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In an investigation of a physics problem, I ran into the following equation:
d^2(y)/(dt)^2 = k * y * (y^2 + c)^-1.5
I know how to solve separable first order differential equations but this one seems to be beyond me. Assistance?
CPL.Luke Dec12-06 01:30 PM
hmm I don't think that one can be solved analytically, can you settle for a numeric answer?
Matthew Rodman Dec12-06 03:57 PM
Well, one thing you can do is multiply by y prime
[tex]y^{\prime} y^{\prime \prime} = \frac{k y y^{\prime}}{(y^2 + c)^\frac{3}{2}} [/tex]
and then integrate to get
[tex] \frac{1}{2} y^{\prime 2} = - \frac{k}{\sqrt{y^2 + c}} + A [/tex]
where A is a constant of integration.
You can then square root the y prime square, pull over all the y stuff on one side (and integrate again) to get x as some horrendous integral in y.
[tex]x = \int{\frac{dy}{\sqrt{2(A- \frac{k}{\sqrt{y^2 + c}})}}} [/tex]
or rather
[tex]x = \frac{1}{\sqrt{2}} \int{\sqrt{\frac{\sqrt{y^2 +c}}{A \sqrt{y^2 +c} - k }} dy} [/tex]
Other than that, I dunno.
That looks intractable. I expected there to be a "clean" or closed (or whatever you call it) solution. This equation arose from me trying to plot the position of a point mass in a field generated by
another point mass. The y is the vertical position (the reference point mass is at the origin and is stationary).
Matthew Rodman Dec13-06 10:42 PM
With the use of a clever substitution, it may yet be soluble. You never know.
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dot plot
Definition of Dot plot
● In a Dot Plot, a set of data is represented by using dots over a number line.
More about Dot plot
• The number of dots over the number line tells the value of the data points.
Example of Dot plot
• A dot plot is as shown in the figure.
The frequency of the data points is represented by the dots.
Solved Example on Dot plot
The number of various kinds of snakes found in a zoo is shown in the dot plot. What is the total number of snakes in the zoo?
A. 29
B. 25
C. 30
D. 34
Correct Answer: A
Step 1: The number of snakes of a particular kind in the zoo is indicated by the number of dots above the name of the snake in the dot plot.
Step 2: From the plot, the number of Black cobras, Pythons, Anacondas, Rattlesnakes and Green cobras in the zoo are 4, 5, 2, 10 and 8 respectively. [Count the number of dots over each snake
Step 3: The total number of snakes in the zoo = 4 + 5 + 2 + 10 + 8 = 29 [Add the number of snakes of each kind.]
Step 4: So, there are a total of 29 snakes in the zoo.
Related Terms for Dot plot
• Data
• Data Points
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Re: finding all dominator trees
Martin Ward <martin@gkc.org.uk>
5 Dec 2006 06:20:50 -0500
From comp.compilers
| List of all articles for this month |
From: Martin Ward <martin@gkc.org.uk>
Newsgroups: comp.compilers
Date: 5 Dec 2006 06:20:50 -0500
Organization: Compilers Central
References: 06-11-096 06-12-025 06-12-033
Keywords: analysis
Posted-Date: 05 Dec 2006 06:20:49 EST
On Monday 04 Dec 2006 13:28, diablovision@yahoo.com wrote:
> Because, in truth, the dominance relation doesn't make any sense unless
> you have either or both of the following conditions:
> 1. a unique entry node.
> 2. an acyclic graph.
Several people, including myself, assumed that when Amir
referred to "dominators in a graph", he meant a control
flow graph (with a unique entry node from which the meaning
of dominance can be defined). He actually wants to compute
the set of dominator trees for the set of control flow graphs
which can be constructed from a given graph by taking each node
in turn to be the entry node (and, presumably, ignoring nodes
which are not reachable from that node). Each of these CFGs
has the same nodes and edges, but a different entry node,
and therefore a different dominance relation and dominator tree.
martin@gkc.org.uk http://www.cse.dmu.ac.uk/~mward/ Erdos number: 4
Don't email: d3457f@gkc.org.uk
Post a followup to this message
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Hunting Serial Killers
“Tonight’s the Night”
A large volume of research goes into the psychological and behavioral analysis of criminals. In particular, serial criminals hold a special place in the imagination and nightmares of the general
public (at least, American public). Those criminals with the opportunity to become serial criminals are logical, cool-tempered, methodical, and, of course, dangerous. They walk among us in crowded
city streets, or drive slowly down an avenue looking for their next victim. They are sometimes neurotic sociopaths, and other times amicable, charming models of society and business. But most of all,
they know their craft well. They work slowly enough to not make mistakes, but fast enough to get the job done and feel good about it. Their actions literally change lives.
In other words, they would be good programmers. If only they all hadn’t given up trying to learn C++!
In all seriousness, a serial killer’s rigid methodology sometimes admits itself nicely to mathematical analysis. For an ideal serial criminal (ideal in being analyzable), we have the following two
axioms of criminal behavior:
1. A serial criminal will not commit crimes too close to his base of operation.
2. A serial criminal will not travel farther than necessary to find victims.
The first axiom is reasonable because a good serial criminal does not want to arouse suspicion from his neighbors. The second axiom roughly describes an effort/reward ratio that keeps serial
offenders from travelling too far away from their homes.
These axioms have a large amount of criminological research behind them. While there is little unifying evidence (the real world is far too messy), there are many bits and pieces supporting these
claims. For example, the frequency of burglaries peak about a block from the offender’s residence, while almost none occur closer than a block (Turner, 1969, “Delinquency and distance”). Further,
many serial rapists commit subsequent rapes (or rather, abductions preceding rape) within a half mile from the previous (LeBeau, 1987, “Patterns of stranger and serial rape offending”). There are
tons of examples of these axioms in action in criminology literature.
On the other hand, there are many types of methodical criminals who do not agree with these axioms. Some killers murder while traveling the country, while others pick victims with such specific
characteristics that they must hunt in a single location. So we take the following models with a grain of salt, in that they only apply to a certain class of criminal behavior.
With these ideas in mind, if we knew a criminal’s base of operation, we could construct a mathematical model of his “buffer zone,” inside of which he does not commit crime. With high probability,
most of his crimes will lie just outside the buffer zone. This in itself is not useful in the grand scheme of crime-fighting. If we know a criminal’s residence, we need not look any further. The key
to this model’s usefulness is working in reverse: we want to extrapolate a criminal’s residence from the locations of his crimes. Then, after witnessing a number of crimes we believe to be committed
by the same person, we may optimize a search for the offender’s residence. We will use the geographic locations of a criminal’s activity to accurately profile him, hence the name, geoprofiling.
Murder, She Coded
Historically, the first geoprofiling model was crafted by a criminologist named Dr. Kim Rossmo. Initially, he overlaid the crime locations on a sufficiently fine $n \times m$ grid of cells. Then, he
uses his model to calculate the probability of the criminal’s residence lying within each cell. Rossmo’s formula is displayed below, and explained subsequently.
$\displaystyle P(x) = \sum \limits_{\textup{crime locations } c} \frac{\varphi}{d(x,c)^f} + \frac{(1-\varphi)B^{g-f}}{(2B-d(x,c))^g}$,
where $\varphi = 1$ if $d(x,c) > B, 0$ otherwise.
Here, $x$ is an arbitrary cell in the grid, $d(x,c)$ is the distance from a cell to a crime location, with some fixed metric $d$. The variable $\varphi$ determines which of the two summands to
nullify based on whether the cell in question is in the buffer zone. $B$ is the radius of the buffer zone, and $f,g$ are formal empirically tuned parameters. Variations in $f$ and $g$ change the
steepness of the decay curve before and after the buffer radius. We admit to have no idea why they need to be related, and cannot find a good explanation in Rossmo’s novel of a dissertation. Instead,
Rossmo claims both parameters should be equal. For the purposes of this blog we find their exact values irrelevant, and put them somewhere between a half and two thirds.
This model reflects the inherent symmetry in the problem. If we may say that an offender commits a crime outside a buffer of some radius $B$ surrounding his residence, then we may also say that the
residence is likely outside a buffer of the same radius surrounding each crime! For a fixed location, we may compute the probability of the offender’s residence being there with respect to each
individual crime, and just sum them up.
This equation, while complete, has a better description for programmers, which is decidedly easier to chew in small bites:
Let d = d(x,c)
if d > B:
P(x) += 1/d^f
P(x) += B^(f-g)/(2B-d)^g
Then we may simply loop this routine over over all such $c$ for a fixed $x$, and get our probability. Here we see the ideas clearly, that outside the buffer zone of the crime the probability of
residence decreases with a power-law, and within the buffer zone it increases approaching the buffer.
Now, note that these “probabilities” are not, strictly speaking, probabilities, because they are not normalized in the unit interval $[0,1]$. We may normalize them if we wish, but all we really care
about are the relative cell values to guide our search for the perpetrator. So we abuse the term “high probability” to mean “relatively high value.”
Finally, the distance metric we actually use in the model is the so-called taxicab metric. Since this model is supposed to be relevant to urban serial criminals (indeed, where the majority of cases
occur), the taxicab metric more accurately describes a person’s mental model of distance within a city, because it accounts for roadways. Note that in order for this to work as desired, the map used
must be rotated so that its streets lie parallel to the $x,y$ axes. We will assume for the rest of this post that the maps are rotated appropriately, as this is a problem with implementation and not
the model itself or our prototype.
Rossmo’s model is very easy to implement in any language, but probably easiest to view and animate in Mathematica. As usual, the entire program for the examples presented here is available on this
blog’s Github page. The decay function is just a direct translation of the pseudocode:
rossmoDecay[p1_, p2_, bufferLength_, f_, g_, distance_] :=
With[{d = distance[p1, p2]},
If[d > bufferLength,
(bufferLength^(g - f))/(2 bufferLength - d)^g]];
We then construct a function which computes the decay from a fixed cell for each crime site:
makeRossmoFunction[sites_, buffer_, f_, g_] :=
Function[{x, y},
Map[rossmoDecay[#,{x,y},buffer,f,g,ManhattanDistance] &,
Now we may construct a “Rossmo function,” (initializing the parameters of the model), and map the resulting function over each cell in our grid:
Array[makeRossmoFunction[sites, 14, 1/3, 2/3], {60, 50}];
Here the Array function accepts a function $f$, and a specification of the dimensions of the array. Then each array index tuple is fed to $f$, and the resulting number is stored in the $i,j$ entry of
the array. Here $f: \mathbb{Z}_{60} \times \mathbb{Z}_{50} \to \mathbb{R}^+$. We use as a test the following three fake crime sites:
sites = {{20, 25}, {47, 10}, {55, 40}};
Upon plotting the resulting array, we have the following pretty picture:
Here, the crime locations are at the centers of each of the diamonds, and cells with more reddish colors have higher values. Specifically, the “hot spot” for the criminal’s residence is in the
darkest red spot in the bottom center of the image.
As usual, in order to better visualize the varying parameters, we have the following two animations:
Variation in the $B$ parameter simply increases or decreases the size of the buffer zone. In both animations above we have it fixed at 14 units.
Despite the pretty pictures, a mathematical model is nothing without empirical evidence to support it. Now, we turn to an analysis of this model on real cases.
“Excellent!” I cried. “Elementary mathematics,” said he.
The first serial killer we investigate is Richard Chase, also known as the Vampire of Sacramento. One of the creepiest murderers in recent history, Richard Chase believed he had to drink the blood of
his victims in order to live. In the month of January 1978, Chase killed five people, dumping their mutilated bodies in locations near his home.
Before we continue with the geographic locations of this particular case, we need to determine which locations are admissible. For instance, we could analyze abduction sites, body drop sites,
locations of weapons caches or even where the perpetrator’s car was kept. Unfortunately, many of these locations are not known during an investigation. At best only approximate abduction sites can be
used, and stash locations are usually uncovered after an offender is caught.
For the sake of the Chase case, and subsequent cases, we will stick to the most objective data points: the body drop sites. We found this particular data in Rossmo’s dissertation, page 272 of the pdf
document. Overlaid on a 30 by 30 grid, they are:
richardChaseSites =
{{3, 17}, {15, 3}, {19, 27}, {21, 22}, {25, 18}};
richardChaseResidence = {19,17};
Then, computing the respective maps, we have the following probability map:
If we overlay the location of Chase’s residence in purple, we see that it is very close to the hottest cell, and well-within the hot zone. In addition, we compare this with another kind of
geoprofile: the center of gravity of the five sites. We color the center of gravity in black, and see that it is farther from Chase’s residence than the hot zone. In addition, we make the crime sites
easy to see by coloring them green.
This is a great result for the model! Let us see how it fares on another case: Albert DeSalvo, the Boston strangler.
With a total of 13 murders and being suspected of over 300 sexual assault charges, DeSalvo is a prime specimen for analysis. DeSalvo entered his victim’s homes with a repertoire of lies, including
being a maintenance worker, the building plumber, or a motorist with a broken-down car. He then proceeded to tie his victims to a bed, sexually assault them, and then strangle them with articles of
clothing. Sometimes he tied a bow to the cords he strangled his victims with.
We again use the body drop sites, which in this case are equivalent to encounter sites. They are:
deSalvoSites = {{10, 48}, {13, 8}, {15, 11}, {17, 8}, {18, 7},
{18, 9}, {19, 4}, {19, 8}, {20, 9}, {20, 10}, {20, 11},
{29, 23}, {33, 28}};
deSalvoResidence = {19,18};
Running Rossmo’s model again, including the same extra coloring as for the Chase murders, we get the following picture:
Again, we win. DeSalvo’s residence falls right in the darker of our two main hot zones. With this information, the authorities would certainly apprehend him in a jiffy. On the other hand, the large
frequency of murders in the left-hand side pulls the center of gravity too close. In this way we see that the center of gravity is not a good “measure of center” for murder cases. Indeed, it violates
the buffer principle, which holds strong in these two cases.
Finally, we investigate Peter Sutcliffe, more infamously known as the Yorkshire Ripper. Sutcliffe murdered 13 women and attacked at least 6 others between 1975 and 1980 in the county of West
Yorkshire, UK. He often targeted prostitutes, hitting them over the head with a hammer and proceeding to sexually molest and mutilate their bodies. He was finally caught with a prostitute in his car,
but was not initially thought to be the Yorkshire Ripper until after police returned to the scene of his arrest to look for additional evidence. They found his murder weapons, and promptly prosecuted
We list his crime locations below. Note that these include body drop sites and the attack sites for non-murders, which were later reported to the police.
sutcliffeSites = {{5, 1}, {8, 7}, {50, 99}, {53, 68}, {56, 72},
{59, 59}, {62, 57}, {63, 85}, {63, 87}, {64, 83}, {69, 82},
{73, 88}, {80, 88}, {81, 89}, {83, 88}, {83, 87}, {85, 85},
{85, 83}, {90, 90}};
sutcliffeResidences = {{60, 88}, {58, 81}};
Notice that over the course of his five-year spree, he lived in two residences. One of these he moved to with his wife of three years (he started murdering after marrying his wife). It is unclear
whether this changed his choice of body drop locations.
Unfortunately, our attempts to pinpoint Sutcliffe’s residence with Rossmo’s model fail miserably. With one static image, guessing at the buffer radius, we have the following probability map:
As we see, both the center of gravity and the hot zones are far from either of Sutcliffe’s residences. Indeed, even with a varying buffer radius, we are still led to search in unfruitful locations:
Even with all of the axioms, all of the parameters, all of the gosh-darn work we went through! Our model is useless here. This raises the obvious question, exactly how applicable is Rossmo’s model?
The Crippling Issues
The real world is admittedly more complex than we make it out to be. Whether the criminal is misclassified, bad data is attributed, or the killer has some special, perhaps deranged motivation, there
are far too many opportunities for confounding variables to tamper with our results. Rossmo’s model even requires that the killer live in a more or less central urban location, for if he must travel
in a specific direction to find victims, he may necessarily produce a skewed distribution of crime locations.
Indeed, we have to have some metric by which to judge the accuracy of Rossmo’s model. While one might propose the distance between the offender’s residence and the highest-probability area produced
on the map, there are many others. In particular, since the point of geographic profiling is to prioritize the search for a criminal’s residence, the best metric is likely the area searched before
finding the residence. We call this metric search area. In other words, search area is the amount of area on the map which has probability greater or equal to the cell containing the actual
residence. Indeed Rossmo touts this metric as the only useful metric.
However, according to his own tests, the amount of area searched on the Sutcliffe case would be over a hundred square miles! In addition, Rossmo neither provides an idea of what amount of area is
feasibly searchable, nor any global statistics on what percentage of cases in his study resulted in an area that was feasibly searchable. We postulate our own analysis here.
In a count of Rossmo’s data tables, out of the fifteen individual cases he studied, the average search area was 395 square kilometers, or 152.5 square miles, while the median was about 87 square
kilometers, or 33.6 square miles. The maximum is 1829 square kilometers, while the min is 0.2 square kilometers. The complete table is contained in the Mathematica notebook on this blog’s Github page
From the 1991 census data for Vancouver, we see that a low density neighborhood has an average population of 2,380 individuals per square kilometer, or about 6,000 per square mile. Applying this to
our numbers from the previous paragraph, we have a mean of 940,000 people investigated before the criminal is found, a median of 200,000, a max of four million (!), and a min of 309.
Even basing our measurements on the median values, this method appears to be unfeasible as a sole means of search prioritization. Of course, real investigations go on a lot more data, including
hunches, to focus search. At best this could be a useful tool for police, but on the median, we believe it would be marginally helpful to authorities prioritize their search efforts. For now, at
least, good ol’ experience will likely prevail in hunting serial killers.
In addition, other researchers have tested human intuition at doing the same geographic profiling analysis, and they found that with a small bit of training (certainly no more than reading this blog
post), humans showed no significant difference from computers at computing this model. (English, 2008) Of course, for the average human the “computing” process (via pencil and paper) was speedy and
more variable, but for experienced professionals the margin of error would likely disappear. As interesting as this model may be, it seems the average case is more like Sutcliffe than Chase; Rossmo’s
model is effectively a mathematical curiosity.
It appears, for now, that our friend Dexter Morgan is safe from the threat of discovery by computer search.
Alternative Models
The idea of a decay function is not limited to Rossmo’s particular equation. Indeed, one might naturally first expect the decay function to be logarithmic, normal, or even exponential. Indeed, such
models do exist, and they are all deemed to be roughly equivalent in accuracy for appropriately tuned parameters. (English, 2008) Furthermore, we include an implementation of a normal growth/decay
function in the Mathematica notebook on this blog’s Github page. After reading that all of these models are roughly equivalent, we did not conduct an explicit analysis of the normal model. We leave
that as an exercise to the reader, in order to become familiar with the code provided.
In addition, one could augment this model with other kinds of data. If the serial offender targets a specific demographic, then this model could be combined with demographic data to predict the sites
of future attacks. It could be (and in some cases has been) weighted according to major roadways and freeways, which reduce a criminals mental model of distance to a hunting ground. In other words,
we could use the Google Maps “shortest trip” metric between any two points as its distance metric. To our knowledge, this has not been implemented with any established mapping software. We imagine
that such an implementation would be slow; but then again, a distributed network of computers computing the values for each cell in parallel would be quick.
Other Uses for the Model
In addition to profiling serial murders, we have read of other uses for this sort of geographic profiling model.
First, there is an established paper on the use of geographic profiling to describe the hunting patterns of great white sharks. Briefly, we recognize that such a model would switch from a taxicab
metric to a standard Euclidean metric, since the movement space of the ocean is locally homeomorphic to three-dimensional Euclidean space. Indeed, we might also require a three-dimensional
probability map for shark predation, since sharks may swim up or down to find prey. Furthermore, shark swimming patterns are likely not uniformly random in any direction, so this model is weighted to
consider that.
Finally, we haphazardly propose additional uses for this model: pinpointing the location of stationary artillery, locating terrorist base camps, finding the source of disease outbreaks, and profiling
other minor serial-type criminals, like graffiti vandalists.
Data! Data! My Kingdom for Some Data!
As recent as 2000, one researcher noted that the best source of geographic criminal data was newspaper archives. In the age of information, and given the recent popularity of geographic profiling
research, this is a sad state of being. As far as we know, there are no publicly available indexes of geographic crime location data. As of the writing of this post, an inquiry to a group of machine
learning specialists has produced no results. There doesn’t seem to be such a forum for criminology experts.
If any readers have information to crime series data that is publicly available on the internet (likely used in some professor’s research and posted on their website), please don’t hesitate to leave
a comment with a link. It would be greatly appreciated.
6 thoughts on “Hunting Serial Killers”
1. Fascinating post!
Thank you for explaining it so well and for giving me more to read up on!
□ Any time! I’m proud to have attracted someone from a non-mathematical community to my humble blog :)
2. What a great post! I had fun reading it, thanks for that. :)
3. Very interesting, and I was looking for this! Except I don’t know how to use it, is there any way I can find out how to apply this?
□ So say that you have a serial case you want to solve, and it’s in some region like a city or county. Take a map and mark the locations of interest on it (murder locations, body drop sites,
weapons cache locations, etc.), and then overlay a grid of squares on it. Label the bottom left square (0,0), the square above (0,1), the square to the right (1,0), etc., as with the usual
Cartesian plane. Then whichever squares the locations of interest fall into are the coordinates of that location. My program takes as input a list of such locations, and outputs the pictures
you see in the article. Is that what you were asking?
4. Pingback: Matemáticas y programación | CyberHades | {"url":"http://jeremykun.com/2011/07/20/serial-killers/","timestamp":"2014-04-16T10:46:31Z","content_type":null,"content_length":"106456","record_id":"<urn:uuid:09099aed-ebe9-4689-9db3-adf90f1f53fe>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00073-ip-10-147-4-33.ec2.internal.warc.gz"} |
algebra related word problem
April 24th 2009, 09:40 AM #1
Nov 2008
algebra related word problem
A club consists of 8 seniors, 7 juniors, and 4 sophomores. A subcommittee consisting of 3 seniors, 2 juniors, and 1 sophomore will be chosen. How many different such subcommittees are there?
Order does not appear to matter here, so you'll be looking for combinations, not permutations.
In how many ways can you choose three of eight seniors?
In how many ways can you choose two of seven juniors?
In how many ways can you choose one of four sophomores?
Where can you go from this...?
April 24th 2009, 10:32 AM #2
MHF Contributor
Mar 2007 | {"url":"http://mathhelpforum.com/algebra/85448-algebra-related-word-problem.html","timestamp":"2014-04-19T14:08:14Z","content_type":null,"content_length":"32678","record_id":"<urn:uuid:1cf1cc61-fa35-4af9-a816-d960423c3822>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00520-ip-10-147-4-33.ec2.internal.warc.gz"} |
Three Houses, Three Utilities
Date: 07/15/99 at 01:43:02
From: Chris
Subject: Lines, etc.
I know that you have answered this before: the question about the
three houses and the three utilities (gas, electricity, water).
Well, the guy who gave me this puzzle says there is a way of solving
it in 2D, without any tricks. He says that it is simple, once you
figure it out. I don't get it. Everywhere, it says that it can only be
done using 3 dimensions. Can you solve it using 2 dimensions? How?
Thank you very much :)
Date: 07/15/99 at 12:38:34
From: Doctor Rob
Subject: Re: Lines, etc.
Thanks for writing to Ask Dr. Math!
You can only solve this if you allow one of the utility lines to run
through someone else's house, or through one of the other utility
companies, which I suppose is possible, but is usually forbidden by
the conditions of the puzzle.
- Doctor Rob, The Math Forum
Date: 07/15/99 at 12:46:43
From: Doctor Peterson
Subject: Re: lines, etc.
Hi, Chris.
He may not call it a trick, but any solution that's really 2D (that
is, done just by drawing non-intersecting curves on a flat sheet of
paper) has to twist the rules somehow. He might, for example, draw the
houses as rectangles and say that it's legal to open the front and
back doors of one house and pass a pipe through. I call that a trick.
Another trick is to solve it on the surface of a donut (a torus) and
point out that any surface is itself 2-dimensional, even though it
exists in a 3-dimensional space. Or you can allow going around to the
other side of the paper through a hole, which is essentially the same
thing, as this answer points out:
When the problem is stated carefully in mathematical terms (continuous
non-intersecting curves from each of three points to each of three
other points), there's no solution; but presented in terms of houses
and utilities (which are inherently three-dimensional), there are lots
of ways to get around it.
I'd like to hear what his answer is.
- Doctor Peterson, The Math Forum | {"url":"http://mathforum.org/library/drmath/view/55165.html","timestamp":"2014-04-19T17:20:56Z","content_type":null,"content_length":"7098","record_id":"<urn:uuid:fdaf6674-7aa3-434e-9ce1-b9b963c92f82>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00439-ip-10-147-4-33.ec2.internal.warc.gz"} |
more CHEM proofreading
Posted by jerson on Sunday, April 27, 2008 at 8:06pm.
for this question A sample of air has a volume of 140.0 mL at 67 degrees celcius. At what temperature would its volume 50.0 mL at constant pressure
I did
140.0 ml... 67+273=340
T2... 50.0
This question i don't get because of the scientific notation. A sample of oxygen that occupies 1.00 x 10^6 mL at 575 mm Hg is subjected to a pressure of 1.25 atm. What will the final volume of the
sample be if the temp is held constant?
• more CHEM proofreading - DrBob222, Sunday, April 27, 2008 at 8:18pm
The first one is ok EXCEPT that if your prof is a stickler for significant figures, you will get points counted off if not the entire question. The smallest number of significant figures is 3
(50.0, 340) so the most s.f. you are allowed in the answer is 3; therefore, you would round the answer to 121. ALSO, some profs will count off if you don't have units; therefore, the complete
answer would be 121 Kelvin.
For the second problem, 10^6 just means to add 6 zeros to 1 (which would be 1,000,000). Or you can key in the number with scientific notation to your calculator and let it keep track of the
decimal. USUALLY, a number expressed as 1.00 x 10^6 mL means your prof looks at s.f.
• more CHEM proofreading - DrBob222, Sunday, April 27, 2008 at 8:39pm
Sure. It says the temperature is held constant. This is a pressure/volume problem. Done the same way except
P1V1 = P2V2.
By the way, do you know how to keep all these formulas straight? Do it this way.
The general formula is
(P1V1)/T1 = (P2V2)/T2
If T is constant, just cover up T1 and T2 with your fingers (or mentally) and you have P1V1 = P2V2 which is Boyle's Law.
If P is held constant, cover up P1 and P2 with your fingers (or mentally), and you have V1/T1 = V2/T2 which is Charles' Law.
If V is held constant, covert up V1 and V2 with yur fingers (or mentally) and you have P1/T1 = P2/V2. Easy, huh?
You only need to memorize the general formula that contains all the variable adn tailor it to fit the problem. And, of course, you must remember that T always goes in with Kelvin.
□ more CHEM proofreading - DrBob222, Sunday, April 27, 2008 at 8:42pm
If V is held constant, covert up V1 and V2 with yur fingers (or mentally) and you have P1/T1 = P2/V2. Easy, huh?
I made a goof here. This should read, if V is held constant, cover up V1 and V2 with your fingers (or mentally) and you have P1/T1 = P2/T2
My fingers sometimes get ahead of my brain.
• more CHEM proofreading - jerson, Sunday, April 27, 2008 at 9:09pm
well im sure u can tell chemistry is my weakest subject because im still confused :/...
so would it be 1000000= 575
1.25= t2?
• more CHEM proofreading - DrBob222, Sunday, April 27, 2008 at 9:37pm
Why do you want to solve for t? It SAYS t is constant so we don't care what it is. And you haven't used the pressure at all.
You didn't do what I said. Just follow the guide lines.
P1V1/T1 = P2V2/T2.
Now, since T is constant, cover T1 and T2 with your fingers, or mentally, and we are left with
P1V1 = P2V2
Now look at the problem.
V1 = 1 x 10^6 mL
P1 = 575 mm Hg
V2 = ?? mL.
P2 = 1.25 atm.
Right away you see that the units on pressure don't match. You must change mm Hg to atmospheres OR change atmospheres to mm Hg. The conversion factor is 760 mm = 1 atm. The easy one to change is
P1 so 575/760 = 0.756 atm.
Then P1V1 = P2V2
0.756*1x10^6 = 1.25*V2 and solve for V2.
• more CHEM proofreading - Anonymous, Tuesday, February 2, 2010 at 10:22pm
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Spring Valley, CA Algebra Tutor
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User Abhinav Kumar
bio website web.mit.edu/abhinavk/www
visits member for 4 years, 4 months
seen Apr 9 at 22:20
stats profile views 1,021
28 reviewed Approve suggested edit on On a particular case of the ``Tumura-Hayman" theorem :
30 reviewed Reject suggested edit on Cohomology after completion
Jan Commutativity of convex hulls and closed balls
16 comment I believe I now have a counterexample - see above.
Jan Commutativity of convex hulls and closed balls
16 revised added 1351 characters in body
15 answered Commutativity of convex hulls and closed balls
14 awarded Good Question
12 awarded Custodian
12 reviewed Approve suggested edit on Distribution of moduli of quadratic residues
Jan Determinant and eigenvalues of a specific matrix
2 comment If you let $e^{-c}$ be $x$, then the matrix has polynomial entries in $x$, and experiment seems to indicate that the determinant is a product of cyclotomic polynomials (for instance, if
$n = 6$, we get $-(x-1)^{15}(x+1)^{15}(x^2+1)^6 (x^2-x+1)^3(x^2+x+1)^3(x^4+1)^2(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$. (In particular, the power of $(x \pm 1)$ seems to be $n$ choose $2$.)
19 awarded Yearling
Nov Lattice points and convex bodies
20 comment @AntonPetrunin: Good point!
Nov Lattice points and convex bodies
20 comment I would be surprised if this were true even for integer polytopes - that the Erhart polynomial determines the polytope (though I can't seem to find an immediate counterexample by
searching online ...). You can certainly do $GL_n(\mathbb{Z})$ transformations without changing the number of integer points.
Nov Nefness on a K3 surface
12 comment Not if the divisor $D$ is reducible - else take $D = C + f$ on a Hirzebruch surface with $C^2 = -2, f^2 = 0, C \cdot f = 1$, and notice $D \cdot C = -1$. If you don't assume $D$ is
effective, then Jason's comment shows you that there's no way to distinguish $D$ from its negative.
12 answered Nefness on a K3 surface
1 awarded Nice Answer
Nov Rationality of curve does not depend on base change
1 revised deleted 1 characters in body
Nov Rationality of curve does not depend on base change
1 revised added 725 characters in body
1 answered Rationality of curve does not depend on base change
Oct Lattice-point-free buffers around circles
30 comment You apply it with $x = r^2$. It gives a lattice point $(a,b)$ with $n = a^2 + b^2$, where $n$ is between $r$ and $\sqrt{r^2 + C\sqrt{r}}$. Therefore the distance to the circle is $\sqrt
{n} - r$, but $r = \sqrt{r^2} \leq \sqrt{n} \leq \sqrt{r^2 + Cr^{1/2}} = r(1 + r^{-3/2}/2 ) = r + r^{-1/2}/2$.
Oct answered Lattice-point-free buffers around circles | {"url":"http://mathoverflow.net/users/2698/abhinav-kumar?tab=activity","timestamp":"2014-04-20T13:42:55Z","content_type":null,"content_length":"45920","record_id":"<urn:uuid:99bf03ab-c790-4d02-b06a-467983e3e622>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00214-ip-10-147-4-33.ec2.internal.warc.gz"} |
ANOVA Diagnostics Module. Stats Design & Analysis Web Guide (DAWG), University of Tennessee
Check for outliers, equal variance, normality, and model validity
You just ran your SAS analysis; you are almost done.
Now you have to examine your dataset for three potential problems in your data. If one or more of these occur, you will have to make corrections and then rerun your SAS analysis. It is uncommon for
datasets to have any of these problems, but if they occur, your conclusions can be altered. So you have to check.
Go to Step 10 of your analysis and click on your example in the gray Example output table to display the output in a popup window. It will be helpful to have the analysis example ouput for your
particular design available, to refer to as you move through this diagnostics module. In the example output, yellow, clickable arrows
The ANOVA tests depend on data that are normally distributed with equal variance. You use probability values to make decisions about treatment differences. These probabilities are incorrect if the
original data are not normal. Equal variance is needed because variability within each treatment is pooled to create an error term. Clearly, if variances are not equal, this one pooled error term
will be too large for some treatments and too small for others. Again, this produces incorrect probabilities. Generally, these two problems can be corrected by transforming your data. And finally,
outliers are a primary cause of unequal variance and non-normality. Since by definition outliers are bad data points, these should be removed before making final conclusions.
If your analysis includes block or covariate effects (if you chose something besides CRD in column 1 and/or chose Covariate in column 3 of Choose Design), there is an additional check that needs to
be done for each. These will be covered in Step D. These steps will not be used if you determine your model does not include these terms.
Therefore, this module will lead you through identifying and correcting these problems. Examples will be shown to illustrate each of the problems. You can also click the Examples tab to view real
datasets having these problems. If you are running your own dataset, compare its diagnostics to the illustrations.
You check for outliers first, in Step A, because they affect all other diagnostics. Then you continue with Steps B and C, to determine if you need to transform your data to correct unequal variance
and/or non-normality (and possibly also apparent outliers). The module will show you how to transform your data. You will finish with Step E, which will summarize what you should do next. You are
sent to Step F only if necessary.
The diagnostics checking process that you will learn here is complicated by each diagnostic potentially affecting others. In this module, unlike other modules in which we can lead you more clearly
through sequential steps, you will bounce back and forth among the various steps. For example, outliers may cause non-normality, but also, non-normality may cause apparent outliers. So you will first
check all diagnostics before fixing any one problem.
Click next >> at the bottom of each page to move through this module. | {"url":"http://dawg.utk.edu/diag/diagnostics.htm","timestamp":"2014-04-18T20:51:25Z","content_type":null,"content_length":"19585","record_id":"<urn:uuid:18a321d6-11b5-4b96-b43a-d895aba96da4>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00206-ip-10-147-4-33.ec2.internal.warc.gz"} |
September 26th 2006, 11:58 PM
Can someone help me on the following questions
If loge (a/b) = kl/2c make a the subject of the formula. Also calculate the value of a when b = 4.16, k = 3.68, 1 = 11.41 and c = 9.23
Thanks Alison
September 27th 2006, 02:44 AM
The symbol for "log base e" is "ln" so I'm going to use that.
ln(a/b) = kl/(2c) (I'm assuming the denominator is this.)
a/b = exp[kl/(2c)] where exp is the exponential function e^
a = b*exp[kl/(2c)]
For the given values I get a = 40.451266618 | {"url":"http://mathhelpforum.com/algebra/5866-logritham-print.html","timestamp":"2014-04-18T04:35:33Z","content_type":null,"content_length":"4353","record_id":"<urn:uuid:42be0bd9-8be3-40a2-a9be-b3417733fa56>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00260-ip-10-147-4-33.ec2.internal.warc.gz"} |
9th grade math word problems
Posted by Angel on Thursday, April 7, 2011 at 4:51pm.
i never can do word problems....please help... i got 7 of them.....these are word for word
Larry is 8 yrs older than his sister.In 3 yrs, hr will be twice as old as she will be then.how old is each now?
Jennifer is 6 yrs older than Sue. in 4 yrs, she will be twice as old as Sue was 5 yrs ago. Find their ages now.
Adam is 5 yrs younger than Eve. In 1 yr, Eve will be three times as old as Adam was 4 yrs ago. Find their ages now.
4)jack is twice as old as Jill. In 2 yrs, Jack will be 4 times as old as Jill was ( uears ago. How old are they now?
four yrs ago, Katie was twice as old as Anne was then. In 6 yrs, Anne will be the same age that Katie is now How old is each now?
five yrs ago, Tom was one third as old as his father was then. In 5 yrs Tom will be half as old as his father will be then. Find their ages now.
Barry is 8 yrs older than sue. In 4 yrs, she will be twice as old as Sue was 5 yrs ago. find their ages now.
• 9th grade math word problems - SraJMcGin, Thursday, April 7, 2011 at 8:09pm
1. If Larry is 13 today and his sister is 5, add 3 years and Larry will be 146 and his sister will be 8.
Here are some word problem tutorials. Look through them to see if you can find something to help you:
That should get you started.
• 9th grade math word problems - f(10)=1/2x-5, Monday, March 26, 2012 at 7:18pm
• 9th grade math word problems - Anonymous, Wednesday, April 25, 2012 at 4:00pm
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steepest descent - Numerical Recipes Forum
Originally Posted by
What is behind the smilies...
For people trying to read code that was posted without code tags:
Pretend you are going to respond to the post (click the
button at the lower right).
When the response edit window opens, copy/paste the problematic lines to your text editor. Then, of course, unless you are really going to respond to the post, you click the
Here's what was in that post:
function g= grad(f,x,h)
%grad.m to get the gradient of f(x) at x.
if nargin<3, h=.00001; end
h2= 2*h; N= length(x);
I= eye(N);
for n=1:N
g(n)= (feval(f,x+I(n,:)*h)-feval(f,x-I(n,:)*h))/h2;
For people posting code: The board will preserve your indentation and the post will show up in a monospace font (like the stuff in the
box, above) if you highlight your lines of code (just the code; not your other narrative), and click the
icon at the top of the edit window. This puts code tags around the highlighted stuff.
When posting, If you click the
Preview Post
button before submitting the reply, you can see if you got it right.
Originally Posted by cloyst3r
Cuz it makes no sense replacing the smilies with : )
Actually, it
make sense to Matlab, I'm thinking. I mean, in general,
designates the
th row of matrix
, doesn't it? | {"url":"http://www.nr.com/forum/showthread.php?t=745","timestamp":"2014-04-19T06:53:16Z","content_type":null,"content_length":"51829","record_id":"<urn:uuid:dec4217e-a3c9-4abc-bcc0-942a63ae17d4>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00039-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Midpoint and Distance Formulas in 3D - Problem 3
The distance formula for vectors can be used to derive equations. Let's take a look at an example. It says find an equation that must be satisfied by the points equidistant from O, the origin, and
point A; 3, 2, 6.
So I've drawn a little picture here. Here is the origin, here is point 3, 2, 6, and here is the midpoint between them. It's probably the first one you think about that's equidistant from the two
points. In space, or even in two dimensions, there are lots of other points that are equidistant from A, and O.
So how do we find all of them? Well I'm going to start by calling this vector AP, and this vector OP. I'm going to start with the condition that defines the set. That vector OP's length has to equal
vector AP's length.
Now in order to analyze that a little bit further, I need to come up with components for these two vectors. So first OP, note that that's a position vector, because O is the origin, so vector OP is
going to be x, y, z. What about vector AP? AP will be x minus 3, y minus 2, and z minus 6. So now I have to take these two vectors, and set their lengths equal to each other.
So the length of OP is the square root of x² plus y² plus z². The length of vector AP, x minus 3², plus y minus 2², plus z minus 6². Now to make any sense out of this equation, I'm going to have to
square both sides. Let me do that up here. I have x², plus y², plus z² on the left and on the right x minus 3², plus 1 minus 2² plus z minus 6².
Now let me expand each of these binomials, because there's going to be a lot of cancellation here. These guys are all going to cancel. X², plus y² plus z² equals x² minus 6x plus 9. Y² minus 4x plus
4, 4y plus 4. Z² minus 12z plus 36. And now the fun part, the cancellation. So you can see that x² will cancel, y² will cancel and z² cancels. Now what are we left with? Well we're left with 0 in the
left, and on the right we've got -6x. Let me pull all the variable coefficients together, variable terms; minus 4y minus 12z. Then I have 9 plus 4 plus 36.
Now just to neaten things up, I'll pull all the variable terms to the left. I have 6x plus 4y plus 12z, and then 40 plus 9, 49. That's a rectangular equation for the set of points that are
equidistant from point O, and point A. It turns out to be a plane. This is the equation of a plane in space.
So the set of all points equidistant from these two points is a plane. In fact in space, the set of all points equidistant from any two points, is a plane that perpendicularly bisects the segment
connecting the two points.
position vectors the vector from point a to point b length magnitude plane perpendicular bisector of a segment | {"url":"https://www.brightstorm.com/math/precalculus/vectors-and-parametric-equations/the-midpoint-and-distance-formulas-in-3d-problem-3/","timestamp":"2014-04-17T04:15:49Z","content_type":null,"content_length":"77556","record_id":"<urn:uuid:fbdfacc9-8674-4a53-bcba-4117b3d39067>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00320-ip-10-147-4-33.ec2.internal.warc.gz"} |
normal distribution
August 4th 2010, 05:24 PM #1
Mar 2010
normal distribution
The lengths of 100 pipes have a normal distribution with a mean of 102.4 inches and a standard deviation of 0.2 inch. If one of the pipes measures exactly 102.1 inches, what percentile does its
length lie in?
Thanks for the help
Calculate Pr(X > 102.1) and link the answer to the definition of percentile.
August 4th 2010, 05:26 PM #2 | {"url":"http://mathhelpforum.com/statistics/152807-normal-distribution.html","timestamp":"2014-04-20T14:43:25Z","content_type":null,"content_length":"33787","record_id":"<urn:uuid:982be9ab-4dc8-43a5-bbdb-9b8f9155db4f>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00220-ip-10-147-4-33.ec2.internal.warc.gz"} |
Vinnings, GA Math Tutor
Find a Vinnings, GA Math Tutor
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Puzzles For Kids
- Lots of Puzzle Fun!
Try these fun puzzles for kids - you'll find the answers at the end of the page.
Puzzles for kids - Puzzle # 1
Dime Disappointment
A girl named Tracy collected old coins. She knew that some old coins were worth a lot of money. But she did not know if any of her coins were valuable.
One day, her friend Megan told her she had heard that 1950 dimes were worth almost two hundred dollars. Tracy had two 1950 dimes!
So, she rushed down to turn in the two coins and collect almost four hundred dollars. However, at the bank she was told that while 1950 dimes were worth almost two hundred dollars, her dimes were
only worth twenty cents. Why?
You will find the answer below.
Puzzles for kids - Puzzle # 2
A Sailor's Puzzle
Long ago, a ship was anchored in the port of Amsterdam. A rope ladder hung over the ship's side. There were sixteen rungs in the ladder. The bottom rung of the ladder just touched the water.
Every half hour, the tide rose exactly one-half the distance between two rungs of the ladder. How long did it take for the water to cover the rungs of the ladder?
You will find the answer below.
Puzzles for kids - Puzzle # 3
Hungry Horses
If five horses can eat five bags of oats in five minutes, how long will it take a hundred horses to eat a hundred bags of oats?
You will find the answer below.
Puzzles for kids - Puzzle # 4
The Bus Driver
The bus driver was going down a street. He went right past a stop sign without stopping. He turned left where there was a "No Left Turn" sign. Then he turned the wrong way into a one-way street. And
yet, he didn't break a single traffic law. Why not?
You will find the answer below.
Puzzles for kids - Puzzle # 5
Leftover Sandwiches
Mrs. Miller made twenty-four sandwiches for a picnic. All but seven were eaten. How many were left?
You will find the answer below.
Puzzles for kids - Puzzle # 6
Even Money
Two mothers and two daughters decided to go shopping. They found that they had twenty-seven dollars, all in one-dollar bills. They divided up the money evenly, without making any of the dollars into
change, so they each had exactly the same amount. How was this possible?
You will find the answer below.
Puzzles for kids - Puzzle # 7
The Nut Collectors
Six squirrels began to gather hickory nuts and put them into a large basket. The squirrels worked so fast that the number of nuts in the basket was doubled at the end of every minute. The basket was
completely full at the end of ten minutes. How many minutes had it taken the squirrels to get the basket half full?
You will find the answer below.
Puzzles for kids - Puzzle # 8
The Tennis Player
Once there was a player in a tennis game.
She played very well.
And won great fame.
Jane was her first name,
What was her last name.
The tennis player's name is hidden in the poem. Can you find it?
You will find the answer below.
Puzzles for kids - Puzzle # 9
The Scientist's Brother
Mr. Jones proudly told everyone that he is the brother of a famous scientist. However, Mr. Jones doesn't have a brother. Even so, he was telling the truth. How is this possible?
You will find the answer below.
Puzzles for kids - Puzzle # 10
The Amazing Bunny
Two bunnies were nibbling clover in a meadow. They were facing in opposite directions. Suddenly, one cried out to the other: "Look out! There's a fox sneaking up behind you!"
The bunny did not hear or smell the fox. How did it know the fox was sneaking up behind its friend?
You will find the answer below.
Spelling Bee
PUZZLES FOR KIDS - ANSWERS
Answer #1 - Dime Disappointment
If you add up 1950 dimes - that is, one thousand, nine hundred and fifty dimes - you'll find they amount to $195.00, which is almost two-hundred dollars. However, a dime with the year 1950 on it is
only worth ten cents, like most other dimes.
Answer #2 - A Sailor's Puzzle
The water never covered any of the rungs. A ship, of course, floats on water. So, as the tide rose, the ship rose with it. And the rope ladder, which was attached to the ship, also rose.
Answer #3 - Hungry Horses
If five horses can eat five bags of oats in five minutes, then it takes each horse five minutes to eat a bag of oats. Therefore, it will take only five minutes for a hundred horses to eat a hundred
bags of oats.
Answer #4 - The Bus Driver
The bus driver didn't break any traffic laws because it was his day off and he was walking.
Answer #5 - Leftover Sandwiches
If all but seven sandwiches were eaten, then seven sandwiches were left, of course!
Answer #6 - Even Money
The two mothers and two daughters were actually only three people - Sally, her mother, and her grandmother. Sally was her mother's daughter, of course, and her mother was the grandmother's daughter -
that's two daughters. Sally's mother was one mother, and her mother, the grandmother, was the other mother. Thus, they divided the twenty-seven dollars three ways, each taking nine dollars.
Answer #7 - The Nut Collectors
If the number of nuts in the basket doubled at the end of every minute, the basket must have been half full in nine minutes. Then, after one more minute, the half would be doubled, thus filling the
other half of the basket.
Answer #8 - The Tennis Player
The tennis player's first and last names are given in the last two lines of the poem.
Jane was her first name,
What was her last name.
The two lines make one complete sentence. And there is a period at the end of the sentence, not a question mark. It tells you that Jane was the player's first name and What was her last name. Her
name was Jane What.
Answer #9 - The Scientist's Brother
The famous scientist was a woman. She was Mr Jones' sister.
Answer #10 - The Amazing Bunny
Although the bunnies were facing in opposite directions, they were facing each other. Thus, the bunny could see the fox sneaking toward his friend.
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Calculate Mixture
[calculate mixture ] [search_but]
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Calculate Mixture
259.0 KB |
Category: Mathematics
You want to make a mixture. You have ingredients composed of the same components.The proportions (percentage,content) of the components are different in each ingredient.How much of each ingredient
you need, the Mix Finder can find for you. You want to make a mixture. You have ingredients composed of the same components. The proportions (percentage,content) of the components is different in
each ingredient. The composition of the mixture will depend on how much of each ingredient you...
As the name suggests, an embellished dialog.
OS: Windows
Software Terms: Calculate Mixture, Chemical Calculator, Holger Buick
Percentage Calculator 1.1
1.5 MB |
Freeware |
Category: Mathematics
Calculate the value of a percentage of a number or enter two numbers to calculate the percentage the first is of the second. It is not difficult to learn how to calculate the percentage of one
number vs. another number. All you have to do is divide one number by another number. Then, then take that number and move the decimal place two spaces to the right. This way you...
OS: Windows
Software Terms: Calculate Percentage, Percentage Calculator
Calculate the kt factor within a cylinder. Calculate the kt factor within a cylinder. Kt Calc is an software that runs under Windows, written in Visual Basic 2005 Express, to calculate the kt factor
useful to calculate stress in mechanical engeneering with the model of a cilinder with an...
389.1 KB |
Freeware |
Category: Skins
"Calculate the number of days between given dates. "Calculate the number of days between given dates." EditByBSEditor: Calculate the number of days between given dates. This is a nifty little
program that will perform date arithmetic. Simply supply any two of the three fields, and the program will...
OS: Windows
Software Terms: Dave, David, Grund, Software, Sr
2.3 MB |
Shareware |
US$19.95 |
Category: Miscellaneous
Allows you to calculate and save the information about the bicycle gears Bicycle Gear Calculator is a very powerful application for serious cyclists. It will not only calculate a bicycle's gearing,
but will also save the information for later retrieval. Measure the wheels, enter the gear numbers, and Bicycle Gear...
OS: Windows
Software Terms: Bicycle, Bicycle Gear Calculator, Calculate, Calculate Gear Ratio, Gear, Wheel Measurement
529.0 KB |
Freeware |
Category: Mathematics
Z-Calculate is an innovative calculator for students, scientists etc. Z-Calculate is an innovative calculator for students, scientists etc. It combines mathematical power and cool features with a
clear and user-friendly interface. Both real and complex numbers are supported. And, best of all, it is completely...
OS: Windows
Calculate resonant home theater modes. Calculate resonant home theater modes. Calculate Room Acoustic Measurements is an automated Excel spreadsheet designed to help you to calculate room acoustic
measurements for home theaters. The only cells in which you need to enter data are in...
0 |
Shareware |
US$29.95 |
Category: Miscellaneous
The NEW Scuba Diving Calculators now comes with a Universal Multi Dive Table Planner providing you with the ability to calculate and log all your dives. Designed easy enough for the beginner scuba
diver to being advanced enough for Scuba Instructors use. Calculate your Depth, Residual Nitrogen, Time in and Out, PSI used, ABT, TBT, Average Depth and more. Each one of your dives is logged into
a database so you will always be able to retrieve...
OS: Windows, XP, 2000, 98, Me
Software Terms: Gas Consumption, Maximum Operating Depth, Mod, Nitrox, Sac, Scuba Diving Calculators, Scuba Tank Volume, Trimix
Warpulator you calculate warp speed. Warpulator you calculate warp speed. This software is dedicatedto all Trekkies and uses the newest formula for your calculation. Features: Allows to calculate
velocity to Warp Scale and Vice Versa.
Easily calculate your electricity bill. Easily calculate your electricity bill. Twin Tigers Watt Hour Meter can calculate your electricity bill. Maximum 10 types of Instruments can be added. The
working hour in a day can be entered in Hours, Minutes and Seconds. After entering the power...
Install this Ebay/Paypal calculate to quickly calculate fees for your ebay listings. Install this Ebay/Paypal calculate to quickly calculate fees for your ebay listings. There are two modes advance,
which is detailed and basic, less detailed. Many options to choose from giving better calculations and more accuracy.
872.0 KB |
Shareware |
US$12 |
Category: Mathematics
The calculator can calculate and plot formulas with variables. The calculator can calculate and plot formulas with variables. Tooltip-Help for all functions of the calculator are available. The
plotter can calculate null, intersection of functions, minima and maxima. The statisic can calculate average and...
OS: Windows
Software Terms: Calculator, Formula, Formula-calculator, Lucent, Lucent Calculator, Mathematic, Plotter, Statistics, Variable, Windows-calculator
Calculate age with this utility. Calculate age with this utility. AGE Calculator can calculate your age. You just enter the needed parameters and this software will give you the answer in a matter
of seconds.
5.0 MB |
Freeware |
Category: Mathematics
Calculate the percentage of any number out of any number. Add percentage to an amount or calculate a discount percentage. For tip calculation. Figure the selling price after a discount was applied.
Percent lets you quickly calculate percentages such as tips, added taxes and price discounts. Easily find a percentage from an amount entered. Calculate the percentage of any number out of any
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OS: Windows
Software Terms: Calculate, Calculator, Discount Calculator, Educate, Education, Find Percent, Learn, Math, Percent, Percentage | {"url":"http://www.bluechillies.com/software/calculate-mixture.html","timestamp":"2014-04-19T09:26:07Z","content_type":null,"content_length":"57710","record_id":"<urn:uuid:f175b602-cc2f-4369-ae33-55b80f98121d>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00385-ip-10-147-4-33.ec2.internal.warc.gz"} |
Steam (Vapor) Power Cycle (Rankine Cycle) Examples - TEST Tutorial
Launch the open cycle daemon located at the page TEST. Daemons. Systems. Open. Steady. Specific. PowerCycles. PhaseChange.
Let us set up the cycle as follows: Device-A: isentropic pumping from State-1 to State-2 ; Device-B : constant pressure boiler with State-2 and State-4 as inlets and State-3 and State-5 as exits;
Device-C : high pressure isentropic turbine from State-3 to State-4 ; Device-D : low pressure isentropic turbine from State-4 to State-5 ; Device-E : constant pressure heat rejection from State-4 to
State-1 .
State-1: Enter mdot1 (assume 1 kg/s), p1 (15 kPa) and x1 (0%). Calculate .
State-2: Enter p2 (15 MPa), s2 ('=s1'), and Calculate.
State-3: Enter p3 ('=p2'), T3 (620C) and Calculate.
State-4: Enter p4 ('=p5'), s4 ('=s3'), and Calculate. Note that p5 is not yet known.
State-5: Enter T5 ('=T3'), s5 ('=s6') and Calculate . Note that s6 is not yet known.
State-6: Choose 'More...' from the state selector to add more states to the menu. Choose State-6. Enter p6 ('=p1'), x6 (90%) and Calculate.
Super-Calculate to propagates s6 back to State-5 and then p5 back to State-4, thus completing evaluation of all states. It is always a good practice to draw a T-s or some other thermodynamic plot to
visualize the calculated states before proceeding to other panels. | {"url":"http://www1.ceit.es/asignaturas/termo/test/testcenter/testhome/Test/intro/exOpenVaporCycleP.html","timestamp":"2014-04-20T16:03:03Z","content_type":null,"content_length":"26860","record_id":"<urn:uuid:c2a48399-49b8-4e23-a7a0-d3f87c0ca3b3>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00495-ip-10-147-4-33.ec2.internal.warc.gz"} |
Embedding Leveled Hypercube Algorithms into Hypercubes
When a leveled hypercube algorithm (one dimension used at a time) is mapped in the straightforward way into a hypercube in which all edges are useable at once, most of the host machine's bandwidth is
unused. This paper shows how to construct embeddings which better utilize the host's edges. In particular, I show how to map an n-dimensional leveled hypercube algorithm into an n-dimensional host
hypercube so that the communication throughput of every guest edge is Theta((n/log n)^log_6 2) = omega(n^0.386) times the communication throughput of a host edge. Furthermore, the routing can be done
on edge-disjoint paths of length at most n. This result can be applied to other algorithms that are run on hypercubes. For example, if an algorithm runs on a mesh with a axes each of length 2^l, but
uses only one axis at a time, then it can be embedded in an la-dimensional hypercube so that each mesh edge has throughput Theta(l(a/log a)^log_6 2).
SPAA '92, 4th Annual ACM Symposium on Parallel Algorithms and Architectures , pp. 264-270, 1992.
PostScript version | {"url":"http://research.microsoft.com/en-us/um/people/dbwilson/embed/","timestamp":"2014-04-19T18:57:03Z","content_type":null,"content_length":"3026","record_id":"<urn:uuid:db260c77-7456-434a-961b-1084b6386a4e>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00185-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Wikipedia Poker page has numbers on the various standard poker hands, but I've been tasked with scoring odd ball hands and hands with wild cards, and furthermore, with validating my scoring
logic. I have test runs, representing all 52-choose-5 [footnote 1] possible hands run through the hand recognition logic (wild cards are 2s).
I was able to predict how many hands would score as most things but the wild card straight flush has me stumped. Here's what I think the number should be and why. Tell me why I'm wrong.
I'm doing two numbers and adding them together. Hands with a logical high card (which might be a wildcard) of A K Q J 10 9 8 or 7 don't use a 2 as a 2 so they have potentially all four wild cards
available (but in reality, four wilds is another hand that takes priority, so we have to subtract that out later). That's 8 possible logical high card values, times 4 suits. Then there's the exact
mixture of actual non-wild cards and wild cards. The five cards composing that whatever-high straight flush can be any mixture of the non-wild normal values and the four wild cards. That's 9 cards
those 5 cards are chosen from, or 9-choose-5. So 8 * 4 * 9-choose-5 which is 4032.
Then the same thing over again, but for 6 and 5 high hands, where one of the wild cards has to stand in for the actual 2: 2 high card values, 4 suits for whichever of the cards aren't wild, then a
mixture of five non-wild cards and the three remaining wilds, or 8-choose-5, which is 448.
Add those for 4480. Subtract out 48+4+480 for the hands four 2s, royal flush, and wild royal flush. And it's still way off what the actual Perl scoring logic came up with: 2,068.
So, what did I do wrong, and how do I fix it?
Footnote 1:
See http://en.wikipedia.org/wiki/Binomial_coefficient for the 52-choose-5 thing. It's normally written as one number over another both surrounded by (). It's the nCr button on your calculator. For
example, 52-choose-5 is 2,598,960.
The Fine Print: The following comments are owned by whoever posted them. We are not responsible for them in any way.
Loading... please wait.
Duplicates... (Score:1)
Alright, I see a problem. Duplicates. Eg,
K 2 2 2 2 and 2 K 2 2 2 are both the same hand, but the above considers one king-high and the other ace-high. (2s are wild.)
Not sure how to represent the problem now.
• Re: (Score:2)
When wild cards are used in poker doesn't five of a kind become the highest hand? At least that's what I vaguely recall from Hoyle.
A brief skimming of wikipedia and the wikibooks for poker doesn't show that but then again I would think the odds of 5 of a kind shouldn't be much different than four of a kind except that it's a
specific 4 cards that are needed.
Not sure how it really affects your calculations though.
Well, this isn't real poker, but instead video poker, where the rules are made up and people like it that way. But five of a kind is the highest paying hand in one of these. The other, five
3-5's, five 6-k, and five aces are all separate hands, and flush comes before four 2s (four wilds). So the number of card combinations that have a chance to match as a hand does vary by game.
Simply subtracting out things that overlap, like subtracting out non-royal flushes and four 2s seems to work, at least for the
• Here is my solution to the problem.
There are 40 hands that you are trying to emulate. All of the straight flushes for 4 suits from ten low down to ace low. To avoid duplicates I will only count ways of representing the best
possible hand. That means we can immediately ignore the 4 hands with a low card of 2 because the wild card can represent a 7 instead for a better hand. Let's segment possibilities by how many
wild cards there are, whether you are representing one of the 4 royal flushes, the 28 straigh
□ *sigh*. Full houses are easy. You can't have any wild cards because if you did you'd go for 4 of a kind instead. So you have 12 ways to pick the suit you're going to have 3 of a kind in, 4
choices of the cards in that kind, 11 ways to pick the one you have a pair in, and 6 choices for what the pair is. For 9504 possible full houses.
Much harder is 3 of a kind.
☆ Re: (Score:2)
jmm (276)
A full house with a wild card can happen. Two pair plus a wild card gives a full house. (Terminology nit: the three of a kind does not occur in a suit, but at a rank. 3 hearts and 2
spades is not a full house. :-)
○ I noticed that then decided not to reply to myself again just to point that out.
I skimmed this to see that you came up with basically the same final result as me and then hurriedly moved on to other things, and then got stuck again. I'm still waiting for word on whether
something without a big fat E's and ()'s is acceptable.
There's two things here... I mentioned this briefly before... but again...
1. How many combinations of the 52 cards are recgognized as that hand by itself
2. How many combinations of the 52 cards are recognized as that hand when other overlapping hands are tested fir | {"url":"http://use.perl.org/use.perl.org/_scrottie/journal/38497.html","timestamp":"2014-04-21T12:40:21Z","content_type":null,"content_length":"36740","record_id":"<urn:uuid:e9469880-7eb4-4458-86ea-49fa72f89b57>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00276-ip-10-147-4-33.ec2.internal.warc.gz"} |
Stellar Numbers Math Ib Sl Portfolio Investigation Mathematics Essay
Mathematics is a subject with simple purpose, which is to find and understand patterns and try to apply the patterns to real life situation. In this investigation however, we are dealing with
sequences which in this specific case are not applicable to real life as we do not encounter stellar numbers and triangular numbers in everyday situations. The purpose of this investigation is to
find and recognize progressive patterns.
For the general statement of the nth triangular number I have constructed a table of trials and improvements. I started by using the trial and improvement method then looked at it algebraically as
well. This process was a lengthy one however I eventually was able to deduce the final correct general statement based on my trials. This is done by replacing Un(The general term which is substituted
by a number i.e. 1,2,3,4,5,6,7,8) by the triangular number we are trying to find (in this case all positive integers work), and n is the same number as well, which is plugged into each equation.
To construct these stellar numbers manually or on a computer, one must first imagine a regular polygon that will represent the number of vertices. So for a 3 vertices stellar, a triangle, a 4
vertices stellar a square…. and a 20 vertices stellar a icosagon. Once you have imagined this shape, you can plot your stage 1 points, beginning with the shape and then a little further away from
your shape and inbetween the points you add dots to make it star like. As demonstrated below(Thick black lines):
There are limitations to the formula as there cannot be a negative number for p or n as there are no negative stellars or 0 stellar numbers. P can also not take the values of 1and 2. Further
limitations include irrational numbers, imaginary numbers, and fractions. Thus the general statement works in natural numbers above 3. The notation would be such: Sn= >3 where is the set of natural
numbers. Irrational numbers such as π would not work as you cannot have a stellar numbers with π vertices or at stages of π. In this case π has no exact value and stellars have an exact value thus it
is impossible to have a π stellar or an e stellar. Also it is impossible to have stellars which are fractions e.g.: 1/5 stellar: Sn=1/5 (2)2-1/5(2)+1=7/5. A 7/5 stellar cannot exist as stellars have
to have a value which is a natural number greater than or equal to 3. Imaginary numbers in the form (X+I), where X is a complex real number and I is an imaginary number such as √-1.
I arrived at the first general statement for a 6 stellar through trial and error, aided with the use of a calculator to help deduce the final general statement. Furthermore as I understood the
pattern for the 6 stellar, I tried the same type of formula for the 7 stellar, 5 stellar, 4 stellar, and 3 stellar. My reasoning was correct and through the analysis of the trials I was able to
deduce the general formula of: Sn=p(n)2-p(n) +1.
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11 projects tagged "Mathematics"
DigitizeIt digitizes scanned graphs and charts. Graphs can be loaded in nearly all common image formats (including gif, tiff, jpeg, bmp, png, psd, pcx, xbm, xpm, tga, pct), pasted from the clipboard,
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system and can be saved in ASCII format, ready to use in many other applications such as Microcal Origin or Excel. Axes can be linear, logarithmic, or reciprocal scale. Multiple data sets can be
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Re: Comments on 2.6 "Multiple Matches" (CR 6 Apr 2006)
From: Seaborne, Andy <andy.seaborne@hp.com> Date: Mon, 24 Jul 2006 10:03:59 +0100 Message-ID: <44C48CFF.6090600@hp.com> To: Fred Zemke <fred.zemke@oracle.com> CC: public-rdf-dawg@w3.org
Fred Zemke wrote:
> 2.6 Multiple matches
> First sentence: "The results of a query is the set of all pattern
> solutions that match the query pattern, giving all the ways a query
> can match the graph queried." But sets eliminate duplicates,
> and we have the DISTINCT operator as an optional syntactic choice
> about whether to eliminate duplicates. Instead, this should say
> that the result of a query is a sequence of solutions. See
> section 10.1 "Solution sequences and result forms".
rq24 : "2.2 Multiple Matches"
The results of a query is a sequence of solutions, giving the ways in which
the query pattern matches the data. The sequence of solutions is further
modified by the solution sequence modifiers. There may be zero, one or
multiple solutions to a query.
> 2.6 Multiple matches
> The semantics of the empty graph pattern has not been defined.
> I think the following queries are instructive:
> a) SELECT ?a
> FROM graph
> WHERE { }
> b) SELECT ?a ?b
> FROM graph
> WHERE { }
> c) SELECT ?a ?b
> FROM graph
> WHERE { ?a foaf:verb foaf:noun }
> d) SELECT ?a ?b
> FROM graph
> WHERE { ?a foaf:verb foaf:noun .
> OPTIONAL { ?a foaf:verb2 ?b } }
Variables need not be bound in any given solution - and may be bound in some
solutions and not others.
Added (rq24) "4.2 Empty Group Pattern"
> One's initial impulse is that query a) should result in the set of
> all mappings of { ?a } to the scoping set (not the set of all
> total mappings of V to the scoping set; see related comment).
> Or equivalently, the user might view the result as an enumeration
> of the scoping set of the graph.
A solution is the requirements for a match. Suggestions for how to say that,
in a way that does not mandate the logical reduction of the solution set over
the data (too expensive as per a previous discussion), welcome.
Something about required terms is noted as needed to be done (an @@ in the
doc) - and it would nicely allow different capabilities for different engine
and level of entailment. owl:sameAs could be exposed.
> Then query b) would result in the set of all mappings of { ?a, ?b }
> to the scoping set, or, naively, the cross product of the scoping
> set with itself.
> However, I believe that c) and d) should result in a subset of the
> result of b). Now in the case of d) in particular, OPTIONAL is
> intended to allow for a result which is a partial binding, ie, one
> that binds ?a but does not bind ?b. If it happens that there is
> no binding for ?b, then the result would not be a subset of the
> cross product of the scoping set with itself.
> My conclusion is that in order to support OPTIONAL and UNION, we
> have to permit a result that is a partial mapping.
> Coming back to query b), in order for it to contain query d) as a
> subset, the result of b) must be all partial functions from {?a, ?b}
> to the scoping set. Alterantively, a naive view might
> imagine augmenting the scoping set with a single "missing" element,
> distinct from all other elements, in which case the result of b)
> is the cross product of the augmented scoping set.
> And as for a), it seems the result must be the set of all partial
> functions of { ?a } to the scoping set, or equivalently, an
> enumeration of the augmented scoping set.
> As a different approach to this issue, consider these two queries:
> a1) SELECT ?a
> FROM graph
> WHERE { BOUND (?a) }
> a2) SELECT ?a
> FROM graph
> WHERE { !BOUND (?a) }
> I believe the following things:
> -- the result of a) should be the union of the result of a1) and a2)
> -- the result of a1 should be an enumeration of the scoping set,
> -- the result of a2 should be a single solution, in which ?a is
> not bound.
> Fred
Received on Monday, 24 July 2006 09:04:43 GMT
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[Numpy-discussion] numpy type mismatch
Benjamin Root ben.root@ou....
Fri Jun 10 14:50:32 CDT 2011
Came across an odd error while using numpy master. Note, my system is
>>> import numpy as np
>>> type(np.sum([1, 2, 3], dtype=np.int32)) == np.int32
>>> type(np.sum([1, 2, 3], dtype=np.int64)) == np.int64
>>> type(np.sum([1, 2, 3], dtype=np.float32)) == np.float32
>>> type(np.sum([1, 2, 3], dtype=np.float64)) == np.float64
So, only the summation performed with a np.int32 accumulator results in a
type that doesn't match the expected type. Now, for even more strangeness:
>>> type(np.sum([1, 2, 3], dtype=np.int32))
<type 'numpy.int32'>
>>> hex(id(type(np.sum([1, 2, 3], dtype=np.int32))))
>>> hex(id(np.int32))
So, the type from the sum() reports itself as a numpy int, but its memory
address is different from the memory address for np.int32.
Ben Root
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How can I see that the slice of a presheaf category is equivalent to the presheaf category of the category of elements?
up vote 3 down vote favorite
Let $\mathcal{C}$ be a small category and $P$ a presheaf on $\mathcal{C}$. Then there is an equivalence $$\widehat{\mathcal{C}} / P \cong \widehat{\int_{\mathcal{C}}} P$$ Now there are (at least) two
ways to see this (from what I've managed to ascertain.) You might derive the equivalence as it were 2-categorically using the universal property of $\widehat{\mathcal{C}}$ which is the equivalence $\
textbf{Cat}/\widehat{\mathcal{C}} = \textbf{Func }(\widehat{\mathcal{C}}) \cong \textbf{DFib}(\mathcal{C})$ where $A$ is any small category and $\textbf{DFib}$ are the discrete fibrations over $\
mathcal{C}$. You do this by showing that any functor $A \rightarrow \widehat{\mathcal{C}}/P$ corresponds to a functor $A \rightarrow \widehat{\mathcal{C}}$ and a natural transformation $A \rightarrow
1 \overset{P}{\rightarrow} \mathcal{C}$ (where $1$ is terminal in $\textbf{Cat}$) and hence that the corresponding morphism of fibrations corresponds to a discrete fibration of ${\int_{\mathcal{C}}}
P$ and hence that
$$\textbf{Func }(\widehat{\mathcal{C}}/P) \cong \textbf{DFib}(\int_{\mathcal{C}}P) \cong \textbf{Func }(\widehat{\int_{\mathcal{C}}P}) $$
which by universality gives the desired equivalence. Now I have to admit I'm a bit sketchy on the details of this, as it uses many non-obvious properties of fibrations. A more detailed version of
this approach was given as a partial answer to this question.
I am more interested in the more elementary approach, i.e. establishing an explicit equivalence between the two categories, so that I can understand, morally, why this equivalence ought to hold. The
only natural functor I've been able to find between the two is given firstly by factorizing $$\int_{\mathcal{C}}P \overset{\pi_P}{\rightarrow} \mathcal{C} \overset{y}{\rightarrow} \widehat{\mathcal
{C}}$$ through the Yoneda embedding to $\widehat{\int_{\mathcal{C}}P}$ and the canonical colimit functor $L : \widehat{\int_{\mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}$ given since $y \circ \
pi_P$ is a functor to a cocomplete category, and then getting to $\widehat{\mathcal{C}}/P$ by pullback $p$ along $P \rightarrow 1$. So the end result is a functor $$ p \circ L: \widehat{\int_{\
mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}/P$$ Does this functor provide the desired equivalence? There seems to be significant loss of information at $p$, but on the other hand $L$ does have
the property that it makes the two respective Yonneda embeddings commute (i.e. $L \circ y = y \circ \pi_P$), so morally it seems that the information lost at $p$ should be the same amount of
information lost by the projection $\pi_P : \int_{\mathcal{C}}P \rightarrow \mathcal{C}$ (which ought to make $p \circ L$ full and faithful; and I think essential surjectivity is straightforward.)
But I also get the sense I am chatting nonsense. If this functor does nothing of the sort then the question, I suppose, is what is the most elementary functor that provides this equivalence?
Any helpful comments will be greatly appreciated both by myself and my bloodshot eyes.
add comment
1 Answer
active oldest votes
You can verify this equivalence elementarily (without the language of fibrations etc.):
Assume first that $C$ is the category with only one morphism (i.e. the terminal category), so that a presheaf on it is just a set. Then the statement is as follows: If $P$ is a set,
then a set $F$ together with a map $F \to P$ is the same as to give a family of sets indexed by $P$. But this is obvious, right? Given $F \to P$, we may look at its fibers $F_s$, where
$s$ runs through the elements $s \in P$. Since $F$ is the disjoint union of the $F_s$, it is also clear how to write down the inverse.
For general $C$ it works in the same way, $C$ is just a sort of parametrization.
up vote 2
down vote If $F \in \widehat{C} / P$, i.e. $F$ is a presheaf on $C$ together with a morphism $F \to P$, then define a presheaf $G$ on $\int_C P$ as follows: If $(X,s)$ is an element of $P$, i.e
accepted $s \in P(X)$, then let $G(X,s)$ be the fiber of $s$ with respect to $F(X) \to P(X)$.
Conversely, if $G$ is a presheaf on $\int_C P$, then define a presheaf $F$ on $C$ as follows: For $X \in C$ let $F(X) = \coprod_{s \in P(X)} G(X,s)$. We have a natural projection $F(X)
\to P(X)$, which gives rise to a morphism $F \to P$.
Now it is straight forward to check that these assignments actually define functors which are pseudo-inverse to each other.
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excel, data & standard deviation Canceler?
October 18th, 2011, 05:59 PM
Join Date: Apr 2006
Executive Member
Posts: 5,141
Originally Posted by
blackjack avenger
A point graph you plot the days in order of occurrence and then connect the dots?
so.... i wasn't doin it right then, no?
maybe more like this ...........?
but how's that ever gonna look like a bell even after decades of plays?
October 18th, 2011, 06:14 PM
Join Date: Apr 2006
Executive Member
Posts: 5,141
hmm, maybe this way?
.......... ehhmm, that's gonna take me a while, lol................
October 18th, 2011, 07:15 PM
Join Date: Apr 2005
Executive Member Location: Minnesota
Posts: 1,625
Late to the party, and didn't bring much, but...
Obviously, with BJ we can run a sim and get our expected win rate and standard deviation. With other games, like poker, and whatever it is you’re doing, we have to generate our statistics based on
our own experience. As we’ve been reminded, with live play it’s unlikely we will ever have enough data to generate meaningful statistics. But still, we want to have
to look at.
I’ve never seen the use of STDEV recommended as a way to do this. In post #9 of
this thread
, statmanhal describes exactly how to set up a spreadsheet to calculate your hourly standard deviation. If you don’t want it to be hourly you can omit that part, or just plug in “1” for the session
duration to pretend that each session was only one hour.
To get your bell-shaped curve I think you’ll need to look at the number of days you get each kind of result. Say you have these kinds of results: Huge Loss, Moderate Loss, Average Outcome, Moderate
Win, Huge Win. The number of days you have an average outcome should be the bulge in the middle. The number of days you have huge wins or losses should be the tails on the sides.
October 18th, 2011, 08:24 PM
Join Date: Apr 2006
Executive Member
Posts: 5,141
Originally Posted by
Obviously, with BJ we can run a sim and get our expected win rate and standard deviation. With other games, like poker, and whatever it is you’re doing, we have to generate our statistics based on
our own experience. As we’ve been reminded, with live play it’s unlikely we will ever have enough data to generate meaningful statistics. But still, we want to have something to look at.
I’ve never seen the use of STDEV recommended as a way to do this. In post #9 of this thread, statmanhal describes exactly how to set up a spreadsheet to calculate your hourly standard deviation. If
you don’t want it to be hourly you can omit that part, or just plug in “1” for the session duration to pretend that each session was only one hour.
ahh ok, i haven't checked it entirely closely, but i think he's using essentially stdevp a function in excel, which is like the first example ie. (dividing by n) in this link:
stdev in excel would be like the second example (ie. dividing by (n-1), ( i believe) where one doesn't know the entire population but just has a sample. so maybe i'd be best off using stdev.......
To get your bell-shaped curve I think you’ll need to look at the number of days you get each kind of result. Say you have these kinds of results: Huge Loss, Moderate Loss, Average Outcome, Moderate
Win, Huge Win. The number of days you have an average outcome should be the bulge in the middle. The number of days you have huge wins or losses should be the tails on the sides.
there ya go, makes sense...... so technically i'd need one heck of a lot more data. and then would have to know how to do the math right in the first place, lol
edit: but anyway, some rainy day i think i'll try this method:
thank you
Last edited by sagefr0g; October 18th, 2011 at 08:33 PM.
October 18th, 2011, 10:18 PM
Join Date: Apr 2006
Executive Member
Posts: 5,141
Originally Posted by
To get your bell-shaped curve I think you’ll need to look at the number of days you get each kind of result. Say you have these kinds of results: Huge Loss, Moderate Loss, Average Outcome, Moderate
Win, Huge Win. The number of days you have an average outcome should be the bulge in the middle. The number of days you have huge wins or losses should be the tails on the sides.
so, like for the image below, which is results over days.....
i can kind of just conceptualize, Huge Loss, Moderate Loss, Average Outcome, Moderate Win, Huge Win ... ect. , and at least get a 'handle' on how things have been going, sorta thing........ gives one
an idea of what kind of dispersal of results has been happening, and
how it will tend to go in the future, and as time goes on and more data comes in, see how it all stacks up again, at least have a qualitative analysis if not quantitative, sorta thing, at least have
something to look at and get a feel for how things are going, maybe stack that analysis up against what excel thinks is one standard deviation, which in this case was $265.27 and the expected value
which is $65.83 ... to where for the time being at least, i can kind of say to myself, well as far as i know right now, i can expect to be within $200 or so plus or minus of a $65 result about 68% of
the time...... no?
Last edited by sagefr0g; October 18th, 2011 at 10:26 PM.
October 18th, 2011, 10:43 PM
Join Date: Apr 2006
Executive Member
Posts: 5,141
Originally Posted by
blackjack avenger
If you don't know your SD hope your betting conservatively.
I dont know
1/4 Kelly comes to mind
Don't want the misses mad
along this line of reasoning, let me ask you a question, since you seem to know a lot about Kelly stuff.
well, first off, speaking of betting, Kelly stuff and bank roll, well the roll so far has more than doubled.
the question being, when you are 'properly' betting and making positive EV plays, errrhh well, i dunno, is there some significance to reaching the point where the roll is doubled, far as Kelly theory
seems i've seen a lot of talk about the point where advantage players have doubled their roll, in conjunction with Kelly stuff. can you shed any light on that subject, ie. significance of doubling
the bank roll and Kelly stuff?
have anything to do with N0, maybe?
edit: like here's an example where doubling the roll is mentioned:
That's why people advocate 1/2 kelly or 1/4 kelly. If you bet half of the kelly fraction, then you have way less chance of losing 50% of your bankroll. For example, a full Kelly better has 1/3 chance
of losing 50% of his bankroll before doubling, while a 1/2 Kelly better has 1/9 chance of losing 50% of his bankroll before doubling.
from this post:
just one of many examples where "doubling the roll" is emphasized when discussing Kelly stuff
Last edited by sagefr0g; October 18th, 2011 at 11:37 PM.
October 19th, 2011, 12:11 AM
Join Date: Apr 2005
Executive Member Location: Minnesota
Posts: 1,625
I just used the Huge Loss, Moderate Loss, etc. to give you the concept of how to get your bell-shaped curve. Not to imply that that kind of chart would have much value, but you did seem to have your
heart set on it.
Nevertheless, here’s an example of what I had in mind. Being lazy, rather than make up fake data, I just used the outcomes of my last 100 sessions of 2/4 Limit Hold’em. (Astute observers will notice
I was a loser at that game, as most people are, but never mind that!)
The result types are in $25 increments. Result Type A is a loss of $175 to $200. B is a loss of $150 to $174.99. H is a loss between $0 and $25. I is a win between $0 and $25. P is a win of $175 to
The columns represent the number of sessions I had of each result type. With so few data points, the chart is only vaguely bell-shaped, but you get the idea. Again, this type of chart is of little
value. Your SD numbers will be more useful, I think.
October 19th, 2011, 12:38 AM
Join Date: Apr 2006
Executive Member
Posts: 5,141
Originally Posted by
I just used the Huge Loss, Moderate Loss, etc. to give you the concept of how to get your bell-shaped curve. Not to imply that that kind of chart would have much value, but you did seem to have your
heart set on it.
yah, lol, it's virtually all i got, for now, lol
excepting, i do know the EV.
just hoping to get some kind of a handle on the swings and how they may come down, regardless of how hazy that handle may be.
but i get your point.
Nevertheless, here’s an example of what I had in mind. Being lazy, rather than make up fake data, I just used the outcomes of my last 100 sessions of 2/4 Limit Hold’em. (Astute observers will notice
I was a loser at that game, as most people are, but never mind that!)
The result types are in $25 increments. Result Type A is a loss of $175 to $200. B is a loss of $150 to $174.99. H is a loss between $0 and $25. I is a win between $0 and $25. P is a win of $175 to
The columns represent the number of sessions I had of each result type. With so few data points, the chart is only vaguely bell-shaped, but you get the idea. Again, this type of chart is of little
value. Your SD numbers will be more useful, I think.
yup, got it, definitely makes it clear how the bell curve is arrived at, thank you.
October 19th, 2011, 04:07 AM
Join Date: Feb 2007
Executive Member
Posts: 2,267
floating point graph?
Originally Posted by
so.... i wasn't doin it right then, no?
maybe more like this ...........?
but how's that ever gonna look like a bell even after decades of plays?
I am sure there is a real name for this graph, but it escapes me.
Start the graph at starting bank in the middle of the Y verticle axis, then let the results be plotted & added & subtracted at each point. Then connect the dots. At the final dot your bank should be
at its current level. A nice thing, this graph can be done with regular notebook or graph paper. Also, this graph can show results over time, an example if you change strategy.
I started a thread in "other games" you may find interesting.
Last edited by blackjack avenger; October 19th, 2011 at 06:55 AM.
October 19th, 2011, 11:52 AM
Join Date: Feb 2007
Executive Member
Posts: 2,267
[QUOTE=sagefr0g;257339]along this line of reasoning, let me ask you a question, since you seem to know a lot about Kelly stuff.
well, first off, speaking of betting, Kelly stuff and bank roll, well the roll so far has more than doubled.
the question being, when you are 'properly' betting and making positive EV plays, errrhh well, i dunno, is there some significance to reaching the point where the roll is doubled, far as Kelly theory
seems i've seen a lot of talk about the point where advantage players have doubled their roll, in conjunction with Kelly stuff. can you shed any light on that subject, ie. significance of doubling
the bank roll and Kelly stuff?
have anything to do with N0, maybe?
Why talk about doubling bank? Are you retiring? So assuming one is going to play on the risk of drawdown numbers approach the infinite numbers quickly.
Chances of losing % of bank with Kelly resizing:
50% chance of losing 50% of bank with Kelly
80% chance of losing 20% of bank with Kelly
Chances of losing % of bank with 1/4 Kelly:
.08% chance of losing 50% of bank with 1/4 Kelly
21% chance of losing 20% of bank with 1/4 Kelly
Begs the question. Why would anyone play a positive expectation game with an ror? Now, what if one figured the advantage correctly on an unsure game but bet kelly? They face a 50% chance of losing
half, if that happens I would think confidence would suffer.
Kelly & N0? Kelly raises N0 by a factor of 9. However betting 1/4 Kelly does not raise N0 greatly though growth suffers.
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y=f(x) transformation
February 13th 2013, 12:50 PM
y=f(x) transformation
1. describe how y=f(x) is transformed by each of the following:
a) y = -f(x+2) - 5
b) y = f(-x) + 5
if some could please help! Thank you in advanced !
February 13th 2013, 01:00 PM
Re: y=f(x) transformation
What does swapping x for x- 2 do to the graph? It might help to graph y= x- 2 itself first.
Same for swapping x for -x.
What does swapping y for y+ 5 do to the graph? What is the difference between y= x and y= x- 5?
February 14th 2013, 10:35 PM
Re: y=f(x) transformation
the transformation f(x+2) translates the graph 2 points on the left. The transformation -f(x+2) -5 translates the function 2 points to the left and 5 points down then refleccts it over the
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newbie: general overview wanted [Archive] - OpenGL Discussion and Help Forums
i'm just starting out trying to learn ogl but i need to know the grand scheme of things before this stuff clicks for me. the following is my thinking. please help me clear it up:
when you open a window ready for OpenGL to draw in it, you specify a coordinate system, like, "the center of the world shown in the window is (0,0,0)".
you then draw a shape, series of connected lines, etc.. by specifying vertices between glBegin() and glEnd() commands. the coordinates of these vertices are with respect to the world c.s. specified
you then tell ogl: where you're looking at the scene from (x,y,z) and whether it's a wide angle view or a close-up. this gets done after (?) the glBegin()/glEnd() business...
now suppose i want to move the specified object about the screen (err,... about the "world"). does ogl keep special 4x4 matrices around for this purpose? how often are the values of these matrices
changed? suppose a scene with many different moving objects is being rendered; do each of these objects have their own transformation/translation matrices or is the same one matrix constantly changed
and used one-by-one on each of the different objects?
and... i really have no idea how glPushMatrix() and glPopMatrix fit it to this whole thing.
thanks in advance for your comments. please begin sermon now. :-) | {"url":"http://www.opengl.org/discussion_boards/archive/index.php/t-129753.html","timestamp":"2014-04-21T02:52:53Z","content_type":null,"content_length":"6478","record_id":"<urn:uuid:058e0ef4-97e4-4fd8-8aa8-37a56d484307>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00054-ip-10-147-4-33.ec2.internal.warc.gz"} |
A Full Day of Grading
Teachers, TAs, and professors love to complain about the amount of grading they have. But how much is that, exactly? I've seen
that say assessments and assessment-related activities (I imagine this includes creation, proctoring, and grading) take up 1/3 - 1/2 of a teacher's time. I certainly believe it. But I thought it
might be interesting/useful/fun procrastination to consider a rough estimate of how many hours per week it might be reasonable to expect a teacher (in this case, imagine a high school teacher) to
spend just grading. And I found what I came up with to be rather amazing (despite all the time I've personally spent grading).
Here are my basic
• A teacher has about 100 students (imagine 4 classes of 25 students each). These days, many teachers teach 5 classes of 30 each. Some pampered private school teachers (like I was) teach 4 classes
of 15 students each. Still, 100 is a reasonable estimate and a nice round number.
• Each student hands in one page to be graded each day. This will likely be a single sheet of homework. This excludes any tests or quizes, which are generally longer than a sheet, so this is
definitely a low-end estimate.
• Each page takes the teacher 1 minute to grade. Again, this sounds like a short time to me if you're actually grading the work. And if you also consider the time it takes to determine the
full-page grade and enter it into the computer system...
These estimates lead to a need to grade 500 pages per week, which I'm estimating could take 500 minutes. That's 8.33 hours, or
a full work day
. Yes, a teacher could easily spend
a full work day each week just
grading. I find that astonishing.
No comments: | {"url":"http://kdphd.blogspot.com/2011/02/full-day-of-grading.html","timestamp":"2014-04-19T19:53:54Z","content_type":null,"content_length":"76518","record_id":"<urn:uuid:efba5894-6940-4fca-85ac-72afcd8b82f5>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00484-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two Stage Least Squares
August 27th 2011, 10:47 PM #1
Junior Member
Nov 2009
Two Stage Least Squares
You are given a dependent variable y, and independent variable x such that the regression would be ( x and y are endogenous):
$y=\beta_{0} + \beta_{1}x + \epsilon$
There is the possibility that the coefficients estimated from the model is affected by x and y having measurement errors in them.
You are also given the lagged (by one period) dataset of x and y (x_1 and y_1 respectively).
How would you go about fixing this error?
May I ask if my approach is correct?
I would use the lagged data to estimate the true values of x & y, so that the errors are no longer correlated so that the model would look as follows
$y_1=\beta_{0} + \beta_{1}x_1 + \epsilon \\ x = x_1 + u_{1} \\ y = y_{1} + u_{2}$
Then this would be the equivalent of running a two way least squares regression with x_{1} and y_{1} as instrumental variables?
Thank you in advance for any feedback
Last edited by lindah; August 27th 2011 at 11:27 PM.
Re: Two Stage Least Squares
I would have to look it up to confirm your approach, but it sounds like the correct approach in applying a two-stage OLS using the autocorrelation to account for the measurement error. I've just
never seen it used like that (due to my ignorance, not because I can say it isn't used like that). If you don't get additional help here, I'd also query at the talk stats forums.
September 3rd 2011, 07:52 PM #2
May 2011
Sacramento, CA | {"url":"http://mathhelpforum.com/advanced-statistics/186831-two-stage-least-squares.html","timestamp":"2014-04-19T13:11:44Z","content_type":null,"content_length":"32600","record_id":"<urn:uuid:c24d3584-9cce-4be7-a9ce-6bcf1b304deb>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00327-ip-10-147-4-33.ec2.internal.warc.gz"} |
Indefinite integral of (x+(1/x))^4
September 13th 2012, 11:00 PM #1
Sep 2012
Hello, I'm Fahad and I'm new here
I'm here to get help with simple math problems for my IGCSES
Plese help me.
The following math is from the binomial expansions part of my syllabus:
⌠ (x+(1/x))^4
Please show your working and thanks for the help.
Also, please refer to the techniques of integration you are using.
$\int \left(x+\frac{1}{x}\right)^4 \, dx$ = $\int \left(6+\frac{1}{x^4}+\frac{4}{x^2}+4 x^2+x^4\right) \, dx$
Re: Indefinite integral of (x+(1/x))^4
Please show your working throughout and please simplify
Re: Indefinite integral of (x+(1/x))^4
We are given:
$\int \left(x+\frac{1}{x} \right)^4\,dx$
Using the binomial theorem on the integrand, we find:
$\left(x+\frac{1}{x} \right)^4=\left(x+x^{-1} \right)^4=\sum_{k=0}^4{4 \choose k}x^{4-k}x^{-k}$
After simplification:
So then we have:
$I=\int x^4+4x^2+6+4x^{-2}+x^{-4}\,dx$
We will integrate term by term, using the power rule for integration $\int k\cdot x^n\,dx=\frac{k}{n+1}x^{n+1}+C$
where k is a constant and $ne-1$
So we have:
Last edited by MarkFL; September 13th 2012 at 11:28 PM.
September 13th 2012, 11:06 PM #2
September 13th 2012, 11:08 PM #3
Sep 2012
September 13th 2012, 11:12 PM #4 | {"url":"http://mathhelpforum.com/new-users/203431-indefinite-integral-x-1-x-4-a.html","timestamp":"2014-04-19T00:04:31Z","content_type":null,"content_length":"40672","record_id":"<urn:uuid:43d1fc29-3c9f-431b-aa94-b766a5d3966d>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00239-ip-10-147-4-33.ec2.internal.warc.gz"} |
Products of matrices of a certain form
up vote 2 down vote favorite
Are $n \times n$ matrices of the form $$\pmatrix{1&1&1&1 \cr x&1&1&1 \cr x&x&1&1 \cr x&x&x&1}$$ studied anywhere? I am interested in the structure of the matrix obtained by multiplying a bunch of
these together.
matrices linear-algebra co.combinatorics
2 These are special Töplitz matrices. Maybe you should look at their inverses: it has -1/(x-1) on the main diagonal, 1/(x-1) on the superdiagonal and x/(x-1) in the lower left corner. All the other
entries are zero. – Martin Rubey Apr 18 '13 at 16:05
They also happen to be semiseparable. – Federico Poloni Apr 18 '13 at 18:16
Thank you Martin and Federico. $1/(1-x)$ resonates well with my problem! – Rodrigo A. Pérez Apr 18 '13 at 22:11
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Topic: It is shown that ZFC is inconsistent
Replies: 0
byron It is shown that ZFC is inconsistent
Posted: Feb 6, 2013 6:46 AM
Posts: 865
Registered: 3/3/09 It is shown that ZFC is inconsistent
in ZFC IS AN AXIOM CALLED THE Axiom schema of specification
3. Axiom schema of specification (also called the axiom schema of
separation or of restricted comprehension): If z is a set, and \phi\! is
any property which may characterize the elements x of z, then there is a
subset y of z containing those x in z which satisfy the property. The
"restriction" to z is necessary to avoid Russell's paradox and its
the axiom of separation is used to outlaw/block/ban impredicative statements like Russells paradox
the axiom of separation of ZFC is it self impredicative as Solomon Ferferman points out
"in ZF the fundamental source of impredicativity is the seperation axiom
which asserts that for each well formed function p(x)of the language ZF
the existence of the set x : x } a ^ p(x) for any set a Since the formular
p may contain quantifiers ranging over the supposed "totality" of all the
sets this is impredicativity according to the VCP this impredicativity is given teeth by the axiom of infinity
thus it outlaws/blocks/bans itself
thus ZFC contradicts itself and 1)ZFC is inconsistent 2) that the paradoxes it was meant to avoid are now still valid and thus mathematics is inconsistent
Now we have paradoxes like
Russells paradox
Banach-Tarskin paradox
Burili-Forti paradox
Which are now still valid
ZFC is shown to be inconsistent by australias leading erotic poet colin leslie dean | {"url":"http://mathforum.org/kb/thread.jspa?threadID=2433204","timestamp":"2014-04-17T09:51:37Z","content_type":null,"content_length":"15522","record_id":"<urn:uuid:cdc3baa1-a151-4f4e-9a51-4d67735623e0>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00060-ip-10-147-4-33.ec2.internal.warc.gz"} |
On Preserving: Essays on Preservationism and Paraconsistent Logic
This volume is an announcement that an important school of philosophical logic is flourishing in Canada. The essays in On Preserving offer an original conception of logic called preservationism and
studies of an accompanying inference relation called forcing (not to be confused with Cohen's eponymous methods in set theory). Peter Schotch and Raymond Jennings pioneered these ideas circa 1980,
and work is now carried on by their former student Bryson Brown, logical polymath Alasdair Urquhart, and a third generation: Dorian Nicholson, Kim Sing Leung, and Gillman Payette. And so the volume
marks an arrival -- the Canadian school of paraconsistency. I came to this book hoping for a self-contained and accessible introduction to preservationism, which has otherwise only been available in
scattered papers. The book will give an attentive reader a solid understanding of the preservationist program, from its foundations to its cutting edge. The book is a landmark contribution to the
projects of modal and paraconsistent logic, as well as philosophical logic more generally.
As a non-classical logic, the basic idea behind preservationism is that something is wrong with the classical inference relation. If we ever find ourselves with beliefs and commitments that are, as a
whole, inconsistent, the classical rule of ex falso quodlibet reduces everything to triviality. This seems neither descriptively nor prescriptively adequate. The preservationist response is not to
replace the rules of classical logic, though, but to graft on a new piece of machinery, called the forcing relation, defined in terms of classical logic. Forcing, when possible, avoids ex falso
quodlibet by being sensitive to the consistency of subsets of premises, not the set of premises overall. To paraphrase (from p. 98):
Question: How do you reason from an inconsistent set of premises?
Answer: You don't, since every formula follows in that case. You reason from consistent subsets of premises.
Preservationism is presented as a contingency plan for when we are stuck with bad data (p. 30). This is an unusual and interesting tack amongst non-classical logics, others of which are commonly
thought of as rivals, replacements, or alternatives to standard logic. It is unusual as a paraconsistent system, too, since at the global level ex falso quodlibet still holds.
In Schotch and Jennings' semi-autobiographical introduction, we learn that preservationism was born out of concerns with modal logic. Wanting to break the validity of certain theorems in deontic
logic, the authors hit on the idea (independently discovered by Tarski and Jonsson in 1951) of generalizing the access relation between worlds; instead of modalities defined in terms of Rxy, the
authors study Rxy[0] … y[n]. Schotch and Jennings presented this work in 1978, not then aware of or concerned with paraconsistency. An audience member pointed out that the n-frames looked to her like
a model of non-trivial reasoning from inconsistent data. With n worlds on hand, there are in effect n separate compartments for storing data. For example, if we must store three formulae in two
worlds, then by 'pigeonhole' reasoning, any two of the formula must appear together at a world, but there need not be a world with all three formula together. Inconsistent information in particular
can be kept separate, and ex falso quodlibet subverted, as long as there isn't too much inconsistency relative to the number of worlds on hand. Thus, as if by accident, Schotch and Jennings found
themselves with the seeds of a paraconsistent logic.
What is the forcing relation, and what does it preserve? The classical inference relation preserves truth between a set of premises and a conclusion. In their important chapter, Schotch and Payette
argue that the classical relation is dubious, since a "true set of premises" is either a meaningless expression (how can a set be true?) or else presupposes that a set of data {A, … , B} can be
incontrovertibly gathered up into a single sentence A& … &B, and the latter is, in effect, just what preservationism controverts. The forcing relation instead relates sets of sentences, and preserves
how finely the sets must be divided up if all the resulting cells are to be internally consistent. In jargon, forcing preserves level. Again, the notion of forcing depends on an already-defined
underlying logic X; the X-level of a set of sentences is n iff n is the smallest partition where each cell is consistent with respect to X. (If no such n exists, as is the classical case for a
singleton like {A&~A}, then the level is said to be infinite.) Where Γ is a set of formulae, A a formula, and X a logic,
Γ forces A,
for any division of Γ into n cells, with n the least possible consistent partitioning, at least one cell X-proves A.
Forcing is a relation between sets, and preserves level, in the sense that the level of Γ is identical to the level of the closure of the forcing relation on Γ. Less tersely, forcing seems to
preserve the coherence, or incoherence, of a set of premises.
The strength of the book throughout is in its attractive technical developments. The layout is effective, with many boxed definitions and clearly marked theorems.
Three of the early essays (chapters 3, 4, and 5) focus on the modal origins of preservationism. The main purpose of the early chapters is to provide proofs, including a lovely arrangement by
Urquhart, that the n-ary frame conditions match up with the Schotch/Jennings condition for necessity, ◻:
◻A holds at world x
for all y[0], … , y[n], Rxy[0] … y[n] and A holds at one or more one of y[0] … y[n].
This is the way that ◻A and □~A can obtain, without ◻(A & ~A). Nicholson's chapter includes some illuminating diagrammatic representations of the otherwise mysterious n-frames, and Leung and
Jennings' chapter is a detailed treatment of the relationship between frames and modal axioms.
How all that relates to forcing and preservation, it must be said, is not immediately clear. Someone not already conversant with preservationism will need to take some time in settling in. On this
count, the arrangement of chapters is not ideal. Faced with 50 pages of algebraic completeness proofs of various stripes, firstly, a non-negligible amount of mathematical sophistication is needed (I
doubt only one semester of formal logic would do), and secondly, while completeness is important, it is informative only after someone fully understands and wants to study preservationism.
I would recommend non-initiate readers to skip quickly to focus on the central chapters by Gillman Payette, starting with "Preserving What?" (co-authored with Schotch). This is where the main
definitions are clearly laid out and some core theorems stated and proved. These helpful chapters detail the mechanics of forcing, and make precise just what can and cannot be preserved. They show,
for example, that a logic can be completely characterized by the concept of level. Results like this are an indication that the preservationist approach has tapped a deep vein.
In his standout chapter "Preserving Logical Structure", Payette investigates which properties of the underlying logic X are retained by the X-level forcing relation. Many properties are.
Monotonicity, though, is not, at least not in full generality. That is, just because Γ forces A, it does not follow that the set theoretic union Γ∪Δ will force A. This is interesting both technically
and philosophically, since failure of monotonicity has elsewhere been thought to model belief revision. Payette shows that preservationists only have the conditioned principle that if adding
information to the premises does not make the premises more inconsistent, then forcing preserves level (p. 129). Other structural properties of the underlying logic X are fully preserved; in fact,
forcing is so well behaved that even logics without a cut rule can inherit a sort of cut rule, which is striking. With his work Payette has done a service to his school and given lucid attention to
an interesting new technology.
Brown and Nicholson's useful chapter produces a syntactic representation of forcing. They develop too the general failure of monotonicity and give it a philosophical basis. The intuition against ex
falso quodlibet is that an inference from (maybe inconsistent) premises to absolutely everything just isn't an inference at all. There is no sense in which an arbitrary formula is derived from
inconsistent premises. What, then, about the dual, the inference from no premises to a conclusion? Usually we call such a conclusion a theorem or logical truth. The authors point out that if ex falso
is bad reasoning, then its dual is bad, too. It was at this point I saw clearly that preservationism is about reasoning in the medium-sized realm of human experience -- filtering out what happens at
the fringes of both contradiction and tautology. Using a multiple conclusion logic, Brown and Nicholson succeed at characterizing an X-forcing relation that captures X-level forcing. In the related
final chapter, Brown compares forcing to relevant and dialetheic logics. He shows in effect how preservationist ideas about ambiguity can lead naturally to Dunn's truth tables for first-degree
entailment FDE. This makes sense in the context of blocking "trivial" inferences -- since as a consequence of its semantics, FDE has no theorems at all.
Motivational philosophical topics such as ethical reasoning and practical inference are introduced by Schotch and Jennings in chapter 2 and treated at length by Schotch in chapter 9. Schotch develops
some generalizations and variations on forcing to model more closely what happens in real life cases of belief revision. He makes a good case that science works by holding various commitments that,
nevertheless, can be overturned, that this involves some degree of inconsistency, and that this process can be modeled by forcing. There are also repeated suggestions, intriguing if rather hastily
stated, that deontic logic ought to take paraconsistency seriously.
This is a slim volume, and, especially as the book is the first official presentation of the school, I would have liked to see more articulation and discussion of this neat philosophy of logic. In
fiction, a collection of interconnected short stories can often signal a novel that never quite came together, and there is some of that feel here. The introductory essay provides some frame for the
whole, but the chapters are not integrated. Some material is repeated from chapter to chapter, while other material is never really explained at all. I think the book could have been a single
co-authored monograph -- but it is not. Let me just register a few loose ends. Apart from a nice diagram by Nicholson on page 56, I did not see any suggestions for how to understand, or even read, n
-ary access relations on worlds, nor justification for the somewhat esoteric Schotch/Jennings ◻ condition. Similarly, the baseline motivations for generalizing the access relation, and/or for
studying a paraconsistent logic, are sketchy. A recurring theme is that we, human reasoners, are prone to errors, and that subsequently, our obligations are prone to be in conflict, since obligations
are human conventions -- but this remains a gesture. It would have helped me to see more carefully worked, concrete examples of inconsistency, deontic or otherwise, and more responsible referencing
and footnoting, to specify literature the (sometimes rather curt) authors seem to have in mind. Instead, a robust strain of rhetoric runs through the book. The words "chafe", "balk", "awestruck",
"jaundiced", "gladdened", "inviting", "dirty", "unpalatable", "browbeaten" and "scornfully" are used -- all between page 85 and 88. A more reasoned invitation to preservationism could better serve
the project.
On a basic level, for instance, there is throughout an important but largely implicit distinction drawn between a set {A, ~A} and a formula A&~A. The authors take it as basically obvious that the
former can be a perfectly sensible (if unfortunate) bit of data, and basically obvious that the latter is always absurd. The latter claim will find many adherents, to be sure, but most of those
adherents I think would "balk" at the former, too. Since preservationists stake out some controversial territory, they owe some explanation as to why other nearby territory is off limits -- beyond
conjecture about what "even the angels" want (p. 10).
There are a few typos. There is a box where there shouldn't be at the bottom of p. 63; there is a capital sigma where there should be a capital gamma on p. 163.
Preservationism cuts a fascinating middle path. On the one hand, we can still say that any change of logic is, in a sense, a change of subject -- for instance, {A&~A} still forces arbitrary B and
allowing A&~A to be true would be to change the meaning of the connectives. The preservationists do not want to replace classical logic. If nothing else, they take it as hopelessly entrenched, and in
this way the program is very conservative (p. 19). On the other hand, we have in the forcing relation a flexible and perspicuous new attachment for our old logical machinery that brings it up to date
and prevents misfiring that cannot be explained away. "Forcing allows us to keep within sight of the familiar", Payette writes, "even while straying into unexplored logical territory" (p. 143).
Preservationism is related to the first proposed paraconsistent logic of Jaskowski and that later presented by Rescher and Brandom. It is also related to relevance logic, in that the semantics
feature a generalized access relation on worlds and that the logic turns out to be paraconsistent as a byproduct. Because it is a sort of meta-logic, deriving a new inference relation over a
pre-existing pool of logics, I think it may be possible that forcing can provide the tools for philosophical discussion of logical pluralism, a topic that is today gaining momentum.
"On Preserving" will suit as part of an advanced undergraduate (or, more likely, graduate) course, and will generously repay independent study. The Canadian school is open and taking enrollments. | {"url":"http://ndpr.nd.edu/news/24157/?id=17405","timestamp":"2014-04-20T03:09:49Z","content_type":null,"content_length":"35544","record_id":"<urn:uuid:441fe2e5-e02f-4a84-a4de-f53c2fb9c660>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00074-ip-10-147-4-33.ec2.internal.warc.gz"} |
PredictionSIGNIFICANCE OF THE PROJECT
IDENTIFYING FORESHOCKS BY THEIR SENSITIVITY TO REMOTE TRIGGERING
Grant Number: 1434-HQ-97-GR-03078
Terry E. Tullis
Brown University
Department of Geological Sciences
Providence, RI 02912-1846
Phone: (401) 863-3829
FAX: (401) 863-2058
E-mail: Terry_Tullis@brown.edu
Program Element IV: Providing Real-time Hazard and Risk Assessment
Key Words: Earthquake forecasting
INVESTIGATIONS UNDERTAKEN
The purpose of this research is to evaluate a proposed earthquake prediction scheme briefly presented by Whiteside and Ben-Zion (1995). They orally reported results suggesting that free oscillations
of the earth were able to trigger earthquakes and that this triggering occurred preferentially in time and space prior to impending earthquakes. The underlying physical idea was that in the
hypocentral region of an impending earthquake the stress was close enough to the failure strength of the rock, or the frictional strength of minor faults. Their suggestion was that the stresses
involved in normal mode vibrations of the earth are high enough to trigger microseismicity, perhaps only by advancing their time of occurrence by a few minutes. Thus the proposed method can be
thought of as a way of distinguishing foreshocks from normal background microseismicity by their sensitivity to triggering. Although the idea has some appeal, the magnitudes of free oscillation
stresses are much smaller than tidal stresses; since tides apparently seldom trigger earthquakes, the likelihood that free oscillations could trigger earthquakes seems low.
However, the data presented by Whiteside and Ben-Zion (1995) suggested that before several major earthquakes, notably the Loma Prieta earthquake of 1989, there was a significant increase in
free-oscillation-triggered earthquakes in the epicentral area. Thus although the physical plausibility of the method seemed low, the suggestive results presented warranted an independent study of the
The results of our evaluation of the method of Whiteside and Ben-Zion (1995) indicate that the method is not a viable method of earthquake forecasting. The triggering reported is in fact not real and
results from random chance. Comparisons of results generated with real data and with random data show virtually no difference. Furthermore, the temporal and spatial variation previously reported can
be shown to result from variations in the numbers of earthquakes and consequent variations in the opportunities for apparent triggering by random chance. We have devised a triggering parameter that
takes into account the opportunities for such random apparent triggering. We find that this triggering parameter is never significantly different from zero in any of the examples for which Whiteside
and Ben-Zion (1995) suggested free-oscillation triggering occurred.
The Method Being Evaluated
The method of Whiteside and Ben-Zion (1995) considers that an earthquake may have been triggered by a free oscillation mode if the time interval between the earthquake and any other earthquake in the
region is equal to the period of that mode, or a multiple thereof. Both earthquakes of the pair are considered to have been triggered. Only certain modes are used, and only as many multiples are
considered as will fall within a specified time interval. A match between the inter-event time and the free oscillation period is considered to occur if they are within a specified tolerance of one
another. The modes used are 0S0, 0S2, 0S3, 1S2, 0T2, 0T3, and 1T2 and take mode splitting into account. The tolerance allowed for a match is plus or minus 9 seconds.
Since one earthquake may be a member of several pairs, an earthquake that is multiply paired is, by definition, multiply triggered. The study area is subdivided into 0.1-degree bins and the number of
earthquake triggers (greater or equal to the number of triggered events) is counted in each bin. The ratio of the number of triggers to the number of total events is calculated and cast as a
percentage. Because many events in a single bin may have been multiply triggered, a bin may have a percentage in excess of 100%.
Plots of percent triggering for a region over several years allows the establishment of a baseline value and therefore detection of times of greater sensitivity of earthquakes to triggering which may
be foreshocks to a large event.
Recreating the Proposed Method and What it Appears to Show
In order to verify that we have successfully recreated the Whiteside method, we used our independently written computer program to examine a data set also examined by Whiteside. For the ten-month
span from 8/1/88 to 5/31/89, the results of this comparison are shown in Figure 1.
Figure1. Percentage of earthquakes that appear to be triggered in 0.1 degree bins, using the method described above in the text. The epicenter of the Loma Prieta earthquake is shown as a black star
on these and all other maps. (a-left) These results, computed and kindly provided by Lowell Whiteside, have been transferred into our map base. (b-right) These results have been computed by us using
the same data and same method, but using our own independently written computer program.
The virtually identical results in Figure 1a and 1b show that we have successfully duplicated the method of Whiteside. We have also prepared a difference map between our results and those computed by
Whiteside and it shows very small differences that are related primarily to different bin boundaries.
In Figure 1 (and more clearly on Figure 2a) note the region of high triggering percentage near the future location of the Loma Prieta earthquake that may indicate foreshocks to the large event. Also
of note is the very high level in the Mammoth Mountain region. Several regions in the study area exhibit similar magnitude of triggering percentage to that seen in the Loma Prieta location.
On the maps presented by Whiteside and Ben-Zion (1995) the triggering shown in Figure 1 around the location of the Loma Prieta earthquake was significantly larger than the observed background level
of prior years. This formed the basis for their claim that sensitivity to triggering increases in the time and space window where a future large earthquake will occur.
Tests vs. Random Data
The apparent earthquake triggering shown in Figure 1 could be due to matches occurring by random chance between the inter-event times for "triggered" pairs of earthquakes and the periods of the
Earth's normal modes. In order to test this we have randomized the data in various ways to see if the triggering disappears, as is should if it is due to physical triggering by the normal modes. We
have randomized the earthquake inter-event times, the times of the earthquakes themselves, and the periods of the normal modes. In every test the randomized data is very similar to the real data,
showing that the triggering is only apparent and is occurring by random chance. This is shown in Figure 2 for the case in which the earthquake inter-event times have been randomized.
Figure 2. Comparison between (a-left) real data and (b-right) data for which the earthquake inter-event times have been randomized. The two results are nearly identical. The percentages are slightly
higher in these maps than in Figure 1. This is because we have corrected an error of Whiteside's incorporated into both Figures 1a and 1b, in which he had used a smaller number of period multiples
than he originally intended and a 2 day time window for matching pairs of events to period multiples. After this was fixed, we switched to using arithmetic multiples of periods and a two hour time
window. The result of the two changes kept the percentage triggering levels similar to those observed previously, but somewhat higher.
Improved Method for Analyzing Data
The use of triggering percentage as employed by Whiteside is not a suitable parameter for quantitative analysis. If there are a large number of earthquakes in the study region, each event in a
particular bin has an increased chance of being in a "triggered" pair of events. Also, if either the plus or minus 9 second tolerance or the time window used for matching is lengthened, each
earthquake may be paired with many more events. To remove this bias, we form a null hypothesis that there is not a statistically greater number of matches than should be observed by random chance.
More explicitly, for each earthquake the number of triggers relative to the number of pairs available for matching (fractional triggers) should equal the ratio of the time available for matching in
the window to the full window width (fractional window). The ratio of fractional triggers to fractional window should equal unity for random data. The triggering parameter is then defined as,
Values of T[r] greater than zero indicate that normal mode forcing enhances triggering of earthquakes while a value less than zero indicates that triggering is inhibited by normal mode forcing. We
find that the statistical distribution of T[r] is essentially Gaussian.
The means of the distributions for both real and random data were found to have slightly negative values. For clarity, plots of T[r] were made by subtracting the mean of the distribution for
randomized inter-event times, thus showing the deviations from random data. Such plots of T[r] for four time periods leading up to the Loma Prieta earthquake are shown in Figure 3.
Figure 3. Maps of Triggering Parameter T[r ]for four time periods leading up to the Loma Prieta earthquake.
Figure 3c (lower left) is the map of T[r] for the same ten-month interval used for Figures 1 and 2. For this and all the time periods we have examined most values are very close to zero, and almost
no value in this binned data is beyond one standard deviation from zero. Thus not only is the apparent triggering in Figure 1 and 2a due to random chance, there is no significant difference in the
triggering parameter as one approaches the time of the Loma Prieta earthquake.
The earthquake prediction method proposed by Whiteside and Ben-Zion (1995) is found upon closer inspection not to be a viable means of earthquake prediction. Results using random data are
indistinguishable from those using real data. Casting the results in terms of a newly defined triggering parameter that takes into account opportunities for random apparent triggering, our results
support the null hypothesis that the apparent triggering by free oscillations is not real.
References Cited
Whiteside, L. S., and Y. Ben-Zion, Universal triggering patterns in earthquake sequences and the use of triggering for earthquake prediction, Eos. Trans. Am. Geophys, Union, Fall Meeting Suppl., 76,
F532, 1995.
NON-TECHNICAL SUMMARY
We have evaluated a previously proposed earthquake prediction method and found that it does not work. The method envisioned that very small stresses associated with vibrations of the earth called
free oscillations could cause small earthquakes to be triggered in areas where the stress was already high because the area was getting ready for a major earthquake. Several examples were presented
that seemed to show that an unusually large number of small earthquakes were triggered before previous major earthquakes. Unfortunately, we have found that the apparent triggering was not real and
was due merely to random chance.
Reports Published
Costello, S. W., and T. E. Tullis, Investigation of a proposed earthquake prediction method that envisions foreshock triggering by free oscillations, Eos. Trans. Am. Geophys, Union, Fall Meeting
Suppl., 78, 490, 1997.
Costello, S., and T.E. Tullis, Can free oscillations trigger foreshocks that allow earthquake prediction?, Geophys. Res. Lett., 26, 891-894, 1999. | {"url":"http://www.geo.brown.edu/faculty/ttullis/prediction_fy98.htm","timestamp":"2014-04-19T11:57:37Z","content_type":null,"content_length":"18645","record_id":"<urn:uuid:4da3e124-aa5b-4b2f-9f5d-3e8a3ff5144b>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00444-ip-10-147-4-33.ec2.internal.warc.gz"} |
sdrt – Macros for Segmented Discourse Representation Theory
The package provides macros to produce the ‘Box notation’ of SDRT (and DRT), to draw trees representing discourse relations, and finally to have an easy access to various
mathematical symbols used in that theory, mostly with automatic mathematics mode, so they work the same in formulae and in text.
Sources /macros/latex/contrib/sdrt
Documentation Readme
Version 1.0
License The LaTeX Project Public License
Maintainer Paul Isambert
Contained in TeXLive as sdrt
MiKTeX as sdrt
Topics support for linguistics
Download the contents of this package in one zip archive (179.3k). | {"url":"http://www.ctan.org/pkg/sdrt","timestamp":"2014-04-18T23:49:24Z","content_type":null,"content_length":"5141","record_id":"<urn:uuid:218c1396-d1ae-4a37-a7e9-a413feeaae75>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00377-ip-10-147-4-33.ec2.internal.warc.gz"} |
Equirectangular Panini
From PanoTools.org Wiki
(Difference between revisions)
(Created page with 'The projection labelled "equirectangular Panini" in libpano13 is the standard Newer edit →
Pannini projection, which is the cylindrical analogue of the stereographic projection. The "Panini
Line 1: Line 1:
The projection labelled "equirectangular Panini" in libpano13 is the standard Pannini projection, The projection labelled "equirectangular Panini" in libpano13 is the standard Pannini projection,
− which is the cylindrical analogue of the stereographic projection. The "Panini general" + which is the cylindrical analogue of the stereographic projection. The "Panini general"
projection, available since early 2010, is an adjustable projection that is identical to projection, available since early 2010, is an adjustable projection that is identical to
"equirectangular Panini" at its default parameter settings. "equirectangular Panini" at its default parameter settings.
Revision as of 01:55, 18 May 2010
The projection labelled "equirectangular Panini" in libpano13 is the standard Pannini projection, which is the cylindrical analogue of the stereographic projection. The "Panini general" projection,
available since early 2010, is an adjustable projection that is identical to "equirectangular Panini" at its default parameter settings. | {"url":"http://wiki.panotools.org/index.php?title=Equirectangular_Panini&diff=12386&oldid=12385","timestamp":"2014-04-23T09:42:36Z","content_type":null,"content_length":"18217","record_id":"<urn:uuid:5443fa6b-d24f-4f8c-9a77-492b0dc5d8fc>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00183-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bell's derivation; socks and Jaynes
I'm pretty much a spectator in these discussions, but I'd like to point out that there was a long thread here about three years ago, about Jaynes's objections to Bell:
This was before you joined PF, so you may not have seen this. It may or may not fit in with the direction you were planning to go.
It was split off from another thread, by the way, which is why it appears to start in the middle of a discussion.
I now had a better look at it, and I think that in particular posts #26 and #31 are important. Anyway I'll give a short summary of how I now see it.
If Jaynes' criticism focuses on Bell's equation no.11 in his "socks" paper, it was perhaps due to a misunderstanding about what Bell meant (his comments were based on an earlier paper).
P(AB|a,b,x) = P(A|a,x) P(B|b,x) (Bel 11)
Here x stands for Bell's lambda, which corresponds to the circumstances that lead to a single pair correlation (in contrast to my earlier X, which causes the overall correlation for many pairs).
According to Jaynes it should be instead, for example:
P(AB|a,b,x) = P(A|B,a,b,x) P(B|a,b,x)
Perhaps Jaynes thought that Bell meant:
P(AB|a,b,X) = P(A|a,X) P(B|b,X)
in which case Jaynes claimed that:
P(AB|a,b,X) = P(A|B,a,b,X) P(B|a,b,X)
This is really tricky.
However, he really was disagreeing with the integral equation.
According to him, it should not be:
P(AB|a,b) = ∫ P(A|a,x) P(B|b,x) p(x) dx
P(AB|a,b) = ∫ P(AB|a,b,x) P(x|a,b) dx
and thus:
P(AB|a,b) = ∫ P(AB|a,b,x) p(x) dx = ∫ P(A|B,a,b,x) P(B|a,b,x) p(x) dx
Is my summary of the disagreement correct?
What is the significance of little p(x) instead of P(x)? | {"url":"http://www.physicsforums.com/showthread.php?p=3785565","timestamp":"2014-04-20T08:33:09Z","content_type":null,"content_length":"92693","record_id":"<urn:uuid:b47ce418-2c55-48a4-9f4b-6d37df99084f>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00067-ip-10-147-4-33.ec2.internal.warc.gz"} |
Discrete Math
On-line Resources: Discrete Math
Think about our world and name some things that can be counted; the number of pixels in an image, the votes cast in an election, the number of fingers on your hand, the number of different ways in
which a network can be traversed. From a mathematical perspective, in each of these examples we are talking about sets of discrete objects. The study of discrete objects and how they can be counted
is fundamental to the study of discrete mathematics.
Now consider things that we measure; the speed of a cannon ball when it leaves the cannon, the distance from the earth to moon, the volume of blood in your body. In these cases the results of the
measurement -the number that you get- depends on the accuracy and the precision of the measurement. The speed of a car could be 35 mph, 35.1 mph, 35.111 mph,and so on. Given a better measuring device
you could always add another digit of precision to your measurement. For this reason we say that measurement usually gives us a “continuous” result. In terms of our study of mathematics, algebra and
calculus are among the mathematical tools that we use to analyze and understand questions that involve continuous quantities.
Discrete math is also commonly associated with computers. That’s because, believe it or not, computers cannot handle those messy imprecise numbers that go along with continuous math. Sure you
probably use your computer to work with numbers containing decimal points all the time, but at the lowest level your computer sees 31.111 as five integers with a decimal point after the first two.
That’s a discrete quantity. The messy part comes in when we use our computers to do calculations with irrational numbers like Pi. As you probably know, Pi cannot be defined as a exact value; the
decimal places go on forever. When a computer does math with numbers like that it has to decide how many digits of precision will be used for each calculation. And when lots of these “rounded
off”numbers come together in some sort of calculation problems can occur.
Another area where discrete math is used is in the definition of the recipes that are used to write computer programs; called algorithms.Students who pursue degrees in information technology or
computer science are usually required to take one or more courses in discrete mathematics.
Image Analysis Using Maple may sound like an arcane topic, but if you own a digital camera you already have an advanced digital image analysis system. Image analysis is central to how digital
imaging works. Some cameras even give you access to some of the image processing features allowing you to shift the contrast, the color balance, or the luminosity of an already captured image.
For Empire State College students studying discrete math, and using Maple math software, you can apply these sorts of image transforms using Maple. Maple provides the basic image processing
functions found in your camera and more advanced ones not typically found in graphics software.
You can download a worksheet from the Maple Applications Center (free registration required) that demonstrates these features. On the Application Center home page search for “image” and select
“image tools” from the list. You can download the worksheet for a demonstration of the different ways that Maple can be used to manipulate images. | {"url":"http://commons.esc.edu/smatresources/category/math/discrete-math/","timestamp":"2014-04-17T21:32:05Z","content_type":null,"content_length":"55707","record_id":"<urn:uuid:458e7497-018d-4e9a-9e9f-bc876bd6922f>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00361-ip-10-147-4-33.ec2.internal.warc.gz"} |
A few homework questions I need help on =)
January 24th 2010, 08:10 AM
A few homework questions I need help on =)
I'm not very good at math and these questions are either stumping me or I don't remember what the teacher said. I don't need the answer spelled out or anything just some instructions on how to
actually do them. Thanks =)
1. Given a triangle with b = 7, c = 3 , and A = 37° what is the length of a? Round to the nearest tenth.
2. Use SOLVTRI to find the missing sides and angles of triangle ABC where angle A= 51 degrees angle C= 76.3 and side C = 50. --- I have no idea what SOLVTRI is one of the questions in the book
said to use it and I don't think the teacher mentioned it in class.
3.The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the
phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models
the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals https://
kcdistancelearning.blackboar...es/es041-1.jpg illuminated.]
4. Eastport, Maine, has among the highest tides in the United States. Write a cosine function that models the oscillation of Eastport’s tides if their amplitude is 9 feet 8 inches, the
equilibrium point is 12 feet 3 inches, the phase shift is -2.68 hours, and the period is 12.34 hours.
5. Find the area of a sector with a central angle of 32° and a radius of 8.5 millimeters. Round to the nearest tenth.
6. For a circle of radius 8 feet, find the arc length s subtended by a central angle of https://kcdistancelearning.blackboar...es/mc020-1.jpg
7. Change 290° to radian measure in terms of p.
Thanks in advance for the help.. I really need to do well in this class!
January 24th 2010, 12:41 PM
I'm not very good at math and these questions are either stumping me or I don't remember what the teacher said. I don't need the answer spelled out or anything just some instructions on how to
actually do them. Thanks =)
1. Given a triangle with b = 7, c = 3 , and A = 37° what is the length of a? Round to the nearest tenth.
2. Use SOLVTRI to find the missing sides and angles of triangle ABC where angle A= 51 degrees angle C= 76.3 and side C = 50. --- I have no idea what SOLVTRI is one of the questions in the book
said to use it and I don't think the teacher mentioned it in class.
3.The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the
phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models
the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals https://
kcdistancelearning.blackboar...es/es041-1.jpg illuminated.]
4. Eastport, Maine, has among the highest tides in the United States. Write a cosine function that models the oscillation of Eastport’s tides if their amplitude is 9 feet 8 inches, the
equilibrium point is 12 feet 3 inches, the phase shift is -2.68 hours, and the period is 12.34 hours.
5. Find the area of a sector with a central angle of 32° and a radius of 8.5 millimeters. Round to the nearest tenth.
6. For a circle of radius 8 feet, find the arc length s subtended by a central angle of https://kcdistancelearning.blackboar...es/mc020-1.jpg
7. Change 290° to radian measure in terms of p.
Thanks in advance for the help.. I really need to do well in this class!
you've posted quite a list of laundry here ... most of it is very basic and at least one problem is rather difficult.
1. use the law of cosines
2. SOLVTRI must be a calculator/computer program ... check with your instructor. Otherwise, use the law of sines and be prepared to deal with the ambiguous case if applicable.
3. not an easy problem. I'd have to think about it.
4. you should already know what the values A, B, C, and D tell you about the graph of $y = A\cos[B(x \pm C)] + D$.
5. look up the formula for the area of a sector.
6. look up the formula for arc length.
7. very basic ... look up how to change degrees to radians and radians back to degrees. you should know how to go both ways.
January 24th 2010, 01:31 PM
I figured out all but two... 3 - the one about the moon.... and the one using SOLVTRI. Thanks though!
January 30th 2010, 03:18 PM
well, i hope you still hear this reply, ladden, because i might be able to help on the SOLVTRI problem, to en extent.
~try going to:
Plane Triangle Solver
it helped my on that problem. | {"url":"http://mathhelpforum.com/trigonometry/125174-few-homework-questions-i-need-help-print.html","timestamp":"2014-04-24T10:11:30Z","content_type":null,"content_length":"11451","record_id":"<urn:uuid:a271a632-3fb2-4bf6-bc80-852fe69265c9>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00549-ip-10-147-4-33.ec2.internal.warc.gz"} |
The main module, exporting types, utility functions, and fuse and connect operators.
The three core types to this package are Source (the data producer), Sink (the data consumer), and Conduit (the data transformer). For all three types, a result will provide the next value to be
used. For example, the Open constructor includes a new Source in it. This leads to the main invariant for all conduit code: these three types may never be reused. While some specific values may work
fine with reuse, the result is generally unpredictable and should no be relied upon.
The user-facing API provided by the connect and fuse operators automatically addresses the low level details of pulling, pushing, and closing, and there should rarely be need to perform these actions
in user code.
data Source m a Source
A Source has two operations on it: pull some data, and close the Source. A Source should free any resources it allocated when either it returns Closed or when it is explicitly closed (the second
record on either the Open or SourceM constructors).
Since 0.3.0
Open (Source m a) (m ()) a A Source providing more data. Provides records for the next Source in the stream, a close action, and the data provided.
Closed A Source which has no more data available.
SourceM (m (Source m a)) (m ()) Requires a monadic action to retrieve the next Source in the stream. Second record allows you to close the Source.
Monad m => IsSource Source m
Monad m => Functor (Source m)
Monad m => Monoid (Source m a)
data BufferedSource m a Source
When actually interacting with Sources, we sometimes want to be able to buffer the output, in case any intermediate steps return leftover data. A BufferedSource allows for such buffering.
A BufferedSource, unlike a Source, is resumable, meaning it can be passed to multiple Sinks without restarting. Therefore, a BufferedSource relaxes the main invariant of this package: the same value
may be used multiple times.
The intention of a BufferedSource is to be used internally by an application or library, not to be part of its user-facing API. For example, the Warp webserver uses a BufferedSource internally for
parsing the request headers, but then passes a normal Source to the web application for reading the request body.
One caveat: while the types will allow you to use the buffered source in multiple threads, there is no guarantee that all BufferedSources will handle this correctly.
Since 0.3.0
MonadIO m => IsSource BufferedSource m
bufferSource :: MonadIO m => Source m a -> m (BufferedSource m a)Source
Places the given Source and a buffer into a mutable variable. Note that you should manually call bsourceClose when the BufferedSource is no longer in use.
Since 0.3.0
unbufferSource :: MonadIO m => BufferedSource m a -> Source m aSource
Turn a BufferedSource into a Source. Note that in general this will mean your original BufferedSource will be closed. Additionally, all leftover data from usage of the returned Source will be
discarded. In other words: this is a no-going-back move.
Note: bufferSource . unbufferSource is not the identity function.
Since 0.3.0
bsourceClose :: MonadIO m => BufferedSource m a -> m ()Source
Close the underlying Source for the given BufferedSource. Note that this function can safely be called multiple times, as it will first check if the Source was previously closed.
Since 0.3.0
class IsSource src m Source
Monad m => IsSource Source m
MonadIO m => IsSource BufferedSource m
data Sink input m output Source
In general, a sink will consume data and eventually produce an output when it has consumed "enough" data. There are two caveats to that statement:
• Some sinks do not actually require any data to produce an output. This is included with a sink in order to allow for a Monad instance.
• Some sinks will consume all available data and only produce a result at the "end" of a data stream (e.g., sum).
Note that you can indicate any leftover data from processing via the Maybe input field of the Done constructor. However, it is a violation of the Sink invariants to return leftover data when no input
has been consumed. Concrete, that means that a function like yield is invalid:
yield :: input -> Sink input m ()
yield input = Done (Just input) ()
A Sink should clean up any resources it has allocated when it returns a value.
Since 0.3.0
Processing (SinkPush input m output) (SinkClose m output) Awaiting more input.
Done (Maybe input) output Processing complete.
SinkM (m (Sink input m output)) Perform some monadic action to continue.
MonadBase base m => MonadBase base (Sink input m)
MonadTrans (Sink input)
Monad m => Monad (Sink input m)
Monad m => Functor (Sink input m)
Monad m => Applicative (Sink input m)
MonadIO m => MonadIO (Sink input m)
type SinkPush input m output = input -> Sink input m outputSource
Push a value into a Sink and get a new Sink as a result.
Since 0.3.0
type SinkClose m output = m outputSource
Closing a Sink returns the final output.
Since 0.3.0
data Conduit input m output Source
A Conduit allows data to be pushed to it, and for each new input, can produce a stream of output values (possibly an empty stream). It can be considered a hybrid of a Sink and a Source.
A Conduit has four constructors, corresponding to four distinct states of operation.
Since 0.3.0
NeedInput (ConduitPush input m Indicates that the Conduit needs more input in order to produce output. It also provides an action to close the Conduit early, for cases when there is no more input
output) (ConduitClose m output) available, or when no more output is requested. Closing at this point returns a Source to allow for either consuming or ignoring the new stream.
Indicates that the Conduit has more output available. It has three records: the next Conduit to continue the stream, a close action for early termination, and the
HaveOutput (Conduit input m output currently available. Note that, unlike NeedInput, the close action here returns () instead of Source. The reasoning is that HaveOutput will only be closed early
output) (m ()) output if no more output is requested, since no input is required.
Indicates that no more output is available, and no more input may be sent. It provides an optional leftover input record. Note: It is a violation of Conduit's
Finished (Maybe input) invariants to return leftover output that was never consumed, similar to the invariants of a Sink.
ConduitM (m (Conduit input m Indicates that a monadic action must be taken to determine the next Conduit. It also provides an early close action. Like HaveOutput, this action returns (), since it
output)) (m ()) should only be used when no more output is requested.
Monad m => Functor (Conduit input m)
type ConduitPush input m output = input -> Conduit input m outputSource
Pushing new data to a Conduit produces a new Conduit.
Since 0.3.0
type ConduitClose m output = Source m outputSource
When closing a Conduit, it can produce a final stream of values.
Since 0.3.0
Connect/fuse operators
($$) :: IsSource src m => src m a -> Sink a m b -> m bSource
The connect operator, which pulls data from a source and pushes to a sink. There are two ways this process can terminate:
1. If the Sink is a Done constructor, the Source is closed.
2. If the Source is a Closed constructor, the Sink is closed.
This function will automatically close any Sources, but will not close any BufferedSources, allowing them to be reused. Also, leftover data will be discarded when connecting a Source, but will be
buffered when using a BufferedSource.
Since 0.3.0
($=) :: IsSource src m => src m a -> Conduit a m b -> Source m bSource
Left fuse, combining a source and a conduit together into a new source.
Any Source passed in will be automatically closed, while a BufferedSource will be left open. Leftover input will be discarded for a Source, and buffered for a BufferedSource.
Since 0.3.0
(=$) :: Monad m => Conduit a m b -> Sink b m c -> Sink a m cSource
Right fuse, combining a conduit and a sink together into a new sink.
Any leftover data returns from the Sink will be discarded.
Since 0.3.0
(=$=) :: Monad m => Conduit a m b -> Conduit b m c -> Conduit a m cSource
Middle fuse, combining two conduits together into a new conduit.
Any leftovers provided by the inner Conduit will be discarded.
Since 0.3.0
Utility functions
:: Monad m
=> state Initial state
-> (state -> m (SourceStateResult state output)) Pull function
-> Source m output
Construct a Source with some stateful functions. This function addresses threading the state value for you.
Since 0.3.0
:: MonadResource m
=> IO state resource and/or state allocation
-> (state -> IO ()) resource and/or state cleanup
-> (state -> m (SourceStateResult state output)) Pull function. Note that this need not explicitly perform any cleanup.
-> Source m output
data SourceStateResult state output Source
The return value when pulling in the sourceState function. Either indicates no more data, or the next value and an updated state.
Since 0.3.0
StateOpen state output
:: MonadResource m
=> IO state resource and/or state allocation
-> (state -> IO ()) resource and/or state cleanup
-> (state -> m (SourceIOResult output)) Pull function. Note that this should not perform any cleanup.
-> Source m output
Construct a Source based on some IO actions for alloc/release.
Since 0.3.0
data SourceIOResult output Source
The return value when pulling in the sourceIO function. Either indicates no more data, or the next value.
Since 0.3.0
transSource :: Monad m => (forall a. m a -> n a) -> Source m output -> Source n outputSource
Transform the monad a Source lives in.
Note that this will not thread the individual monads together, meaning side effects will be lost. This function is most useful for transformers only providing context and not producing side-effects,
such as ReaderT.
Since 0.3.0
:: Monad m
=> state initial state
-> (state -> input -> m (SinkStateResult state input output)) push
-> (state -> m output) Close. Note that the state is not returned, as it is not needed.
-> Sink input m output
Construct a Sink with some stateful functions. This function addresses threading the state value for you.
Since 0.3.0
data SinkStateResult state input output Source
A helper type for sinkState, indicating the result of being pushed to. It can either indicate that processing is done, or to continue with the updated state.
Since 0.3.0
StateDone (Maybe input) output
StateProcessing state
:: MonadResource m
=> IO state resource and/or state allocation
-> (state -> IO ()) resource and/or state cleanup
-> (state -> input -> m (SinkIOResult input output)) push
-> (state -> m output) close
-> Sink input m output
Construct a Sink. Note that your push and close functions need not explicitly perform any cleanup.
Since 0.3.0
data SinkIOResult input output Source
A helper type for sinkIO, indicating the result of being pushed to. It can either indicate that processing is done, or to continue.
Since 0.3.0
IODone (Maybe input) output
transSink :: Monad m => (forall a. m a -> n a) -> Sink input m output -> Sink input n outputSource
Transform the monad a Sink lives in.
See transSource for more information.
Since 0.3.0
sinkClose :: Monad m => Sink input m output -> m ()Source
Close a Sink if it is still open, discarding any output it produces.
Since 0.3.0
:: Conduit a m b The next Conduit to return after the list has been exhausted.
-> m () A close action for early termination.
-> [b] The values to send down the stream.
-> Conduit a m b
A helper function for returning a list of values from a Conduit.
Since 0.3.0
:: Monad m
=> state initial state
-> (state -> input -> m (ConduitStateResult state input output)) Push function.
-> (state -> m [output]) Close function. The state need not be returned, since it will not be used again.
-> Conduit input m output
Construct a Conduit with some stateful functions. This function addresses threading the state value for you.
Since 0.3.0
data ConduitStateResult state input output Source
A helper type for conduitState, indicating the result of being pushed to. It can either indicate that processing is done, or to continue with the updated state.
Since 0.3.0
StateFinished (Maybe input) [output]
StateProducing state [output]
Functor (ConduitStateResult state input)
:: MonadResource m
=> IO state resource and/or state allocation
-> (state -> IO ()) resource and/or state cleanup
-> (state -> input -> m (ConduitIOResult input output)) Push function. Note that this need not explicitly perform any cleanup.
-> (state -> m [output]) Close function. Note that this need not explicitly perform any cleanup.
-> Conduit input m output
data ConduitIOResult input output Source
A helper type for conduitIO, indicating the result of being pushed to. It can either indicate that processing is done, or to continue.
Since 0.3.0
IOFinished (Maybe input) [output]
IOProducing [output]
Functor (ConduitIOResult input)
transConduit :: Monad m => (forall a. m a -> n a) -> Conduit input m output -> Conduit input n outputSource
Transform the monad a Conduit lives in.
See transSource for more information.
Since 0.3.0
type SequencedSink state input m output = state -> Sink input m (SequencedSinkResponse state input m output)Source
Helper type for constructing a Conduit based on Sinks. This allows you to write higher-level code that takes advantage of existing conduits and sinks, and leverages a sink's monadic interface.
Since 0.3.0
:: Monad m
=> state initial state
-> SequencedSink state input m output
-> Conduit input m output
sequence :: Monad m => Sink input m output -> Conduit input m outputSource
Specialised version of sequenceSink
Note that this function will return an infinite stream if provided a SinkNoData constructor. In other words, you probably don't want to do sequence . return.
Since 0.3.0
data SequencedSinkResponse state input m output Source
Emit state [output] Set a new state, and emit some new output.
Stop End the conduit.
StartConduit (Conduit input m output) Pass control to a new conduit.
data Flush a Source
Provide for a stream of data that can be flushed.
A number of Conduits (e.g., zlib compression) need the ability to flush the stream at some point. This provides a single wrapper datatype to be used in all such circumstances.
Since 0.3.0
Functor Flush
Eq a => Eq (Flush a)
Ord a => Ord (Flush a)
Show a => Show (Flush a)
Convenience re-exports
data ResourceT m a
The Resource transformer. This transformer keeps track of all registered actions, and calls them upon exit (via runResourceT). Actions may be registered via register, or resources may be allocated
atomically via allocate. allocate corresponds closely to bracket.
Releasing may be performed before exit via the release function. This is a highly recommended optimization, as it will ensure that scarce resources are freed early. Note that calling release will
deregister the action, so that a release action will only ever be called once.
Since 0.3.0
MonadTrans ResourceT
MonadTransControl ResourceT
MonadBase b m => MonadBase b (ResourceT m)
MonadBaseControl b m => MonadBaseControl b (ResourceT m)
Monad m => Monad (ResourceT m)
Functor m => Functor (ResourceT m)
Typeable1 m => Typeable1 (ResourceT m)
Applicative m => Applicative (ResourceT m)
(MonadThrow m, MonadUnsafeIO m, MonadIO m) => MonadResource (ResourceT m)
MonadThrow m => MonadThrow (ResourceT m)
(MonadIO m, MonadActive m) => MonadActive (ResourceT m)
MonadIO m => MonadIO (ResourceT m)
class (MonadThrow m, MonadUnsafeIO m, MonadIO m) => MonadResource m
A Monad which allows for safe resource allocation. In theory, any monad transformer stack included a ResourceT can be an instance of MonadResource.
Note: runResourceT has a requirement for a MonadBaseControl IO m monad, which allows control operations to be lifted. A MonadResource does not have this requirement. This means that transformers such
as ContT can be an instance of MonadResource. However, the ContT wrapper will need to be unwrapped before calling runResourceT.
Since 0.3.0
(MonadThrow m, MonadUnsafeIO m, MonadIO m) => MonadResource (ResourceT m)
MonadResource m => MonadResource (MaybeT m)
MonadResource m => MonadResource (ListT m)
MonadResource m => MonadResource (IdentityT m)
(Monoid w, MonadResource m) => MonadResource (WriterT w m)
(Monoid w, MonadResource m) => MonadResource (WriterT w m)
MonadResource m => MonadResource (StateT s m)
MonadResource m => MonadResource (StateT s m)
MonadResource m => MonadResource (ReaderT r m)
(Error e, MonadResource m) => MonadResource (ErrorT e m)
MonadResource m => MonadResource (ContT r m)
(Monoid w, MonadResource m) => MonadResource (RWST r w s m)
(Monoid w, MonadResource m) => MonadResource (RWST r w s m)
class Monad m => MonadThrow m where
A Monad which can throw exceptions. Note that this does not work in a vanilla ST or Identity monad. Instead, you should use the ExceptionT transformer in your stack if you are dealing with a non-IO
base monad.
Since 0.3.0
MonadThrow IO
MonadThrow m => MonadThrow (ResourceT m)
Monad m => MonadThrow (ExceptionT m)
MonadThrow m => MonadThrow (MaybeT m)
MonadThrow m => MonadThrow (ListT m)
MonadThrow m => MonadThrow (IdentityT m)
(Monoid w, MonadThrow m) => MonadThrow (WriterT w m)
(Monoid w, MonadThrow m) => MonadThrow (WriterT w m)
MonadThrow m => MonadThrow (StateT s m)
MonadThrow m => MonadThrow (StateT s m)
MonadThrow m => MonadThrow (ReaderT r m)
(Error e, MonadThrow m) => MonadThrow (ErrorT e m)
MonadThrow m => MonadThrow (ContT r m)
(Monoid w, MonadThrow m) => MonadThrow (RWST r w s m)
(Monoid w, MonadThrow m) => MonadThrow (RWST r w s m)
class Monad m => MonadUnsafeIO m where
A Monad based on some monad which allows running of some IO actions, via unsafe calls. This applies to IO and ST, for instance.
Since 0.3.0
MonadUnsafeIO IO
(MonadTrans t, MonadUnsafeIO m, Monad (t m)) => MonadUnsafeIO (t m)
MonadUnsafeIO (ST s)
MonadUnsafeIO (ST s)
runResourceT :: MonadBaseControl IO m => ResourceT m a -> m a
Unwrap a ResourceT transformer, and call all registered release actions.
Note that there is some reference counting involved due to resourceForkIO. If multiple threads are sharing the same collection of resources, only the last call to runResourceT will deallocate the
Since 0.3.0 | {"url":"http://hackage.haskell.org/package/conduit-0.3.0/docs/Data-Conduit.html","timestamp":"2014-04-19T22:06:36Z","content_type":null,"content_length":"85415","record_id":"<urn:uuid:3d275e76-8872-4d61-ab02-74f7952b5cab>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00064-ip-10-147-4-33.ec2.internal.warc.gz"} |
Richard A. Mollin, Fundamental Number Theory with Applications, Second Edition
Richard A. Mollin, Fundamental Number Theory with Applications, Second Edition
C..n & H..C 2/ed (2-2008) | PDF | 384 pages | ISBN:
| 6.2Mb
An update of the most accessible introductory number theory text available, Fundamental Number Theory with Applications, Second Edition presents a mathematically rigorous yet easy-to-follow treatment
of the fundamentals and applications of the subject. The substantial amount of reorganizing makes this edition clearer and more elementary in its coverage.
New to the Second Edition
* Removal of all advanced material to be even more accessible in scope
* New fundamental material, including partition theory, generating functions, and combinatorial number theory
* Expanded coverage of random number generation, Diophantine analysis, and additive number theory
* More applications to cryptography, primality testing, and factoring
* An appendix on the recently discovered unconditional deterministic polynomial-time algorithm for primality testing
Taking a truly elementary approach to number theory, this text supplies the essential material for a first course on the subject. Placed in highlighted boxes to reduce distraction from the main text,
nearly 70 biographies focus on major contributors to the field. The presentation of over 1,300 entries in the index maximizes cross-referencing so students can find data with ease.
Click on the link below to start downloading this free ebook:
Richard A. Mollin, Fundamental Number Theory with Applications, Second Edition download link 1Richard A. Mollin, Fundamental Number Theory with Applications, Second Edition download link 2 | {"url":"http://www.ebookl.com/708838-richard-mollin-fundamental-number-theory-applications-second-edition","timestamp":"2014-04-19T05:40:46Z","content_type":null,"content_length":"43256","record_id":"<urn:uuid:8cb93066-b164-46f8-adcf-887c9e9ec951>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00052-ip-10-147-4-33.ec2.internal.warc.gz"} |
Trigonometric Functions and Their Graphs:
The Sine and Cosine (page 1 of 3)
Sections: The sine and cosine, The tangent, The co-functions
At first, trig ratios related only to right triangles. Then you learned how to find ratios for any angle, using all four quadrants. Then you learned about the unit circle, in which the value of the
hypotenuse was always r = 1 so that sin(θ) = y and cos(θ) = x. In other words, you progressed from geometrical figures to a situation in which there was just one input (one angle measure, instead of
three sides and an angle) leading to one output (the value of the trig ratio). And this kind of relationship can be turned into a function.
Looking at the sine ratio in the four quadrants, we can take the input (the angle measure θ), "unwind" this from the unit circle, and put it on the horizontal axis of a standard graph in the x,y
-plane. Then we can take the output (the value of sin(θ) = y) and use this value as the height of the function. The result looks like this:
As you can see, the height of the red line, being the value of sin(θ) = y, is the same in each graph. In the unit circle on the left, the angle is indicated by the green line. On the "regular" graph
on the right, the angle is indicated by the scale on the horizontal axis.
If the green angle line had gone backwards, counting into negative angle measures, the horizontal graph on the right would have extended back to the left of zero. If, instead of starting over again
at zero for every revolution on the unit circle, we'd counted up higher angles, then the horizontal graph on the right would have continued, up and down, over and over again, past 2π.
The Sine Wave
From the above graph, showing the sine function from –3π to +5π, you can probably guess why this graph is called the sine "wave": the circle's angles repeat themselves with every revolution, so the
sine's values repeat themselves with every length of 2π, and the resulting curve is a wave, forever repeating the same up-and-down wave. (My horizontal axis is labelled with decimal approximations of
π because that's all my grapher can handle. When you hand-draw graphs, use the exact values: π, 2π, π/2, etc.)
When you do your sine graphs, don't try to plot loads of points. Instead, note the "important" points. The sine wave is at zero (that is, on the x-axis) at x = 0, π, and 2π; it is at 1 when x = π/2;
it is at –1 when x = 3π/2. Plot these five points, and then fill in the curve.
We can do the same sort of function conversion with the cosine ratio:
The relationship is a little harder to see here, because the unit circle's line is horizontal while the standard graph's line is vertical, but you can see how those two purple lines are the same
length, while the angle measure is moving from zero to 2π. And just as with the sine graph, the cosine graph can be extended outside the interval from zero to 2π:
The Cosine Wave
As you can see from the extended sine and cosine graphs, each curve repeats itself regularly. This trait is called "periodicity", because there is a "period" over which the curve repeats itself over
and over. The length of the period for the sine and cosine curves is clearly 2π: "once around" a circle. Also, each of sine and cosine vary back and forth between –1 and +1. The curves go one unit
above and below their midlines (here, the x-axis). This value of "1" is called the "amplitude".
When you graph, don't try to plot loads of points. Note that the cosine is at y = 1 when x = 0 and 2π; at y = 0 for x = π/2 and 3π/2, and at y = –1 for x = π. Plot these five "interesting" points,
and then fill in the curve.
Original URL: http://www.purplemath.com/modules/triggrph.htm
Copyright 2009 Elizabeth Stapel; All Rights Reserved.
Terms of Use: http://www.purplemath.com/terms.htm | {"url":"http://www.purplemath.com/modules/printtrig.cgi?path=/modules/triggrph.htm","timestamp":"2014-04-17T09:35:34Z","content_type":null,"content_length":"10596","record_id":"<urn:uuid:8402259d-a96b-44f9-bb43-8308488f4559>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00120-ip-10-147-4-33.ec2.internal.warc.gz"} |
Course Information
Text Book
Main course book: An Introduction to Mechanics, by Daniel Kleppner and Robert J. Kolenkow, McGraw-Hill International Editions (Sections of the book that are part of the course are indicated below)
Supplementary text book for additional reading: Classical Mechanics, by John R. Taylor
Old Examinations
Two recent solved exams are: May 2005 , Aug 2005 . Two earlier examination papers are found here . (Copies of still older exams with solutions have been distributed in the class. If you have not got
one, please contact me).
Suggested Problems
Suggested problems from the book and from the yellow "Övnningsproblem i mekanik" can be found here (Consider only chapters that have been covered in this course and note that the description in line
2 and 3 does not apply to this course)
Homework Problems
Homework problem set I pdf file (Due date: May 2, 2006)
Course Content
Chapter 1: Vectors and Kinematics
Topics Covered:
• Vectors, addition of vectors, unit vectors, components of a vector (in Cartesian coordinates), Scalar (dot) product and vector (cross) product of two vectors.
• Basic description of motion: position vector, displacement, velocity, momentum, acceleration, description of motion with constant acceleration.
• Uniform circular motion (angular velocity and radial acceleration), polar coordinates in 2 dimensions, general 2-dimensional motion in polar coordinates, radial and angular components of velocity
and acceleration.
Relevant Sections in the Book: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.9 (important, upto page 37)
Chapter 2: Newton's Laws of Motion
Topics Covered:
• The need for reference frames, Newton's first law of motion, the concept of inertial and non-inertial reference frames, isolated bodies.
• Newton's second law of motion, unit of force, principle of superposition, interaction as the origin of force, non-inertial frames and the concept of fictitious forces.
• Newton's third law of motion, action and reaction
• Some common examples of force: Newton's law of gravitation, gravity due to empty and solid spherical shells (no derivations), weight, acceleration due to gravity.
• Linear restoring force (Hooke's law), Motion under a linear restoring force (Simple Harmonic Motion): general solution of the simple harmonic equation, the concept of initial conditions, time
period and frequency of the oscillations.
• Contact forces: tension on a string, normal force (perpendicular to a surface), friction
Relevant Sections in the Book: 2.1, 2.2, 2.4, 2.5 (topics on pulleys and viscocity not included)
Relevant Examples: 2.3, 2.5, 2.6, 2.8, 2.10, 2.11, 2.12, 2.17, 2.18
Chapter 3: Momentum
Topics Covered:
• Newton's second law in terms of momentum, Dynamics of a system of particles: total momentum and total external force, center of mass, motion of center of mass, conservation of momentum and its
importance, impulse and the significance of interaction time.
Relevant Sections in the Book: 3.1, 3.2, 3.3, 3.4
Relevant Examples: 3.2, 3.6, 3.8 (no need to use the CoM coordinates), 3.10
Chapter 4: Work and Energy
Topics Covered:
• Work, Integrating the equation of motion in 1 dimension, Work-Energy theorem in 1 dim, Calculation of work for some forces: a) uniform gravitational field, b) linear restoring force, c) inverse
square force (escape velocity)
• Integrating equation of motion in 3 dimensions, Work-Energy theorem, Work done by uniform and central forces, Conservative forces and potential energy, Total mechanical energy and its
conservation, Potential in a uniform gravitational field, harmonic oscillator potential, gravitational potential, Shape of potential energy curve and stability.
Relevant Sections in the Book: 4.1, 4.2, 4.3, 4.4, 4.5, 4.7, 4.8, (recommended reading: 4.9, 4.10)
Relevant Examples: 4.1, 4.2, 4.3, 4.4, 4.5, 4.7, 4.8, 4.11, 4.12, 4.13,
Chapter 6: Angular Momentum and Fixed Axis Rotation
Topics Covered:
• Angular momentum of a particle, Torque and the conservation of angular momentum, Central force motion and Kepler's law of equal areas, Angular momentum of an extended body and moment of inertia,
Calculation of moment of inertia in some simple cases, The parallel axis theorem, Dynamics of pure rotation about an axis, kinetic energy of a rotating body, axis of gyration, The physical
• Motion Involving Both Translations and Rotations, analysis for angular momentum, torque and kinetic energy.
Relevant Sections in the Book: 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, 6.7
Relevant Examples: 6.1, 6.2, 6.3, 6.5, 6.6, 6.7, 6.8, 6.9, 6.10, 6.14, 6.15, 6.16, 6.17
Chapter 8: Non-inertial Systems and Fictitious Forces
Topics covered:
• Transformations between reference frames, Galilean transformations and inertial frames, Uniformly accelerated systems and the equivalence principle, Physics in rotating coordinate systems,
centrifugal and Coriolis forces, The case of the rotating earth, Foucault's pendulum
Relevant Sections in the Book: 8.1, 8.2, 8.3, 8.4, 8.5
Relevant Examples: 8.1, 8.2, 8.3, 8.6, 8.7, 8.8, 8.9, 8.11,
Chapter 10: The Harmonic Oscillator
Topics covered:
• Review of simple harmonic motion, Damped harmonic oscillator, Forced harmonic oscillator, Forced damped harmonic oscillator, resonance and energy considerations
Relevant Sections in the Book: 10.1, 10.2 (leave out the Q-factor), 10.3, 10.4
Relevant Examples: 10.1, 10.3, 10.4, 10.5,
Chapter 11: The Special Theory of Relativity
Topics covered:
• Galilean transformations and Galilean addition of velocities, Qualitative description of inconsistency with experiments and with the theory of electromagnetism (constancy of the speed of light),
The postulates of Special Relativity, Derivation of Lorentz transformations.
Relevant Sections in the Book: 11.1, 11.3,
Relevant Examples: 11.1, 11.2
Chapter 12: Relativistic Kinematics
Topics covered:
• Implications of Lorentz Tansformations: observer dependence of simultaneity, Lorentz contraction, Time dilation, Relativistic transformation of velocity.
Relevant Sections in the Book: 12.1, 12.2, 12.3, 12.4
Relevant Examples: 12.1, 12.2, 12.3, 12.4, 12.5, 12.6, 12.7
Chapter 13: Relativistic Momentum and Energy
Topics covered:
• Relativistic momentum and its conservation, the concept of velocity dependent mass, Relativistic kinetic energy,Total energy and mass-energy equivalence, energy-momentum relationship, the notion
of massless particles.
Relevant Sections in the Book: 13.1, 13.2, 13.3
Relevant Examples: 13.2, 13.3, 13.5
Space-time Intervals and Diagrams
Topics covered:
• Space intervals, Time intervals and Space-time intervals, Invariance of space-time intervals under Lorentz transformations, Types of Space-time intervals: space-like, time-like and light-like,
The notion of causality and its invariance under Lorentz transformations, Space-time interval as a distance between two events in Minkowski space.
• Space-time diagrams, the light-cone and causality | {"url":"http://www.fysik.su.se/~fawad/mechanics.html","timestamp":"2014-04-19T11:58:55Z","content_type":null,"content_length":"9969","record_id":"<urn:uuid:5693bc91-2f34-4858-a829-99ce96b6227f>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00571-ip-10-147-4-33.ec2.internal.warc.gz"} |
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General covariance
Yes, I think the quantum spin-2 way is the best, since if one formulates GR as field on flat spacetime, then there is a flat spacetime which is clearly not background independent.
Again, it depends on definitions. If you are going by the definitions that Ben Niehoff for instance is using above, then the above *is* background independant.
Note the difference between the two following definitions...
1) Background dependance is tantamount to using the background field method for gravity, and ONLY for gravity (eg the metric tensor is split into a classical but arbitrary fixed background metric + a
small perturbation). The approximation is valid up to some cutoff, whereupon the backreaction of the pertubation on the background can no longer be ignored.
2) Background independance is like asking whether the metric field is dynamical or not in the Lagrangian of the theory. In the sense that if you look at the variation in the action and consider (d/d&
G), then you look for something that vanishes. So for instance, coupling a topological field theory to a theory with curvature invariants is clearly background independant in this definition. The
terms with curvature invariants, owing to their general covariance, will integrate out any metric dependance, and terms that are topological have no metric dependancy at all. Contrast that with
something like a Maxwell term, which when acted with the operator, will instantly pull out the nondynamical and absolute fixed structure.
Both definitions (as well as anyone that you can think off) are not going to generalize universally, or serve as a theory 'filter'. The first problem is that the word 'background' is often
generalized in the literature to mean something more than just a classical solution of Einstein's equations. Second, its a little bit unclear what physical principle you are trying to capture that is
so damn important, considering that even classical GR can be written in ways that make it look background dependant. (Consider writing GR like field theorists for the first case and consider the pure
connection formalism for the 2nd) | {"url":"http://www.physicsforums.com/showthread.php?p=4200198","timestamp":"2014-04-19T07:28:25Z","content_type":null,"content_length":"55916","record_id":"<urn:uuid:ba136456-308d-4579-82aa-aef43bca5f62>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00146-ip-10-147-4-33.ec2.internal.warc.gz"} |
On most systems, programmers of high level languages have an advantage over assembly language programmers in that a lot of the nitty gritty detail work has been done for them by what ever compiler or
interpretter they re using. An example of this is in the outputting of numbers. High level language programmers usually take it for granted that they can output a number in just about any format they
want to. Pascal s writeln and C s printf make it a trivial task to change the field width of both integers and reals. While it s true that assembly programmers can make use of _NumToString or the
SANE formatter, we still need to play with the resulting strings to get nice formatted numbers. Until now.
The basic idea when converting an integer into its string equivalent is to first break it down into its digits and then convert each digit into its ASCII equivalent. The only tricky part is that a
number is represented as binary internally and we need its decimal equivalent.
There are at least 2 different ways to approach the problem of extracting base 10 digits from a binary number. The _NumToString routine that Apple provides in the system file (see listing 1) uses
binary coded decimal (BCD) arithmetic to calculate the digits and then calls a separate subroutine to build the string two digits at a time (the least significant bytes of registers D1-D5 are used to
store 2 BCD digits each). The main problem with this algorithm is that it runs in more or less constant time (the number 1 takes as long to convert as 2147483648 = 2^31) and isn t very efficient
except for very large numbers (about 7 or more digits).
Another way to get digits one by one is to subtract by successively smaller powers of 10, starting with 10^9 (since it is the largest power of 10 that can be represented by a 32 bit integer). Count
how many times you can subtract each power of 10 from the number and then convert that number (which will be in the range 0..9) to its ASCII equivalent. If you use multi-word compares and subtracts,
this method can be used to convert any size integers (like 64 or 128 bits) into strings.
My own NumToString routine (Shown in the program listing) uses the subtract method. How does the subtract method compare to the real _NumToString? As for size, _NumToString is 118 bytes (after all of
the trap related instructions are removed) and my routine is 120 bytes. Pretty close. As for speed, time trials on 10,000 random 32 bit signed integers showed my routine to be faster by about 2% --
no big deal. But for time trials on 10,000 16 bit signed integers my routine was faster by 32% and for 10,000 8 bit signed integers my routine was faster by 45% (to be fair, _NumToString was timed
after it had been isolated so there was no overhead for trap calling). The reason the smaller numbers showed so much improvement is because my routine doesn t spend much time on little digits
(0,1,2...) and virtually no time on leading zeros. If you have a time critical application that makes a lot of calls to _NumToString, it may be worthwhile to use my routine instead. Of course, you
could just patch my routine over the existing _NumToString in the system file to speed up all applications that use _NumToString -- no, wait, I didn t just say that (and I m not responsible for the
consequences. Is there any reason that wouldn t work?).
When you use DIVS or DIVU, the 32 bit result is made up of 16 bits of quotient and 16 bits of remainder. The FixPtToString routine uses these two pieces to convert a fixed point number to a string.
Since the quotient is just an integer, we can use NumToString to convert it. Then add a decimal point and convert the remainder. But in order to convert the remainder into a fractional number, we
need to know the original divisor. We also need to specify how many digits we want after the decimal point.
For each digit you want after the decimal point, the routine multiplies the remainder by 10 and then divides the it by the divisor, getting a new remainder in the process. It adds the digit to the
string and then repeats the process for the next digit. However, there is a limited number of times that this can be done accurately (since we only have 16 bits of remainder to begin with). The
number of digits that can be accurately calculated depends on the magnitude of the divisor, but 5 or 6 digits should be safe.
Note that the number you pass to FixPtToString doesn t have to be the result of a divide instruction. You can output any arbitrary fixed point number you want by using the same basic idea. If you
wanted an integer part bigger than 16 bits, you could output the number in a two part process. FixPtToString could be modified to be FractionToString by eliminating instructions 3 (EXT.L D0) thru 17
(BEQ.S @6) inclusive. Then it will tack on whatever fraction you pass it to what ever string you pass it. For instance, to output the number 1864723135.24226 (.24226 = 47/194):
MOVE.L stringPtr(A5),A0 MOVE.L #1864723135,D0 ;quotient JSR NumToString MOVE #47,D0 ;remainder SWAP D0;no need to set quot MOVE #194,D1 ;divisor MOVE #5,D2 ;5 digits accuracy JSR FractionToString
MOVE.L A0,-(SP) _DrawString
Now that we can get integers and reals into strings we need to be able to set field widths. Also, an option to add commas would be nice. The FormNumString routine does both of these. You pass it a
pointer to a format string of the form: [ , ][q[ . [r]]] where [ ] denotes something optional and . and , denote a constant character. Q and r are string variables in the range [ 0 .. 99'] and
represent how many spaces should be allocated for the quotient (including sign and commas) and remainder (not including decimal point). Notice that everything is optional, so the empty string is a
legal one, but one that won t do much (i.e. any) formatting. Also, the word string here does not mean a pascal string; i.e. there is no length byte. The nested brackets mean that you can t have a
. if a q was not provided and you can t have an r if a . wasn t provided. If q is too small to contain the number, space is used as needed. If it is too big, the number will be padded with
spaces. If r is bigger than any existing remainder the number might have, zeros are appended after the decimal point (which doesn t do much for the accuracy of the extra digits). If r is smaller than
any existing remainder, then digits are just dropped. No attempt at rounding is made.
Another use for this routine is to reformat the output string from the SANE formatter. The SANE formatter can format any SANE data type into a fixed sytle number (see the Apple Numerics Manual for
everything you ever wanted to know about SANE). So, you could use SANE to do all of your calculations in floating point and have your result formatted into a fixed style string and then use
FormNumString to add commas, pad with spaces and delete decimal places (or add zeros) to make all your numbers uniform.
The 3 routines NumToString, FixPtToString and FormNumString have been pieced together to form DrawNumsInAString which can be used to output complete sentences containing any number of formatted
numbers. For instance, you could pass it the string result = \f,8.4. along with the fixed point number 123456.789 and it will output result = 12,3456.7890. Each number you want formatted in the
input string will begin with a \ character and be followed by an i (for longints) or an f (for fixed points). After that comes the FormNumString style format codes for the number. The only
peculiar part of using the routine is that the numbers you want formatted have to be pushed on the stack in reverse of their occuring order in the string. For instance, if your input string looks
like x = \i4 y = \i4 z = \i4 then you would do this:
MOVE.L z,-(SP) ;3rd parameter MOVE.L y,-(SP) ;2nd MOVE.L x,-(SP) ;1st PEA inputString JSR DrawNumsInAString
The reason for passing the parameters that way is (1) because we don t know how many will be present in advance, and (2) so that they can be taken off the stack in the order needed.
The FormatDemo program shows how all of this comes together. It is a bare bones application that demonstrates how the routines presented here might be used. A click in the window will generate
another set of random numbers to format. Desk accessories are supported and you can see a simple technique for providing window updating without an update event. Whenever a content click is detected,
a set of addition and division problems are displayed in formatted output using the formatting routines discussed. After displaying the numbers, a quickdraw picture is taken of the window output
using copybits on the portRect of the window. The appropriate field in the window record is set with the pointer to this picture so that when the window needs updating, it is updated automatically
from the picture. Figures 1 and 2 show the demo program in operation with a desk accessory showing this easy update method. A good reference book for this and all assembly language techniques is Dan
Weston s classic The Complete Book of Macintosh Assembly Language Programming, vol. 1 and 2, from Scott, Foresman and Company.
{1} Listing 1. The Apple Way ; NOTE: This code is Apple Computer s. Only ; the comments are mine. ;========== NumToString ;========== ; convert a 32 bit longint into a pascal string ; input: D0
longint ; A0 points to a space of at least 12 bytes ; output: A0 points to pascal string MOVEM.LD0-D6/A1,-(SP) MOVEQ #0,D1 ;init digits MOVEQ #0,D2 MOVEQ #0,D3 MOVEQ #0,D4 MOVEQ #0,D5 MOVEQ #31,D6 ;
loop counter LEA 1(A0),A1 ;skip length byte TST.L D0;is num zero? BGT.S @2 BMI.S @1 MOVE.B # 0',(A1)+ ;special case for zero BRA.S @3 @1 MOVE.B # - ,(A1)+ ;give string a minus sign NEG.L D0;make
number positive @2 ADD.LD0,D0 ;shift a bit into extend flag ABCD D5,D5 ;tens & ones digits ABCD D4,D4 ;thous & hunds digits ABCD D3,D3 ;hund thous & ten thous digits ABCD D2,D2 ;ten mils &
mils digits ABCD D1,D1 ;bils & hund mils digits DBRA D6,@2 ;do next bit ; NOTE: at this point D6 = -1. It is used as a flag to ; kill leading zeros. BSR.S Do2Digits MOVE.B D2,D1 BSR.S Do2Digits
MOVE.B D3,D1 BSR.S Do2Digits MOVE.B D4,D1 BSR.S Do2Digits MOVE.B D5,D1 BSR.S Do2Digits ;calculate length of string that was created @3 MOVE A1,D0 ;end of string + 1 SUB A0,D0 ;beginning of string
SUBQ.B #1,D0 ;minus 1 for length byte MOVE.B D0,(A0) MOVEM.L(SP)+,A1/D0-D6 RTS Do2Digits ;convert BCD byte in D1 into 2 ASCII digits. ;do most significant digit ROR #4,D1 BSR.S DoADigit ;do least
significant digit ROL #4,D1 DoADigit TST D6;have we had a non-zero digit yet? BPL.S @6 TST.B D1;is this a leading zero? BEQ.S @7 MOVEQ #0,D6 ;print all zeros from now on @6 ORI.B#$30,D1 ;covert BCD
digit to ASCII MOVE.B D1,(A1)+ ;add it to the string SUB.B D1,D1 @7 RTS ; DrawNumsInAString.asm ;---------------------- ; by Mike Scanlin Include Traps.D Xref
DrawNumsInAString,NumToString,FixPtToString,FormNumString Xref GetANumber ;================ DrawNumsInAString ;================
Draw a string that may contain implicit formatted numbers. input: stack contains numbers that will be needed and a pointer a string. The numbers should be pushed on the stack in order from last to
first (so they can be poped off in order from first to last). The last thing to be pushed on the stack is the string pointer. Each formatted number within the input string begins with a \ and then
either an i (for longints) or a f (for a fixed point number). For fixed points, first push the fixed point number, then the divisor to be used to calculate the remainder. The formatting after the
i or f is the same as for the FormNumString routine. output: a string is drawn A0,D0 are trashed.
{2} MOVEM.LA0-A3/D1-D2,-(SP) LEA 28(SP),A3;point to first parameter MOVE.L A3,-(SP) ;save initial position MOVE.L (A3)+,A2 ;string pointer @1 MOVEQ#0,D0 MOVE.B (A2)+,D0 BEQ.S @10 ;end of string found
CMPI.B # \ ,D0 ;is it a number? BEQ.S @2 MOVE D0,-(SP) _DrawChar BRA.S @1 ;we got a number to format @2 LEA scratch,A0 MOVE.B (A2)+,D0 CMPI.B # i ,D0 ;is it a longint? BNE.S @4 ;handle integers
MOVE.L (A3)+,D0 ;get longint JSR NumToString BRA.S @5;go format it @4 CMPI.B # f ,D0;is it a fixed point? BNE.S @1;if not, ignore it ;handle fixed point MOVEA.LA2,A1 ;find out how many decimal places
should be passed to ; FixPtToString MOVE.B (A1)+,D0 ;skip comma, if present CMPI.B # , ,D0 BEQ.S @4.1 SUBA #1,A1 @4.1 JSR GetANumber BMI.S @1;no quotient present MOVE.B (A1)+,D0 CMPI.B # . ,D0 BNE.S
@4.2 JSR GetANumber MOVE D0,D2 BPL.S @4.3 @4.2 MOVEQ #0,D2;no remainder @4.3 MOVE(A3)+,D1 ;get divisor MOVE.L (A3)+,D0 ;get fixed point num JSR FixPtToString ;do the formatting @5 MOVEA.LA2,A1 ;addr
of format string JSR FormNumString MOVEA.LA1,A2 ;point past format string MOVE.L A0,-(SP) _DrawString BRA.S @1 @10SUBA.L (SP)+,A3 ;calc len of params MOVE A3,D0 MOVEM.L(SP)+,D1-D2/A0-A3 MOVE.L (SP)
+,A0 ;get return addr ADDA D0,SP ;length of parameters JMP (A0) scratch DCB.B 40,0 ; FixPtToString.asm ;------------------ ; by Mike Scanlin Xref FixPtToString,NumToString ;============
FixPtToString ;============ ; convert a 32 bit fixed point number into a pascal ; string. ; input: D0 fixed point number ; D1 16 bit divisor used when D0 was calculated ; D2 # of digits after decimal
point (D2=0 for no ; dec point) ; A0 points to a space of at least (8 + D2) bytes ; output: A0 points to pascal string MOVEM.LD0-D3/A1,-(SP) MOVE.L D0,D3 ;save quotient & remainder EXT.L D0;sign
extend quotient JSR NumToString ;if q = 0 and the result should be < 0, we ll have to ; add a minus sign. ;(NumToString won t know about it, since all it sees is ; a zero quotient) TST D3;q = 0?
BNE.S @0 TST.L D3;check remainder BEQ.S @0;q & r both zero ;if r & divisor have the same sign, then result will ; be > 0 EXT.L D1 MOVE.L D1,D0 EOR.L D3,D0 BPL.S @0 MOVE.B # - ,1(A0) MOVE.B # 0',2(A0)
MOVE.B #2,(A0) ;new length @0 TST D2;do we want a decimal point? BEQ.S @6 MOVEQ #0,D0 MOVE.B (A0),D0 ;length of quotient LEA 1(A0,D0),A1;end of string + 1 MOVE.B # . ,(A1)+ TST.L D1;make divisor
positive BPL.S @1 NEG.L D1 @1 TST.LD3;make remainder positive BPL.S @2 NEG.L D3 @2 SWAP D3 ANDI.L #$FFFF,D3;isolate remainder SUBQ #1,D2 ;loop control @3 ADD.LD3,D3 ;mult r by 10 MOVE.L D3,D0 ADD.L
D0,D0 ;4x ADD.L D0,D0 ;8x ADD.L D0,D3 ;10x = 8x + 2x MOVEQ # 0',D0 ;init digit @4 CMP.LD1,D3 ;is 10r > divisor? BLT.S @5 ADDQ #1,D0 ;increase digit SUB.L D1,D3 ;subtract divisor BNE.S @4 @5 MOVE.B
D0,(A1)+ ;add to string DBRA D2,@3 MOVE A1,D0 ;calc length of new string SUB A0,D0 SUBQ.B #1,D0 ;minus 1 for length byte MOVE.B D0,(A0) @6 MOVEM.L(SP)+,A1/D0-D3 RTS ; FormNumString.asm ;
------------------ ; by Mike Scanlin Xref FormNumString,GetANumber ;============ FormNumString ;============ ; format the string representation of a number (integer ; or real) according to a format
string. ; input: A0 points to string of a number (which should ; be in a space big enough for formatting result) ; A1 points to format string ; syntax of format string is [ , ]d[d][ . [d[d]]] ; where
denote a constant char and d denotes ; a digit 0 .. 9' ; there is no length byte for these strings. ; valid strings invalid strings ;3 230. (230 too big [99 max]) ; 4.0 or 4. .0(need a d before
. ) ;,3.4 ,.4 (need a d between ,. ) ;,12.20 0.123 (123 too big) ; output: A0 points to formatted string of a number ; A1 points to first byte after format string MOVEM.LD0-D5/A2,-(SP) CMPI.B # , ,
(A1) BNE.S @6 ADDA #1,A1 ;do commas ;first find out if a decimal point in the string MOVEQ #0,D0 MOVE.B (A0),D0 ;length of string MOVE D0,D2 ;save in case it s and int SUBQ #1,D0 ;loop control ; D0
counts how many digits from the end of the string ; to the decimal point (including the decimal ; point -- which is why it starts out as 1 and not 0). ; If the number is an int, then D0=0 MOVEQ #1,D1
@1 CMPI.B # . ,1(A0,D0) ;dec point, it s a real BEQ.S @2 ADDQ #1,D1 DBRA D0,@1 ;it s an integer MOVE D2,D0 MOVEQ #0,D1 ;now add some commas @2 MOVE D1,D5 ;save length of fraction MOVEQ #3,D2 ;# of
digits until next comma SUBQ #1,D0 @3 ADDQ #1,D1 ;total len, from end to cur pos SUBQ #1,D2 BNE.S @5 MOVE D0,D3 MOVE.B (A0,D0),D0 BSR IsItADigit ;test next digit BNE.S @6 MOVE D3,D0 ;move some chars,
add a comma, increase string length. MOVE D1,D4 ;loop control SUBQ #1,D4 LEA 1(A0,D0),A2 @4 MOVE.B (A2,D4),1(A2,D4) DBRA D4,@4 MOVE.B # , ,(A2) ADDI.B #1,(A0) ADDQ #1,D1 ;for the comma MOVEQ #3,D2 ;
reset counter @5 DBRA D0,@3 ;get next byte of format string @6 BSR.SGetANumber ;D0 = q BMI.S @16 ;no q provided -- leave MOVEQ #0,D2 MOVE.B (A0),D2 ;length of string MOVE D2,D1 SUB D5,D1 ;D1=len of
quotient now SUB D1,D0 BMI.S @10 BEQ.S @10 ADD.B D0,(A0) ;increase length by D0 spaces ;add D0 # of preceeding spaces @7 LEA 1(A0,D0),A2 SUBQ #1,D2 @8 MOVE.B 1(A0,D2),(A2,D2) DBRA D2,@8 SUBQ #1,D0 @9
MOVE.B # ,1(A0,D0) DBRA D0,@9 ;do remainder @10CMPI.B # . ,(A1) BNE.S @16 ADDA #1,A1 ;make sure there is a decimal point already LEA 1(A0),A2 MOVEQ #0,D0 MOVE.B (A0),D0 SUBQ #1,D0 @11CMP.B# . ,(A2)
+ BEQ.S @12 DBRA D0,@11 MOVE.B # . ,(A2)+ ADD.B #1,(A0) @12BSR.SGetANumber BMI.S @16 ;no r provided -- leave ; D0 is how many digits they want. D5 is ; what we ve already got. @13SUBQ #1,D5 @14CMP
D5,D0 BEQ.S @16 BPL.S @15 SUB.B #1,(A0) BRA.S @13 @15MOVE.B # 0',(A2,D5) ;add a zero ADD.B #1,(A0) ADDQ #1,D5 BRA.S @14 @16MOVEM.L(SP)+,A2/D0-D5 RTS ;========= GetANumber ;========= ; convert a one
or two digit ASCII integer into ; its numerical form ; input: A1 points to digit(s) ; output: D0 is the decimal equivalent (-1 if A1 ; didn t point to a digit) ; A1 points to byte after digit(s) ; Z
and N flags reflect the value of D0 MOVEM.LD1-D2,-(SP) MOVEQ #0,D0 MOVE.B (A1),D0 @1 BSR.SIsItADigit BEQ.S @2 MOVEQ #-1,D0 BRA.S @4 @2 ADDA #1,A1 MOVE D0,D1 ;save first digit MOVE.B (A1),D0 BSR.S
IsItADigit BEQ.S @3 MOVE D1,D0 BRA.S @4 @3 ADDA #1,A1 ADD D1,D1 ;multiply first digit by 10 MOVE D1,D2 ADD D2,D2 ADD D2,D2 ADD D2,D1 ADD D1,D0 ;add to second digit @4 MOVEM.L(SP)+,D1-D2 TST D0 RTS ;=
======== IsItADigit ;========= ; If the ASCII byte in D0 is in 0 .. 9' it s ; value [0..9] ; is returned in D0. If D0 is a digit, all ; flags are set. CMPI.B # 0',D0 BLT.S @1 CMPI.B # 9',D0 BGT.S @1
SUBI.B # 0',D0 MOVE #-1,CCR ;set all flags RTS @1 MOVE #0,CCR ;clear all flags RTS ; NumToString.asm ;---------------- ; by Mike Scanlin ; an alternative to _NumToString Xref NumToString ;==========
NumToString ;========== ; convert a 32 bit integer into a pascal string ; input: D0 longint ; A0 points to a space of at least 12 bytes ; output: A0 points to pascal string MOVEM.LD0-D4/A1-A2,-(SP)
LEA 1(A0),A1 ;skip length byte TST.L D0;is number zero? BGT.S @2 BMI.S @1 MOVE.B # 0',(A1)+ ;special case for zero BRA.S @8 @1 MOVE.B # - ,(A1)+ ;give string a minus sign NEG.L D0;make number
positive @2 LEA PowersTable,A2 MOVEQ #1,D3 ;set leading zeros flag MOVEQ #9,D4 ;loop counter @3 MOVE.L (A2)+,D2 ;get a power of 10 MOVEQ # 0',D1 ;init digit @4 CMP.LD2,D0 ;is # > power of 10? BLT.S
@5 ADDQ #1,D1 ;increase digit SUB.L D2,D0 ;subtract power of 10 BNE.S @4 @5 TST D3;have we had a non-zero digit yet? BEQ.S @6 CMP.B # 0',D1 ;is this a leading zero? BEQ.S @7 MOVEQ #0,D3 ;print all
zeros from now on @6 MOVE.B D1,(A1)+ @7 DBRA D4,@3 @8 MOVE A1,D0 ;calc length of new string SUB A0,D0 SUBQ.B #1,D0 ;minus 1 for length byte MOVE.B D0,(A0) MOVEM.L(SP)+,A1-A2/D0-D4 RTS PowersTable:
DC.L 1000000000 DC.L 100000000 DC.L 10000000 DC.L 1000000 DC.L 100000 DC.L 10000 DC.L 1000 DC.L 100 DC.L 10 DC.L 1 ; Resource File RESOURCE FRED 0 IDENTIFICATION DC.B 14, Format Program .ALIGN
2 RESOURCE BNDL 128 BUNDLE DC.L FRED ;name DC.W 0,1 ;data DC.L ICN# ;icon map DC.W 0 ;mapping-1 DC.W 0,128 ;map 0 to 128 DC.L FREF ;file ref DC.W 0 ;maps-1 DC.W 0,128 ;map 0 to 128 RESOURCE
FREF 128 FREF 1 DC.B APPL ,0,0,0 .ALIGN 2 RESOURCE ICN# 128 MY ICON DC.L $00000000, $00000000, $07FFFFC0, $18000030 DC.L $202038E8, $40604514, $80204412, $80204422 DC.L $80204441, $80204481,
$802139F1, $80000001 DC.L $80000001, $80F81021, $80103061, $802010A1 DC.L $80701121, $800811F1, $80881021, $80711021 DC.L $80000001, $80201801, $80602001, $80204001 DC.L $9E207802, $80204402,
$40204404, $20213808 DC.L $18000030, $07FFFFC0, $00000000, $00000000 DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF,
$FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF,
$FFFFFFFF, $FFFFFFFF DC.L $FFFFFFFF, $FFFFFFFF, $FFFFFFFF, $FFFFFFFF .ALIGN 2 RESOURCE MENU 1 APPLE DC.W 1 ;menu id DC.W 0 ;width holder DC.W 0 ;height holder DC.L 0 ;resource id holder DC.L
$1FB;flags enable all except 2 DC.B 1 ;title length DC.B 20;title apple symbol DC.B 22;menu item length DC.B About this program... DC.B 0 ;no icon DC.B 0 ;no keyboard equiv DC.B 0 ;marking char
DC.B 0 ;style of item text DC.B 2 ;menu item length DC.B -- DC.B 0 ;no icon DC.B 0 ;no keyboard equiv DC.B 0 ;marking char DC.B 0 ;style of item text DC.B 0 ;end of menu items .ALIGN 2 RESOURCE
MENU 2 FILE DC.W 2 ;menu id DC.W 0 ;width holder DC.W 0 ;height holder DC.L 0 ;resource id DC.L $1FF;flags DC.B 4 ;title length DC.B File ;title DC.B 6 ;menu item length DC.B Quit/Q DC.B 0 ;no
icon DC.B 0 ;no keyboard DC.B 0 ;no marking DC.B 0 ;style DC.B 0 ;end of menu items .ALIGN 2 RESOURCE MENU 3 MESSAGE DC.W 3 ;menu id DC.W 0 ;width holder DC.W 0 ;height holder DC.L 0 ;resource id
DC.L $1F8;flags - DISABLE DC.B 16;title length DC.B Click in Window ;title DC.B 0 ;menu item length DC.B DC.B 0 ;no icon DC.B 0 ;no keyboard DC.B 0 ;no marking DC.B 0 ;style DC.B 0 ;end of menu
Have a Special Dead Trigger 2 Easter Basket Full of Goodies, Courtesy of Madfinger Games Posted by Rob Rich on April 18th, 2014 [ permalink ] Dead Trigger 2 | Read more »
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The 61st Carnival of Mathematics
Today is a big day! Not only is it the first day of the new year but it’s also the first day of a new decade! In addition to all of that it is also time for the 61st Carnival of Mathematics and
this one has shaped up to be a great one thanks to the growing army of carnival contributors out there. So, put off joining the gym for one more day; Sit back, relax and enjoy this feast of
pulchritudinous mathematics.
First off, as per long standing carnival tradition, let’s look at some interesting properties of the number 61. Well, it’s prime for a start but so are a lot of numbers so maybe that isn’t so
interesting. However, 61 is the smallest multidigit prime p such that the sum of digits of p^p is a square (pop-quiz – what is the next one?). While on the subject of primes, 61 is the smallest prime
who’s digit reversal is square! It also turns out that the 61st Fibonacci number (2504730781961) is the smallest Fibonacci number which contains all the digits from 0 to 9. (Thanks to Number Gossip
for these by the way).
So 61 is a lot more interesting than you thought huh? If it had a name then it would be Keith and he would be Australian.
Puzzles, games and problems.
Let’s kick things off with a few puzzles. Sam Shah has submitted a problem for you to try which was originally created by his sister (A physics teacher) in A stubborn equilateral triangle.
If you are in the market for some online math games and lessons then head over to TutorFi.com and see what Meaghan Montrose has found for you.
Jonathan has a very interesting puzzle over at his blog, jd2718, called Who Am I (Teacher Edition) which should keep you thinking while you recover from the new year festivities.
Finally, Erich Friedman has prepared a set of holiday puzzles for 2009 for you all to try.
Explorations, discussions and messing about with maths
Pi is irrational right? Have you ever seen the proof? If not then you need to check out Brent Yorgey’s three part series on the irrationality of Pi over at The Math Less Travelled. Part 3 of this
series forms Brent’s submission to today’s carnival.
Pat Ballew has had his math class working on some maximization problems recently and his article Exploring an Isoperimetric Theme discusses a discovery about a “rule of thumb” for some maximization
problems. In a later post he wonders if there is a relation between the shape of a polygon and the maxium length of the diagonals (for a fixed perimeter).
Something that has kept potamologists awake over the years is the geometry of meandering rivers. If you’ve ever wondered about the same thing then head over to Division by Zero to see what Dave
Richeson has to say on the subject.
Terry Tao has been getting into the holiday season with a ‘more frivolous post than usual’ in A demonstration of the non-commutativity of the English language while Qiaochu Yuan of Annoying Precision
gets more serious and considers the combinatorics of words in The cyclotomic identity and Lyndon words.
Matters of a statistical nature
John D. Cook has been contemplating questions involving rare diseases and counterfeit coins over at The Endeavour.
Over at An Ergodic Walk the author has been discussing a statistical problem of how to estimate a probability distribution from samples when you don’t know e.g. how many possible values there are.
An example application is estimating the number of different butterfly species from a sample containing many unique species.
I don’t know about you but I like to have a game of cards with my mates from time to time (Poker is our usual game of choice and I am fantastically bad at it). Every now and then a small
‘discussion’ breaks out concerning how shuffled the deck of cards is which is usually solved by somone reshuffling them ‘properly’. But how many suffles are necessary to randomize a deck of cards?
Mathematically, card-shuffling can be viewed as a random walk on a finite group and, thus, it can be modeled by a Markov chain. Rod Carvalho has the details.
Techno Techno Techno….the technological side of mathematics
Sage is one of the best free mathematical software packages you can get at the moment and the project is led by William Stein, an associate professor at The University of Washington. In his post
Mathematical Software and Me he discusses his past experiences with mathematical software and recounts the series of events that led him to start the development of Sage. If you are interested in
Sage and can program in Python and Javascript then you may want to consider my Sage Bounty Hunt.
Wolfram Alpha has been a big hit among mathematics bloggers this year and, since it was launched back in May, Wolfram Research have added a lot of new features to it. For a list of some of the more
recent features check out the latest post from the Wolfram Alpha Blog – New Features in Wolfram|Alpha: Year-End Update.
Visualisation of volume data is getting easier every day thanks to products such as MATLAB and Patrick Kalita recently gave an internal talk to Mathworks engineers explaining how to do it. This
talk was recorded and turned into a series of 9 blog posts by Doug Hull over at Doug’s MATLAB Video Tutorials and the final part was posted early in December.
Teaching, learning and testing
Explaining mathematics can be hard and there are many different ways of teaching it. In How We Teach, Joel Feinstein shares some of his methodologies and includes a screencast of a talk of his
entitled “Using a tablet PC and screencasts when teaching mathematics.”
Every year, many hundreds of mathematics graduate students take language exams where they translate some technical writing in French,German or Russian into English. These translations are then
graded and thrown away which seems like a waste of effort when you think about it. David Speyer wonders if there is a more useful way to administer language exams in his post Let’s make language
exams useful.
Finally, Eric Mazur has posted a ‘video confession’ on YouTube saying “I thought I was a good teacher until I discovered my students were just memorizing information rather than learning to
understand the material. Who was to blame?”
Happy new Year – Math Carnival Style!
Now that it is officially 2010 you will be in need of a new calendar which is where Ron Doerfler of Dead Reckonings comes in. He has created a great looking calendar called The Age of Graphical
Computing and has made it all available for free. Just download, print and away you go. I have to confess that I am too lazy to build them myself and so only wish that he could make them available
for sale somehow. What about it Ron?
So, that’s it for this edition of the carnival – I hope you enjoyed it. The next one will be published on February 4th and I am still looking for someone to host it. So, if you have blog about
mathematics and would like a traffic boost then drop me a line and we’ll discuss it.
January 1st, 2010 at 23:44
Reply | Quote | #1
Pub-quiz next primes are:
313, 463, 739, 37957, 54667, 66571, 80809, 98041…
I checked until 125003 (which has 637131 digits).
Mathematica code:
January 2nd, 2010 at 00:19
Reply | Quote | #2
Great Site keep up the good work.
January 2nd, 2010 at 00:20
Reply | Quote | #3
I really enjoyed the stubborn triangle question! Although it was solved in a couple of lines ;-)
forces=Total[MapThread[#2/Norm[{x,y}-#1]^3 ({x,y}-#1)&,{p,m}]];
sol1 = {x, y} /. NMinimize[{forcemagnitude, x^2 + y^2 < 1 \[And] x < 0 \[And] y > 0}, {x, y}, MaxIterations -> 10^6][[2]]
sol2 = {x, y} /. NMinimize[{forcemagnitude, x^2 + y^2 < 1 \[And] x > 0 \[And] y < 0}, {x, y}, MaxIterations -> 10^6][[2]]
After which you can plot:
ContourPlot[Log[forcemagnitude], {x, -1.1, 1.1}, {y, -1.1, 1.1}, PlotPoints -> 10, PlotRangePadding -> 0,
Contours -> 20, Epilog -> {Red, PointSize[Large], Point[{sol1, sol2}], Orange,Point[p]},
RegionFunction -> (#1^2 + #2^2 < 1 &), FrameLabel -> {"x", "y"}]
Solutions are (we can have two solutions because the fields decay with distance squared):
{-0.477646475033636079575008490756, 0.174709836993090262676835042780}
{0.601438719335870730289962396423, -0.116809119162900477536841860600}
January 6th, 2010 at 00:08
Reply | Quote | #4
Some more pub-quiz:
the last one having 2484755 digits! | {"url":"http://www.walkingrandomly.com/?p=2113","timestamp":"2014-04-19T01:48:38Z","content_type":null,"content_length":"54923","record_id":"<urn:uuid:711d968a-498b-4d64-945b-d5aa582da419>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00298-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Solving the Cubic with Cardano - Aspects of Abbaco Mathematics
Aspects of Abbaco Mathematics
To understand the abbaco mathematics used by Cardano, we have to step back and look at the medieval tradition of abbaco schools and their masters. Though the subject is a fascinating and deep one,
there is one particular aspect of this tradition that is crucial in the following account: abbaco masters thought in terms of canonical problems, and one particular canonical problem, the “Problem of
Ten,” arises in the solution of the cubic that we will examine.
Throughout the Middle Ages and the Renaissance, city-states like Florence or Milan hired abbaco masters and established state-supported abbaco schools for children aged about nine to eleven. Students
were introduced to Hindu-Arabic numerals; the basic arithmetic operations of addition, subtraction, multiplication, and division; proportions; some practical geometry; and often algebra up to the
quadratic equation. The mathematical content was derived from Islamic texts, brought to Italy by writers such as Fibonacci. Abbaco masters often served their cities in capacities outside of the
abbaco school itself, as engineers or accountants, roles in which their mastery of abbaco mathematics made them indispensable [Note 1].
Abbaco mathematics was rhetorical—in Cardano’s time, most of the algebraic symbols with which we are so familiar were either recently invented, concurrent with the Ars Magna, or were well in the
future. For example, ‘\(+\)’ and ‘\(–\)’ were first recorded in the 1480s, and were not in common use in 1545, when the Ars Magna was published. Robert Recorde would not invent the equals sign until
1557, and the use of letters and exponential notation would have to await Francois Viete in the 1590s and the Geometrie of Rene Descartes of 1637 [Note 2]. What Descartes would write as \(x^3=ax+b,\)
Cardano wrote as “cubus aequalis rebus & numero” [Cardano 1662, Chapter 12, p. 251].
Rhetorical formulas can be difficult to remember, so algebraic rules were presented with canonical examples, which encoded the rules as algorithms within the examples. Thus, the mind of the abbaco
master was a storehouse of such canonical examples, to which he compared the new problems that he came across in his work. When he recognized a parallel structure between the new problem and a
canonical problem, he could solve the new problem by making appropriate substitutions into the canonical example.
Such canonical examples occurred even in the foundational texts of abbaco mathematics, including the Algebra of al-Khwarizmi. An important example for us, one that occurs implicitly in Cardano’s
solution to the cubic, is the “problem of ten” [Note 3]. Most abbaco texts had such problems, and one from Robert of Chester’s 1215 translation of al-Khwarizmi’s Algebra into Latin [al-Khwarizmi, p.
111] ran as follows:
Denarium numerum sic in duo diuido, vt vna parte cum altera multiplicata, productum multiplicationis in 21 terminetur. Iam ergo vnam partem, rem proponimus quam cum 10 sine re, quae alteram partem
habent, multiplicamus...
In his translation of this passage into English, Louis Karpinski used \(x\) for ‘rem’ (thing), and so I offer my own translation, without symbols [Note 4]:
Ten numbers in two parts I divide in such a way, in order that one part with the other multiplied has the product of the multiplication conclude with 21. Now therefore one part we declare the thing,
and then, with 10 without the thing, which the other part is, we multiply...
Al-Khwarizmi here first divided ten into two parts with product 21. He then named the two parts, one being 'thing' and the other '10 without the thing.' Continuing al-Khwarizmi’s solution
symbolically, he distributed \(x\) over \(10-x\) to get \(10x-x^2=21.\) By applying ‘al-jabr’ or restoration [Note 5], he transformed the latter to \(x^2+21=10x\) or, in rhetorical style, to the
problem of “square and number equal to thing.” Al-Khwarizmi had a solution for such a problem, complete with a geometric justification [al-Khwarizmi, p. 83]. Karpinski translated his solution as:
Take ½ of the unknown, that is 5, and multiply this by itself, giving 25. From this, subtract 21, giving 4. Take the root of this, 2, and subtract it from half of the root, leaving 3, which
represents one of the parts.
The other solution is, of course, 7. There are neither symbols nor symbolic formulas in either al-Khwarizmi’s statement of the problem, or in his solution.
The structure of the “problem of ten” was that of a number \(a\) broken into two parts \(x\) and \(y,\) with a condition on the parts; symbolically: \[x+y=a\,\,{\rm and}\,\,f(x,y)=b\] for some
function \(f(x,y)\) and number \(b.\) The usual method of solution was to express the two parts as “thing” and “number minus thing” and then to substitute into the condition, as al-Khwarizmi did
above. The “problem of ten” was canonical for quadratic problems, and served as a way to remember the rules for solving such problems.
As we shall see, Cardano’s solution to the cubic rested upon a “problem of ten.”
Notes for Aspects of Abbaco Mathematics
1. See Jens Høyrup, Jacopo da Firenze’s Tractatus Algorismi and Early Italian Abbacus Culture, for more details on abbaco mathematics and culture. Paul Grendler, Schooling in Renaissance Italy, is
also useful.
2. See Florian Cajori, A History of Mathematical Notations, for much more information about the history of mathematical symbolism.
3. The first example in Chapter 3 of the Ars Magna is an explicit working of a “problem of 10,” demonstrating Cardano’s familiarity with this problem [Cardano 1662, pp. 227-228].
4. A fairly literal translation, to catch the rhetorical style of non-symbolic algebraic writing.
5. For a discussion of the Arabic terms ‘al-jabr’ and ‘al-muqabala’, see [Katz, p. 244]. | {"url":"http://www.maa.org/publications/periodicals/convergence/solving-the-cubic-with-cardano-aspects-of-abbaco-mathematics?device=mobile","timestamp":"2014-04-19T23:16:35Z","content_type":null,"content_length":"27933","record_id":"<urn:uuid:c24b02aa-41cf-49ff-8a1c-6e8f188ce5a4>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00653-ip-10-147-4-33.ec2.internal.warc.gz"} |
George S. Fishman
These C-languange programs can be used to solve exercises in this book. cndev.c,
Bruec.c, and Pruec.c all use the pseudorandom number generator in rng_afm.c.
rng_afm.c For the pseudorandom number generator known as the Mersenne twister,
MT19937, this C-language program provides a means for loading seeds
at the beginning of a run and saving the final numbers in the sub-sequence
at the end of the run. This allows for non-overlapping sub-sequences
on successive runs.
cndev.c C-language code from Moro (1995) to generate a sample from the normal
distribution with 0 mean and variance 1 by the inverse-transform method.
The following are taken from WinRand at the Institute of Statistics at the University
of Graz at
Bruec.c C-language code to sample from the Binomial distribution using the
ratio-of- uniforms and inverse-transform methods.
Pruec.c C-language code to sample from the Poisson distribution using the ratio-of-
uniforms and inverse-transform methods. | {"url":"http://www.unc.edu/~gfish/fcmc.html","timestamp":"2014-04-21T04:57:13Z","content_type":null,"content_length":"4422","record_id":"<urn:uuid:fc6647b6-535f-4a2f-8008-eaef6aa3aeb7>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00605-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: Interesting trivia - anybody has a different answer than I keep getting - zero
Replies: 6 Last Post: Nov 29, 2012 10:39 AM
Messages: [ Previous | Next ] Topics: [ Previous | Next ]
Re: Interesting trivia - anybody has a different answer than I keep getting - zero
Posted: Nov 29, 2012 2:49 AM
In article <kpkab8tdbc77kkrg9mjeahi0hkrsb8rml3@4ax.com>,
Stone Bacchus <x@x.com> wrote:
> Thanks Jussi, yours and Mike's answer have me thinking on those lines.
> Thanks for your input. They will help for a nice evening of more
> analysis on those lines. appreciated.
> s
> On 27 Nov 2012 23:21:37 +0200, Jussi Piitulainen
> <jpiitula@ling.helsinki.fi> wrote:
> >Stone Bacchus writes:
> >
> >> My daughter and I were solving a math trivia and I could not come up
> >> with any answer other than zero. Would be interesting to see if
> >> somebody has a different opinion. The problem follows:
> >>
> >> You are at the start of a 1000 mile road with 3000 gummybears and a
> >> donkey. At the end of the road is a supermarket. You want to find
> >> the greatest number of gummy bears you can sell. Unfortunately, your
> >> donkey has a disease and can only carry 1000 gummybears at 1 time.
> >> Also, the donkey must eat 1 gummybear per mile.
> >>
> >> - You can drop off gummybears anywhere on the road
> >> - You can't carry gummybears while walking
> >> - No loopholes
> >>
> >> Again, this was a math trivia question and I could not ask anybody for
> >> clarification about what some the caveats meant or what the "no
> >> loopholes" meant, therefore I got zero.
> >
> >Take 900 bears on the donkey, walk it 300 miles and back, leaving 300
> >bears at that milepost. The donkey will have eaten 600 bears. You and
> >the donkey and 2100 bears are standing where you started.
> >
> >Take 1000 bears, walk the 300 miles. The donkey will have eaten
> >another 300 bears and has room for the pile of 300 bears that are
> >waiting there. (Take them.)
> >
> >The donkey is again carrying 1000 bears. Walk the remaining 700 miles.
> >You will have taken 300 bears to the market (and left 1100 behind).
> >
> >Therefore, the answer is at least 300. Probably more, of course.
Using a depot:
1. Stock up the depot: Load 1000 GBs; walk 333 1/3 miles;
leave 333 1/3 GBs at the depot; walk back 333 1/3 miles.
2. Final leg: Load 1000 GBs; walk 333 1/3 miles; load the 333 1/3 GBs
from the depot.
You have now effectively moved the origin for your final leg (with 1000
GBs) ahead 333 1/3 miles. Walk the remaining 667 2/3 miles to the
market. You have 333 1/3 GBs remaining on the donkey, and can sell 333
of them.
Note that you have used only 2000 of the original 3000 GBs, having left
1000 GBs behind at the origin. Instead of wasting them, you can use
them to push your depot (call it Z) forward some amount by using
another, previous depot (call it Y).
Y will need to have 2000 GBs, in order to be the new origin for stocking
depot Z. You can not stock Y in 1.5 roundtrips like depot Z, since the
donkey cannot carry enough. With < 1000 GBs moveable per trip, you need
at least 2.5 roundtrips. So, dividing the 2000 GBs by 4 stocking loads
(2 roundtrips and twice the load for the final one-way trip[1]) gives
500 GBs to stock per trip.
The donkey can carry 1000 GBs total, leaving 500 for donkey food on each
roundtrip. The donkey uses half of this for each leg of a roundtrip, or
250 GBs. Thus depot Y can be 250 miles from the origin.
Except, you run out of GBs. 250 GBs * 5 legs = 1250 GBs, but there are
only 1000 GBs left behind. So, the maximum donkey fuel per leg is only
1000 / 5 = 200 GBs. Thus depot Y can only be 200 miles from the origin.
So depot Z will be 200 miles closer to the market, so you will have 200
more GBs remaining from the 1000 you start out with on the final leg.
So the total remaining at the market will be 333 1/3 + 200 = 533 1/3 GBs.
> >(I don't see how to get the donkey back, though.)
Hopefully there is some suitable long-distance donkey food available at
the market which is much cheaper per mile than GBs, and which the donkey
can carry 1000 miles' worth. And besides, if you only feed the donkey
GBs it is sure to get diabetes and other health problems.
Alternatively, sell the diabetic donkey.
[1] Exercise: why twice?
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| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb | | {"url":"http://mathforum.org/kb/thread.jspa?messageID=7929748&tstart=0","timestamp":"2014-04-21T03:11:29Z","content_type":null,"content_length":"28820","record_id":"<urn:uuid:60f17597-f6ba-45bb-98dc-c8a5bc2eb2ca>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00280-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Renewal theory
From Encyclopedia of Mathematics
A branch of probability theory describing a large range of problems connected with the rejection and renewal of the elements of some system. The principal concepts in renewal theory are those of a
renewal process and renewal equation. A renewal process may be described using the classical scheme of sums of independent random variables in the following manner. Let
If the
For Poisson process, in which
The renewal process queueing theory, reliability theory, storage theory, the theory of branching processes (cf. Branching process), etc. A large number of results in renewal theory is connected with
the study of the asymptotic properties of the renewal function
where [1]), that if the distribution
Numerous results are available which generalize and precisize equations (3) and (4) in several respects. Results of the kind presented in (3) and (4) are used to study the asymptotic behaviour of the
in which the free term
The relation
follows from definition (1). Since limit theorems for sums
There exists a large number of generalizations of the scheme just described. One such generalization, connected with semi-Markov processes (cf. Semi-Markov process), yields the so-called Markov
renewal process, in which the system has some number of states, and the lifetimes of the individual elements are random elements which depend on the states of the system before and after the moment
of renewal.
[1] D.R. Cox, "Renewal theory" , Methuen (1962)
How to Cite This Entry:
Renewal theory. B.A. Sevast'yanov (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Renewal_theory&oldid=19000
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098 | {"url":"http://www.encyclopediaofmath.org/index.php/Renewal_theory","timestamp":"2014-04-19T09:38:28Z","content_type":null,"content_length":"21942","record_id":"<urn:uuid:b6f97b82-8f10-4c00-80cf-b5c7e1e3f706>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00129-ip-10-147-4-33.ec2.internal.warc.gz"} |
Average parameterization and partial kernelization for computing medians
- In Proceedings of the 27th International Symposium on Theoretical Aspects of Computer Science (STACS ’10), volume 5 of LIPIcs
"... Abstract. Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into
the field of multivariate algorithmics. Correspondingly, we describe how to perform a systematic investiga ..."
Cited by 24 (19 self)
Add to MetaCart
Abstract. Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into the
field of multivariate algorithmics. Correspondingly, we describe how to perform a systematic investigation and exploitation of the “parameter space ” of computationally hard problems.
, 2010
"... To make a joint decision, agents (or voters) are often required to provide their preferences as linear orders. To determine a winner, the given linear orders can be aggregated according to a
voting protocol. However, in realistic settings, the voters may often only provide partial orders. This direc ..."
Cited by 12 (0 self)
Add to MetaCart
To make a joint decision, agents (or voters) are often required to provide their preferences as linear orders. To determine a winner, the given linear orders can be aggregated according to a voting
protocol. However, in realistic settings, the voters may often only provide partial orders. This directly leads to the POSSIBLE WINNER problem that asks, given a set of partial votes, whether a
distinguished candidate can still become a winner. In this work, we consider the computational complexity of POSSIBLE WINNER for the broad class of voting protocols defined by scoring rules. A
scoring rule provides a score value for every position which a candidate can have in a linear order. Prominent examples include plurality, k-approval, and Borda. Generalizing previous NP-hardness
results for some special cases, we settle the computational complexity for all but one scoring rule. More precisely, for an unbounded number of candidates and unweighted voters, we show that POSSIBLE
WINNER is NP-complete for all pure scoring rules except plurality, veto, and the scoring rule defined by the scoring vector (2, 1,...,1, 0), while it is solvable in polynomial time for plurality and
"... Abstract. RANK AGGREGATION is important in many areas ranging from web search over databases to bioinformatics. The underlying decision problem KE-MENY SCORE is NP-complete even in case of four
input rankings to be aggregated into a “median ranking”. We study efficient polynomial-time data reduction ..."
Cited by 1 (0 self)
Add to MetaCart
Abstract. RANK AGGREGATION is important in many areas ranging from web search over databases to bioinformatics. The underlying decision problem KE-MENY SCORE is NP-complete even in case of four input
rankings to be aggregated into a “median ranking”. We study efficient polynomial-time data reduction rules that allow us to find optimal median rankings. On the theoretical side, we improve a result
for a “partial problem kernel ” from quadratic to linear size. On the practical side, we provide encouraging experimental results with data based on web search and sport competitions, e.g., computing
optimal median rankings for real-world instances with more than 100 candidates within milliseconds. 1
"... Abstract. Given a collection C of partitions of a base set S, the NP-hard Consensus Clustering problem asks for a partition of S which has a total Mirkin distance of at most t to the partitions
in C, where t is a nonnegative integer. We present a parameterized algorithm for Consensus Clustering with ..."
Cited by 1 (1 self)
Add to MetaCart
Abstract. Given a collection C of partitions of a base set S, the NP-hard Consensus Clustering problem asks for a partition of S which has a total Mirkin distance of at most t to the partitions in C,
where t is a nonnegative integer. We present a parameterized algorithm for Consensus Clustering with running time O(4.24 k ·k 3 +|C|·|S | 2), where k: = t/|C| is the average Mirkin distance of the
solution partition to the partitions of C. Furthermore, we strengthen previous hardness results for Consensus Clustering, showing that Consensus Clustering remains NP-hard even when all input
partitions contain at most two subsets. Finally, we study a local search variant of Consensus Clustering, showing W[1]-hardness for the parameter “radius of the Mirkin-distance neighborhood”. In the
process, we also consider a local search variant of the related Cluster Editing problem, showing W[1]-hardness for the parameter “radius of the edge modification neighborhood”. 1
- PROC. 27TH STACS , 2010
"... Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into the field
of multivariate algorithmics. Correspondingly, we describe how to perform a systematic investigation and ..."
Add to MetaCart
Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into the field of
multivariate algorithmics. Correspondingly, we describe how to perform a systematic investigation and exploitation of the “parameter space” of computationally hard problems.
"... Abstract. Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into
the field of multivariate algorithmics. Correspondingly, we describe how to perform a systematic investiga ..."
Add to MetaCart
Abstract. Research on parameterized algorithmics for NP-hard problems has steadily grown over the last years. We survey and discuss how parameterized complexity analysis naturally develops into the
field of multivariate algorithmics. Correspondingly, we describe how to perform a systematic investigation and exploitation of the “parameter space ” of computationally hard problems.
"... Abstract. We review NP-hard voting problems together with their status in terms of parameterized complexity results. In addition, we survey standard techniques for achieving fixed-parameter (in)
tractability results in voting. 1 ..."
Add to MetaCart
Abstract. We review NP-hard voting problems together with their status in terms of parameterized complexity results. In addition, we survey standard techniques for achieving fixed-parameter (in)
tractability results in voting. 1
"... In a typical covering problem we are given a universe U of size n, a family S (S could be given implicitly) of size m and an integer k and the objective is to check whether there exists a
subfamily S ′ ⊆ S of size at most k satisfying some desired properties. If S ′ is required to contain all the e ..."
Add to MetaCart
In a typical covering problem we are given a universe U of size n, a family S (S could be given implicitly) of size m and an integer k and the objective is to check whether there exists a subfamily S
′ ⊆ S of size at most k satisfying some desired properties. If S ′ is required to contain all the elements of U then it corresponds to the classical Set Cover problem. On the other hand if we require
S ′ to satisfy the property that for every pair of elements x, y ∈ U there exists a set S ∈ S ′ such that |S ∩ {x, y} | = 1 then it corresponds to the Test Cover problem. In this paper we consider a
natural parameterization of Set Cover and Test Cover. More precisely, we study the (n − k)-Set Cover and (n − k)-Test Cover problems, where the objective is to find a subfamily S ′ of size at most n
− k satisfying the respective properties, from the kernelization perspective. It is known in the literature that both (n−k)-Set Cover and (n−k)-Test Cover do not admit polynomial kernels (under some
well known complexity theoretic assumptions). However, in this paper we show that they do admit “partially polynomial kernels”. More precisely, we give polynomial time algorithms that take as input
an instance (U, S, k) of (n − k)-Set Cover ((n − k)-Test Cover) and return an equivalent instance ( Ũ, ˜ S, ˜ k) of (n − k)- | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=10190270","timestamp":"2014-04-17T06:07:23Z","content_type":null,"content_length":"32428","record_id":"<urn:uuid:e832aaa7-38af-4aa4-91a6-0c57ff0f03e2>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00409-ip-10-147-4-33.ec2.internal.warc.gz"} |
A to Z Kids Stuff | Preschool Shapes Theme
Ricky Rectangle
Ricky Rectangle is my name.
My four sides are not the same.
2 are short and 2 are long.
Count my sides, come right along.
1, 2, 3, 4.
Sponge Painting
Need: construction paper rectangle, paint, rectangular sponges
Place paint in a shallow container. Have children dip the sponges into
the paint and press onto their construction paper to create a rectangle
shapes mural.
Rectangle Mural
Need: variety of rectangles
Cut rectangles from many different types of paper; construction, gift wrap... Give each child a sheet of construction paper. Place the pre-cut rectangles on the table with glue. The children pick
out and glue onto thier paper the rectangles to create a rectangle collage.
Train -Finish The Train
Sammy Snake Rectangle
Celery sticks, carrot sticks, rectangular crackers, or granola bars
Draw a Square
A to Z Kids Stuff eBook
Draw a square, draw a square
Shaped like a tile floor.
Draw a square, draw a square
All with corners four.
Sammy Square
Sammy Square is my name.
My four sides are just the same.
Turn me around, I don't care.
I'm always the same.
I'm Sammy Square.
Friendly Shapes Two
Make a circle, make a circle, (draw in the air)
Draw it in the sky
Use your finger, use your finger,
Make it round as a pie
Draw a square, draw a square,
Make the lines so straight.
Make a square, make a square,
Draw a box in the air.
Square College
Need: Various sized squares cut from construction paper, one 8-inch construction paper square per child.
Give each child one 8 inch square cut from construction paper. Have an assortment of various size squares in piles from which the children may choose. Have the children glue various sizes of
squares onto their large square paper.
When the children are finished make a wall decoration by taping each square side by side onto a wall or bulletin board. Retain the basic square shape by taping each paper in a equal array of rows
and columns.
Shiny Stars
Cut a star shape from poster board. Glue aluminum foil onto the shape. Punch a hole near the top and using yarn or string hang the stars from the ceiling.
Star pattern
Twinkle, twinkle, little star
Print and color
Sammy Snake Star
Dora's Star Catching Storybook
Dora's Collect A Star coloring book
Draw a Triangle
Draw a triangle, draw a triangle
With corners three.
Draw a triangle, draw a triangle
Draw it just for me.
Tommy Triangle
I'm Tommy Triangle.
Look at me!
Count my sides.
One, two, three.
Tissue Triangle Art
Need: Tissue paper triangles, diluted glue, wax paper, paint brushes
Cut triangles out of different colors of tissue paper. Set out brushes and diluted glue. Give each child a piece of waxed paper. Have the children brush the glue on their papers and place the
triangles on top of the glue.
Encourage them to work on small areas at a time and to overlap their triangles to create new colors. For a shiny effect, brush more glue over the children's papers when they have finished. Attach
construction paper frames. Then punch a hole in one corner of each frame and hang the papers from the ceiling or in a window.
Cut triangles of different sizes from green construction paper. Have the children glue the green triangles on blue paper to create Christmas tree "forests." Different shades of green
make the pictures more attractive. Try brushing some glue on the construction paper and sprinkling with salt to make snow.
Ollie Oval
I am Ollie Oval
A football shape is mine
Some people think that I'm an egg
But I think I look fine!
*Other Sites
Which Shape is Different activity sheet from Bry-Back Manor
Connect the dots to find Zini'a favorite shapes
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WSU Libraries homepage
WSU Libraries Online Catalog--locate books and materials owned by the Libraries
Electronic Journal List--a growing number of the libraries' journals are available online
WSU Mathematics & Statistics Department
Full list of Databases by Subject
Applied Science Full Text--One of the WilsonWeb databases. Indexes journal articles in the applied sciences and all engineering fields from 1983 to the present and select full text from 1997. Broader
scope makes this a good source for undergraduate research .
General Science Full Text --One of the WilsonWeb databases. Very broad coverage of science. Indexes journal articles from 1984 to the present. Selected full text available from 1993.
Mathscinet --Mathsci is the Mathematical Sciences database produced by the American Mathematical Society (AMS). Mathsci provides comprehensive coverage of international research in mathematics and
mathematically related research in statistics, computer science, physics, operations research, engineering, biology, and other disciplines. It contains abstracts from approximately 2,000 journals,
conference proceedings and proceedings papers, books or chapters, advanced-level textbooks, and published dissertations. Dates of coverage are: bibliographic entries from 1940, and bibliographic data
and reviews from 1980 to present.
CRC Standard Mathematical Tables and Formulae, Latest edition in reference (call no. QA47.M315).
Encyclopedia of Mathematics, 10 volumes updated periodically by supplements (call no. QA5. M3713 1988)
Encyclopedia of Statistical Sciences, 9 volume set plus 3 volume Update (call no. QA276.14. E5)
McGraw-Hill Encyclopedia of Science & Technology--New 2007 Edition available on Index Table 3. Good general resource in science, applied science and engineering (call no. Q121.M3 2007).
Standard Probability and Statistics Tables and Formulae, published by CRC (call no. QA273.3 .Z95 2000).
AMS Math on the Web--American Mathematical Society's directory to mathematical resources.
AWM Association for Women in Mathematics--organization dedicated to encouraging women and girls in the mathematical sciences. Site includes biographies of women in the field, scholarship and grant
opportunities, links for students, parents, and educators
MAA Online--The Mathematical Association of America
MacTutor History of Mathematics--searchable index of math history from the School of Mathematics and Statistics University of St. Andrews, Scotland. Indexes include women mathematicians, famous
curves, quotations, etc.
The Mathematics ArXiv --the UC Davis front end for the mathematics arXiv, maintained at Cornell University. Searchable preprint server.
Mathematics WWW virtual Library--searchable index of math resources available on the Web maintained by the Florida State University Department of Mathematics.
Math Forum @ Drexel--math resources for students and educators K-12 and beyond.
MathWorld--Eric Weisstein's World of Mathematics is a searchable, interactive, math encyclopedia.
NSF Division of Mathematical Sciences--(National Science Foundation). This site includes funding opportunities, research highlights, programs in mathematics, etc.
On-Line Encyclopedia of Integer Sequences--searchable database of number sequences with links to additional information, authors, and extensions.
Zentralblatt MATH--searchable abstracting and reviewing service in pure and applied mathematics.
ASEE Presentation--MS Powerpoint slides, printout as handouts.
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Pair correlation functions around a pulled A particle at T = 0.14 for different forces f and at equilibrium (dashed lines). The solid lines in the plots correspond to the forces f = 2.5, 6.0, 8.0,
10.0, and 20.0 (from top to bottom). Note that the latter curves are shifted with respect to the equilibrium curves by multiples of −0.5 on the ordinate. g AA(x) in (a) and g AB(x) in (b) are the
pair correlation functions in force direction, g AA(r) in (c) and g AB(r) in (d) those perpendicular to the force.
The same as Fig. 1 , but now for a pulled B particle. g BA(x) in (a) and g BB(x) in (b) are the pair correlation functions in force direction, g BA(r) in (c) and g BB(r) in (d) those perpendicular to
the force.
Time dependence of displacement Δx of a pulled A particle for the forces f = 0.5. 1.0, 1.5, 2.0, 2.5, and 5.0 (from right to left) at temperature T = 0.14. Solid lines correspond to the transient
state from switch-on of the force at t = 0 to the steady-state at long times. Dashed lines correspond to the steady state where Δx displays a linear time dependence.
(a) Steady state velocity of the pulled A particle as a function of f for different temperatures T, as indicated. The bold dashed line is a linear fit to the data for T = 1.0. (b) Peclet number Pe*
as a function of (see text).
(a) Mean-squared displacement ⟨Δx 2(t)⟩ − ⟨Δx(t)⟩2 for pulled A particle at T = 0.14. The curves correspond to the forces f = 0.0, 0.5, 1.0, 1.5, 2.5, 4.0, 6.0, and 10.0 (from right to left). (b)
Effective exponents α as a function of f for different temperatures, as indicated.
Mean-squared displacement ⟨Δx 2(t)⟩ − ⟨Δx(t)⟩2 for pulled A particle at T = 0.14 and f = 2.0 for different thermostats (see text). The inset shows the corresponding mean-squared displacement
displacements divided by t.
(a) Typical trajectories, x(t), of pulled A particles at T = 0.14 and f = 1.0. (b) Typical trajectories, x(t), of pulled A particles at f = 1.0 and different temperatures, as indicated.
van Hove correlation function of pulled A particle at T = 0.14 for the indicated times, (a) f = 1.0, (b) f = 2.5.
Dependence of self-diffusion constant on inverse temperature for pulled A particles for the forces f = 0.0 (equilibrium), 0.5, 1.0, 1.5, 2.0, 2.5, and 5.0. The insets show the same data, but now
using the scaled effective temperatures T eff(f) instead of T.
Time dependence of the incoherent scattering functions, F s(q, t) in perpendicular direction for A particles at f = 1.0 and q = 6.0 for the temperatures T = 0.14, 0.15, 0.16, 0.17, 0.18, 0.21, 0.25,
0.30, and 0.34 (from right to left).
Dependence of (a) α relaxation time and (b) friction coefficient on inverse temperature for pulled A particles for the same forces as in Fig. 9 . The insets show the data as function of the effective
temperatures T eff(f).
T eff/T − 1 as function of f, as obtained from the scaling of the self-diffusion constant D orth and the friction coefficient ξ. Results for A and B particles are shown, as indicated. The dotted and
dashed lines are fits with T eff/T − 1 = C × f 2 yielding the fit parameters C = 0.0326 and C = 0.0181 for D orth and C = 0.0266 and C = 0.0134 for ξ of A and B particles, respectively. | {"url":"http://scitation.aip.org/content/aip/journal/jcp/138/12/10.1063/1.4770335","timestamp":"2014-04-18T14:21:11Z","content_type":null,"content_length":"85578","record_id":"<urn:uuid:b4935483-1901-449b-8fd2-be37411b816e>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00310-ip-10-147-4-33.ec2.internal.warc.gz"} |
Kalman filter vs Complementary filter
Note: At the bottom of the post the complete source code
The use of accelerometer and gyroscope to build little robots, such as the self-balancing, requires a math filter in order to merge the signals returned by the sensors.
The gyroscope has a drift and in a few time the values returned are completely wrong. The accelerometer, from the other side, returns a true value when the acceleration is progressive but it suffers
much the vibrations, returning values of the angle wrong.
Usually a math filter is used to mix and merge the two values, in order to have a correct value: the Kalman filter . This is the best filter you can use, even from a theoretical point of view, since
it is one that minimizes the errors from the true signal value. However it is very difficult (see here) to understand. In fact, the filter needs to be able to calculate the coefficients of the
matrices, the process-based error, measurement error, etc. that are not trivial.
In the hobbistic world, recently are emerging other filters, called complementary filters. In fact, they manage both high-pass and low-pass filters simultaneously. The low pass filter filters high
frequency signals (such as the accelerometer in the case of vibration) and low pass filters that filter low frequency signals (such as the drift of the gyroscope). By combining these filters, you get
a good signal, without the complications of the Kalman filter.
Making a study from a theoretical point of view, the discussion is complicated and is beyond the scope of this tutorial. The complementary filters can be have different ‘orders’. Here I speak about
the so-called first-order filter that filter already well, and the second-order filter which filter even better. Clearly, going from first to second order, the algorithm is more complicated to use
and perhaps there is no gain so obvious to justify the increase in complexity.
A great introduction to the first order complementary filters applied to the an accelerometer and a gyroscope, comes from MIT (here). It introduces the filter in a a very simple mode. On this
document is based the first Arduino algorithm:
// a=tau / (tau + loop time)
// newAngle = angle measured with atan2 using the accelerometer
// newRate = angle measured using the gyro
// looptime = loop time in millis()
float tau=0.075;
float a=0.0;
float Complementary(float newAngle, float newRate,int looptime) {
float dtC = float(looptime)/1000.0;
x_angleC= a* (x_angleC + newRate * dtC) + (1-a) * (newAngle);
return x_angleC;
It ’enough to choose the response time of tau, to send the arguments, ie the angle measured with the accelerometer and the gyroscope, the time of the loop and you get in two lines, the angle
calculated by the filter.
The algorithm at the base of the second order complementary filter is described here. Indeed it is not described at all, but now we’ve figured out how the filter works by the MIT’s documentation.
The principle is the same, the algorithm is more complicated. The translation of this algorithm for the Arduino:
// newAngle = angle measured with atan2 using the accelerometer
// newRate = angle measured using the gyro
// looptime = loop time in millis()
float Complementary2(float newAngle, float newRate,int looptime) {
float k=10;
float dtc2=float(looptime)/1000.0;
x1 = (newAngle - x_angle2C)*k*k;
y1 = dtc2*x1 + y1;
x2 = y1 + (newAngle - x_angle2C)*2*k + newRate;
x_angle2C = dtc2*x2 + x_angle2C;
return x_angle2C;
Here too we just have to set the k and magically we get the angle.
If we want to apply the Kalman filter, we can re-use one of the codes already present in internet. This is the code that I copied from the Arduino forum (here):
// KasBot V1 - Kalman filter module
float Q_angle = 0.01; //0.001
float Q_gyro = 0.0003; //0.003
float R_angle = 0.01; //0.03
float x_bias = 0;
float P_00 = 0, P_01 = 0, P_10 = 0, P_11 = 0;
float y, S;
float K_0, K_1;
// newAngle = angle measured with atan2 using the accelerometer
// newRate = angle measured using the gyro
// looptime = loop time in millis()
float kalmanCalculate(float newAngle, float newRate,int looptime)
float dt = float(looptime)/1000;
x_angle += dt * (newRate - x_bias);
P_00 += - dt * (P_10 + P_01) + Q_angle * dt;
P_01 += - dt * P_11;
P_10 += - dt * P_11;
P_11 += + Q_gyro * dt;
y = newAngle - x_angle;
S = P_00 + R_angle;
K_0 = P_00 / S;
K_1 = P_10 / S;
x_angle += K_0 * y;
x_bias += K_1 * y;
P_00 -= K_0 * P_00;
P_01 -= K_0 * P_01;
P_10 -= K_1 * P_00;
P_11 -= K_1 * P_01;
return x_angle;
To get the answer, you have to set 3 parameters: Q_angle, R_angle,R_gyro. The activity is a bit complicated .
But what happens with these algorithms? Similar curves are obtained? Here’s a comparison:
There are 5 curves:
Color lines:
• Red - accelerometer
• Green - Gyro
• Blue - Kalman filter
• Black - complementary filter
• Yellow - the second order complementary filter
As you can see the signals filtered are very similarly. Note that in the presence of vibrations, the accelerometer (red) generally go crazy. The gyro (green) has a very strong drift increasing int
the time.
Now let’s see a comparison only between a filtered signal. That kalman (green), complementary (black) and complementary second-order (yellow). You can see how the Kalman is a bit late vs
complementary filters, but it is more responsive to the vibration. In this case the second order filter does not return an ideal curve, probably I have to work a bit on the coefficients.
In conclusion I think that the complementary filter, in this case the first order, can be used in place of the Kalman filter. The smoothing is good and the algorithm is much simpler than Kalman.
The hardware I used was composed of:
- Arduino 2009
- 6-axis IMU SparkFun Razor 6 DOF
This is the complete source code:
#1 by Iwan on 21 November 2012 - 05:40
how do you display the graph the results of Kalman and complementary?
This chart comes from the hardware or the results of the simulation matlab?
Tell me please,
#2 by robottini on 21 November 2012 - 06:02
No Matlab simulation. They are the hardware results. The code is included in the post.
#3 by Iwan on 3 December 2012 - 19:55
can you give me a complete source code? I want to try it..
#4 by robottini on 3 December 2012 - 21:27
I sent you an email with the code
#5 by xw on 6 December 2012 - 22:35
can you give me a complete source code? I want to try it..
#6 by ducati on 7 December 2012 - 00:54
nice post!
I, too, am interested in how you collect the data for the graph.
if possible, please send me the code.
ciao e grazie mille!
#7 by tantun on 7 December 2012 - 15:14
can you send to me complete source code?
#8 by Izbert on 11 December 2012 - 03:56
Can i also get a copy of this source code?
#9 by phuongnut on 12 December 2012 - 16:21
I want to try it.
Send to me complete source code pls!
#10 by robottini on 12 December 2012 - 17:50
FOR ALL: I put the source code in the post. please take it if you need
#11 by jj on 1 January 2013 - 11:34
Are the codes that you post are all complete source codes? If not can I have them? Thanks
#12 by robottini on 1 January 2013 - 13:03
Yes, i put the code in the post.
#13 by kamal on 19 January 2013 - 09:57
Dear , can u send me to code
#14 by robottini on 19 January 2013 - 10:16
At the bottom of the post the complete source code
#15 by Someone on 30 January 2013 - 05:47
Can you send a completed code to me? I want to try it since I do not have any experience about that. Thanks
#16 by robottini on 30 January 2013 - 08:39
At the bottom of the post the complete source code
#17 by Someone on 30 January 2013 - 09:13
Okay, I can see that. Thanks
#18 by Morten on 2 May 2013 - 18:25
Hi, when I click on Filters1 for downloading the source code I just get redirected to this page.
Is it possible to repost the source code?
#19 by robottini on 3 May 2013 - 07:48
Thanks Morten, now the link works.
#20 by Pablopaolus on 27 May 2013 - 16:29
Thank you for sharing your job
I’m using PIC18F46J50 as MCU along with Sparkfun IMU – 6DOF ITG3200/ADXL345, and I’m trying to combine accelerometer and giroscope data using the first order complementary filter. I’ve ported your
code (to CCS v4.140). It seems to work well enough within an angle range. However, if I tilt for instance pitch angle to 90 or -90 deg, roll angle goes crazy. The same occurs to pitch if I tilt roll
to +-90. I don’t know if I’ve explained myself clearly, so I’ve made a video:
Here is my code:
float tau=0.075;
float a=0.0;
float rollAcc=0, rollGyro=0;
float pitchAcc=0, pitchGyro=0;
//RwAcc[x] in g
//Gyro[x] in deg/s
void ComplementaryFilter2() {
interval = (0.250*(float)timerCount250us)/1000.0; //Units: sec
timerCount250us = 0; //Restart the counters
rollAcc=atan2(RwAcc[1], RwAcc[2]) * 180 / PI;
pitchAcc=atan2(RwAcc[0], RwAcc[2]) * 180 / PI;
roll = a * (roll + Gyro[0] * interval) + (1-a) * (rollAcc);
pitch = a * (pitch + Gyro[1] * interval) + (1-a) * (pitchAcc);
I use an internal interruption so as to calculate the interval. Timer0 overflows every 250us:
void TIMER0_isr(void)
Sorry for the long post.
I would greatly appreciate if you could help me.
Thank you.
#21 by Tjaart on 10 June 2013 - 21:17
Hello, I’m currently trying to implement a Kalman filter using the code above. After I plotted the accelerometer angle vs the Kalman angle, they seemed to be about the same. After looking through the
code I found that :
x_angle += K_0 * y;
x_angle += K_0 * (newAngle – x_angle);
x_angle += (P_00 / S) * (newAngle – x_angle);
x_angle += (P_00 / (P_00 + R_angle)) * (newAngle – x_angle);
In most cases my P_00>>R_angle (sometimes maxed at about +-1800) whis basically means K_0~1 and then
x_angle = x_angle + newAngle – x_angle;
x_angle = newAngle;
Am I calculating P_00 wrong? Any help would be appreciated!
#22 by irina on 13 June 2013 - 21:37
Hi, I’m trying to use a ADXL345 and ITG 3200 for a platform with 3 analog servos. I need to filtrate the data from gyro and accel and using your complementary filter implementations. Hoe would the
filter look like for all three axis active? thanks
#23 by siddharth on 28 October 2013 - 12:43
Dear Sir,
plz send me source code of 2 wheel balancing robot using gyro-521 and kalman filter.
I will be very thankfulto you.
#24 by siddharth on 28 October 2013 - 12:44
Dear Sir,
plz send me the code of two wheel balancing robot using gyro and kalman filter.
#25 by robottini on 28 October 2013 - 12:48
Sorry the code is complex, I can’t work on it. You can find many implementations on internet about it.
#26 by kevan on 7 November 2013 - 14:55
sir, can you help me. why result the source code can’t look in serial monitor arduino?
#27 by robottini on 7 November 2013 - 15:03
Are you using the right speed in the serial monitor (the same of Serial.begin in the code)?
#28 by kevan on 12 November 2013 - 09:13
no sir, i’m using serial 1200.
it’s make value of the sensor just 0?
#29 by robottini on 12 November 2013 - 10:31
You have to use the same speed in the serial monitor and in the Serial.begin in the code. Please use 57600 as a value in the Serial.beglin (57600) and in the Serial monitor. This is the first point.
After, we can see if there are other problems.
#30 by kevan on 12 November 2013 - 17:24
sir, i was changed serial begin become 57600 but the result constant 0..
i’m using accelerometer (ADXL345) and Gyroscope (ITG3200), how setting and read this sensor sir?
#31 by Eric on 23 December 2013 - 10:02
我想回覆#20 by Pablopaolus
Here is the equation:
#32 by Ali Hamza on 2 February 2014 - 17:21
Hi, I was looking at your code and I kinda get hold of it and understand it. However, I have a question. This code gives all the 6 axis readings. like the 3 axis of gyro and 3 axis of accelerometer.
Because I am trying to implement it on my quadroter. So wanted to know if I can implement it on it.
Sorry if I dnt make too much sense as I am new to all this.
#33 by TARIKU w/TSADIK on 23 March 2014 - 15:44
Hi, I am tried to implement Kalman filter for noisey Gyro-accelerometer data in matlab. Is there anyone who could help me ,please?
Tags: complementary filter, Kalman filter
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Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 791.05037
Autor: Erdös, Paul; Tuza, Zsolt
Title: Rainbow subgraphs in edge-colorings of complete graphs. (In English)
Source: Gimbel, John (ed.) et al., Quo vadis, graph theory? A source book for challenges and directions. Amsterdam: North-Holland, Ann. Discrete Math. 55, 81-88 (1993).
Review: We raise the following problem. Let F be a given graph with e edges. Consider the edge colorings of K[n] (n large) with e colors, such that every vertex has degree at least d in each color (d
< n/e). For which values of d does every such edge coloring contain a subgraph isomorphic to F, all of whose edges have distinct colors? The case when F is the triangle K[3] is well-understood, but
for other graphs F many interesting questions remain open, even for d-regular colorings when n = de+1.
Classif.: * 05C15 Chromatic theory of graphs and maps
Keywords: rainbow subgraphs; complete graphs; edge colorings
© 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 │
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October 31st 2008, 08:11 PM
The scale of a map is 1:300 000
Find the actual area of the village, in km^2, if the area on the map is 3cm^2
November 1st 2008, 07:25 PM
Prove It
It tells us that 1cm on the map gives 300,000 cm in real life.
If we had a square of side length 1cm, it would correspond to a square of side lengths 300,000cm. So a square with area $1cm \times 1cm = 1cm^2$ would correspond to a square of area $300,000cm \
times 300,000cm = 90,000,000,000cm^2$.
So if the area on the map is $3cm^2$ then it would correspond to an area of $3 \times 90,000,000,000cm^2 = 270,000,000,000cm^2$.
Since there are 100cm in 1 metre, there are $100^2cm^2=10,000cm^2$ in $1m^2$.
So that means that $270,000,000,000cm^2 = 27,000,000m^2$.
Since there are 1000m in 1km, there are $1000^2m^2 = 1,000,000m^2$ in $1km^2$.
So $27,000,000m^2 = 27km^2$.
Therefore the actual area that is represented by this map is $27km^2$.
November 1st 2008, 07:29 PM
OH I SEE! thank you very much. I have been very stupid. i understand now | {"url":"http://mathhelpforum.com/geometry/56839-maps-print.html","timestamp":"2014-04-16T08:21:39Z","content_type":null,"content_length":"6951","record_id":"<urn:uuid:0d09e5bf-3a46-42e0-b333-887e45ac36ee>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00074-ip-10-147-4-33.ec2.internal.warc.gz"} |
Contour integral calculation?
October 18th 2011, 12:24 PM #1
Contour integral calculation?
So basically I need to calculate this
where L is defined as
How should I approach it? "Wolfram Mathematica" or "Wolfram Alpha" would be preferred, however "Mathcad" or "Maple" input is also fine as long as one does the job... Thanks.
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Karl Pearson
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File:Karl Pearson.gif
Karl Pearson (pencil sketch in notebook; there is some see-through of writing on next page)
Karl Pearson (March 27, 1857 – April 27, 1936) was a major contributor to the early development of statistics as a serious scientific discipline in its own right. He founded the Department of Applied
Statistics at University College London in 1911; it was the first university statistics department in the world.
Karl Pearson was born in Islington, London on March 27, 1857, the son of barrister William Pearson and Fanny, née Smith. He was educated privately at University College School, after which he went to
King's College, Cambridge to study mathematics. He then spent part of 1879 and 1880 studying medieval and 16th century German literature at the universities of Berlin and Heidelberg – in fact, he
became sufficiently knowledgeable in this field that he was offered a post in the German department at Cambridge University.
His next career move was to Lincoln's Inn, where he read law until 1881 (although he never practised). After this, he returned to mathematics, deputising for the mathematics professor at King's
College London in 1881 and for the professor at University College London in 1883. In 1884, he was appointed to the Goldsmid Chair of Applied Mathematics and Mechanics at University College London.
1891 saw him also appointed to the professorship of Geometry at Gresham College; here he met Walter Frank Raphael Weldon, a zoologist who had some interesting problems requiring quantitative
solutions. The collaboration, in biometry and evolutionary theory, was a fruitful one and lasted until Weldon died in 1906. Weldon introduced Pearson to Charles Darwin's cousin Francis Galton, who
was interested in aspects of evolution such as heredity and eugenics. Pearson became Galton's protégé — his "statistical heir" as some have put it — at times to the verge of hero worship. After
Galton's death in 1911, Pearson embarked on producing his definitive biography—a three-volume tome of narrative, letters, genealogies, commentaries, and photographs—published in 1914, 1924, and 1930,
with much of Pearson's own financing paying for their print runs. The biography, done "to satisfy myself and without regard to traditional standards, to the needs of publishers or to the tastes of
the reading public", triumphed Galton's life, work, and personal heredity. He predicted that Galton, rather than Charles Darwin, would be remembered as the most prodigious grandson of Erasmus Darwin.
When Galton died, he left the residue of his estate to the University of London for a Chair in Eugenics. Pearson was the first holder of this chair, in accordance with Galton's wishes. He formed the
Department of Applied Statistics (with financial support from the Drapers' Company), into which he incorporated the Biometric and Galton laboratories. He remained with the department until his
retirement in 1933, and continued to work until his death in 1936.
Pearson married Maria Sharpe in 1890, and between them they had two daughters and a son. The son, Egon Sharpe Pearson, succeeded him as head of the Applied Statistics Department at University
When the 23 year-old Albert Einstein started a study group, the Olympia Academy, with his two younger friends, Solovine and Habicht, he suggested that the first book to be read was Pearson's The
Grammar of Science. This book covered several themes that were later to become part of the theories of Einstein and other scientists. Pearson asserted that the laws of nature are relative to the
perceptive ability of the observer. Irreversibility of natural processes, he claimed, is a purely relative conception. An observer who travels at the exact velocity of light would see an eternal now,
or an absence of motion. He speculated that an observer who traveled faster than light would see time reversal, similar to a cinema film being run backwards. Pearson also discussed antimatter, the
fourth dimension, and wrinkles in time.
Pearson's relativity was based on idealism, in the sense of ideas or pictures in a mind. "There are many signs," he wrote, "that a sound idealism is surely replacing, as a basis for natural
philosophy, the crude materialism of the older physicists." (Preface to 2nd Ed., The Grammar of Science) Further, he stated, "...science is in reality a classification and analysis of the contents of
the mind...." "In truth, the field of science is much more consciousness than an external world." (Ibid., Ch. II, § 6) "Law in the scientific sense is thus essentially a product of the human mind and
has no meaning apart from man." (Ibid., Ch. III, § 4)
Aside from his professional life, Pearson was active as a prominent freethinker and socialist. He gave lectures on such issues as "the woman's question" (this was the era of the suffragist movement
in the UK) and upon Karl Marx. His commitment to socialism and its ideals led him to refuse the offer of being created an OBE (Officer of the Order of the British Empire) in 1920, and also to refuse
a Knighthood in 1935. In the 1930s he had a protracted feud with R.A. Fisher over a statistical disagreement, which continued after his death through his son.
Pearson's views on eugenics, however, would be considered deeply racist today. According to a BBC report on the history of genetics, "Pearson was a fanatic – a cold, calculating measurer of man who
claimed to be a socialist, but loathed the working class." Pearson openly advocated "war" against "inferior races," and saw this as a logical implication of his scientific work on human measurement:
"My view – and I think it may be called the scientific view of a nation," he wrote, "– is that of an organized whole, kept up to a high pitch of internal efficiency by insuring that its numbers are
substantially recruited from the better stocks, and kept up to a high pitch of external efficiency by contest, chiefly by way of war with inferior races." He reasoned that, if August Weismann's
theory of germ plasm is correct, then the nation is wasting money when it tries to improve people who come from poor stock. Weismann claimed that acquired characteristics could not be inherited.
Therefore, training benefits only the trained generation. Their children will not exhibit the learned improvements and, in turn, will need to be improved. "No degenerate and feeble stock will ever be
converted into healthy and sound stock by the accumulated effects of education, good laws, and sanitary surroundings. Such means may render the individual members of a stock passable if not strong
members of society, but the same process will have to be gone through again and again with their offspring, and this in ever-widening circles, if the stock, owing to the conditions in which society
has placed it, is able to increase its numbers." (Introduction, The Grammar of Science).
Awards from professional bodies
Pearson achieved widespread recognition across a range of disciplines and his membership of, and awards from, various professional bodies reflects this:
• 1896: elected Fellow of the Royal Society
• 1898: awarded the Darwin Medal
• 1911: awarded the honorary degree of LLD from St Andrews University
• 1911: awarded a DSc from University of London
• 1920: offered (and refused) the OBE
• 1932: awarded the Rudolf Virchow medal by the Berliner Anthropologische Gesellschaft
• 1935: offered (and refused) a knighthood
He was also elected an Honorary Fellow of King's College Cambridge, the Royal Society of Edinburgh, University College London and the Royal Society of Medicine, and a Member of the Actuaries' Club.
Contributions to statistics
Pearson's work was all-embracing in the wide application and development of mathematical statistics, and encompassed the fields of biology, epidemiology, anthropometry, medicine and social history.
In 1901, with Weldon and Galton, he founded the journal Biometrika whose object was the development of statistical theory. He edited this journal until his death. He also founded the journal Annals
of Eugenics (now Annals of Human Genetics) in 1925. He published the Drapers' Company Research Memoirs largely to provide a record of the output of the Department of Applied Statistics not published
Pearson's thinking underpins many of the 'classical' statistical methods which are in common use today. Some of his main contributions are:
1. Linear regression and correlation. Pearson was instrumental in the development of this theory. One of his classic data sets (originally collected by Galton) involves the regression of sons'
height upon that of their fathers'. Pearson built a 3-dimensional model of this data set (which remains in the care of the Statistical Science Department) to illustrate the ideas. The Pearson
product-moment correlation coefficient is named after him, and it was the first important effect size to be introduced into statistics.
2. Classification of distributions. Pearson's work on classifying probability distributions forms the basis for a lot of modern statistical theory; in particular, the exponential family of
distributions underlies the theory of generalized linear models.
3. Pearson's chi-square test. A particular kind of chi-square test, a statistical test of significance.
See also
External links
• The MacTutor History of Mathematics archive at St. Andrews University includes biographies of mathematicians and statisticians (including Pearson), as well as general information on the history
of mathematics.
• John Aldrich's Karl Pearson: a Reader's Guide contains many useful links to further sources of information.
• Gavan Tredoux's Francis Galton site, galton.org, contains Pearson's biography of Francis Galton, and several other papers - in addition to nearly all of Galton's own published works.
Further reading
Most of the biographical information above is taken from A list of the papers and correspondence of Karl Pearson (1857-1936) held in the Manuscripts Room, University College London Library, compiled
by M.Merrington, B.Blundell, S.Burrough, J.Golden and J.Hogarth and published by the Publications Office, University College London, 1983. See http://www.ucl.ac.uk/stats/history/pearson.html.
Further references which may be of use are:
• Porter, Theodore M. (2004): Karl Pearson: The Scientific Life in a Statistical Age, Princeton University Press.
• Eisenhart, Churchill (1974): Dictionary of Scientific Biography, pp. 447–73. New York, 1974.
• Filon, L. N. G. and Yule, G. U. (1936): Obituary Notices of the Royal Society of London, Vol. ii, No. 5, pp. 73–110.
• Pearson, E. S. (1938): Karl Pearson: An appreciation of some aspects of his life and work. Cambridge University Press. | {"url":"http://psychology.wikia.com/wiki/Karl_Pearson?oldid=26064","timestamp":"2014-04-24T12:50:08Z","content_type":null,"content_length":"78647","record_id":"<urn:uuid:72f6686c-52d3-4e3a-934c-df0b6f84f725>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00253-ip-10-147-4-33.ec2.internal.warc.gz"} |
Post a reply
I am used to doing this problem in reverse so I am hoping that I did it correctly.
Find the system of equations that is equivalent to the matrix equation given below.
[-2 5] [x] = [3]
[ 1 4] [y] = [6]
The answer that I got was {-2x+5y=3
{ x+4y=6
Did I do this correctly? | {"url":"http://www.mathisfunforum.com/post.php?tid=19513","timestamp":"2014-04-17T07:02:07Z","content_type":null,"content_length":"15671","record_id":"<urn:uuid:20b51c8d-c1a1-4256-ad13-01dcf09f471e>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00644-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Kendriya Vidyalaya HOTS Questions for Class 10
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The Abstract Format
6 The Abstract Format
This document describes the standard representation of parse trees for Erlang programs as Erlang terms. This representation is known as the abstract format. Functions dealing with such parse trees
are compile:forms/[1,2] and functions in the modules epp, erl_eval, erl_lint, erl_pp, erl_parse, and io. They are also used as input and output for parse transforms (see the module compile).
We use the function Rep to denote the mapping from an Erlang source construct C to its abstract format representation R, and write R = Rep(C).
The word LINE below represents an integer, and denotes the number of the line in the source file where the construction occurred. Several instances of LINE in the same construction may denote
different lines.
Since operators are not terms in their own right, when operators are mentioned below, the representation of an operator should be taken to be the atom with a printname consisting of the same
characters as the operator.
A module declaration consists of a sequence of forms that are either function declarations or attributes.
• If D is a module declaration consisting of the forms F_1, ..., F_k, then Rep(D) = [Rep(F_1), ..., Rep(F_k)].
• If F is an attribute -module(Mod), then Rep(F) = {attribute,LINE,module,Mod}.
• If F is an attribute -export([Fun_1/A_1, ..., Fun_k/A_k]), then Rep(F) = {attribute,LINE,export,[{Fun_1,A_1}, ..., {Fun_k,A_k}]}.
• If F is an attribute -import(Mod,[Fun_1/A_1, ..., Fun_k/A_k]), then Rep(F) = {attribute,LINE,import,{Mod,[{Fun_1,A_1}, ..., {Fun_k,A_k}]}}.
• If F is an attribute -compile(Options), then Rep(F) = {attribute,LINE,compile,Options}.
• If F is an attribute -file(File,Line), then Rep(F) = {attribute,LINE,file,{File,Line}}.
• If F is a record declaration -record(Name,{V_1, ..., V_k}), then Rep(F) = {attribute,LINE,record,{Name,[Rep(V_1), ..., Rep(V_k)]}}. For Rep(V), see below.
• If F is a wild attribute -A(T), then Rep(F) = {attribute,LINE,A,T}.
• If F is a function declaration Name Fc_1 ; ... ; Name Fc_k, where each Fc_i is a function clause with a pattern sequence of the same length Arity, then Rep(F) = {function,LINE,Name,Arity,[Rep
(Fc_1), ...,Rep(Fc_k)]}.
Record fields
Each field in a record declaration may have an optional explicit default initializer expression
• If V is A, then Rep(V) = {record_field,LINE,Rep(A)}.
• If V is A = E, then Rep(V) = {record_field,LINE,Rep(A),Rep(E)}.
Representation of parse errors and end of file
In addition to the representations of forms, the list that represents a module declaration (as returned by functions in erl_parse and epp) may contain tuples {error,E} and {warning,W}, denoting
syntactically incorrect forms and warnings, and {eof,LINE}, denoting an end of stream encountered before a complete form had been parsed.
There are five kinds of atomic literals, which are represented in the same way in patterns, expressions and guards:
• If L is an integer or character literal, then Rep(L) = {integer,LINE,L}.
• If L is a float literal, then Rep(L) = {float,LINE,L}.
• If L is a string literal consisting of the characters C_1, ..., C_k, then Rep(L) = {string,LINE,[C_1, ..., C_k]}.
• If L is an atom literal, then Rep(L) = {atom,LINE,L}.
Note that negative integer and float literals do not occur as such; they are parsed as an application of the unary negation operator.
If Ps is a sequence of patterns P_1, ..., P_k, then Rep(Ps) = [Rep(P_1), ..., Rep(P_k)]. Such sequences occur as the list of arguments to a function or fun.
Individual patterns are represented as follows:
• If P is an atomic literal L, then Rep(P) = Rep(L).
• If P is a compound pattern P_1 = P_2, then Rep(P) = {match,LINE,Rep(P_1),Rep(P_2)}.
• If P is a variable pattern V, then Rep(P) = {var,LINE,A}, where A is an atom with a printname consisting of the same characters as V.
• If P is a universal pattern _, then Rep(P) = {var,LINE,'_'}.
• If P is a tuple pattern {P_1, ..., P_k}, then Rep(P) = {tuple,LINE,[Rep(P_1), ..., Rep(P_k)]}.
• If P is a nil pattern [], then Rep(P) = {nil,LINE}.
• If P is a cons pattern [P_h | P_t], then Rep(P) = {cons,LINE,Rep(P_h),Rep(P_t)}.
• If E is a binary pattern <<P_1:Size_1/TSL_1, ..., P_k:Size_k/TSL_k>>, then Rep(E) = {bin,LINE,[{bin_element,LINE,Rep(P_1),Rep(Size_1),Rep(TSL_1)}, ..., {bin_element,LINE,Rep(P_k),Rep(Size_k),Rep
(TSL_k)}]}. For Rep(TSL), see below. An omitted Size is represented by default. An omitted TSL (type specifier list) is represented by default.
• If P is P_1 Op P_2, where Op is a binary operator (this is either an occurrence of ++ applied to a literal string or character list, or an occurrence of an expression that can be evaluated to a
number at compile time), then Rep(P) = {op,LINE,Op,Rep(P_1),Rep(P_2)}.
• If P is Op P_0, where Op is a unary operator (this is an occurrence of an expression that can be evaluated to a number at compile time), then Rep(P) = {op,LINE,Op,Rep(P_0)}.
• If P is a record pattern #Name{Field_1=P_1, ..., Field_k=P_k}, then Rep(P) = {record,LINE,Name, [{record_field,LINE,Rep(Field_1),Rep(P_1)}, ..., {record_field,LINE,Rep(Field_k),Rep(P_k)}]}.
• If P is #Name.Field, then Rep(P) = {record_index,LINE,Name,Rep(Field)}.
• If P is ( P_0 ), then Rep(P) = Rep(P_0), i.e., patterns cannot be distinguished from their bodies.
Note that every pattern has the same source form as some expression, and is represented the same way as the corresponding expression.
A body B is a sequence of expressions E_1, ..., E_k, and Rep(B) = [Rep(E_1), ..., Rep(E_k)].
An expression E is one of the following alternatives:
• If P is an atomic literal L, then Rep(P) = Rep(L).
• If E is P = E_0, then Rep(E) = {match,LINE,Rep(P),Rep(E_0)}.
• If E is a variable V, then Rep(E) = {var,LINE,A}, where A is an atom with a printname consisting of the same characters as V.
• If E is a tuple skeleton {E_1, ..., E_k}, then Rep(E) = {tuple,LINE,[Rep(E_1), ..., Rep(E_k)]}.
• If E is [], then Rep(E) = {nil,LINE}.
• If E is a cons skeleton [E_h | E_t], then Rep(E) = {cons,LINE,Rep(E_h),Rep(E_t)}.
• If E is a binary constructor <<V_1:Size_1/TSL_1, ..., V_k:Size_k/TSL_k>>, then Rep(E) = {bin,LINE,[{bin_element,LINE,Rep(V_1),Rep(Size_1),Rep(TSL_1)}, ..., {bin_element,LINE,Rep(V_k),Rep
(Size_k),Rep(TSL_k)}]}. For Rep(TSL), see below. An omitted Size is represented by default. An omitted TSL (type specifier list) is represented by default.
• If E is E_1 Op E_2, where Op is a binary operator, then Rep(E) = {op,LINE,Op,Rep(E_1),Rep(E_2)}.
• If E is Op E_0, where Op is a unary operator, then Rep(E) = {op,LINE,Op,Rep(E_0)}.
• If E is #Name{Field_1=E_1, ..., Field_k=E_k}, then Rep(E) = {record,LINE,Name, [{record_field,LINE,Rep(Field_1),Rep(E_1)}, ..., {record_field,LINE,Rep(Field_k),Rep(E_k)}]}.
• If E is E_0#Name{Field_1=E_1, ..., Field_k=E_k}, then Rep(E) = {record,LINE,Rep(E_0),Name, [{record_field,LINE,Rep(Field_1),Rep(E_1)}, ..., {record_field,LINE,Rep(Field_k),Rep(E_k)}]}.
• If E is #Name.Field, then Rep(E) = {record_index,LINE,Name,Rep(Field)}.
• If E is E_0#Name.Field, then Rep(E) = {record_field,LINE,Rep(E_0),Name,Rep(Field)}.
• If E is #{W_1, ..., W_k} where each W_i is a map assoc or exact field, then Rep(E) = {map,LINE,[Rep(W_1), ..., Rep(W_k)]}. For Rep(W), see below.
• If E is E_0#{W_1, ..., W_k} where W_i is a map assoc or exact field, then Rep(E) = {map,LINE,Rep(E_0),[Rep(W_1), ..., Rep(W_k)]}. For Rep(W), see below.
• If E is catch E_0, then Rep(E) = {'catch',LINE,Rep(E_0)}.
• If E is E_0(E_1, ..., E_k), then Rep(E) = {call,LINE,Rep(E_0),[Rep(E_1), ..., Rep(E_k)]}.
• If E is E_m:E_0(E_1, ..., E_k), then Rep(E) = {call,LINE,{remote,LINE,Rep(E_m),Rep(E_0)},[Rep(E_1), ..., Rep(E_k)]}.
• If E is a list comprehension [E_0 || W_1, ..., W_k], where each W_i is a generator or a filter, then Rep(E) = {lc,LINE,Rep(E_0),[Rep(W_1), ..., Rep(W_k)]}. For Rep(W), see below.
• If E is a binary comprehension <<E_0 || W_1, ..., W_k>>, where each W_i is a generator or a filter, then Rep(E) = {bc,LINE,Rep(E_0),[Rep(W_1), ..., Rep(W_k)]}. For Rep(W), see below.
• If E is begin B end, where B is a body, then Rep(E) = {block,LINE,Rep(B)}.
• If E is if Ic_1 ; ... ; Ic_k end, where each Ic_i is an if clause then Rep(E) = {'if',LINE,[Rep(Ic_1), ..., Rep(Ic_k)]}.
• If E is case E_0 of Cc_1 ; ... ; Cc_k end, where E_0 is an expression and each Cc_i is a case clause then Rep(E) = {'case',LINE,Rep(E_0),[Rep(Cc_1), ..., Rep(Cc_k)]}.
• If E is try B catch Tc_1 ; ... ; Tc_k end, where B is a body and each Tc_i is a catch clause then Rep(E) = {'try',LINE,Rep(B),[],[Rep(Tc_1), ..., Rep(Tc_k)],[]}.
• If E is try B of Cc_1 ; ... ; Cc_k catch Tc_1 ; ... ; Tc_n end, where B is a body, each Cc_i is a case clause and each Tc_j is a catch clause then Rep(E) = {'try',LINE,Rep(B),[Rep(Cc_1), ..., Rep
(Cc_k)],[Rep(Tc_1), ..., Rep(Tc_n)],[]}.
• If E is try B after A end, where B and A are bodies then Rep(E) = {'try',LINE,Rep(B),[],[],Rep(A)}.
• If E is try B of Cc_1 ; ... ; Cc_k after A end, where B and A are a bodies and each Cc_i is a case clause then Rep(E) = {'try',LINE,Rep(B),[Rep(Cc_1), ..., Rep(Cc_k)],[],Rep(A)}.
• If E is try B catch Tc_1 ; ... ; Tc_k after A end, where B and A are bodies and each Tc_i is a catch clause then Rep(E) = {'try',LINE,Rep(B),[],[Rep(Tc_1), ..., Rep(Tc_k)],Rep(A)}.
• If E is try B of Cc_1 ; ... ; Cc_k catch Tc_1 ; ... ; Tc_n after A end, where B and A are a bodies, each Cc_i is a case clause and each Tc_j is a catch clause then Rep(E) = {'try',LINE,Rep(B),
[Rep(Cc_1), ..., Rep(Cc_k)],[Rep(Tc_1), ..., Rep(Tc_n)],Rep(A)}.
• If E is receive Cc_1 ; ... ; Cc_k end, where each Cc_i is a case clause then Rep(E) = {'receive',LINE,[Rep(Cc_1), ..., Rep(Cc_k)]}.
• If E is receive Cc_1 ; ... ; Cc_k after E_0 -> B_t end, where each Cc_i is a case clause, E_0 is an expression and B_t is a body, then Rep(E) = {'receive',LINE,[Rep(Cc_1), ..., Rep(Cc_k)],Rep
• If E is fun Name / Arity, then Rep(E) = {'fun',LINE,{function,Name,Arity}}.
• If E is fun Module:Name/Arity, then Rep(E) = {'fun',LINE,{function,Rep(Module),Rep(Name),Rep(Arity)}}. (Before the R15 release: Rep(E) = {'fun',LINE,{function,Module,Name,Arity}}.)
• If E is fun Fc_1 ; ... ; Fc_k end where each Fc_i is a function clause then Rep(E) = {'fun',LINE,{clauses,[Rep(Fc_1), ..., Rep(Fc_k)]}}.
• If E is fun Name Fc_1 ; ... ; Name Fc_k end where Name is a variable and each Fc_i is a function clause then Rep(E) = {named_fun,LINE,Name,[Rep(Fc_1), ..., Rep(Fc_k)]}.
• If E is query [E_0 || W_1, ..., W_k] end, where each W_i is a generator or a filter, then Rep(E) = {'query',LINE,{lc,LINE,Rep(E_0),[Rep(W_1), ..., Rep(W_k)]}}. For Rep(W), see below.
• If E is E_0.Field, a Mnesia record access inside a query, then Rep(E) = {record_field,LINE,Rep(E_0),Rep(Field)}.
• If E is ( E_0 ), then Rep(E) = Rep(E_0), i.e., parenthesized expressions cannot be distinguished from their bodies.
Generators and filters
When W is a generator or a filter (in the body of a list or binary comprehension), then:
• If W is a generator P <- E, where P is a pattern and E is an expression, then Rep(W) = {generate,LINE,Rep(P),Rep(E)}.
• If W is a generator P <= E, where P is a pattern and E is an expression, then Rep(W) = {b_generate,LINE,Rep(P),Rep(E)}.
• If W is a filter E, which is an expression, then Rep(W) = Rep(E).
Binary element type specifiers
A type specifier list TSL for a binary element is a sequence of type specifiers TS_1 - ... - TS_k. Rep(TSL) = [Rep(TS_1), ..., Rep(TS_k)].
When TS is a type specifier for a binary element, then:
• If TS is an atom A, Rep(TS) = A.
• If TS is a couple A:Value where A is an atom and Value is an integer, Rep(TS) = {A, Value}.
Map assoc and exact fields
When W is an assoc or exact field (in the body of a map), then:
• If W is an assoc field K => V, where K and V are both expressions, then Rep(W) = {map_field_assoc,LINE,Rep(K),Rep(V)}.
• If W is an exact field K := V, where K and V are both expressions, then Rep(W) = {map_field_exact,LINE,Rep(K),Rep(V)}.
There are function clauses, if clauses, case clauses and catch clauses.
A clause C is one of the following alternatives:
• If C is a function clause ( Ps ) -> B where Ps is a pattern sequence and B is a body, then Rep(C) = {clause,LINE,Rep(Ps),[],Rep(B)}.
• If C is a function clause ( Ps ) when Gs -> B where Ps is a pattern sequence, Gs is a guard sequence and B is a body, then Rep(C) = {clause,LINE,Rep(Ps),Rep(Gs),Rep(B)}.
• If C is an if clause Gs -> B where Gs is a guard sequence and B is a body, then Rep(C) = {clause,LINE,[],Rep(Gs),Rep(B)}.
• If C is a case clause P -> B where P is a pattern and B is a body, then Rep(C) = {clause,LINE,[Rep(P)],[],Rep(B)}.
• If C is a case clause P when Gs -> B where P is a pattern, Gs is a guard sequence and B is a body, then Rep(C) = {clause,LINE,[Rep(P)],Rep(Gs),Rep(B)}.
• If C is a catch clause P -> B where P is a pattern and B is a body, then Rep(C) = {clause,LINE,[Rep({throw,P,_})],[],Rep(B)}.
• If C is a catch clause X : P -> B where X is an atomic literal or a variable pattern, P is a pattern and B is a body, then Rep(C) = {clause,LINE,[Rep({X,P,_})],[],Rep(B)}.
• If C is a catch clause P when Gs -> B where P is a pattern, Gs is a guard sequence and B is a body, then Rep(C) = {clause,LINE,[Rep({throw,P,_})],Rep(Gs),Rep(B)}.
• If C is a catch clause X : P when Gs -> B where X is an atomic literal or a variable pattern, P is a pattern, Gs is a guard sequence and B is a body, then Rep(C) = {clause,LINE,[Rep({X,P,_})],Rep
A guard sequence Gs is a sequence of guards G_1; ...; G_k, and Rep(Gs) = [Rep(G_1), ..., Rep(G_k)]. If the guard sequence is empty, Rep(Gs) = [].
A guard G is a nonempty sequence of guard tests Gt_1, ..., Gt_k, and Rep(G) = [Rep(Gt_1), ..., Rep(Gt_k)].
A guard test Gt is one of the following alternatives:
• If Gt is an atomic literal L, then Rep(Gt) = Rep(L).
• If Gt is a variable pattern V, then Rep(Gt) = {var,LINE,A}, where A is an atom with a printname consisting of the same characters as V.
• If Gt is a tuple skeleton {Gt_1, ..., Gt_k}, then Rep(Gt) = {tuple,LINE,[Rep(Gt_1), ..., Rep(Gt_k)]}.
• If Gt is [], then Rep(Gt) = {nil,LINE}.
• If Gt is a cons skeleton [Gt_h | Gt_t], then Rep(Gt) = {cons,LINE,Rep(Gt_h),Rep(Gt_t)}.
• If Gt is a binary constructor <<Gt_1:Size_1/TSL_1, ..., Gt_k:Size_k/TSL_k>>, then Rep(Gt) = {bin,LINE,[{bin_element,LINE,Rep(Gt_1),Rep(Size_1),Rep(TSL_1)}, ..., {bin_element,LINE,Rep(Gt_k),Rep
(Size_k),Rep(TSL_k)}]}. For Rep(TSL), see above. An omitted Size is represented by default. An omitted TSL (type specifier list) is represented by default.
• If Gt is Gt_1 Op Gt_2, where Op is a binary operator, then Rep(Gt) = {op,LINE,Op,Rep(Gt_1),Rep(Gt_2)}.
• If Gt is Op Gt_0, where Op is a unary operator, then Rep(Gt) = {op,LINE,Op,Rep(Gt_0)}.
• If Gt is #Name{Field_1=Gt_1, ..., Field_k=Gt_k}, then Rep(E) = {record,LINE,Name, [{record_field,LINE,Rep(Field_1),Rep(Gt_1)}, ..., {record_field,LINE,Rep(Field_k),Rep(Gt_k)}]}.
• If Gt is #Name.Field, then Rep(Gt) = {record_index,LINE,Name,Rep(Field)}.
• If Gt is Gt_0#Name.Field, then Rep(Gt) = {record_field,LINE,Rep(Gt_0),Name,Rep(Field)}.
• If Gt is A(Gt_1, ..., Gt_k), where A is an atom, then Rep(Gt) = {call,LINE,Rep(A),[Rep(Gt_1), ..., Rep(Gt_k)]}.
• If Gt is A_m:A(Gt_1, ..., Gt_k), where A_m is the atom erlang and A is an atom or an operator, then Rep(Gt) = {call,LINE,{remote,LINE,Rep(A_m),Rep(A)},[Rep(Gt_1), ..., Rep(Gt_k)]}.
• If Gt is {A_m,A}(Gt_1, ..., Gt_k), where A_m is the atom erlang and A is an atom or an operator, then Rep(Gt) = {call,LINE,Rep({A_m,A}),[Rep(Gt_1), ..., Rep(Gt_k)]}.
• If Gt is ( Gt_0 ), then Rep(Gt) = Rep(Gt_0), i.e., parenthesized guard tests cannot be distinguished from their bodies.
Note that every guard test has the same source form as some expression, and is represented the same way as the corresponding expression.
The compilation option debug_info can be given to the compiler to have the abstract code stored in the abstract_code chunk in the BEAM file (for debugging purposes).
In OTP R9C and later, the abstract_code chunk will contain
where AbstractCode is the abstract code as described in this document.
In releases of OTP prior to R9C, the abstract code after some more processing was stored in the BEAM file. The first element of the tuple would be either abstract_v1 (R7B) or abstract_v2 (R8B). | {"url":"http://www.erlang.org/doc/apps/erts/absform.html","timestamp":"2014-04-16T10:11:33Z","content_type":null,"content_length":"42731","record_id":"<urn:uuid:eeea21c1-1b89-41d5-83ca-7a98499c3e5c>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00614-ip-10-147-4-33.ec2.internal.warc.gz"} |
[FOM] interesting real number
Martin Davis martin at eipye.com
Sat Apr 15 16:30:51 EDT 2006
Replying to a query by Bob Solovay, Ron Graham wrote:
> This real is well known to be e^(1/e)
Just to comment that this number has a nostalgic interest for me. When I
was a freshman I managed to prove that the sequence defined by s_1=x,
s_{n+1} = x^(s_n) converges precisely for real numbers x satisfying (1/e)^e
<= x <= e^{1/e).
More information about the FOM mailing list | {"url":"http://www.cs.nyu.edu/pipermail/fom/2006-April/010416.html","timestamp":"2014-04-18T16:10:26Z","content_type":null,"content_length":"2641","record_id":"<urn:uuid:82785d88-5e57-4e37-a889-7a24f31fb18d>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00031-ip-10-147-4-33.ec2.internal.warc.gz"} |
What is the minimum set of combinations C(p,n) required to guarantee q<p matches with a target combination (pT,n)
up vote 1 down vote favorite
A state lottery draws p numbers out of a grid of n numbers. Players participate by filling in p numbers into a grid, at unit cost. They can sumbit as many grids as they like.
The lottery pays out when a player's grid matches at least q numbers with the outcome of the drawing, the "winning combination" p[T]. How many grids does a player need to submit, to ensure a payout ?
Obviously, a player could cover the whole space of C(p/n) = n!/p!(p-n)! possible combinations, thereby guaranteeing every possible match of q, q+1, .., p numbers with the target set p[T]. That would
certainly cover the C(q/p).C(p-q / n-p) combinations that match exactly q elements of the target set p[T]. But only one such match is required. How does one construct a minimum subset of C(p/n) to
guarantee at least one grid matching at least q elements of the target set p[T] ?
I have previously submitted this question as a Project Euler challenge, but it wasn't selected.
2 You seem to be asking for a minimum block design? en.wikipedia.org/wiki/Block_design – András Salamon Aug 9 '10 at 14:59
add comment
4 Answers
active oldest votes
Hmmm... I recently saw a paper dealing exactly with this. You may like to learn that this problem is known in the litterature as the "lottery number".
I was not able to find this paper back, though I found one from 2008 whose introduction contains several interesting references :
up vote 0 down vote A note on a symmetrical set covering problem: The lottery problem
Hoping this is useful...
Thanks Nathann. The paper you referred to, has numerical results for small cases. No closed solution for the general case seems yet to exist. I understand now, why my question
was not used on Project Euler. – Walter Baeck Aug 10 '10 at 16:05
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You may find the following interesting.
I think what you're asking for is the minimum cardinality of a code of length $p$ over the alphabet $Z_n$ (the integers mod $n$, assuming) the possible numbers are $1,2,\ldots,n-1$
with covering radius $p-q$. This is in general a very hard NP-complete problem.
Given two vectors $x,y$ in $Z_n^p$, their Hamming distance $d(x,y)$ is defined as the number of coordinates on which they differ. Given a subset $C$ of $Z_n^p$ which corresponds to the
collection of lottery selections, the covering radius $R(C)$ is
$$ R(C) = \max d(x,C) $$
up vote 0 down
vote where the maximum is taken as $x$ ranges over $Z_n^p$ and where $d(x,C)=\min d(x,c)$ with the minimum taken over $c \in C.$ When you think about it, you're asking for the minimal
cardinality code with covering radius $p-q,$ i.e., with $p-q$ wrong numbers out of $p$.
In general there are bounds on $R(C)$ and the case $n=3$ is of interest in football [soccer] pools. The Golay code is relevant here since it is a perfect ternary code with good
covering radius.
There was a nice article MAA monthly years ago entitled something like "football pools, a problem for mathematicians".
add comment
I'm a programmer, not a mathematician, but knowing that the probability of any payout match on a single grid is given by
$p = _pC_r / _nC_r$, where $r$ is the number of matches required for a payout and
$(r <= q)$,
intuitively, having $1/p$ grids covering all r-tuples will result in a probability of $1$; a guarantee.
up vote 0 However (unfortunately) I don't believe there exists a design for a $1/p$ set of grids covering all r-tuples uniquely (my brute force computer trials have all failed), but I don't have the
down vote math skills to prove it. Such sets exist, but with many more than $1/p$ members.
As a side note, this concept was implemented successfully by a group who purchased several hundred grids weekly with a guaranteed payout that covered about 60% of the cost. Since their
grids covered 100% of the smallest r-tuples thus covering a significant portion of (r+1)-tuples which paid out substantially more, frequently their payout was out enough to turn a "profit"
thus sustataining the pool without additional funds. Over a couple of years, the scheme won many larger payouts (but not a jackpot) until the grid price was doubled after which the group
add comment
As I had pointed out earlier, this is a difficult problem to solve in general.
up vote 0 down vote I have managed to find a nice description of the problem in sections VI.8 and V.8 in the CRC Handbook of Combinatorial Designs by Colbourn and Dinitz.
add comment
Not the answer you're looking for? Browse other questions tagged co.combinatorics or ask your own question. | {"url":"http://mathoverflow.net/questions/35003/what-is-the-minimum-set-of-combinations-cp-n-required-to-guarantee-qp-matches/35018","timestamp":"2014-04-20T08:34:52Z","content_type":null,"content_length":"65018","record_id":"<urn:uuid:a7bdf7e3-4699-47c9-92bd-a01c5bef8940>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00203-ip-10-147-4-33.ec2.internal.warc.gz"} |
Structs pointing to each other, stack overflow
I'm trying to build a simply binary tree (made of node structs which each have two "child" slots and one "parent" slot)
(defstruct node child-a child-b parent)
I can for example create two new nodes like so:
(defparameter a (make-node))
(defparameter b (make-node))
Now I want to create their parent like so:
(defparameter c (make-node :child-a a :child-b b))
Fine, good so far.
Now I need to let the children nodes (a and b) link back to their parent c. (so the binary tree can be traversed in 2 directions)
HOWEVER I get an immediate error when trying to set the parent value for a and b:
(setf (node-parent a) c)
*** - Program stack overflow. RESET
Does CL not allow structs pointing to each other?
Re: Structs pointing to each other, stack overflow
DEFSTRUCT is too simple-minded to handle inheritance properly, you would need to write some extra functions to keep track of all possible inheritance issues. In Paul Graham's ANSI Common Lisp are
some examples how this would work.
(setf (node-parent a) c) creates an infinite loop because a already contains a reference to c, where a, b, c are instances of one and the the same STRUCTURE-CLASS, so CLOS can't resolve the
precedence properly.
DEFCLASS and MAKE-INSTANCE give much better options to define and control inheritance dependencies.
Nick Levine's Fundamentals of CLOS tutorial describes on one single page pretty well how to implement single and multiple inheritance with CLOS.
If you want to read books:
• Sonya E. Keene Object-oriented programming in Common LISP - easy to understand
• Gregor Kiczales The Art of the Metaobject Protocol - hardcore OOP (don't read this first)
- edgar
Re: Structs pointing to each other, stack overflow
Some time later...
Here is a readable version of the code from ANSI Common Lisp, using DEFSTRUCT to create doubly-linked binary trees:
Adding a new node to the tree:
The trick is to define the precedence by a sort-function. If the tree contains only numbers, the sort-function would be #'< or #'>, like here:
Finding nodes in the tree:
Here is how it works:
The original code, including functions to remove elements and other stuff, can be found here:
Look for function names starting with "bst-..." (binary search tree).
But I still think that using DEFCLASS and MAKE-INSTANCE is the better idea.
- edgar
Re: Structs pointing to each other, stack overflow
Thanks so much for the information.
I was trying to play C with CL and I think I didn't realize the simple nature of the structs.
Re: Structs pointing to each other, stack overflow
You do all ok, stack overflow is just as lisp tries to print your circular graph
of structures.
Set *print-circle* to t and all would work.
Answer by edgar-rft is wrong.
Re: Structs pointing to each other, stack overflow
budden73 wrote:You do all ok, stack overflow is just as lisp tries to print your circular graph
of structures.
Set *print-circle* to t and all would work.
Answer by edgar-rft is wrong.
Yep. budden73 is perfectly right. The problem is the "printing". Nothing to do with the structures setup. Doing
Code: Select all
cl-prompt> (progn (setf (node-parent c) a) t)
would just print T. Of course, setting *print-circle* to NIL will also solve the problem.
Finally, let's not forget the print-object/print-function facilities in DEFSTRUCT.
Marco Antoniotti | {"url":"http://www.lispforum.com/viewtopic.php?p=11495","timestamp":"2014-04-21T12:10:37Z","content_type":null,"content_length":"26770","record_id":"<urn:uuid:9cb60ebf-e103-4c87-b076-0da5bbab911e>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00619-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Lapack] A old problem arises again -- symmetric matrix triple product?
LAPACK Archives
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[Lapack] A old problem arises again -- symmetric matrix triple product?
From: John G. Lewis
Date: Thu, 13 Apr 2006 16:43:03 -0700
Over the years I've repeatedly run into a symmetric matrix modification
operation that's not captured in the BLAS. Following Beresford's
symmetric letter convention, the operation is
A = A + C^T W C
where A and W are real symmetric (or complex Hermitian) and C is
rectangular. This occurs as a rank 2 modification in symmetric
indefinite factorizations. You've heard me before on that. Now I've
encountered it in a very general form in computational chemistry. Here C
and W are full rank matrices, of considerably large rank.
Obviously this could be done with one intermediate step, as A + (C^T M)
C or A + C^T (M C). Storing the intermediate product creates a storage
penalty that's not particularly important at the moment. The major
problem is that the only place where BLAS or the PBLAS recognize that a
matrix-matrix product X*Y is symmetric is when Y = X^T. (The symmetric
matrix argument to PDSYMM is X or Y, but the product is general.) This
means that we have only PDGEMM available to do this, which requires
twice as much work as we really need to do.
Has anyone else ever faced this problem? Does anyone know of an
implementation of either
* the elegant solution (modifying PDSYRK to accept the extra
symmetric matrix argument and to do the intermediate products on
the fly), or
* the inelegant solution, computing the intermediate product as a
separate matrix and then modifying PDGEMM to avoid the work in one
or the other triangle?
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• [Lapack] A old problem arises again -- symmetric matrix triple product?, John G. Lewis <=
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Algorithm 410 (partial sorting
- In Proc. 8th Workshop on Algorithm Engineering and Experiments (ALENEX , 2006
"... Let A be a set of size m. Obtaining the first k ≤ m elements of A in ascending order can be done in optimal O(m+k log k) time. We present an algorithm (online on k) which incrementally gives the
next smallest element of the set, so that the first k elements are obtained in optimal time for any k. We ..."
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Let A be a set of size m. Obtaining the first k ≤ m elements of A in ascending order can be done in optimal O(m+k log k) time. We present an algorithm (online on k) which incrementally gives the next
smallest element of the set, so that the first k elements are obtained in optimal time for any k. We also give a practical algorithm with the same complexity on average, which improves in practice
the existing online algorithm. As a direct application, we use our technique to implement Kruskal’s Minimum Spanning Tree algorithm, where our solution is competitive with the best current
implementations. We finally show that our technique can be applied to several other problems, such as obtaining an interval of the sorted sequence and implementing heaps. 1 | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=4838984","timestamp":"2014-04-19T03:05:56Z","content_type":null,"content_length":"12342","record_id":"<urn:uuid:80f2b542-c133-45ba-a2c4-b7444787170e>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00469-ip-10-147-4-33.ec2.internal.warc.gz"} |
Greatest Common Denominators - Definition and Uniqueness
October 13th 2012, 01:00 AM #1
Greatest Common Denominators - Definition and Uniqueness
On page 274 in Chaper 8 - Euclidean, Principal Ideal and Unique Factorization Domains, Dummit and Foote state Proposition 3 as follows: (see attachment - Proposition 3 ... )
================================================== ====================
Proposition 3
Let R be an integral domain. If two elements d and d' of R generate the same pricipal ideal, i.e. (d) = (d'), then d' = ud for some unit u in R.
In particular if d and d' are both greatest common divisors of a and b, then d' = ud for some unit u
================================================== =====================
So we can apply Proposition 3 to $\mathbb{Z}$ yeilding two gcds for each pair of integers a, b.
For example the gcds of 12 and 18 would be 6 and -6
So why do D&F make a special definition on page 4 of a unique gcd for $\mathbb{Z} - \{0\}$ by stipulating that the gcd must be positive - see attachment GCD - Properties of the Integers ... for
the definition.
This seems inconsistent .. also why deal with $\mathbb{Z} - \{0\}$ instead of dealing with $\mathbb{Z}$. Why do we need a separate definition from Proposition 3. Is it something to do with the
fact the the Euclidean Algorithm also gives a positive gcd? I cannot see the motivation for the definition (3) on page 4 of D&F [See attachement GCD - Properties of the Integers ...]
Can someone with more knowledge please explain why the two definitions are necessary and what the motivation for D&F could be in this matter ... Would appreciate the help.
Last edited by Bernhard; October 13th 2012 at 01:04 AM.
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Bisymmetric Matrix, solving set of linear equations.
up vote 4 down vote favorite
A bisymmetric matrix is a square matrix that is symmetric about both of its main diagonals.
If $A$ is a bisymmetric matrix and I'm interested in solving $Ax=b$.
Are there techniques used to exploit this structure when solving the system of linear equations?
Note: I'm looking for techniques which exploit more than just the fact that the matrix is symmetric.
matrices linear-algebra
add comment
1 Answer
active oldest votes
The condition of symmetry about the antidiagonal says that $A$ commutes with reversal of coordinates. Call this operation $R$, so $R^2 = 1$ and $AR = RA$.
$R$ has a $+1$ eigenspace and a $-1$ eigenspace.
up vote 5 For any solution, you can project both $x$ and $b$ to the two eigenspaces, by averaging them with either their reversals or $-$ the reversals. You can get the induced action of $A$ on
down vote these (roughly if in odd dimension) half-size eigenspaces similarly. The two halves of $A$ are still symmetric, so you're left with the easier problem of solving two symmetric systems of
accepted equations in half the number of variables.
Excellent! This is a little surprising. The operation count by this method is $~2\frac{1}{3}(\frac{n}{2})^3 = \frac{1}{12}n^3$ (not including the memory savings). I was expecting at
least $~\frac{1}{6}n^3$. Can you explain why this is? – alext87 Sep 23 '10 at 14:58
It's almost always simpler if you can separate variables cheaply. Basically you're eliminating the need to consider interactions between them. This is part of a more general method
1 that usually helps if there is known symmetry. A compact (e.g. finite) group $G$ acting on a vector space canonically splits into subspaces according to the irreducible
representations of $G$: the sum of all copies of a particular irreducible representation. Anything that commutes with the action of $G$ preserves this decomposition, so questions
split into a usually easier collection of smaller subquestions. – Bill Thurston Sep 23 '10 at 15:21
add comment
Not the answer you're looking for? Browse other questions tagged matrices linear-algebra or ask your own question. | {"url":"http://mathoverflow.net/questions/39618/bisymmetric-matrix-solving-set-of-linear-equations","timestamp":"2014-04-23T23:48:31Z","content_type":null,"content_length":"53328","record_id":"<urn:uuid:46f86b41-6902-4ebb-a26f-375055f854df>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00233-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cryptology ePrint Archive: Report 2011/118
New Fully Homomorphic Encryption over the IntegersGu ChunshengAbstract: We first present a fully homomorphic encryption scheme over the integers, which modifies the fully homomorphic encryption
scheme in [vDGHV10]. The security of our scheme is merely based on the hardness of finding an approximate-GCD problem over the integers, which is given a list of integers perturbed by the small error
noises, removing the assumption of the sparse subset sum problem in the origin scheme [vDGHV10]. Then, we construct a new fully homomorphic encryption scheme, which extends the above scheme from
approximate GCD over the ring of integers to approximate principal ideal lattice over the polynomial integer ring. The security of our scheme depends on the hardness of the decisional approximate
principle ideal lattice polynomial (APIP), given a list of approximate multiples of a principal ideal lattice. At the same time, we also provide APIP-based fully homomorphic encryption by introducing
the sparse subset sum problem. Finally, we design a new fully homomorphic encryption scheme, whose security is based on the hardness assumption of approximate lattice problem and the decisional SSSP.
Category / Keywords: Fully Homomorphic Encryption, Approximate Lattice Problem, Approximate Principal Ideal Lattice, Approximate GCD, BDDP, SSSPDate: received 9 Mar 2011, last revised 8 Jul 2011
Contact author: guchunsheng at gmail comAvailable format(s): PDF | BibTeX Citation Version: 20110709:020809 (All versions of this report) Discussion forum: Show discussion | Start new discussion[
Cryptology ePrint archive ] | {"url":"http://eprint.iacr.org/2011/118","timestamp":"2014-04-16T22:22:06Z","content_type":null,"content_length":"2857","record_id":"<urn:uuid:01ef3dd9-5d91-4f58-ac2a-9efa1f7d9da7>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00148-ip-10-147-4-33.ec2.internal.warc.gz"} |
MathGroup Archive: September 2011 [00415]
[Date Index] [Thread Index] [Author Index]
Re: Plot axis length and size ratio (TwoPlot revive)
• To: mathgroup at smc.vnet.net
• Subject: [mg121571] Re: Plot axis length and size ratio (TwoPlot revive)
• From: "David Park" <djmpark at comcast.net>
• Date: Tue, 20 Sep 2011 06:09:40 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <27663898.180702.1316428706866.JavaMail.root@m06>
This is how I would do it with the Presentations application. I would use a
Frame plot rather than an Axes plot. I think if you look in scientific
journals you will seldom see an axes plot where an axis cuts across the
<< Presentations`
yticks1 = CustomTicks[2.5 # &, {-8, 8, 4, 4}];
yticks2 =
CustomTicks[Identity, {-30, 30, 10, 5},
CTNumberFunction -> (Style[#, Red] &)];
{Draw[2.5 x Sin[x], {x, 0, 10}],
Draw[x (x - 10) Cos[x], {x, 0, 10}]},
AspectRatio -> 0.6,
Frame -> True,
FrameTicks -> {{yticks1, yticks2}, {Automatic, Automatic}},
FrameLabel -> {x, y1, None, Style[y2, Red]}, RotateLabel -> False,
PlotLabel -> "Two Axis Plot",
BaseStyle -> {FontSize -> 12}]
First we define custom ticks for the two y axes. Then in a single Draw2D
drawing statement we draw the two curves, scaling y1 by a factor of 2.5.
Then we specify the overall look of the plot with options using the custom
ticks defined initially.
A notebook and PDF of this result will appear on the archive kept for me by
Peter Lindsay at St. Andrews University Mathematics Department.
David Park
djmpark at comcast.net
From: matyigtm [mailto:matyigtm at gmail.com]
I want to combine two plots with the same "x" axis into one graphic.
Because the two plots have different magnitude, I choosed the "Inset"
command inside of one of the two plots. I want to put the two "y" axes
on left and right for the two plots, resp. (aka. TwoPlot)
My problem is that the two plots take different sizes when the numbers
written at the ticks on the "y" axis get different length, or when
AxesLabel is used. I've understood the the area of the plot is
extended by PlotRangePadding and with ImagePadding, but I cannot
retrieve the latter from the plots to get the right sizes.
The Inset command help a lot to position the origins of the two plots,
but its "size" parameter looses any relation to the coordinates used
in the plots.
The following example illustrates the problem, the hardcoded number
1.045 is to be programmed to give good results.
plt2 = Plot[x (x - 10) Cos[x], {x, 0, 10}, PlotStyle -> Red,
AxesOrigin -> {10, 0}, AxesStyle -> Red];
Plot[x Sin[x], {x, 0, 10}, Epilog -> Inset[plt2, {0, 0}, {0, 0}, 10 *
Any idea to align the two plots is appreciated, | {"url":"http://forums.wolfram.com/mathgroup/archive/2011/Sep/msg00415.html","timestamp":"2014-04-16T16:30:14Z","content_type":null,"content_length":"27940","record_id":"<urn:uuid:9d1ae900-9e19-4dc5-af84-849a4fae8228>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00608-ip-10-147-4-33.ec2.internal.warc.gz"} |
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This is a terminal mathematics course for liberal arts and social science majors. Topics include sets and counting, probability, linear systems, linear programming, statistics, and mathematics of
finance, with emphasis on applications.
2670 8:00a-9:20a TTh MC 83 Moassessi M
2671 8:00a-11:05a F MC 66 Harjuno T
Above section 2671 requires that students have internet access.
2672 9:30a-10:50a MW LS 205 Jahani F
2673 10:45a-12:05p TTh BUNDY 156 Gharamanians J
Above section 2673 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2674 11:15a-12:35p MW MC 82 Lee P H
2675 12:45p-2:05p MW PAC 103 Karasik P
Above 2675 section meets at the Performing Arts Center, 1310 11th Street.
2676 2:00p-3:20p TTh MC 66 Carty B
2677 2:15p-3:35p MW LA 228 Lai I
Above section 2677 requires that students have internet access.
4327 5:15p-6:35p MW LS 203 Nikolaychuk A M
4328 7:35p-8:55p TTh MC 71 Harjuno T
MATH 26, Functions and Modeling for Business and Social Science 3 units
Transfer: UC*, CSU • IGETC AREA 2 (Mathematical Concepts) • Prerequisite: Mathematics 20.
This course is a preparatory course for students anticipating enrollment in Math 28 (Calculus 1 for Business and Social Science). Topics include algebraic, exponential and logartihmic functions and
their graphical representations, and using these functions to model applications in business and social science.
Math 26 is not recommended as a terminal course to satisfy transfer requirements. *Maximum UC credit for Math 2, 22 and 26 is one course.
2678 9:30a-10:50a TTh MC 70 Rodas B G
Above section 2678 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2679 11:15a-12:35p MW LA 231 Gromova I M
2680 12:45p-3:50p W LA 231 Garcia E E
Above section 2680 requires that students have internet access.
2681 2:00p-3:20p MW LS 203 Baskauskas V
2682 3:30p-4:50p TTh MC 66 Garcia E E
Above section 2682 requires that students have internet access.
4329 6:00p-7:20p TTh MC 83 Kaviani K
MATH 28, Calculus 1 for Business and Social Science 5 units
Transfer: UC, CSU • IGETC AREA 2 (Mathematical Concepts) • Prerequisite: Math 26.
This class was formerly Math 23.
This course is intended for students majoring in business or social sciences. It is a survey of differential and integral calculus with business and social science applications. Topics include
limits, differential calculus of one variable, including exponential and logarithmic functions, introduction to integral calculus, and mathematics of finance.
2683 9:30a-10:35a MTWTh MC 10 Wong B L
2684 12:45p-1:50p MTWTh MC 71 McGraw C K
4330 7:35p-10:00p TTh LA 231 Kaviani K
MATH 29, Calculus 2 for Business and Social Science 3 units
Transfer: UC, CSU • IGETC AREA 2 (Mathematical Concepts) • Prerequisite: Math 28.
Formerly Math 24.
Topics include techniques and applications of integration, improper integrals, functions of several variables, partial derivatives, method of least squares, maxima and minima of functions of several
variables with and without constraints, methods of LaGrange Multipliers, double integrals and their application, elementary differential equations with applications, probability and calculus.
Maximum UC credit is allowed for only one series, either Math 7,8 or 28, 29.
2685 9:30a-10:50a MW MC 73 Rodas B G
Above section 2685 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
MATH 31, Elementary Algebra 5 units
• Prerequisite: Math 84 or Math 85.
Topics include: Arithmetic operations with real numbers, polynomials, rational expressions, and radicals; factoring polynomials; linear equations and inequalities in one and two variables; systems of
linear equations and inequalities in two variables; application problems; equations with rational expressions; equations with radicals; introduction to quadratic equations in one variable.
This course is equivalent to one year high school algebra. Course credit may not be applied toward satisfaction of Associate Degree requirements. Students enrolled in this course are required to
spend 16 documented supplemental learning hours outside of class during the semester. This can be accomplished in the Math Lab on the main campus, BUNDY 116, or electronically (purchase of an access
code required).
2686 6:45a-7:50a MTWTh MC 82 Phung Q T
Arrange-1 Hour
Above section 2686 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2687 7:00a-8:05a MTWTh MC 70 Fanelli D
Arrange-1 Hour
2688 8:15a-9:20a MTWTh LA 228 Rahnavard M
Arrange-1 Hour
2689 9:30a-10:35a MTWTh MC 74 Soury S
Arrange-1 Hour
2690 9:30a-10:35a MTWTh LS 201 Jimenez B S
Arrange-1 Hour
Above section 2690 is part of the Black Collegians Program. See Special Programs section of schedule for program information. Above section is part of the Latino Center Adelante Program. See Special
Programs section of schedule for program information.
2691 9:30a-11:55a MW BUNDY 153 Ward J E
Arrange-1 Hour
Above section 2691 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2692 11:00a-12:30p MWF MC 2 Cho M
Arrange-1 Hour
2693 12:45p-1:50p MTWTh BUNDY 217 Foreman N
Arrange-1 Hour
Above section 2693 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2694 12:45p-1:50p MTWTh MC 66 Nguyen D T
Arrange-1 Hour
2695 12:45p-2:15p MWF MC 12 Bresloff J L
Arrange-1 Hour
2696 12:45p-3:10p MTWTh MC 73 Lee P H
Arrange-2 Hours
Above section 2696 meets for 8 weeks, Aug 26 to Oct 17.
2697 12:45p-3:10p TTh MC 70 Garcia E E
Arrange-1 Hour
Above section 2697 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2698 2:15p-3:20p MTWTh LS 201 Pachas-Flores W
Arrange-1 Hour
2699 2:15p-3:20p MTWTh MC 82 Allen C A
Arrange-1 Hour
2700 2:15p-4:40p MW BUNDY 156 King W S
Arrange-1 Hour
Above section 2700 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 2700 requires that students have internet access. Students are required to purchase an
access code in order to complete online homework.
2701 2:15p-4:40p TTh MC 74 Emerson A J
Arrange-1 Hour
2702 3:15p-4:20p MTWTh BUNDY 221 Bellin E
Arrange-1 Hour
Above section 2702 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2703 3:15p-5:40p MW LS 103 Bronie B L
Arrange-1 Hour
2704 3:30p-4:35p MTWTh MC 82 Lopez Ma
Arrange-1 Hour
Above section 2704 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2705 3:30p-5:55p MW MC 74 Mazorow M M
Arrange-1 Hour
4331 5:00p-7:25p MW LS 201 Saakian L
Arrange-1 Hour
4332 5:00p-7:25p TTh MC 66 Schlenker J D
Arrange-1 Hour
4333 6:45p-9:10p W LS 203 Owaka E A
12:30p-2:55p Sat LS 203 Owaka E A
Arrange-1 Hour
4334 7:35p-10:00p MW LS 103 Ghahramanyan A
Arrange-1 Hour
4335 7:35p-10:00p MW MC 70 Sheynshteyn A S
Arrange-1 Hour
Above section 4335 requires that students have internet access.Students are required to purchase an access code in order to complete online homework.
4336 7:35p-10:00p TTh MC 73 Evinyan Z
Arrange-1 Hour
4337 7:35p-10:00p TTh MC 70 Bojkov A
Arrange-1 Hour
MATH 32, Plane Geometry 3 units
• Prerequisite: Math 31. • Advisory: Math 20.
This is an introductory course in geometry whose goal is to increase student’s mathematical maturity and reasoning skills. Topics include elementary logical reasoning, properties of geometric
figures, congruence, similarity, and right triangle relationships using trigonometric properties. Formal proof is introduced and used within the course.
2706 8:00a-9:20a TTh BUS 201 London J S
Above section 2706 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2707 9:30a-10:50a MW MC 66 Gizaw A
2708 11:15a-12:35p MW LS 101 London J S
Above section 2708 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2709 11:15a-12:35p TTh MC 71 Karasik P
2710 2:00p-3:20p MW MC 10 McDonnell P
Above section 2710 requires that students have internet access.
2711 3:30p-4:50p TTh LS 201 Chau E
4338 5:15p-6:35p MW MC 83 Lai I
Above section 4338 requires that students have internet access.
4339 5:15p-6:35p TTh LA 228 Mardirosian R
4340 6:00p-7:20p MW LA 228 Perez R E
MATH 41, Mathematics for Elementary School Teachers 3 units
Transfer: CSU • Prerequisite: Math 20.
This course is designed for preservice elementary school teachers. The course will examine five content areas: Numeration (historical development of numeration system); Set Theory (descriptions of
sets, operations of sets, Venn Diagrams); Number Theory (divisibility, primes and composites, greatest common divisor, least common multiple); Properties of Numbers (whole numbers, integers, rational
numbers and models for teaching binary operations); and Problem Solving (strategies, models to solve problems, inductive and deductive reasoning).
Math 41 fulfills the mathematics requirement for the Liberal Arts major at CSU campuses but does not meet the mathematics admission requirement at any of the CSU campuses. Please contact a counselor
if you have questions.
4341 5:15p-8:20p T BUNDY 221 Perez L
Above section 4341 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
MATH 54, Elementary Statistics 4 units
Transfer: UC, CSU • Prerequisite: Math 20 or Math 18 with a grade of C or better. • IGETC AREA 2 (Mathematical Concepts).
It is recommended that students who were planning to take Math 52 to fulfill requirements should take Math 54.
This course covers concepts and procedures of descriptive statistics, elementary probability theory and inferential statistics. Course material includes: summarizing data in tables and graphs;
computation of descriptive statistics; measures of central tendency; variation; percentiles; sample spaces; classical probability theory; rules of probability; probability distributions; binomial,
normal, T, Chi-square and F distributions; making inferences; decisions and predictions. This course develops confidence intervals for population parameters, hypothesis testing for both one and two
populations, correlation and regression, ANOVA, test for independence and non-parametric method. This course develops statistical thinking through the study of applications in a variety of
disciplines. The use of a statistical/graphing calculator or statistical analysis software is integrated into the course.
2712 7:00a-9:05a TTh BUNDY 221 Gharamanians J
Above section 2712 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2713 8:00a-10:05a MW MC 67 McGraw C K
2714 8:00a-10:05a TTh MC 67 McGraw C K
Above section 2714 is part of the Scholars Program and enrollment is limited to program participants. See Special Programs section of class schedule or www.smc.edu/scholars for additional
2715 8:00a-12:05p F MC 70 Zilberbrand M
Above section 2715 requires that students have internet access.
2716 9:00a-1:05p Sat HSS 152 Zilberbrand M
Above section 2716 requires that students have internet access.
2717 10:15a-12:20p MW MC 67 Jahangard E
Above section 2717 requires that students have internet access.
2718 10:15a-12:20p MW BUNDY 221 Miao W
Above section 2718 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 2718 requires that students have internet access. Students are required to purchase an
access code in order to complete online homework.
2719 10:15a-12:20p TTh MC 67 Soleymani S
Above section 2719 requires that students have internet access.
2721 12:45p-2:50p MW MC 67 Edinger G C
Above section 2721 requires that students have internet access.
2722 12:45p-2:50p MW BUNDY 221 Foster M
Above section 2722 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 2722 requires that students have internet access.
2723 12:45p-2:50p TTh BUNDY 156 Yankey K
Above section 2723 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2724 12:45p-2:50p TTh BUNDY 221 Foster M
Above section 2724 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 2724 requires that students have internet access.
2725 12:45p-2:50p TTh MC 67 Edinger G C
Above section 2725 requires that students have internet access.
2726 2:00p-4:05p MW BUNDY 153 Miao W
Above section 2726 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 2726 requires that students have internet access. Students are required to purchase an
access code in order to complete online homework.
2727 2:15p-4:20p MW MC 1 Nguyen D T
2728 3:00p-5:05p MW MC 67 Kamin G
2729 3:00p-5:05p TTh MC 67 Jahangard E
Above section 2729 requires that students have internet access.
2730 3:30p-5:35p TTh PAC 103 Kaush A
Above 2730 section meets at the Performing Arts Center, 1310 11th Street.
2731 3:45p-5:50p MW MC 10 McDonnell P L
Above section 2731 requires that students have internet access.
4342 5:15p-7:20p MW MC 67 Martinez M G
4343 5:15p-7:20p TTh MC 67 Martinez M G
4344 5:15p-9:20p Th BUNDY 221 Walker C W
Above section 4344 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
4345 6:45p-8:50p MW BUNDY 221 Yee D K
Above section 4345 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
4346 6:45p-8:50p MW MC 10 Pachas-Flores W
4347 7:35p-9:40p MW MC 67 Perez R E
Above section 4347 requires that students have internet access.
4348 7:35p-9:40p TTh MC 67 Malakar S R
MATH 81, Basic Arithmetic 3 units
• Prerequisite: None.
The aim of this course is to develop number and operation sense with regard to whole numbers, fractions, decimals and percents; as well as measurement and problem solving skills. Course content also
includes ratios, proportions, and practical applications of the arithmetic material.
*Course credit may not be applied toward satisfaction of Associate in Arts Degree requirements. Students enrolled in this course are required to spend 16 documented supplemental learning hours
outside of class during the session. This can be accomplished in the Math Lab on the main campus, in Bundy 116, or electronically (purchase of an access code required).
2732 6:30a-7:50a TTh MC 74 Chau E
Arrange-1 Hour
2733 8:00a-9:20a MTWTh MC 82 Phung Q T
Arrange-2 Hours
Above section 2733 meets for 8 weeks, Aug 26 to Oct 17. Above section 2733 requires that students have internet access. Students are required to purchase an access code in order to complete online
2734 8:00a-9:20a MW LS 152 Quevedo J M
Arrange-1 Hour
2735 8:00a-11:05a F MC 74 Lai I
Arrange-1 Hour
Above section 2735 requires that students have internet access.
2736 9:00a-12:05p Sat LS 103 Ulrich J W
Arrange-1 Hour
Above section 2736 requires that students have internet access.
2737 11:15a-12:35p MW MC 74 Phung Q T
Arrange-1 Hour
Above section 2737 is part of the Black Collegians Program. See Special Programs section of schedule for program information. Above section 2737 requires that students have internet access. Students
are required to purchase an access code in order to complete online homework.
2738 1:00-2:20 p.m. T Th LA 231 Lopez Ma
Arrange-1 Hour
Above section 2738 is part of the Latino Center Adelante Program. See Special Programs section of schedule for program information.
2739 2:15p-3:35p MW MC 11 Bayssa B T
Arrange-1 Hour
2740 2:30p-3:50p TTh LA 231 Wang E
Arrange-1 Hour
2741 3:45p-5:05p MW MC 73 Perez L
Arrange-1 Hour
4349 4:30p-5:50p TTh MC 73 Evinyan Z
Arrange-1 Hour
4350 5:00p-6:20p MW BUNDY 156 King W S
Arrange-1 Hour
Above section 4350 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 4350 requires that students have internet access. Students are required to purchase an
access code in order to complete online homework.
4351 5:15p-6:35p MW MC 82 Ghahramanyan A
Arrange-1 Hour
4352 5:15p-6:35p TTh MC 10 Chau E
Arrange-1 Hour
4353 5:15p-6:35p TTh BUNDY 156 Yan S K
Arrange-1 Hour
Above section 4353 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
4354 6:00p-7:20p MW MC 73 Okonjo-Adigwe C E
Arrange-1 Hour
4355 6:00p-7:20p TTh MC 73 Aka D O
Arrange-1 Hour
Above section 4355 requires that students have internet access.
4356 6:00p-7:20p TTh LA 231 Huang C
Arrange-1 Hour
4357 6:45p-8:05p MW MC 74 Lee L S
Arrange-1 Hour
4358 6:45p-9:50p MW HSS 205 Hecht S E
Arrange-2 Hours
Above section 4358 meets for 8 weeks, Aug 26 to Oct 16. Above section 4358 requires that students have internet access. Students are required to purchase an access code in order to complete online
4359 6:45p-9:50p T MC 74 Chitgar M H
Arrange-1 Hour
4360 7:35p-8:55p MW LA 228 Atique N
Arrange-1 Hour
MATH 84, Pre-Algebra 3 units
• Prerequisite: Math 81.
This course prepares the student for Elementary Algebra. It assumes a thorough knowledge of arithmetic. Course content includes integers, signed fractions, signed decimals, grouping symbols, the
order of operations, exponents, and algebraic expressions and formulas. The emphasis is on concepts essential for success in algebra.
*Course credit may not be applied toward satisfaction of Associate In Arts Degree requirements. Students enrolled in this course are required to spend 16 documented supplemental learning hours
outside of class during the session. This can be accomplished in the Math Lab on the main campus, in Bundy 116, or electronically (purchase of an access code required).
2742 6:30a-7:50a MW LS 103 Gizaw A
Arrange-1 Hour
Above section 2742 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2743 6:30a-7:50a TTh MC 9 Chua T D
Arrange-1 Hour
2744 8:00a-9:20a MTWTh MC 82 Phung Q T
Arrange-2 Hours
Above section 2744 meets for 8 weeks, Oct 21 to Dec 12. Above section 2744 requires that students have internet access. Students are required to purchase an access code in order to complete online
2745 8:00a-9:20a MW LS 205 Staff
Arrange-1 Hour
2746 8:00a-11:05a F LA 228 Chen C
Arrange-1 Hour
2747 9:00a-12:05p Sat LS 203 Owaka E
Arrange-1 Hour
2748 9:30a-10:50a MW MC 70 Cho M
Arrange-1 Hour
2749 11:15a-12:35p MW MC 70 Tsvikyan A
Arrange-1 Hour
Above section 2749 requires that students have internet access.
2750 11:15a-12:35p TTh MC 83 Esmaeili F
Arrange-1 Hour
2751 12:45p-2:05p MW MC 70 Chan Hy
Arrange-1 Hour
2752 2:15p-3:35p MW MC 83 Ward J E
Arrange-1 Hour
Above section 2752 is part of the Black Collegians Program. See Special Programs section of schedule for program information. Above section is part of the Latino Center Adelante Program. See Special
Programs section of schedule for program information.
2753 2:15p-3:35p TTh MC 14 Chan Hy
Arrange-1 Hour
2754 3:45p-5:05p MW MC 83 Lee L S
Arrange-1 Hour
2755 3:45p-5:05p TTh LS 103 Halaka E F
Arrange-1 Hour
2756 4:15p-5:35p MW BUNDY 153 Jiang J
Arrange-1 Hour
Above section 2756 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2757 4:15p-5:35p TTh MC 83 Chan Hy
Arrange-1 Hour
4361 5:00p-6:20p TTh BUNDY 153 Owens D J
Arrange-1 Hour
Above section 4361 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066. Above section 4361 requires that students have internet access.
4362 6:00p-7:20p MW LS 103 Man S
Arrange-1 Hour
4363 6:00p-7:20p TTh MC 82 England A
Arrange-1 Hour
4364 6:00p-7:20p TTh LS 103 Bronie B L
Arrange-1 Hour
4365 6:45p-9:50p MW HSS 205 Hecht S E
Arrange-2 Hours
Above section 4365 meets for 8 weeks, Oct 21 to Dec 11. Above section 4365 requires that students have internet access. Students are required to purchase an access code in order to complete online
4366 7:35p-8:55p TTh LS 201 Mozafari R
Arrange-1 Hour
MATH 85, Arithmetic and Prealgebra 5 units
• Prerequisite: None.
This course offers an accelerated option for preparation for Elementary Algebra. The material covered is equivalent to that covered separately in Math 81 (Basic Arithmetic) and Math 84 (Prealgebra).
This course develops number and operation sense with regard to whole numbers, integers, rational numbers, mixed numbers, and decimals. Grouping symbols, order of operations, estimation and
approximation, scientific notation, ratios, percents, proportions, geometric figures, and units of measurement with conversions are included. An introduction to algebraic topics, including simple
linear equations, algebraic expressions and formulas, and practical applications of the material also are covered. All topics will be covered without the use of a calculating device.
This course is fast-paced and intensive. Course credit may not be applied toward satisfaction of Associate degree requirements.
2758 6:50a-9:15a MW MC 73 Tsvikyan A
Above section 2758 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
2759 7:00a-8:05a MTWTh MC 71 Rahnavard M
2760 8:00a-9:05a MTWTh BUNDY 153 Raffel C
Above section 2760 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2761 9:30a-10:35a MTWTh LA 231 London J S
Above section 2761 is part of the Black Collegians Program. See Special Programs section of schedule for program information. Above section is part of the Latino Center Adelante Program. See Special
Programs section of schedule for program information. Above section 2761 requires that students have internet access. Students are required to purchase an access code in order to complete online
2762 9:30a-10:35a MTWTh LS 101 Esmaeili F
2763 11:15a-12:20p MTWTh LA 228 Baskauskas V A
2764 11:15a-12:20p MTWTh MC 73 Soury S
2765 12:45p-1:50p MTWTh BUNDY 153 Petikyan G
Above section 2765 meets at the Bundy Campus, 3171 South Bundy Drive, Los Angeles, CA 90066.
2766 2:00p-3:05p MTWTh LS 103 Halaka E F
2767 3:30p-5:55p TTh LS 203 Mozafari R
2768 4:00p-5:05p MTWTh LA 231 Pachas-Flores W
4367 5:00p-7:25p TTh LS 201 Lopez Ma
Above section 4367 requires that students have internet access. Students are required to purchase an access code in order to complete online homework.
4368 7:35p-10:00p MW MC 5 Bateman M
MATH 88A, Independent Studies in Mathematics 1 unit
Transfer: CSU
Please see “Independent Studies” section.
2769 Arrange-1 Hour MC 26 Emerson A J
│ IMPORTANT! Many SMC classes require the use of a computer with Internet access to reach class resources and/or to complete assignments and/or take exams. To locate a computer lab on campus go to │
│ www.smc.edu/acadcomp and click on the “Labs” link. │ | {"url":"http://www2.smc.edu/schedules/2013/fall/071_133_schedule.htm","timestamp":"2014-04-16T18:57:08Z","content_type":null,"content_length":"75185","record_id":"<urn:uuid:7371bbf9-5dd5-4b36-809b-a02bc3db36a4>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00549-ip-10-147-4-33.ec2.internal.warc.gz"} |
Brentwood, CA Algebra 2 Tutor
Find a Brentwood, CA Algebra 2 Tutor
...One way that I make this easier for students is by giving you a step-by-step recipe of how to do the task while allowing room for variations so you know how to solve any type of problem. I can
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30 Subjects: including algebra 2, English, reading, Spanish
...I prepare students for the following SAT tests: SAT RT Math (the “big” SAT) and SAT Subject Math Level 2. I began tutoring for these tests as an instructor for a couple of premier Test Prep
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14 Subjects: including algebra 2, calculus, geometry, statistics
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Millbury, MA ACT Tutor
Find a Millbury, MA ACT Tutor
...My name is Francesca and I am a recent graduate from Clark University, where I received a bachelor's degree in Art History and Studio Art. During my career at Clark, I studied a variety of
subjects from British literature to management. I was a PLA (Peer Learning Assistant) for an art history survey course and I have experience working with children of all ages.
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I am an experience math tutor for students in middle school through college. I have an PhD in Applied Math from UC Berkeley and have been tutoring students part time in the last four years. I
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1997-98 UAF Catalog
Degrees and Programs
Course Description Index
College of Science, Engineering and Mathematics
Department of Mathematical Sciences (907) 474-7332
Minimum Requirements for Degree: 120 credits
Statistics is a collection of methods for making decisions or estimating unknown quantities from incomplete information. Statistical techniques are useful, for example, in estimating plant, animal
and mineral abundances; forecasting social, political and economic trends; planning field plot experiments in agriculture; performing clinical trials in medical research; and maintaining quality
control in industry. Employment opportunities are excellent for statisticians in many of these areas of application.
The curriculum for the B.S. in statistics provides a strong mathematics and statistics background and integrates this with an area of application. The program allows considerable flexibility in the
choice of the area of application.
The statistics program is administered by the Department of Mathematical Sciences. In addition to the B.S. in statistics, the department offers a bachelor's degree in mathematics with an emphasis in
statistics. A minor in statistics is also available.
Undergraduate Degree Requirements
Statistics -- B.S. Degree
A student may declare Statistics as a major only when she/he is eligible to enroll into MATH 200, Calculus I.
1. Complete the general university requirements and B.S. degree requirements. The mathematics requirements should be met with MATH 200-201. ENGL 314 is recommended to fulfill one of the writing
intensive course requirements.
2. Complete the following major requirements:
A. Statistics Core (26 Credits)
MATH 202X -- Calculus (4 credits)
MATH 371 -- Probability (3 credits)
MATH 408 -- Mathematical Statistics (3 credits)
CS 103 -- Intro. to Computer Programming
or any higher level CS course (3 credits)
STAT 200 -- Elementary Probability and Statistics
or STAT 300 -- Statistics (3 credits)
STAT 401 -- Regression and Analysis of Variance (4 credits)
STAT 402 -- Scientific Sampling (3 credits)
STAT 498 -- Senior Project (3 credits)
B. Electives in the Major
Choose two of the following:
STAT 461 -- Applied Multivariate Statistics (3 credits)
MATH 307 -- Discrete Mathematics (3 credits)
MATH 310 -- Numerical Analysis (3 credits)
MATH 314 -- Linear Algebra (3 credits)
MATH 401 -- Advanced Calculus I (3 credits)
MATH 402 -- Advanced Calculus II (3 credits)
MATH 460 -- Mathematical Modeling (3 credits)
STAT, MATH or statistical discipline oriented course approved by the statistics program chairperson (3 credits)
C. Area of Application* (24 Credits)
Complete a minimum of 24 credits, including at least 6 upper division, in a single discipline in which a UAF Bachelor's Degree is offered. Joint approval in writing is required from the
department head in the area of application and the statistics advisor.**
Minimum credits required (120 credits)
* Credits received in the area of application may reduce the number of required credits in the general distribution requirements of humanities/social science and science.
** Examples of programs for areas of application for computer science, biology, wildlife, geology, natural resource management, and economics are available. Other areas of application are available.
A mathematics minor is completed by all statistics majors.
A Statistics/Math double major may be obtained by taking the following in addition to Items 1 and 2A above: MATH 215, 308, 314, 401, 492 and complete 12 additional credits in upper division math or
statistics. A math elective package is MATH 371 and 408, STAT 401 and 402 plus 8 credits upper division MATH or STAT. The statistics elective package is MATH 314 and 401. Total credit hours 60
including MATH 200-201. Other double majors are available.
Minor in Statistics:
Complete the following: Credits
STAT 200 -- Elementary Probability and Statistics
or STAT 300 -- Statistics (3 credits)
STAT 401 -- Regression and Analysis of Variance (3 credits)
MATH 371 -- Probability (3 credits)
MATH 408 -- Mathematical Statistics (3 credits)
In addition, complete three (3) credits of approved MATH, STAT or STAT related coursework (e.g., BA 360, GEOS 430, ANTH 424, MATH 460, etc.) (3 credits)
Fisheries majors selecting the research option need only complete MATH 371 and 408 in addition to their fisheries requirements to obtain a minor in statistics.
*MATH 371 requires MATH 200-201-202 as prerequisites. These courses can be used to simultaneously satisfy other major or general distribution requirements.
Catalog Index | Admissions | UAF Home | UAF Search | News and Events
Send comments or questions to the UAF Admissions Office.
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Homework Help
Posted by Helga on Thursday, March 1, 2012 at 2:53am.
How do I calculate 4^n=1/(2(n-n^2))? Thanks.
• Algebra - MathMate, Thursday, March 1, 2012 at 9:18am
cross multiply to get:
Since there is no more division and the right-hand side is less than 1, we can start searching for n where n<1, say 1/2.
Try n=1/2 and see if the equation is satisfied.
Hint: one of the laws of exponents tells us that
• Algebra - Helga, Thursday, March 1, 2012 at 10:49am
Is there a more elegant solution than trial and error?
• Algebra - MathMate, Thursday, March 1, 2012 at 2:37pm
You have an excellent question!
In other words, your question was more like:
Find all solutions for n such that
I do not see an explicit solution to the equation. Perhaps someone else can find one.
An explicit solution is expressed as
n=expression where expression does not contain n.
Lack of an explicit solution, I proceed as follows:
1. first bound the solution.
We can conclude that for n outside of [-1,1], we cannot have the right hand side equal to -1.
2. Find approximate solutions by graphing or otherwise. By graphing, there are two solutions, at 0.5 and 0.8, approximately.
3. proceed to refine the solutions by iterations (a glamorous name for trial and error).
Here are the details of iteration using Newton's method:
For the case of n=0.5, it is exact and so requires no further iteration.
For the case of n=0.8, we can refine the solution by setting up Newton's iteration equation:
let f(x)=4^x-1/(2*(x-x^2))
and find the derivative
and finally calculate a better approximation of x as
and proceed to calculate
which means that 0.786097643010236 is our (approximate) solution.
Hope I have answered your question.
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Continuous Markov Processes
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Here's the question you clicked on:
The recipe for a loafcake calls for 3/4 pound of jelly beans in assorted colors. If 16 cakes are needed how many pounds of jelly beans will be used.
• one year ago
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over 9000!
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thats what i was gonna write was 12. thanks fnuuosh
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No problem.
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can u help me with another problem
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A manufacture of bird feeders cute spacers from a tube that is 6 1/9inches long. How many spacers can be cut from the tube if each spacer must be 5/9inches thick? The estimate answer is _____?
The exact number of spacers is_____?
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First it would be much easier if the values are changed into decimals. 6 1/9=6.11 5/9=.55, So to estimate you would want to say 6/.5, so 12. To fine the exact dumber you will divide 6.11 by .55
If it is .anything you will round down.
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estimate is 12. exact is 11.1090 i have too type a whole number or a simplified fraction
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yes. :)
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No. 180: Pi
Today, let's talk about
School-children can always be found competing to see who knows the most digits of pi -- the ratio of the circumference of a circle to its diameter. My interest and patience usually ran out at
3.14159256. But finding out what those digits were has been a major mathematical challenge ever since the invention of the wheel first stirred a real interest in circles. The earliest recorded
values of pi were Phoenician and Egyptian. They were 3 1/8 and 3 13/81. Both values are accurate within a half a percent.
Where did these values came from? From measurements? Well, try measuring pi with a piece of string. You won't come this close. The Hebrew peoples used a rough empirical value of pi in the Bible.
It was three, and it's found in a text that shows up in both the 1st Book of Kings and the 2nd Book of Chronicles:
And he made a molten sea, ten cubits from the one brim to
the other: It was round ... and a line of thirty cubits
did compass it ... about.
From time to time you hear stories about legislative bodies that've tried to make pi = 3 into law on the basis of this text. Science writer Petr Beckmann was unable to verify any of these
stories, but he does report a remarkable event in the 1897 Indiana State Legislature. An Indiana doctor thought he'd solved the classical problem of squaring the circle. That means specifying the
size of a square with the same area as a circle. If you could do that, you'd also be able to get an exact value of pi. This fellow tried to get his proof enacted as law. But the text of his bill
was muddled. It would've made pi greater than nine.
The House had trouble finding anyone to review the bill. They finally gave it to the Committee on Swamp Lands, who said it looked okay to them. When it cleared the House, the Senate gave it to
their Committee on Temperance. Temperance could no more figure it out than Swamps could, so it got preliminary approval. After that, local academics heard of what Congress was up to and started
questioning legislators. The Bill mysteriously disappeared from sight and was never heard from again.
All this happened 15 years after mathematicians had shown it was impossible either to square the circle or to evaluate pi exactly. That's bizzare enough, even if fundamentalists didn't really try
to make pi = 3 into law. But historians have also found out that the accurate Phoenician and Egyptian values of pi didn't come from measurements after all. These ancient engineers actually
deduced them, and they used elegant geometry and logic to do it. It's a sobering fact that they had clear-headed answers to questions that still troubled a lot of people 4000 years later.
I'm John Lienhard, at the University of Houston, where we're interested in the way inventive minds work.
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I have been stuck on this for a bit now
May 2nd 2012, 05:21 AM #1
May 2012
Find all values x such that $(2^x-4)^3+(4^x-2)^3 = (4^x+2^x-6)^3$
I expanded the entire thing:
$2^3^x+2^2^x-38+4^3^x+4^2^x+4^x = (4^3^x)(2^5^x)+(4^2^x)(2^3^x)+4^x+4^x-105+2^x+2^x$
but not quite sure what to do from here. Is there a logarithm I can use?
Re: I have been stuck on this for a bit now
I don't think you can have expanded these out correctly... Remember \displaystyle \begin{align*} (a + b)^3 = a^3 + 3a^2b + 3a\,b^2 + b^3 \end{align*}...
Re: I have been stuck on this for a bit now
Let a=2^x-4 and b=4^x-2 so we are solving a^3+b^3=(a+b)^3
Multiply out the bracket and get 3ab(a+b)=0 So a=0 or b=0 or a+b=0 giving x= 1/2 x=1 and x=2
Re: I have been stuck on this for a bit now
>_< These unknown exponents are really screwing with my head, I can't seem to get it expanded correctly.
I'm going to attempt biffboy's method to see if I can get it to work that way.
Btw, I graphed both sides out and they it the x-axis at 1; but it doesn't show the other values that way either. :/
Thank you both for the help and advice so far. Much appreciated.
Re: I have been stuck on this for a bit now
I then expanded the right side
So I have to set it to 0 so I have to subtract $a^3$ and $b^3$ getting:
and then I factor out $3ab$
I did the following and got your same answers, but don't I have to use logarithms?
$2^x = 4$
$4^x = 2$
$x = 1/2$
problem I am having is if $a+b=0$
I assume I resubstitute the originals back in?
Should I be using Logarithm's to solve these when they get to the form $6^x=6$ or am I to just figure it out since its a simple guess?
Last edited by brokaliv; May 2nd 2012 at 12:11 PM. Reason: figured out the other values
Re: I have been stuck on this for a bit now
This is correct now. You will get the right answers if you use logs but it is not necessary if you can see the answer without using them, as in 2^x=4 and 6^x=6.
For 4^x=2 we see that 2 is the square root of 4 so x=1/2. Alternatively log(4^x)=log2 so xlog4=log2 so x= log2/log4= 0.5
Re: I have been stuck on this for a bit now
Could have sworn I said thank you both for the help...apparently cell phone didn't post it O_o
Thank you both!
May 2nd 2012, 05:32 AM #2
May 2nd 2012, 06:31 AM #3
Senior Member
Mar 2012
Sheffield England
May 2nd 2012, 10:25 AM #4
May 2012
May 2nd 2012, 11:22 AM #5
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May 2nd 2012, 12:28 PM #6
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Sheffield England
May 10th 2012, 09:05 AM #7
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Discrete Mathematics Kenneth H. Rosen
About the Book
to accompany Discrete Mathematics and Its Applications 4/E
This manual contains the full solutions to all of the odd-numbered exercises in the textbook. These solutions explain why a particular method is used, and why it works. For some exercises, one or two
other possible approaches are described to illustrate that a problem may be solved in several different ways. Suggested references for the writing projects, found at the end of each chapter, are also
included in this guide. The manual contains a guide to writing proofs and a list of common errors that students make in discrete mathematics. In order to assist students in preparing for
examinations, sample test questions and answers, and sample crib sheets, are provided for each chapter. Now in its Fourth Edition, students find the Student Solutions Guide very useful. Order your
copy today! The ISBN number is 0072899069.
to accompany Discrete Mathematics and Its Applications 4/E
This manual contains the full solutions to all of the even-numbered exercises in the textbook, and provides suggestions on how to teach the material in each chapter of the book, including suggested
key points to stress in each section and how to put the material in perspective for your students. Sample syllabi are also presented. In addition, the manual includes a printed test bank of over 1300
sample examination questions and answers, as well as sample chapter tests and answers.
to accompany Discrete Mathematics and Its Applications 4/E
Macintosh (0072899085)
Windows (0072899093)
An extensive test bank of over 1300 question is available for use on either Windows or Macintosh operating systems. Instructors can utilize this software to create their own tests. Test questions and
answers may be selected manually or randomly. Instructors are able to add their own headings and instructions to the exams, print scrambled versions of the same test, and edit existing questions or
add their own. A printed version of this test bank (test questions and answers) is included in the Instructor’s Resource Guide.
Supplementary texts:
EXPLORING DISCRETE MATHEMATICS AND ITS APPLICATIONS WITH MAPLE by Kenneth H. Rosen, John S. Devitt, Troy Vasiga, James McCarron, Eithne Murray, and Ed Roskos.
This ancillary is a separate book designed to help students use the MAPLE computer algebra system to do a wide variety of computations in discrete mathematics. For each chapter of the main text, this
ancillary includes the following: a description of relevant MAPLE functions and how they are used, MAPLE programs that carry out relevant computations, suggestions and examples showing how MAPLE can
be used for the computations and explorations at the end of each chapter, and exercises that can be worked using MAPLE.
APPLICATIONS OF DISCRETE MATHEMATICS by John G. Michaels and Kenneth H. Rosen
This ancillary is a separate text that can be used either in conjunction with the main textbook, or independently. It contains more than 20 chapters (each with its own set of exercises) written by
instructors that have used the main text. Following a common format, similar to that of the main text, the chapters in this book can be used as a text for a separate course, for a student seminar, or
for a student doing independent study.
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Posts by Richard Smith
2 posts • joined 20 Sep 2006
"Error correction inadequate"
Fixing up bit flips may seem important, but is hardly sufficient for a "serious HPC center" to trust the alleged answers generated. As is well-known, there are many other sources of error:
floating-point rounding, measurement error, approximate physical constants, use of discrete models in place of continuous models. I wouldn't trust any answer, regardless of the presence of logic
detecting bit-flips, unless the solutions are accompanied by guaranteed bounds on the errors. Now if these new GPUs had hardware implementations of Interval Arithmetic instructions, that might be
something to get excited about.
Interval Arithmetic
One of the common uses of floating point arithmetic is in
modelling real-world phenomena. The techniques described in
the article I believe are inadequate to handle some of the
issues that crop up in trying to solve these models. For
example: we often have measurement error; know physical
constants to varying degrees of accuracy; discretize data
that is supposed to represent a continuum; use algorithms
that may not be numerically stable over the entire domain;
or deal with problems that are inherently stiff. The real
world is nonlinear.
There is an alternative to using fixed point schemes
(whether integers, scaled integers, rationals, or floating
point numbers as approximations to a continuum), which is
to compute with sets of numbers. For reasons of efficiency
and to take advantage of hardware acceleration, we
generally use intervals, defined as the set of all numbers
between a lower bound and upper bound [a,b].
By using intervals we can represent measurement error,
floating point rounding error, and imprecise constants in a
unified and consistent way. For some of us, one of the
greatest strengths of this approach is that when you
compute something, you also obtain an indication of the
quality of the answer i.e. the width of the interval.
Some problems have traditionally been considered
intractable, which may no longer be the case when using IA.
Consider a (large) solution space. By eliminating boxes
(multidimensional intervals) where it can be proved the
solution cannot be, you can iterate towards more and more
accurate approximations of the solution, subject to the
precision of the arithmetic being used. As the size of
boxes shrink, switch to higher precision if required. A
classic example of this technique is an Interval Newton
method for finding all roots of a function.
For all of this to work, you do need an implementation of
interval arithmetic, one that guarantees containment of the
true solution for all operator-operand combinations. In my
own work, I use the implementation that is part of Sun's
Fortran compilers. | {"url":"http://forums.theregister.co.uk/user/382/","timestamp":"2014-04-20T04:49:41Z","content_type":null,"content_length":"31117","record_id":"<urn:uuid:98be8c34-f7e9-4465-a5bf-d4779a863aa3>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00434-ip-10-147-4-33.ec2.internal.warc.gz"} |
Matches for:
Foundations of Analysis is an excellent new text for undergraduate students in real analysis. More than other texts in the subject, it is clear, concise and to the point, without extra bells and
whistles. It also has many good exercises that help illustrate the material. My students were very satisfied with it.
--Nat Smale, University of Utah
I have taught our Foundations of Analysis course (based on Joe Taylor.s book) several times recently, and have enjoyed doing so. The book is well-written, clear, and concise, and supplies the
students with very good introductory discussions of the various topics, correct and well-thought-out proofs, and appropriate, helpful examples. The end-of-chapter problems supplement the body of the
text very well (and range nicely from simple exercises to really challenging problems).
--Robert Brooks, University of Utah
An excellent text for students whose future will include contact with mathematical analysis, whatever their discipline might be. It is content-comprehensive and pedagogically sound. There are
exercises adequate to guarantee thorough grounding in the basic facts, and problems to initiate thought and gain experience in proofs and counterexamples. Moreover, the text takes the reader near
enough to the frontier of analysis at the calculus level that the teacher can challenge the students with questions that are at the ragged edge of research for undergraduate students. I like it a
--Don Tucker, University of Utah
My students appreciate the concise style of the book and the many helpful examples.
--W.M. McGovern, University of Washington
Analysis plays a crucial role in the undergraduate curriculum. Building upon the familiar notions of calculus, analysis introduces the depth and rigor characteristic of higher mathematics courses.
Foundations of Analysis has two main goals. The first is to develop in students the mathematical maturity and sophistication they will need as they move through the upper division curriculum. The
second is to present a rigorous development of both single and several variable calculus, beginning with a study of the properties of the real number system.
The presentation is both thorough and concise, with simple, straightforward explanations. The exercises differ widely in level of abstraction and level of difficulty. They vary from the simple to the
quite difficult and from the computational to the theoretical. Each section contains a number of examples designed to illustrate the material in the section and to teach students how to approach the
exercises for that section.
The list of topics covered is rather standard, although the treatment of some of them is not. The several variable material makes full use of the power of linear algebra, particularly in the
treatment of the differential of a function as the best affine approximation to the function at a given point. The text includes a review of several linear algebra topics in preparation for this
material. In the final chapter, vector calculus is presented from a modern point of view, using differential forms to give a unified treatment of the major theorems relating derivatives and
integrals: Green's, Gauss's, and Stokes's Theorems.
At appropriate points, abstract metric spaces, topological spaces, inner product spaces, and normed linear spaces are introduced, but only as asides. That is, the course is grounded in the concrete
world of Euclidean space, but the students are made aware that there are more exotic worlds in which the concepts they are learning may be studied.
Request an examination or desk copy.
Undergraduate students interested in real analysis. | {"url":"http://ams.org/bookstore?fn=20&arg1=whatsnew&ikey=AMSTEXT-18","timestamp":"2014-04-18T19:28:02Z","content_type":null,"content_length":"18852","record_id":"<urn:uuid:73e26ea2-2d17-45a4-a1e2-8912ba25fa50>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00167-ip-10-147-4-33.ec2.internal.warc.gz"} |
Prime Numbers
An integer is prime if its only positive divisors are itself and one.
Prime numbers have always been a keen interest to mathematicians for various reasons. One being there appears to be no rhyme or reason to the distribution of primes and there is no limit to finding
new primes. The largest known prime was found in November 2001 and it is over 4 million digits long!
Finding Primes
Computers have made finding these large prime numbers possible. The faster computers can crunch numbers, the larger the primes will be found. Here’s an interesting graph showing the largest known
primes by year.
Simple Method of Finding Primes
A prime number is one that has no prime factors. So a simple way of checking if a number is prime is by trying all known primes less than it and seeing if it divides evenly into that number.
Example: Finding the first couple of primes
Starting off with the number 2, it is prime because the only divisors are itself and 1, meaning the only way to multiple two numbers to get 2 is 2 x 1. Likewise for 3.
So that starts us off with two known primes 2 and 3. To check the next number we can check if 4 modulo 2 equals 0. This means when divide 2 into 4 there is no remainder, which means 2 is a factor of
4. Specifically we know 2 x 2 = 4. Thus 4 is not prime, since it has a prime factor.
Moving on to the next number: 5. To check if 5 is prime, try (5 modulo 2) and (5 modulo 3), both of which equals 1. So 5 is prime, since all primes less than it are not factors of 5. So add 5 to our
list of known primes and then continue on checking the next number. This rather tedious process for people is great for a computer.
Speeding up our Search
This method can be a little slow when the number of primes gets large since you have to check every previous prime number. One way to speed this up is to stop checking when the prime squared is
greater than the number.
For example if we are checking if 47 is prime, we can stop our check at 7, since 7 squared (49) is greater. If there were a larger factor that went into 47, it would have to be multiplied by a
smaller prime number to give us 47, and we would have already found the smaller number earlier in our checks.
This method is called the Sieve of Eratosthenes (ca 240 BC), which is stated as:
Make a list of all the integers less than or equal to n (and greater than one). Strike out the multiples of all primes less than or equal to the square root of n, then the numbers that are left
are the primes
My Python Script Implementing Sieve of Eratosthenes
The following python program will print out the primes from 2 to 1,000,000. On my Mac OS X PowerPC G4 500mhz it took approximately 4 minutes when printing to screen, and only 1 minute when output
directed to a file.
If you don’t have python, consider it pseudo-code to review for learning purposes.
Download Python Script: findthem.py
C Source Code
After working on my Krypto problem I found that Python is really slow when compared to C. So I came back and wrote up a simple C version of this script.
Download C Program: findthem.c - in just 12 minutes, I was able to generate the 5,761,453 primes below 100 million, which is a 54mb text file! | {"url":"http://mkaz.com/math/prime-numbers.html","timestamp":"2014-04-18T00:12:56Z","content_type":null,"content_length":"7280","record_id":"<urn:uuid:6aff6b57-32cb-44dc-ba31-e9240bcc7096>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00216-ip-10-147-4-33.ec2.internal.warc.gz"} |
Small letters after tags
Has this happened to everyone.Maybe it is the browser or the Sulfur layout,but after hide and math tags text is of smaller font.
And this here text appears smaller to me.I don't know why.Does this happen in other layouts?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
You should not just write randomtext, if you use \text{randomtext} it can be enlarged.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
The text inside the tags doesn't bother me.It is the text after the tag.In the case of the post above,the text:
"And this here text appears smaller to me.I don't know why.Does this happen in other layouts?"
is smaller for me than:
"Has this happened to everyone.Maybe it is the browser or the Sulfur layout,but after hide and math tags text is of smaller font."
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
They are the same size for me.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
Let me try oxygen.
EDIT:It is probably browser,cause it's the same in Oxygen as well.
Last edited by anonimnystefy (2012-04-24 18:31:49)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
What browser are you using?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
Google Chrome.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
I see, it could be a font setting in that browser.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
Maybe.But it is strange that it happens after tags.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
Do you have a screenshot of it?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
If the pic is alright then it should be right below.
Last edited by anonimnystefy (2012-04-24 19:12:44)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
No picture. Trim it down to size and try again.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
Do you see a picture there?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
It does appear slightly smaller to me to.
Do you have font settings?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
All you can do is play with these if you want to. Maybe a different font or size will help.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
Ok.I will see what I can do.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
Are there other settings for fonts?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
Then you will have to play with those.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
I'll try it.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Small letters after tags
My, what ugly people.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Small letters after tags
Very. But,did you read the text? It is funny.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=213203","timestamp":"2014-04-17T16:10:01Z","content_type":null,"content_length":"40397","record_id":"<urn:uuid:0ea994ea-ed59-42ba-ac86-06f9f4ce21ca>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00454-ip-10-147-4-33.ec2.internal.warc.gz"} |
Factorization of propagator matrices for efficient computation of
ASA 126th Meeting Denver 1993 October 4-8
5aUW10. Factorization of propagator matrices for efficient computation of wave-fields in horizontally layered fluid--solid media.
Sven Ivansson
[Dept. of Hydroacoust. and Seismol., Natl. Defence Res. Establishment FOA 260), S-172 90 Sundbyberg, Sweden]
The wave field is decomposed into its frequency--wave-number components. Compound matrices for solid layers provide a convenient way of computing the boundary values at a fluid--solid interface [M.
B. Porter and E. L. Reiss, J. Acoust. Soc. Am. 77, 1760--1767 1985)], with loss-of-precision control. A certain vector is propagated through a sequence of multiplications with compound matrices, one
for each layer. It is shown that computations of this kind can be performed more efficiently if each compound matrix is decomposed as a product of sparse matrices that are applied in sequence. Two
kinds of compound-matrix factorizations are proposed. In connection with dispersion computations, our first factorization gives a method that is related to the ``fast form'' of Knopoff's method [F.
Schwab et al., Bull. Seismol. Soc. Am. 74, 1555--1578 1984)]. This algorithm is slightly more efficient, however, and its range of applicability is wider. The second compound-matrix factorization
gives a method that is significantly faster than the ``fast form'' of Knopoff's method. Very few arithmetic operations are needed. It also provides a good basis for analyzing the numerical
performance of compound-matrix propagation. Finally, it is shown how propagator-matrix factorization can be used to enhance the efficiency for multi-frequency computations and computation of full
wave-fields, by wave-number integration or modal synthesis. | {"url":"http://www.auditory.org/asamtgs/asa93dnv/5aUW/5aUW10.html","timestamp":"2014-04-19T18:45:16Z","content_type":null,"content_length":"2180","record_id":"<urn:uuid:0835bdc0-e88b-45d2-88ea-7f3806598546>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00044-ip-10-147-4-33.ec2.internal.warc.gz"} |
668pages on
this wiki
Relations are nonnegative ordered pairs.A relation consists of a domain and range.A domain is a set of all first coordinates(x) of a relation, while a range is the set of all second coordinates(y).
E.G. Relation is {(1,10),(2,20),(3,30),(4,40)}
Its domain is {1,2,3,4}
Its range is {10,20,30,40}
Informally, a relation is a rule that describes how elements of a set relate, or interact, with elements of another set. Relations can include, but are not limited to, familial relations (Person A is
Person B's mother; or Person A and Person B have the same last name), geographic relations (State A shares a border with State B), and numerical relations ($A=B$; or $x \leq y$).
A relation from a set A to a set B is any subset of the Cartesian product A×B.
For example, if we let $S$ be the set of all cities, and $T$ the set of all U.S. States, we can define a relation $R$ to be the the set of ordered pairs $(s,t)$ for which the city $s$ is in the state
See also total order.
As a relation $\sim$ from a set $S$ to a set $T$ is formally viewed as a subset of the Cartesian product $S \times T$, the expression $\left(s,t\right)\in \sim$ is a valid mathematical expression.
However, such an expression can be cumbersome to write, and so we may adopt the alternate notation $s \sim t$. | {"url":"http://math.wikia.com/wiki/Relation","timestamp":"2014-04-17T18:42:10Z","content_type":null,"content_length":"54785","record_id":"<urn:uuid:593f4038-afcd-489f-b2d5-e6c3addf4b33>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00539-ip-10-147-4-33.ec2.internal.warc.gz"} |
search results
Expand all Collapse all Results 1 - 7 of 7
1. CJM 2011 (vol 63 pp. 798)
Representing Multipliers of the Fourier Algebra on Non-Commutative $L^p$ Spaces
We show that the multiplier algebra of the Fourier algebra on a locally compact group $G$ can be isometrically represented on a direct sum on non-commutative $L^p$ spaces associated with the right
von Neumann algebra of $G$. The resulting image is the idealiser of the image of the Fourier algebra. If these spaces are given their canonical operator space structure, then we get a completely
isometric representation of the completely bounded multiplier algebra. We make a careful study of the non-commutative $L^p$ spaces we construct and show that they are completely isometric to those
considered recently by Forrest, Lee, and Samei. We improve a result of theirs about module homomorphisms. We suggest a definition of a Figa-Talamanca-Herz algebra built out of these non-commutative
$L^p$ spaces, say $A_p(\widehat G)$. It is shown that $A_2(\widehat G)$ is isometric to $L^1(G)$, generalising the abelian situation.
Keywords:multiplier, Fourier algebra, non-commutative $L^p$ space, complex interpolation
Categories:43A22, 43A30, 46L51, 22D25, 42B15, 46L07, 46L52
2. CJM 2010 (vol 62 pp. 845)
Biflatness and Pseudo-Amenability of Segal Algebras
We investigate generalized amenability and biflatness properties of various (operator) Segal algebras in both the group algebra, $L^1(G)$, and the Fourier algebra, $A(G)$, of a locally compact
Keywords:Segal algebra, pseudo-amenable Banach algebra, biflat Banach algebra
Categories:43A20, 43A30, 46H25, 46H10, 46H20, 46L07
3. CJM 2009 (vol 61 pp. 1262)
On the Local Lifting Properties of Operator Spaces
In this paper, we mainly study operator spaces which have the locally lifting property (LLP). The dual of any ternary ring of operators is shown to satisfy the strongly local reflexivity, and this
is used to prove that strongly local reflexivity holds also for operator spaces which have the LLP. Several homological characterizations of the LLP and weak expectation property are given. We also
prove that for any operator space $V$, $V^{**}$ has the LLP if and only if $V$ has the LLP and $V^{*}$ is exact.
Keywords:operator space, locally lifting property, strongly locally reflexive
4. CJM 2007 (vol 59 pp. 966)
Operator Amenability of the Fourier Algebra in the $\cb$-Multiplier Norm
Let $G$ be a locally compact group, and let $A_{\cb}(G)$ denote the closure of $A(G)$, the Fourier algebra of $G$, in the space of completely bounded multipliers of $A(G)$. If $G$ is a weakly
amenable, discrete group such that $\cstar(G)$ is residually finite-dimensional, we show that $A_{\cb}(G)$ is operator amenable. In particular, $A_{\cb}(\free_2)$ is operator amenable even though $
\free_2$, the free group in two generators, is not an amenable group. Moreover, we show that if $G$ is a discrete group such that $A_{\cb}(G)$ is operator amenable, a closed ideal of $A(G)$ is
weakly completely complemented in $A(G)$ if and only if it has an approximate identity bounded in the $\cb$-multiplier norm.
Keywords:$\cb$-multiplier norm, Fourier algebra, operator amenability, weak amenability
Categories:43A22, 43A30, 46H25, 46J10, 46J40, 46L07, 47L25
5. CJM 2004 (vol 56 pp. 983)
Fubini's Theorem for Ultraproducts \\of Noncommutative $L_p$-Spaces
Let $(\M_i)_{i\in I}$, $(\N_j)_{j\in J}$ be families of von Neumann algebras and $\U$, $\U'$ be ultrafilters in $I$, $J$, respectively. Let $1\le p<\infty$ and $\nen$. Let $x_1$,\dots,$x_n$ in $\
prod L_p(\M_i)$ and $y_1$,\dots,$y_n$ in $\prod L_p(\N_j)$ be bounded families. We show the following equality $$ \lim_{i,\U} \lim_{j,\U'} \Big\| \summ_{k=1}^n x_k(i)\otimes y_k(j)\Big\|_{L_p(\M_i\
otimes \N_j)} = \lim_{j,\U'} \lim_{i,\U} \Big\| \summ_{k=1}^n x_k(i)\otimes y_k(j)\Big\|_{L_p(\M_i\otimes \N_j)} . $$ For $p=1$ this Fubini type result is related to the local reflexivity of duals
of $C^*$-algebras. This fails for $p=\infty$.
Keywords:noncommutative $L_p$-spaces, ultraproducts
Categories:46L52, 46B08, 46L07
6. CJM 2004 (vol 56 pp. 843)
Type Decomposition and the Rectangular AFD Property for $W^*$-TRO's
We study the type decomposition and the rectangular AFD property for $W^*$-TRO's. Like von Neumann algebras, every $W^*$-TRO can be uniquely decomposed into the direct sum of $W^*$-TRO's of type
$I$, type $II$, and type $III$. We may further consider $W^*$-TRO's of type $I_{m, n}$ with cardinal numbers $m$ and $n$, and consider $W^*$-TRO's of type $II_{\lambda, \mu}$ with $\lambda, \mu =
1$ or $\infty$. It is shown that every separable stable $W^*$-TRO (which includes type $I_{\infty,\infty}$, type $II_{\infty, \infty}$ and type $III$) is TRO-isomorphic to a von Neumann algebra. We
also introduce the rectangular version of the approximately finite dimensional property for $W^*$-TRO's. One of our major results is to show that a separable $W^*$-TRO is injective if and only if
it is rectangularly approximately finite dimensional. As a consequence of this result, we show that a dual operator space is injective if and only if its operator predual is a rigid rectangular ${\
OL}_{1, 1^+}$ space (equivalently, a rectangular
Categories:46L07, 46L08, 46L89
7. CJM 2002 (vol 54 pp. 1100)
The Operator Biprojectivity of the Fourier Algebra
In this paper, we investigate projectivity in the category of operator spaces. In particular, we show that the Fourier algebra of a locally compact group $G$ is operator biprojective if and only if
$G$ is discrete.
Keywords:locally compact group, Fourier algebra, operator space, projective
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