it shows a direct, linear correlation between the vertical scale and the horizontal scale. Obviously, you’ve seen tons of graphs that use this
By the way, in case you hadn’t noticed, this second graph illustrates the trick employed by cheap guitar pots to mimic a true log curve. The graph’s vertical scale represents the output of a pot,
from 0 to 100 percent. The pot’s value is irrelevant – this scale shows percentage values, not absolute values. The horizontal scale represents the 300 degrees of rotation for a typical CTS pot as
used in the guitar industry.
The black line shows the plot that a linear taper pot would make. It’s perfectly straight: if you turn the pot 10 percent (30°) then you get 10 percent output. Turn it 70 percent (210°) and you get
70 percent output.
Now look at the blue line. This is what a true logarithmic plot would look like. You can see that you have to turn the pot pretty far before the output starts to rise appreciably. If you had a pot
whose output truly followed this plot, then it would produce a perfect volume increase or decrease. That complaint you may have heard about some pots that act more like a switch than a pot, reducing
volume suddenly and dramatically in a way that makes it difficult to control, wouldn’t apply to this pot at all. It would work
But don’t hold your breath. The gray line is what you actually get in a guitar pot. The trick? Determine the pot’s midpoint value (typically 10 percent – remember last month’s column?) and then print
two separate carbon sections – one on each side of center – to make up the overall trace. Oh, and make the two batches of carbon material that are used for the two sections have different resistive
values. Clever, huh? You can see that the result, which is really two linear sections, does a moderately effective job of following the logarithmic plot that it’s designed to mimic. Of course, it’s
far from perfect, but at $5 it’s the best you’re going to get, and it’s arguably good enough. I say arguably because it works and most people don’t notice that it’s not perfect. If you do, if you
want better and aren’t afraid to pay for it, then it may be time to start rummaging through that audiophile parts bin down at the Wal-Mart.
Happy hunting!
George Ellison
Founder, Acme Guitar Works
acmeguitarworks.com george@acmeguitarworks.com | {"url":"http://www.premierguitar.com/articles/taper-tricks-1","timestamp":"2014-04-18T08:39:19Z","content_type":null,"content_length":"58874","record_id":"<urn:uuid:191efa8f-4440-4aeb-988c-333a5ad38f2c>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00517-ip-10-147-4-33.ec2.internal.warc.gz"} |